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Chapter 1 – Electric Circuit Variables Exercises Exercise 1.2-1 Find the charge that has entered an element by time t when i = 8t2 – 4t A, t ≥ 0. Assume q(t) = 0 for t < 0. 8 Answer: q(t ) = t 3 − 2t 2 C 3 Solution: i (t ) = 8 t 2 − 4 t A q(t ) = t ∫0 i dτ + q(0) = 8 3 8 2 2 t τ − τ τ + = τ − τ = t3 −2 t 2 C (8 4 ) d 0 2 ∫0 0 3 3 t Exercise 1.2-2 The total charge that has entered a circuit element is q(t) = 4 sin 3t C when t ≥ 0 and q(t) = 0 when t < 0. Determine the current in this circuit element for t > 0. d Answer: i (t ) = 4sin 3t = 12 cos 3t A dt Solution: i (t ) = dq d = 4sin 3t = 12 cos 3t dt dt A Exercise 1.3-1 Which of the three currents, i1 = 45 μA, i2 = 0.03 mA, and i3 = 25 × 10–4 A, is largest? Answer: i3 is largest. Solution: i1 = 45 μA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A 1-1 Exercise 1.5-1 Figure E 1.5-1 shows four circuit elements identified by the letters A, B, C, and D. (a) Which of the devices supply 12 W? (b) Which of the devices absorb 12 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device D? Answers: (a) B and C, (b) A and D, (c)–12 W, (c) 12 W, (e)–12 W + 4V – – 2V + + 6V – – 3V 3A 6A 2A 4A (A) (B) (C) (D) + Figure E 1.5-1 Solution: (a) B and C. The element voltage and current do not adhere to the passive convention in B and C so the product of the element voltage and current is the power supplied by these elements. (b) A and D. The element voltage and current adhere to the passive convention in A and D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) −12 W. The element voltage and current do not adhere to the passive convention in B, so the product of the element voltage and current is the power received by this element: (2 V)(6 A) = −12 W. The power supplied by the element is the negative of the power delivered to the element, 12 W. (d) 12 W (e) –12 W. The element voltage and current adhere to the passive convention in D, so the product of the element voltage and current is the power received by this element: (3 V)(4 A) = 12 W. The power supplied by the element is the negative of the power received to the element, −12 W. 1-2 Problems Section 1-2 Electric Circuits and Current Flow P1.2.1 The total charge that has entered a circuit element is q(t) = 1.25(1–e–5t) when t ≥ 0 and q(t) = 0 when t < 0. Determine the current in this circuit element for t ≥ 0. Answer: i(t) = 6.25e–5t A i (t ) = Solution: d 1.25 (1 − e−5t ) = 6.25 e−5t A dt P 1.2-2 The current in a circuit element is i(t) = 4(1–e–5t) A when t ≥ 0 and i(t) = 0 when t < 0. Determine the total charge that has entered a circuit element for t ≥ 0. t 0 −∞ −∞ Hint: q (0) = ∫ i (τ ) = ∫ 0dτ = 0 Answer: q(t) = 4t + 0.8e–5t– 0.8 C for t ≥ 0 Solution: t t t t 4 4 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 (1 − e−5τ ) dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C 0 0 0 0 5 5 P 1.2-3 The current in a circuit element is i(t) = 4 sin 3t A when t ≥ 0 and i(t) = 0 when t < 0. Determine the total charge that has entered a circuit element for t ≥ 0. t 0 −∞ −∞ Hint: q (0) = ∫ i (τ )dτ = ∫ 0 dτ = 0 Solution: t t 0 0 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 5τ dτ + 0 = − 4 4 4 t cos 3τ 0 = − cos 3 t + C 5 5 5 1-3 ⎧0 t < 2 ⎪ ⎪2 2 < t < 4 where the units of current are A P 1.2-4 The current in a circuit element is i (t ) = ⎨ t 1 4 8 − < < ⎪ ⎪⎩0 8 < t and the units of time are s. Determine the total charge that has entered a circuit element for t ≥ 0. Answer: t<2 2<t <4 where the units of charge are C. 4<t <8 8<t ⎧0 ⎪2t − 4 ⎪ q (t ) = ⎨ ⎪8 − t ⎩⎪0 Solution: t t −∞ t −∞ q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0. t t 2 t 2 t 2 4 t 4 t 8 8 q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2τ = 2t−4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C. q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8−t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C. t q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t . P 1.2-5 The total charge q(t), in coulombs, that enters the terminal of an element is ⎧0 ⎪ q (t ) = ⎨2t ⎪3 + e −2t (t −2) ⎩ t<0 0≤t ≤2 t>2 Find the current i(t) and sketch its waveform for t ≥ 0. Solution: dq ( t ) i (t ) = dt ⎧0 ⎪ i (t ) = ⎨2 ⎪ −2( t − 2 ) ⎩−2e t <0 0< t < 2 t >2 1-4 i P 1.2-6 An electroplating bath, as shown in Figure P 1.2-6, is used to plate silver uniformly onto objects such as kitchen ware and plates. A current of 600 A flows for 20 minutes, and each coulomb transports 1.118 mg of silver. What is the weight of silver deposited in grams? i Object to be plated Silver bar Bath Figure P 1.2-6 Solution: i = 450 A = 450 C s C s mg Silver deposited = 450 ×20 min×60 = 6.0372×105 mg=603.72 g ×1.118 s min C 1-5 P1.2-7 Find the charge q(t) and sketch its waveform when the current entering a terminal of ab element is as shown in Figure P1.2-7. Assume that q(t) = 0 for t < 0. Figure P1.2-7 Solution: ⎧1 when 0 < t ≤ 2 i (t ) = ⎨ ⎩ t − 1 when 2 ≤ t t t 0 0 q (t ) = ∫ i (τ )dτ + q (0) = ∫ i (τ )dτ and since q(0) =0. When 0 < t ≤ 2, we have t q = ∫ 1 dτ = t C 0 When t ≥ 2, we have t 2 0 0 t q = ∫ i (τ )dτ = ∫ 1 dτ + ∫ (τ − 1) dτ 2 2 t t = τ 0 + τ2 − τ 2 = 2 2 t2 2 −t + 2 C The sketch of q(t) is shown to the right:. 1-6 Section 1-3 Systems of Units P 1.3-1 A constant current of 3.2 μA flows through an element. What is the charge that has passed through the element in the first millisecond? Answer: 3.2 nC Solution: Δq = i Δt = ( 3.2 × 10 −6 A )(1 × 10 − 3 s ) = 3.2 × 10 − 9 As = 3.2 × 10 − 9 C = 3.2 × 10 − 9 nC P 1.3-2 A charge of 45 nC passes through a circuit element during a particular interval of time that is 5 ms in duration. Determine the average current in this circuit element during that interval of time. Answer: i = 9 μA Solution: Δ q 45 × 10−9 i= = = 9 × 10−6 = 9 μA −3 Δt 5 × 10 P 1.3-3 Ten billion electrons per second pass through a particular circuit element. What is the average current in that circuit element? Answer: i = 1.602 nA Solution electron ⎤ ⎡ C ⎤ ⎡ i = ⎢10 billion 1.602 ×10−19 = ⎥ ⎢ s electron ⎥⎦ ⎣ ⎦⎣ C ⎤ ⎡ 9 electron ⎤ ⎡ 1.602 × 10−19 ⎢⎣10×10 ⎥ ⎢ s electron ⎥⎦ ⎦⎣ electron C = 1010 × 1.602 ×10−19 electron s C = 1.602 × 10−9 = 1.602 nA s 1-7 P1.3-4 The charge flowing in a wire is plotted in Figure P1.3-4. Sketch the corresponding current. Figure P1.3-4 P1.3-4 ⎧ 15 × 10−9 −3 = 7.5 mA when 0 < t < 2μ s ⎪ −6 = 7.5 × 10 2 10 × ⎪ ⎪ 15 × 10−9 d −3 = −5 mA when 4μ s < t < 7 μ s i ( t ) = q ( t ) = the slope of the q versus t plot = ⎨− −6 = −5 × 10 dt ⎪ 3 × 10 0 otherwise ⎪ ⎪ ⎩ 1-8 P1.3-5 The current in a circuit element is plotted in Figure P1.3-5. Sketch the corresponding charge flowing through the element for t > 0. Figure P1.3-5 Solution: t ⎧ 450 μ A dτ when 0 < t < 80 ms ∫ 0 ⎪ t t ⎪ q ( t ) = ∫ i (τ ) dτ = ⎨( 450 × 10−6 )( 80 × 10−3 ) + ∫ ( −600μ A ) dτ when 80 ms < t < 140 ms 0 80 ms ⎪ t ⎪ 450 ×10−6 )( 80 ×10−3 ) + ( −600 × 10−6 )( 60 × 10−3 ) + ∫ 0 dτ ( 140 ms ⎩ ⎧ 450 ×10−6 ) t when 0 < t < 80 ms ( ⎪ ⎪ = ⎨( 36 × 10−6 ) + ( −600 ×10−6 ) t when 80 ms < t < 140 ms ⎪ 0 C when 140 ms < t ⎪⎩ While 0 < t < 80 ms q(t) increases linearly from 0 to 36 μC and while 80 < t < 140 ms q(t) decreases linearly from 36 to 0 μC. Here’s the sketch: 1-9 P1.3-6 The current in a circuit element is plotted in Figure P1.3-6. Determine the total charge that flows through the circuit element between 300 and 1200 μs. Figure P1.3-6 Solution: q (t ) = ∫ 1000 μ s 300 μ s i (τ ) dτ = "area under the curve between 300 μ s and 1000 μ s" ⎛ 360 + 720 ⎞ q (t ) = ⎜ ×10−9 ⎟ (100 × 10−6 ) + ( 720 ×10−9 )( 600 ×10−6 ) = ( 54 + 432 ) × 10−12 = 486 pC 2 ⎝ ⎠ 1-10 Section 1-5 Power and Energy P1.5-1 Figure P1.5-1 shows four circuit elements identified by the letters A, B, C, and D. (a) Which of the devices supply 30 mW? (b) Which of the devices absorb 0.03 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device C? Figure P1.5-1 Solution: (a) A and D. The element voltage and current do not adhere to the passive convention in Figures P1.5- A and D so the product of the element voltage and current is the power supplied by these elements. (b) B and C. The element voltage and current adhere to the passive convention in Figures P1.5-1 B and C so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) 30 mW. The element voltage and current adhere to the passive convention in Figure P1.5-1B, so the product of the element voltage and current is the power received by this element: (5 V)(6 mA) = 30 mW. The power supplied by the element is the negative of the power received to the element, −30 W. (d) −30 mW (e) –30 mW. The element voltage and current adhere to the passive convention in Figure P1.51C, so the product of the element voltage and current is the power received by this element: (5 V)(6 mA) = 30 mW. The power supplied by the element is the negative of the power received to the element, −30 W. 1-11 P 1.5-2 An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours. (a) Find the charge in coulombs that passes through the range. (b) Find the power absorbed by the range. (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours. Solution: a.) q = ∫ i dt = iΔt = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10 4 C b.) P = v i = (110 V )(10 A ) = 1100 W c.) Cost = 0.12$ × 1.1kW × 2 hrs = 0.264 $ kWhr P 1.5-3 A walker’s cassette tape player uses four AA batteries in series to provide 6 V to the player circuit. The four alkaline battery cells store a total of 200 watt-seconds of energy. If the cassette player is drawing a constant 10 mA from the battery pack, how long will the cassette operate at normal power? Solution: P = ( 6 V )(10 mA ) = 0.06 W Δt = 200 W⋅s Δw = = 3.33×103 s P 0.06 W 1-12 P 1.5-4 The current through and voltage across an element vary with time as shown in Figure P 1.5-4. Sketch the power delivered to the element for t > 0. What is the total energy delivered to the element between t = 0 and t = 25 s? The element voltage and current adhere to the passive convention. v (volts) i (amp) 30 30 5 5 0 10 15 25 0 t (s) 10 15 (a) 25 t (s) (b) Figure P 1.5-4 Solution: for 0 ≤ t ≤ 10 s: v = 30 V and i = 30 t = 2t A ∴ P = 30(2t ) = 60t W 15 25 t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V 5 v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W for 10 ≤ t ≤ 15 s: v ( t ) = − 30 t +b A 10 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − i (25) = 0 ∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W Energy = ∫ P dt = = 30t 2 10 0 ∫ 10 0 60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt 15 15 25 25 + 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J 3 10 2 15 1-13 P 1.5-5 An automobile battery is charged with a constant current of 2 A for five hours. The terminal voltage of the battery is v = 11 + 0.5t V for t > 0, where t is in hours. (a) Find the energy delivered to the battery during the five hours. (b) If electric energy costs 15 cents/kWh, find the cost of charging the battery for five hours. Answer: (b) 1.84 cents Solution: a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 5( 3600 ) t 5 ( 3600 ) ⎛ 0.5 τ ⎞ 0.5 2 2 ⎜11 + w = ∫ Pdt = ∫0 vi dτ = ∫0 ⎟ dτ = 22 t + 3600 τ 3600 ⎠ ⎝ 0 = 441× 103 J = 441 kJ b.) Cost = 441kJ × 1 hr 15¢ × = 1.84¢ 3600s kWhr 1-14 P 1.5-6 Find the power, p(t), supplied by the element shown in Figure P 1.5-6 when v(t) = 4 sin 3t V and i(t) = (1/12) sin 3t A. Evaluate p(t) at t=0.5 s and t = 1 s. Observe that the power supplied by this element has a positive value at some times and a negative value at other times. 1 Hint: (sin at )(sin bt ) = (cos(a − b)t − cos(a + b)t ) 2 Answer: p(t) = (1/6)cos(6t) W, p(0.5) = 0.0235 W, p(1) = −0.02466 W Solution: 1 ⎛1 ⎞ 1 p ( t ) = v ( t ) i ( t ) = ( 4 cos 3 t ) ⎜ sin 3 t ⎟ = ( sin 0 + sin 6 t ) = sin 6 t 6 ⎝ 12 ⎠ 6 1 p ( 0.5 ) = sin 3 = 0.0235 W 6 1 p (1) = sin 6 = −0.0466 W 6 W Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % v=4*cos(3*t); i=(1/12)*sin(3*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power initial time final time time increment time plot(t,p) xlabel('time, s'); ylabel('power, W') 1-15 i P 1.5-7 Find the power, p(t), supplied by the element shown in Figure P 1.5-6 when v(t) = 8 sin 3t V and i(t) = 2 sin 3t A. 1 Hint: (sin at )(sin bt ) = (cos(a − b)t − cos(a + b)t ) 2 Answer: p(t) = 8 – 8cos 6t W Solution: + v – Figure P 1.5-7 p ( t ) = v ( t ) i ( t ) = ( 8sin 3 t )( 2sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t W Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=8*sin(3*t); i=2*sin(3*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') 1-16 i P 1.5-8 Find the power, p(t), supplied by the element shown in Figure P 1.5-6. The element voltage is represented as v(t) = 4(1–e–2t)V when t ≥ 0 and v(t) = 0 when t < 0. The element current is represented as i(t) = 2e–2t A when t ≥ 0 and i(t) = 0 when t < 0. –2t –2t Answer: p(t) = 8(1 – e )e Solution: + v – W Figure P 1.5-7 ( ) ( ) p ( t ) = v ( t ) i ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e−2 t W Here is a MATLAB program to plot p(t): clear t0=0; tf=2; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=4*(1-exp(-2*t)); i=2*exp(-2*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') 1-17 P 1.5-9 The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA. What power is absorbed by the bulb? Find the energy absorbed by the bulb in a five-minute period. Solution: The power is P = V I =3 × 0.015=0.045 W . Next, the energy is w = PΔ t = 0.045 × 5 × 60=13.5 J . 1-18 P1.5-10 Medical researchers studying hypertension often use a technique called “2D gel electrophoresis” to analyze the protein content of a tissue sample. An image of a typical “gel” is shown in Figure 1.5-10a. The procedure for preparing the gel uses the electric circuit illustrated in Figure 1.5-10b. The sample consists of a gel and a filter paper containing ionized proteins. A voltage source causes a large, constant voltage, 500 V, across the sample. The large, constant voltage moves the ionized proteins from the filter paper to the gel. The current in the sample is given by Devon Svoboda, Queen’s University (a) sample i ( t ) = 2 + 30 e − at mA + 500 V – i (t) where t is the time elapsed since the beginning of the procedure and the value of the constant a is 1 a = 0.85 hr Determine the energy supplied by the voltage source when the gel preparation procedure lasts 3 hours. 500 V (b) Figure 1.5-10 (a) An image of a gel and (b) the electric circuit used to preparation a gel. Solution: T T 0 0 energy = w ( t ) = ∫ p ( t ) d t = ∫ v ( t ) i ( t ) d t = 500 3 2 + 30 e − 0.85t d t ∫ 0 1000 ( ) 3 3 0 0 ( ) = ∫ d t + 15∫ e − 0.85t d t 15 e − 2.55 − 1 −0.85 = 3 + 16.3 = 19.3 J = (3 − 0) + ( ) 1-19 Section 1.7 How Can We Check…? – 5V + P 1.7-1 Conservation of energy requires that the sum of the power absorbed by all of the elements in a circuit be zero. Figure P 1.7-1 shows a circuit. All of the element voltages and currents are specified. Are these voltage and currents correct? Justify your answer. + 2A –2 V – –5 A + + 4V – 3V 2A – 5A 3A + 1V – Figure P 1.7-1 Hint: Calculate the power absorbed by each element. Add up all of these powers. If the sum is zero, conservation of energy is satisfied and the voltages and currents are probably correct. If the sum is not zero, the element voltages and currents cannot be correct. Solution: Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W =0W The element voltages and currents satisfy conservation of energy and may be correct. 1-20 P 1.7-2 Conservation of energy requires that the sum of the power absorbed by all of the elements in a circuit be zero. Figure P 1.7-2 shows a circuit. All of the element voltages and currents are specified. Are these voltage and currents correct? Justify your answer. Hint: Calculate the power absorbed by each element. Add up all of these powers. If the sum is zero, conservation of energy is satisfied and the voltages and currents are probably correct. If the sum is not zero, the element voltages and currents cannot be correct. + 4V – 3A + + – 3V 3V 3V – 3A – 2A + 2A – –3 A –3 V + 4V – + –3 A Figure P 1.7-2 Solution: Notice that the element voltage and current of some branches do not adhere to the passive convention. The sum of the powers absorbed by each branch are: -(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A) = -9 W + 6 W + 6 W + 12 W + 9 W -12 W ≠0W The element voltages and currents do not satisfy conservation of energy and cannot be correct. 1-21 – 3V + P 1.7-3 The element currents and voltages shown in Figure P 1.7-3 are correct with one exception: the a reference direction of exactly one of the element currents is reversed. Determine + which reference direction has been 5V reversed. – –3A – 1V + 4A 7A – 2V + b c −2A – −6V + 2A – −8V + –5A d Figure P 1.7-3 Solution: Let’s tabulate the power received by each element. We’ll identify each element by its nodes. nodes a c So Power received, W ( 3)( −3) = −9 a b − (1)( 4 ) = −4 b c a d b d c d ( 2 )( −2 ) = −4 − ( 5 )( 7 ) = −35 ( −6 )( 2 ) = −12 ( −8)( −5) = 40 Total power received = − ( 9 + 4 + 4 + 35 + 12 ) + 40 = −24 ≠ 0 Changing the current reference direction for a particular element will change the total power by twice the power of the particular element. Since the element connected between nodes b and d receives -12 W, changing the reference direction of its current will increase the total power received by 24 W, as required. After making that change Total power received = − ( 9 + 4 + 4 + 35 ) + (12 + 40 ) = 0 We conclude that it is the reference direction of the element connected between nodes and b that has been reversed. 1-22 Design Problems DP 1-1 A particular circuit element is available in three grades. Grade A guarantees that the element can safely absorb 1/2 W continuously. Similarly, Grade B guarantees that 1/4 W can be absorbed safely and Grade C guarantees that 1/8 W can be absorbed safely. As a rule, elements that can safely absorb more power are also more expensive and bulkier. The voltage across an element is expected to be about 20 V and the current in the element is expected to be about 8 mA. Both estimates are accurate to within 25 percent. The voltage and current reference adhere to the passive convention. Specify the grade of this element. Safety is the most important consideration, but don’t specify an element that is more expensive than necessary. Solution: The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25) = 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust the estimates of the maximum voltage and current and a Grade A device otherwise. 1-23 DP 1-2 The voltage across a circuit element is v(t) = 20 (1–e–8t)V when t ≥ 0 and v(t) = 0 when t < 0. The current in this element is i(t) = 30e–8t mA when t ≥ 0 and i(t) = 0 when t < 0. The element current and voltage adhere to the passive convention. Specify the power that this device must be able to absorb safely. Hint: Use MATLAB, or a similar program, to plot the power. Solution: ( ) ( ) p ( t ) = 20 1 − e −8t × 0.03 e −8t = 0.6 1 − e −8 t e −8 t Here is a MATLAB program to plot p(t): W Here is the plot: clear t0=0; tf=1; dt=0.02; t=t0:dt:tf; % % % % initial time final time time increment time v=20*(1-exp(-8*t)); i=.030*exp(-8*t); % device voltage % device current for k=1:length(t) p(k)=v(k)*i(k); end % power plot(t,p) xlabel('time, s'); ylabel('power, W') The circuit element must be able to absorb 0.15 W. 1-24 Chapter 2 Circuit Elements Exercises Exercise 2.4-1 Find the power absorbed by a 100-ohm resistor when it is connected directly across a constant 10-V source. Answer: 1-W v 2 (10 ) P= = =1 W R 100 2 Solution: Exercise 2.4-2 A voltage source v = 10 cos t V is connected across a resistor of 10 ohms. Find the power delivered to the resistor. Answer: 10 cos2 t W Solution: v 2 (10 cos t ) 2 P= = = 10 cos 2 t W R 10 Exercise 2.7-1 Find the power absorbed by the CCCS in Figure E 2.7-1. Hint: The controlling element of this dependent source is a short circuit. The voltage across a short circuit is zero. Hence, the power absorbed by the controlling element is zero. How much power is absorbed by the controlled element? Answer:–115.2 watts are absorbed by the CCCS. (The CCCS delivers +115.2 watts to the rest of the circuit.) – 1. 2 0 + 2 4. 0 Ammeter Voltmeter 2Ω 2Ω ic 12 V + – + 4Ω id = 4ic vd – Figure E 2.7-1 Solution: ic = − 1.2 A, vd = 24 id = 4 ( − 1.2) = − 4.8 V A id and vd adhere to the passive convention so P = vd id = (24) (−4.8) = −115.2 is the power received by the dependent source W Exercise 2.8-1 For the potentiometer circuit of Figure 2.8-2, calculate the meter voltage, vm, when θ = 45°, Rp = 20 kΩ, and I = 2 mA. Answer: vm = 5 V (1 – a)Rp Voltmeter + I vm – + vm aRp I Rp – (a) (b) Figure 2.8-2 Solution: θ = 45° , I = 2 mA, R p = 20 kΩ a= θ 45 ⇒ aR = (20 kΩ) = 2.5 kΩ p 360 360 vm = (2 ×10−3 )(2.5 ×103 ) = 5 V Exercise 2.8-2 The voltage and current of an AD590 temperature sensor of Figure 2.8-3 are 10 V and 280 μA, respectively. Determine the measured temperature. i(t) + + v(t) AD590 – v(t) i(t) = kT – Answer: T = 280°K, or approximately 6.85°C (a) (b) Figure 2.8-3 Solution: v = 10 V, i = 280 μA, k = 1 μA °K for AD590 ⎛ °K ⎞ i ⎟ = 280° K i = kT ⇒ T = = (280μA)⎜1 ⎜ μA ⎟ k ⎝ ⎠ t=5s t=3s Exercise 2.9-1 What is the value of the current i in Figure E 2.9-1 at time t = 4 s? Answer: i = 0 amperes at t = 4 s (both switches are open). 12 V + – 3 kΩ 6V + – i Figure E 2.9-1 Solution: At t = 4 s both switches are open, so i = 0 A. t=5s Exercise 2.9-2 What is the value of the voltage v in Figure E 2.9-2 at time t = 4 s? At t = 6 s? Answer: v = 6 volts at t = 4 s, and v = 0 volts at t = 6 s. + 3 kΩ 2 mA i Figure E 2.9-2 Solution: At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V . At t = 6 s the switch is in the down position, so v = 0 V. v – Section 2-2 Engineering and Linear Models P 2.2-1 An element has voltage v and current i as shown in Figure P 2.2-1a. Values of the current i and corresponding voltage v have been tabulated as shown in Figure P 2.2-1b. Determine if the element is linear. i + v – (a) v, V i, A –3 –4 0 12 32 60 –3 –2 0 2 4 6 (b) Figure P 2.2-1 Solution: The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P 2.2-2 A linear element has voltage v and current i as shown in Figure P 2.2-2a. Values of the current i and i corresponding voltage v have been tabulated as shown in + Figure P 2.2-2b. Represent the element by an equation that v expresses v as a function of i. This equation is a model of – the element. (a) Verify that the model is linear. (b) Use the model to predict the value of v corresponding to a current of i = 40 mA. (c) Use the model to predict the value of i (a) v, V i, A –3.6 2.4 6.0 –30 20 50 (b) Figure P 2.2-2a corresponding to a voltage of v = 4 V. Hint: Plot the data. We expect the data points to lie on a straight line. Obtain a linear model of the element by representing that straight line by an equation. Solution: (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed linear. (b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV 3 (c) When v = 3 V, i = = 25 A 0.12 P 2.2-3 A linear element has voltage v and current i as shown in Figure P 2.2-3a. Values of the current i and i corresponding voltage v have been tabulated as shown in + Figure P 2.2-3b. Represent the element by an equation that v v, V i, mA expresses v as a function of i. This equation is a model of – 3.078 5.13 12.825 12 20 50 the element. (a) Verify that the model is linear. (b) Use the model to predict the value of v corresponding to a current of i = 4 mA. (c) Use the model to predict the value of i (a) (b) Figure P 2.2-3 corresponding to a voltage of v = 12 V. Hint: Plot the data. We expect the data points to lie on a straight line. Obtain a linear model of the element by representing that straight line by an equation. Solution: (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is v = 256.5 i . The element is indeed linear. (b) When i = 6 mA, v = (256.5 V/A)×(6 mA) = (256.5 V/A)×(0.006 A) = 1.054 V 12 (c) When v = 12 V, i = = 0.04678 A = 46.78 mA. 256.5 P 2.2-4 An element is represented by the relation between current and voltage as v = 3i + 5 Determine whether the element is linear. Solution: Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence, the property of homogeneity is not satisfied. The element is not linear. P 2.2-5 The circuit shown in Figure P 2.3-5 consists of a current source, a resistor, and element A. + 0.4 A Consider three cases. 10 Ω v A i − Figure P 2.3-5 (a) When element A is a 40-Ω resistor, described by i = v / 40, then the circuit is represented v v by 0.4 = + 10 40 Determine the values of v and i. Notice that the above equation has a unique solution. (b) When element A is a nonlinear resistor described by i = v2 / 2, then the circuit is v v2 0.4 = + represented by 10 2 Determine the values of v and i. In this case there are two solutions of the above equation. Nonlinear circuits exhibit more complicated behavior than linear circuits. v2 When element A is a nonlinear resistor described by i = 0.8 + , then the circuit is 2 2 v v described by 0.4 = + 0.8 + 10 2 Show that this equation has no solution. This result usually indicates a modeling problem. At least one of the three elements in the circuit has not been modeled accurately. (c) Solution: (a) (b) Using the quadratic formula v v v + = ⇒ v = 3.2 V 10 40 8 v i= = 0.08 A 40 v v2 v 0.4 = + ⇒ v 2 + − 0.8 = 0 10 2 5 0.4 = v= −0.2 ± 1.8 = 0.8, − 1.0 V 2 ( −1) = 0.5 A . 0.82 = 0.32 A . When v = -1.0 V then i = When v = 0.8 V then i = 2 2 2 v v v 0.4 = + 0.8 + ⇒ v 2 + + 0.8 = 0 (c) 10 2 5 2 −0.2 ± 0.04 − 3.2 2 So there is no real solution to the equation. Using the quadratic formula v= Section 2-4 Resistors P 2.4-1 A current source and a resistor are connected in series in the circuit shown in Figure P 2.4-1. Elements connected in series have the same current, so i = is in this circuit. Suppose that is = 3 A and R = 7 Ω. Calculate the voltage v across the resistor and the power absorbed by the resistor. is i + R v – Figure P 2.4-1 Answer: v = 21 V and the resistor absorbs 63 W. Solution: i = is = 3 A and v = Ri = 7 × 3 = 21 V v and i adhere to the passive convention ∴ P = v i = 21 × 3 = 63 W is the power absorbed by the resistor. P 2.4-2 A current source and a resistor are connected in series in the circuit shown in Figure P 2.4-1. Elements connected in series have the same current, so i = is in this circuit. Suppose that i = 3 mA and v = 24 V. Calculate the resistance R and the power absorbed by the resistor. Answer: R = 8 kΩ and the resistor absorbs 72 mW. is i + R v – Figure P 2.4-1 Solution: i = i = 3 mA and v = 48 V s 48 v = = 16000 = 16 K Ω R = i 0.003 P = (3×10− 3 )×48 = 144×10− 3 = 144 mW P 2.4-3 A voltage source and a resistor are connected in parallel in the circuit shown in Figure P 2.4-3. Elements connected in parallel have the same voltage, so v = vs in this circuit. Suppose that vs = 10 V and R = 5 Ω. Calculate the current i in the resistor and the power absorbed by the resistor. i vs R + v – Figure P 2.4-3 Answer: i = 2 A and the resistor absorbs 20 W. Solution: + – v = vs =10 V and R = 5 Ω v 10 = =2 A R 5 v and i adhere to the passive convention ∴ p = v i = 2⋅10 = 20 W is the power absorbed by the resistor i = P 2.4-4 A voltage source and a resistor are connected in parallel in the circuit shown in Figure P 2.4-3. Elements connected in parallel have the same voltage, so v = vs in this circuit. Suppose that vs = 24 V and i = 2 A. Calculate the resistance R and the power absorbed by the resistor. i vs Answer: R = 12 Ω and the resistor absorbs 48 W. Solution: v = vs =24 V and i = 3 A v 24 V = = 8Ω i 3A p = v i = 24( 3) = 72 W R = + – R + v – Figure P 2.4-3 P 2.4-5 A voltage source and two resistors are connected in parallel in the circuit shown in Figure P 2.4-5. Elements connected in parallel have the same voltage, so v1 = vs and v2 = vs in this circuit. Suppose that vs = 150 V, R1 = 50 Ω, and R2 = 25 Ω. Calculate the current in each resistor and the power absorbed by each resistor. i1 + – vs i2 + v1 R1 R2 – + v2 – Figure P 2.4-5 Hint: Notice the reference directions of the resistor currents. Answer: i1 = 3 A and i2 = –6 A. R1 absorbs 450 W and R2 absorbs 900 W. Solution: v1 = v 2 = vs = 150 V; R1 = 50 Ω; R2 = 25 Ω v 1 and i1 adhere to the passive convention so v 1 150 = =3 A R 1 50 150 v v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A 25 R2 i1 = The power absorbed by R1 is P1 = v1 i1 = 150 ⋅ 3 = 450 W The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W P 2.4-6 A current source and two resistors are connected in series in the circuit shown in Figure P 2.4-6. Elements connected in series have the same current, so i1 = is and i2 = is in this circuit. Suppose that is = 2 A, R1 = 4 Ω, and R2 = 8 Ω. Calculate the voltage across each resistor and the power absorbed by each resistor. – v1 + i1 is R1 i2 + R2 v2 – Hint: Notice the reference directions of the resistor voltages. Figure P 2.4-6 Answer: v1 = –8 V and v2 = 16 V. R1 absorbs 16 W and R2 absorbs 32 W. Solution: i1 = i2 = is =25 mA and R1 = 4 Ω and R2 = 8 Ω v1 and i1 do not adhere to the passive convention so v1 = − R1 i1 = −4 ( 0.025 ) = −0.1 V. The power absorbed by R1 is P1 =−v1 i1 =−(−0.1)(0.025) = 2.5 mW. v 2 and i 2 do adhere to the passive convention so v 2 = R 2 i 2 = 8 ( 0.025 ) = 0.2 V. The power absorbed by R 2 is P2 = v 2 i 2 =(0.2)(0.025) = 5 mW. P 2.4-7 An electric heater is connected to a constant 250-V source and absorbs 1000 W. Subsequently, this heater is connected to a constant 210-V source. What power does it absorb from the 210-V source? What is the resistance of the heater? Hint: Model the electric heater as a resistor. Solution: v2 v2 (250)2 ⇒ R = = = 62.5 Ω R P 1000 v 2 (220)2 = = 774.4 W P= R 62.5 Model the heater as a resistor, then from P = with a 220 V source P 2.4-8 The portable lighting equipment for a mine is located 100 meters from its dc supply source. The mine lights use a total of 5 kW and operate at 120 V dc. Determine the required cross-sectional area of the copper wires used to connect the source to the mine lights if we require that the power lost in the copper wires be less than or equal to 5 percent of the power required by the mine lights. Hint: Model both the lighting equipment and the wire as resistors. Solution: The current required by the mine lights is: i = P 5000 125 A = = 3 v 120 Power loss in the wire is : i 2 R Thus the maximum resistance of the copper wire allowed is 0.05P 0.05×5000 = = 0.144 Ω (125/3) 2 i2 now since the length of the wire is L = 2×100 = 200 m = 20,000 cm R= thus R = ρ L / A with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1 A= ρL R = 1.7×10−6 ×20,000 = 0.236 cm 2 0.144 *P 2.4-9 The resistance of a practical resistor depends on the nominal resistance and the resistance tolerance as follows: t ⎞ t ⎞ ⎛ ⎛ Rnom ⎜1 − ⎟ ≤ R ≤ Rnom ⎜ 1 + ⎟ ⎝ 100 ⎠ ⎝ 100 ⎠ i R1 + vs where Rnom is the nominal resistance and t is the resistance tolerance expressed as a percentage. For example, a 100-Ω, 2 percent resistor will have a resistance given by + – R2 vo – Figure P 2.4-9 98 Ω ≤ R ≤ 102 Ω The circuit shown in Figure P 2.4-9 has one input, vs, and one output, vo. The gain of this circuit is given by v R2 gain = 0 = vs R1 + R2 Determine the range of possible values of the gain when R1 is the resistance of a 100-Ω, 2 percent resistor and R2 is the resistance of a 400-Ω, 5 percent resistor. Express the gain in terms of a nominal gain and a gain tolerance. Solution: 0.7884 = 380 420 ≤ gain ≤ = 0.8108 102 + 380 98 + 420 nominal gain = gain tolerance = So 0.7884 + 0.8108 = 0.7996 2 0.7996 − 0.7884 0.8108 − 0.7996 × 100 = × 100 = 1.40% 0.7996 0.7996 gain = 0.7996 ± 1.40% + va – P 2.4-10 The voltage source shown in va, V vb, V Figure P 2.4-10 is an adjustable dc voltage 11.75 7.05 source. In other words, the voltage vs is a 40 Ω + 7.5 4.5 constant voltage, but the value of that constant vs +– 5.625 3.375 R vb can be adjusted. The tabulated data were 10 6 – 4.375 2.625 collected as follows. The voltage, vs, was set to some value, and the voltages across the Figure P 2.4-10 resistor, va and vb, were measured and recorded. Next, the value of vs was changed, and the voltages across the resistors were measured again and recorded. This procedure was repeated several times. (The values of vs were not recorded.) Determine the value of the resistance, R. Solution: Label the current i as shown. That current is the element current in both resistors. First va i= 40 va vb ⇒ R = 40 Next v b = R i = R 40 va For example, R = 40 7.05 = 24 Ω 11.75 P2.4-11 Consider the circuit shown in Figure P2.4-11. (a) Suppose the current source supplies 3.125 W of power. Determine the value of the resistance R. Figure P2.4-11 (b) Suppose instead the resistance is R = 12 Ω. Determine the value of the power supplied by the current source. Solution: (a) Suppose the current source supplies 3.125 W of power. 1.25 v 3 = 3.125 ⇒ v 3 = 2.5 V Then, using KVL R= 20 + 2.5 = 18 Ω 1.25 (b) Suppose instead the resistance is R = 12 Ω. From KVL 1.25 (12 ) − v 3 − 20 = 0 ⇒ v 3 = −5 V The value of the power supplied by the current source is 1.25 v 3 = 1.25 ( −5 ) = −6.25 W P2.4-12.We will encounter “ac circuits” in Chapter 10. Frequently we analyze ac circuits using “phasors” and “impedances”. Phasors are complex numbers that represents currents and voltages in an ac circuit. Impedances are complex numbers that describe ac circuit elements. (See Appendix B for a discussion of complex numbers.) Figure P2.4-11 shows a circuit element in an ac circuit. I and V are complex numbers representing the element current and voltage. Z is a complex number describing the element itself. “Ohm’s law for ac circuits” indicates that V=ZI (a) Suppose V = 12∠45° V , I = B∠θ A and Z = 18 + j 8 Ω . Determine the values of B and θ. (b) Suppose V = 48∠135° V , I = 3∠15° A and Z = R + j X Ω . Determine the values of R and X. Figure P2.4-12 Solution: (a) 12∠45° = (18 + j 8 )( B∠θ ) B∠θ = so 12∠45° 12∠45° = = 0.609∠21° 18 + j 8 19.7∠24° B=0.609 A and θ = 21° 48∠135° = ( R + j X )( 3∠15° ) (b) R+ j X = so 48∠75° = 18∠60° = 9 + j 15.6 Ω 3∠15° R = 9 Ω and X = 15.6 Ω Section 2-5 Independent Sources P 2.5-1 A current source and a voltage source are connected in parallel with a resistor as shown in Figure P 2.5-1. All of the elements connected in parallel have the same voltage, vs in this circuit. Suppose that vs = 15 V, is = 3 A, and R = 5 Ω. (a) Calculate the current i in the resistor and the power absorbed by the resistor. (b) Change the current source current to is = 5 A and recalculate the current, i, in the resistor and the power absorbed by the resistor. i vs + is R – Figure P 2.5-1 Answer: i = 3 A and the resistor absorbs 45 W both when is = 3 A and when is = 5 A. Solution: v s 15 2 = = 3 A and P = R i 2 = 5 ( 3 ) = 45 W R 5 (b) i and P do not depend on is . (a) i = P 2.5-2 A current source and a voltage source are connected in series with a resistor as shown in Figure P 2.5-2. All of the elements connected in series have the same current, is, in this circuit. Suppose that vs = 10 V, is = 2 A, and R = 5 Ω. (a) Calculate the voltage v across the resistor and the power absorbed by the resistor. (b) Change the voltage source voltage to vs = 5 V and recalculate the voltage, v, across the resistor and the power absorbed by the resistor. + The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A. – vs is R + v – Figure P 2.5-2 Answer: v = 10 V and the resistor absorbs 20 W both when vs = 10 V and when vs = 5 V. Solution: (a) From Ohm’s law v = R i s = 5 ( 3) = 15 V . (The resistor voltage does not depend on v 2 152 = = 45 W . 5 R (b) Since v and P do not depend on v s the values of v and P are 15 V and 45 W both when vs =10V and when vs =5V . the voltage source voltage.) Next P = P 2.5-3 The current source and voltage source in the circuit shown in Figure P 2.5-3 are connected in parallel so that they both have the same voltage, vs. The current source and voltage source are also connected in series so that they both have the same current, is. Suppose that vs = 12 V and is = 3 A. Calculate the power supplied by each source. Answer: The voltage source supplies –36 W, and the current source supplies 36 W. Solution: is is + vs – vs + – Figure P 2.5-3 Consider the current source: i s and v s do not adhere to the passive convention, so Pcs =i s v s =3⋅12 = 36 W is the power supplied by the current source. Consider the voltage source: i s and v s do adhere to the passive convention, so Pvs = i s vs =3 ⋅12 = 36 W is the power absorbed by the voltage source. ∴ The voltage source supplies −36 W. P 2.5-4 The current source and voltage source in the circuit shown in Figure P 2.5-4 are connected in parallel so that they both have the same voltage, vs. The current source and voltage source are also connected in series so that they both have the same current, is. Suppose that vs = 12 V and is = 3 A. Calculate the power supplied by each source. Answer: The voltage source supplies 36 W, and the current source supplies –36 W. Solution: Consider the current source. iS and vS adhere to the passive convention so PCS = iS vS = 2 (12 ) = 24 W is the received by the current source. The current source supplies −24 W. Consider the voltages source. iS and vS do not adhere to the passive convention so PCS = iS vS = 2 (12 ) = 24 W is the supplied by the voltage source. is + vs – vs + – is Figure P 2.5-4 P 2.5-5 (a) Find the power supplied by the voltage source shown in Figure P 2.5-5 when for t ≥ 0 we have i v v = 2 cos t V and (b) + – i =10 cos t mA Determine the energy supplied by this voltage source for the period 0 ≤ t ≤ 1 s. Figure P 2.5-5 Solution: (a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW ⎛1 1 ⎞ (b) w = ∫ P dt = ∫ 20 cos t dt = 20⎜ t + sin 2t ⎟ 0 0 ⎝2 4 ⎠ 1 1 1 = 10 + 5 sin 2 mJ 2 0 P 2.5-6 Figure P 2.5-6 shows a battery connected to a load. The load in Figure P 2.5-6 might represent automobile headlights, a digital camera, or a cell phone. The energy supplied by the battery to load is given by i + – v R t2 w = ∫ vi dt t1 When the battery voltage is constant and the load resistance is fixed, then the battery current will be constant and battery load Figure P 2.5-6 w = vi (t2 − t1 ) The capacity of a battery is the product of the battery current and time required to discharge the battery. Consequently, the energy stored in a battery is equal to the product of the battery voltage and the battery capacity. The capacity is usually given with the units of Ampere-hours (Ah). A new 12-V battery having a capacity of 800 mAh is connected to a load that draws a current of 50 mA. (a) How long will it take for the load to discharge the battery? (b) How much energy will be supplied to the load during the time required to discharge the battery? Solution: (a) time to discharge = capacity 800 mAh = = 32 hours current 25 mA (b) energy = (12 V) (0.025 A) (32 * 60*60 seconds) =34.56 kJ Section 2-6 Voltmeters and Ammeters P 2.6-1 (a) For the circuit of Figure P 2.6-1: + 5 . 0 – . 5 0 Voltmeter Ammeter What is the value of the resistance R? (b) How much power is delivered by the voltage source? R 12 V + – 1 2 A Figure P 2.6-1 Solution: v 5 (a) R = = = 10 Ω i 0.5 (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W. P 2.6-2 The current source in Figure P 2.6-2 supplies 40 W. What values do the meters in Figure P 2.6-2 read? Ammeter Voltmeter 4Ω + – + v – i 12 V 2A Figure P 2.6-2 Solution: The voltmeter current is zero so the ammeter current is equal to the current source current except for the reference direction: i = -2 A The voltage v is the voltage of the current source. The power supplied by the current source is 40 W so 40 = 2 v ⇒ v = 20 V P 2.6-3 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 2.6-3a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 2.6-3b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. Ideally, there is no current in the 100-Ω resistor and the voltmeter measures vmi = 12 V, the ideal value of vm. In Figure P 2.6-3c the voltmeter is modeled by the resistance Rm. Now the voltage measured by the voltmeter is ⎛ Rm ⎞ vm = ⎜ ⎟12 ⎝ Rm +100 ⎠ As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi = l2 V. When Rm < ∞, the voltmeter is not ideal and vm < vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a) Express the measurement error that occurs when Rm = 900 Ω as a percent of vmi. (b) Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi. 100 Ω 12 V 100 Ω 12 V + – + – Voltmeter + vmi = 12 V − + vm (b) 100 Ω − + (a) 12 V + – Rm vm − (c) Figure P 2.6-3 Solution: ⎛ 900 ⎞ vm = ⎜ ⎟12 = 10.8 V ⎝ 900 + 100 ⎠ (a) 12 − 10.8 = 0.1 = 10% 12 (b) We require ⎛ Rm ⎞ 12 − ⎜ 12 ⎜ R m + 100 ⎟⎟ ⎝ ⎠ 0.02 ≥ 12 ⇒ Rm R m + 100 ≥ 0.98 ⇒ R m ≥ 4900 Ω (checked: LNAP 6/16/04) 2.6-4 An ideal ammeter is modeled as a short circuit. A more realistic model of an ammeter is a small resistance. Figure P 2.6-4a shows a circuit with an ammeter that measures the current im. In Figure P 2.6-4b the ammeter is replaced by the model of an ideal ammeter, a short circuit. Ideally, there is no voltage across the 1-kΩ resistor and the ammeter measures imi = 2 A, the ideal value of im. In Figure P 2.6-4c the ammeter is modeled by the resistance Rm. Now the current measured by the ammeter is ⎛ 1000 ⎞ im = ⎜ ⎟2 ⎝ 1000 + Rm ⎠ As Rm → 0, the ammeter becomes an ideal ammeter and im → imi = 2 A. When Rm > 0, the ammeter is not ideal and im < imi. The difference between im and imi is a measurement error caused by the fact that the ammeter is not ideal. (a) Express the measurement error that occurs when Rm = 10 Ω as a percent of imi. (b) Determine the maximum value of Rm required to ensure that the measurement error is smaller than 5 percent. imi = 2 A im Ammeter 2A 2A 1 kΩ 1 kΩ (b) im (a) 2A Rm 1 kΩ (c) Figure P 2.6-4 Solution: (a) ⎛ 1000 ⎞ im = ⎜ ⎟ 2 = 1.98 A ⎝ 1000 + 10 ⎠ % error = 2 − 1.98 × 100 = 0.99% 2 (b) ⎛ 1000 ⎞ 2−⎜ 2 ⎜ 1000 + R m ⎟⎟ ⎝ ⎠ 0.05 ≥ 2 ⇒ 1000 ≥ 0.95 1000 + R m ⇒ R m ≤ 52.63 Ω (checked: LNAP 6/17/04) P 2.6-5 The voltmeter in Figure P 2.6-5a measures the voltage across the current source. Figure P 2.6-5b shows the circuit after removing the voltmeter and labeling the voltage measured by the voltmeter as vm. Also, the other element voltages and currents are labeled in Figure P 2.6-5b. Given that 25 Ω Voltmeter 12 V + – 2A (a) 12 = vR + vm and − iR = is = 2 A vR = 25 iR and (a) (b) 25 Ω iR + vR – Determine the value of the voltage measured by the meter. Determine the power supplied by each element. 12 V + – 2A is vm – (b) Figure P 2.6-5 Solution: a.) + v R = 25 i R = 25 ( −2 ) = −50 V v m = 12 − v R = 12 − ( −50 ) = 62 V b.) Element voltage source Power supplied current source 62 ( 2 ) = 124 W resistor −v R × i R = − ( −50 )( −2 ) = −100 W total 0 ( ) −12 i s = −12 ( 2 ) = −24 W P 2.6-6 The ammeter in Figure P 2.6-6a measures the current in the voltage source. Figure P 2.6-6b shows the circuit after removing the ammeter and labeling the current measured by the ammeter as im. Also, the other element voltages and currents are labeled in Figure P 2.6-6b. Given that Ammeter 12 V + – 25 Ω 2A 2 + im = iR and vR = vs =12 V vR = 25iR and (a) (b) (a) Determine the value of the current measured by the meter. Determine the power supplied by each element. + im 12 V + – 25 Ω iR vR – (b) Figure P 2.6-6 Solution: a.) iR = vR 25 = 12 = 0.48 A 25 i m = i R − 2 = 0.48 − 2 = −1.52 A b.) Element voltage source Power supplied current source v s ( 2 ) = 12 ( 2 ) =24 W resistor −v R × i R = − (12 )( 0.48 ) = −5.76 W total 0 ( ) 12 i m = 12 ( −1.52 ) = −18.24 W + 2A vs – Section 2-7 Dependent Sources P 2.7-1 The ammeter in the circuit shown in Figure P 2.7-1 indicates that ia = 2 A, and the voltmeter indicates that vb = 8 V. Determine the value of r, the gain of the CCVS. Answer: r = 4 V/A 2 . 0 0 Ammeter 8 . 0 0 Voltmeter R ia + + – vs r ia + – vb – Figure P 2.7-1 Solution: r = vb 8 = =4 Ω ia 2 1 P 2.7-2 The ammeter in the circuit shown in Figure P 2.7-2 indicates that ia = 2 A, and the voltmeter indicates that vb = 8 V. Determine the value of g, the gain of the VCCS. Answer: g = 0.25 A/V 2 . 0 0 Ammeter 8 . 0 0 Voltmeter ia R1 + vs g vb + – R2 vb – Figure P 2.7-2 Solution: vb = 8 V ; g v b = i a = 2 A ; g = P 2.7-3 ia 2 A = = 0.25 vb 8 V The ammeters in the circuit shown in Figure P 2.7-3 indicate that ia = 32 A and ib = 8 A. Determine the value of d, the gain of the CCCS. Answer: d = 4 A/A 3 2 . 0 Ammeter 8 . 0 0 Ammeter ia R d ib vs ib + – Figure P 2.7-3 Solution: i b = 8 A ; d i b = i a = 32A ; d = i a 32 A = =4 ib 8 A 2 The voltmeters in the circuit shown in Figure P 2.7-4 indicate that va = 2 V and vb = 8 P 2.7-4 V. Determine the value of b, the gain of the VCVS. Answer: b = 4 V/V 2 . 0 0 Ammeter 8 . 0 0 Voltmeter R ia + + – vs r ia + – vb – Figure P 2.7-4 Solution: va = 2 V ; b va = vb = 8 V ; b = P 2.7-5 vb 8 V = =4 va 2 V The values of the current and voltage of each circuit element are shown in Figure P 2.7-5. Determine the values of the resistance, R, and of the gain of the dependent source, A. −2V+ ia = −0.5 A A ia = 2 V −4V+ + −2 A 10 V R + – + 1.5 A 2.5 A 14 V − – 3.5 A + – 12 V −4 A Figure P 2.7-5 Solution: R=− 4 2 V = 2 Ω and A = = −4 −2 −0.5 A (checked: LNAP 6/6/04) 3 P 2.7-6 Find the power supplied by the VCCS in Figure P 2.7-6. Answer: 17.6 watts are supplied by the VCCS. (–17.6 watts are absorbed by the VCCS.) – 2. 0 0 Voltmeter + 2. 2 0 + Voltmeter – vc 2Ω + 0.2 Ω –15.8 V + – 6.9 Ω id = 4vc vd – Figure P 2.7-6 Solution: vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V id and vd adhere to the passive convention so P = vd id = (2.2) (−8) = −17.6 W is the power received by the dependent source. The power supplied by the dependent source is 17.6 W. 4 P2.7-7 The circuit shown in Figure P2.7-7 contains a dependent source. Determine the value of the gain k of that dependent source. Figure P2.7-7 Solution: v a = 10 ( 0.2 ) = 2 V k v a = − ( −10 ) = 10 V k= k va va = 10 V =5 2 V P2.7-8 The circuit shown in Figure P2.7-8 contains a dependent source. Determine the value of the gain k of that dependent source. Figure P2.7-8 Solution: ia = − 10 = −0.05 A = −50 mA 200 k i a = −450 mA k= k ia ia = −450 A =9 −50 A 5 P2.7-9 The circuit shown in Figure P2.7-9 contains a dependent source. The gain of that dependent source is k = 25 V A Determine the value of the voltage vb. Figure P2.7-9 Solution: −1 = 0.2 A 5 V k = 25 A ia = − v b = 25 ( 0.2 ) = 5 V P2.7-10 The circuit shown in Figure P2.7-10 contains a dependent source. The gain of that dependent source is k = 90 mA A = 0.09 V V Determine the value of the current ib. Figure P2.7-10 Solution: v a = 100 ( 0.05 ) = 5 V k = 90 mA A = 0.09 V V i b = − ( 0.09 )( 5 ) = −0.45 A = −450 mA 6 Section 2-8 Transducers P 2.8-1 For the potentiometer circuit of Figure 2.8-2, the current source current and potentiometer resistance are 1.1 mA and 100 kΩ, respectively. Calculate the required angle, θ, so that the measured voltage is 23 V. (1 – a)Rp Voltmeter + I – vm + vm aRp I Rp – (a) (b) Figure 2.8-2 Solution: a= θ 360 , θ = 360 v (360)(23 V) m= = 75.27° (100 kΩ)(1.1 mA) R i p P 2.8-2 An AD590 sensor has an associated constant k =1 . The sensor has a voltage v = 20 V; and the μA °K measured current, i(t), as shown in Figure 2.8-3, is i(t) + + v(t) AD590 – v(t) i(t) = kT – 4 μA < i < 13 μA in a laboratory setting. Find the range of measured temperature. (a) (b) Figure 2.8-3 Solution: AD590 : k =1 μA , K v =20 V (voltage condition satisfied) ° 4 μ A < i < 13 μ A ⎫ ⎪ ⎬ i T = ⎪⎭ k ⇒ 4 ° K< T <13° K Section 2-9 Switches t=2s P 2.9-1 Determine the current, i, at t = 1 s and at t = 4 s for the circuit of Figure P 2.9-1. t=3s 15 V + – 5 kΩ + – 10 V i Figure P 2.9-1 Solution: At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. i= v 10 = = 2 mA R 5×103 At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V. i= v 15 = = 3 mA R 5×103 t=3s P 2.9-2 Determine the voltage, v, at t = 1 s and at t = 4 s for the circuit shown in Figure P 2.9-2. + 1 mA 5 kΩ v – Figure P 2.9-2 Solution: At t = 1 s the current in the resistor is 3 mA so v = 15 V. At t = 4 s the current in the resistor is 0 A so v = 0 V. t=2s 2 mA Rs 12 V + + − 100 Ω v 12 V − (a) + − + 100 Ω v − (b) Figure P 2.9-3 P 2.9-3 Ideally an open switch is modeled as an open circuit and a closed switch is modeled as a closed circuit. More realistically, an open switch is modeled as a large resistance and a closed switch is modeled as a small resistance. Figure P 2.9-3a shows a circuit with a switch. In figure P 2.9-3b the switch has been replaced with a resistance. In figure P 2.9-3b the voltage v is given by ⎛ 100 ⎞ v =⎜ ⎟12 R + 100 s ⎝ ⎠ Determine the value of v for each of the following cases. (a) The switch is closed and Rs = 0 (a short circuit). (b) The switch is closed and Rs = 5 Ω. (c) The switch is open and Rs = ∞ (an open circuit). (d) The switch is open and Rs = 10 kΩ. Solution: (a) v = 12 V ⎛ 100 ⎞ (b) v = ⎜ ⎟12 = 11.43 V ⎝ 105 ⎠ (c) v = 0 V ⎛ 100 ⎞ (d) v = ⎜ ⎟12 = 0.1188 0.12 V ⎝ 10100 ⎠ Section 2-10 How Can We Check…? P 2.10-1 The circuit shown in Figure P 2.10-1 is used to test the – 2 . 0 4 0 . 0 Ammeter Voltmeter CCVS. Your lab partner claims that this measurement shows that the gain of the CCVS is –20 V/A R + – vs instead of +20 V/A. Do you agree? Justify your answer. is vo is CCVS = 20 V A + vo – Figure P 2.10-1 Solution: vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .) So vo 40 V = = 20 is 2 A Your lab partner is wrong. P 2.10-2 The circuit of Figure P 2.10-2 is used 0 . 0 0 to measure the current in the resistor. Once this Ammeter current is known, the resistance can be calculated i v R as R = s . The circuit is constructed using a i voltage source with vs = 12 V and a 25-Ω, 1/2-W + v s – resistor. After a puff of smoke and an unpleasant smell, the ammeter indicates that i = 0 A. The resistor must be bad. You have more 25-Ω, 1/2-W resistors. Should you try another resistor? Justify Figure P 2.10-2 your answer. Hint: 1/2-W resistors are able to safely dissipate one 1/2 W of power. These resistors may fail if required to dissipate more than 1/2 watt of power. Solution: vs 12 = = 0.48 A. The power absorbed by R 25 this resistor will be P = i vs = (0.48) (12) = 5.76 W. We expect the resistor current to be i = A half watt resistor can't absorb this much power. You should not try another resistor. Design Problems DP 2-1 Specify the resistance R in Figure DP 2-1 so that both of the following conditions are satisfied: 1. i > 40 mA. 2. The power absorbed by the resistor is less than 0.5 W. i + R 10 V – Figure DP 2-1 Solution: 1.) 10 10 > 0.04 ⇒ R < = 250 Ω R 0.04 2.) 102 1 < ⇒ R > 200 Ω R 2 Therefore 200 < R < 250 Ω. For example, R = 225 Ω. DP 2-2 Specify the resistance R in Figure DP 2-2 so that both of the following conditions are satisfied: 1. v > 40 V. 2. The power absorbed by the resistor is less than 15 W. Hint: There is no guarantee that specifications can always be satisfied. 2A R Figure DP 2-2 Solution: 1.) 2 R > 40 ⇒ R > 20 Ω 15 2.) 2 2 R < 15 ⇒ R < = 3.75 Ω 4 Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously. + v – R1 R2 R3 DP 2-3 Resistors are given a power rating. For example, resistors are available with ratings of 1/8 W, 1/4 W, 1/2 W, and 1 W. A ir = is 1/2-W resistor is able to safely dissipate 1/2 W of power, indefinitely. Resistors with larger power ratings are more expensive and bulkier than resistors with lower power ratings. Good is engineering practice requires that resistor Figure DP 2-3 power ratings be specified to be as large as, but not larger than, necessary. Consider the circuit shown in Figure DP 2-3. The values of the resistances are R1 = 1000 Ω, R2 = 2000 Ω, and R3 = 4000 Ω The value of the current source current is is = 30 mA Specify the power rating for each resistor. Solution:: P1 = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W 2 2 P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W 2 2 P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W 2 2 Chapter 3 Resistive Circuits Exercises Figure E 3.2-1 Exercise 3.2-1 Determine the values of i3, i4, i6, v2, v4, and v6 in Figure E 3.2-1. Answer: i3 = –3 A, i4 = 3 A, i6 = 4 A, v2 = –3 V, v4 = –6 V, v6 = 6 V Solution: Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 A Apply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 A Apply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A Apply KVL to the loop consisting of elements A and B to get -v2 – 3 = 0 ⇒ v2 = -3 V Apply KVL to the loop consisting of elements C, E, D, and A to get 3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V Apply KVL to the loop consisting of elements E and F to get v6 – 6 = 0 ⇒ v6 = 6 V Check: The sum of the power supplied by all branches is -(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0 Exercise 3.3-1 Determine the voltage measured by the voltmeter in the circuit shown in Figure E 3.3-1a. Hint: Figure E3.3-1b shows the circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter, vm. Answer : vm = 2 V Figure E 3.3-1 Solution: 25 From voltage division ⇒ v = (8) = 2 V m 25+75 Exercise 3.3-2 Determine the voltage measured by the voltmeter in the circuit shown in Figure E 3.32a. Hint: Figure E 3.3-2b shows the circuit after the ideal voltmeter has been replaced by the equivalent open circuit and a label has been added to indicate the voltage measured by the voltmeter, vm. Answer: vm = –2 V Figure E 3.3-2 Solution: 25 From voltage division ⇒ v = ( −8 ) = −2 V m 25+75 Exercise 3.4-1 A resistor network consisting of parallel resistors is shown in a package used for printed circuit board electronics in Figure E 3.4-1a. This package is only 2 cm × 0.7 cm, and each resistor is 1 kΩ. The circuit is connected to use four resistors as shown in Figure E 3.4-1b. Find the equivalent circuit for this network. Determine the current in each resistor when is = 1 mA. Answer: Rp = 250 Ω Figure E 3.4-1 Solution: 1 R = eq 1 1 1 1 4 + 3+ 3+ 3= 3 3 10 10 10 10 10 ⇒ R eq = By current division, the current in each resistor = 103 1 = kΩ 4 4 1 -3 1 (10 ) = mA 4 4 Exercise 3.4-2 Determine the current measured by the ammeter in the circuit shown in Figure E 3.4-2a. Hint: Figure E 3.4-2b shows the circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter, im. Answer: im = –1 A Solution: From current division 10 i = ( −5 ) = −1 A m 10+40 Figure E 3.4-2 Exercise 3.6-1 Determine the resistance measured by the ohmmeter in Figure E 3.6-1. Figure E 3.6-1 Answer: (30 + 30) ⋅ 30 + 30 = 50 Ω (30 + 30) + 30 Section 3-2 Kirchhoff’s Laws P 3.2-1 Consider the circuit shown in Figure P 3.2-1. Determine the values of the power supplied by branch B and the power supplied by branch F. Figure P 3.2-1 Solution: Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is power received by element B. The power supplied by element B is 12 W. Apply KVL to the loop consisting of elements D, F, E, and C to get 4 + v + (-5) – 12 = 0 ⇒ v = 13 V The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W is the power supplied by element F. Check: The sum of the power supplied by all branches is -(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0 Determine the values of i2, i4, v2, v3, and v6 in Figure P 3.2-2. P 3.2-2 Solution: Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = −4 A Apply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 A Apply KVL to the loop consisting of elements A and B to get -v2 – 6 = 0 ⇒ v2 = -6 V Apply KVL to the loop consisting of elements C, D, and A to get -v3 – (-2) – 6 = 0 ⇒ v4 = -4 V Apply KVL to the loop consisting of elements E, F and D to get 4 – v6 + (-2) = 0 ⇒ v6 = 2 V Check: The sum of the power supplied by all branches is −(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0 P 3.2-3 Consider the circuit shown in Figure P 3.2-3. (a) Suppose that R1 = 6 Ω and R2 = 3 Ω. Find the current i and the voltage v. (b) Suppose, instead, that i = 1.5 A and v = 2 V. Determine the resistances R1 and R2. (c) Suppose, instead, that the voltage source supplies 24 W of power and that the current source supplies 9 W of power. Determine the current i, the voltage v, and the resistances R1 and R2. Figure P 3.2-3 Solution: KVL : −12 − R 2 (3) + v = 0 (outside loop) v = 12 + 3R 2 or R 2 = KCL i+ 12 − 3 = 0 (top node) R1 i = 3− (a) (b) v = 12 + 3 ( 4 ) = 24 V and i = 3 − R2 = v − 12 3 12 12 or R1 = R1 3−i 12 = 1.5 A 8 42 − 12 12 = 10 Ω ; R1 = = 16 Ω 3 3 − 2.25 (checked using LNAP 7/27/08) (c) 24 = − 12 i, because 12 and i adhere to the passive convention. 12 = 2.4 Ω 3+ 2 because 3 and v do not adhere to the passive convention ∴ i = − 2 A and R1 = 9 = 3v, 3 − 12 = −3 Ω 3 The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative. ∴ v = 3V and R 2 = P 3.2-4 Determine the power absorbed by each of the resistors in the circuit shown in Figure P 3.2-4. Figure P 3.2-4 Solution: Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W 2 Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W 1 Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W 4 12 i = =2A 1 6 20 i = = 5A 2 4 i = 3−i = − 2 A 3 2 i =i +i =3A 4 2 3 (checked using LNAP 8/16/02) P 3.2-5 Determine the power absorbed by each of the resistors in the circuit shown in Figure P 3.2-5. Answer: The 4-Ω resistor absorbs 16 W, the 6-Ω resistor absorbs 24 W, and the 8-Ω resistor absorbs 8 W. Figure P 3.2-5 Solution: v1 = 8 V v2 = −8 + 8 + 12 = 12 V (checked using LNAP 8/16/02) v3 = 2⋅ 4 = 8 V v2 4Ω : P = 3 = 16 W 4 v2 6Ω : P = 2 = 24 W 6 2 v 8Ω : P = 1 = 8 W 8 P 3.2-6 Determine the power supplied by each voltage source in the circuit of Figure P 3.2-6. Answer: The 2-V voltage source supplies 2 mW and the 3-V voltage source supplies –6 mW. Figure P 3.2-6 Solution: P2 V = + ⎡⎣ 2 × (1× 10−3 ) ⎤⎦ = 2 × 10−3 = 2 mW P3 V = + ⎡⎣3 × ( −2 × 10−3 ) ⎤⎦ = −6 × 10−3 = −6 mW (checked using LNAP 8/16/02) P 3.2-7 What is the value of the resistance R in Figure P 3.2-7? Hint: Assume an ideal ammeter. An ideal ammeter is equivalent to a short circuit. Answer: R = 4 Ω Figure P 3.2-7 Solution: KCL: iR = 2 + 1 ⇒ iR = 3 A KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V ∴ R= vR 12 = =4Ω iR 3 (checked using LNAP 8/16/02) P 3.2-8 The voltmeter in Figure P 3.2-8 measures the value of the voltage across the current source to be 56 V. What is the value of the resistance R? Hint: Assume an ideal voltmeter. An ideal voltmeter is equivalent to an open circuit. Answer: R = 10 Ω Figure P 3.2-8 Solution: KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V KCL: iR + 8 = 0 ⇒ iR = −8 A ∴ R= vR −80 = = 10 Ω iR −8 (checked using LNAP 8/16/02) P 3.2-9 Determine the values of the resistances R1 and R2 in Figure P 3.2-9. Figure P 3.2-9 Solution: KCL at node b: 5.61 3.71 − 5.61 12 − 5.61 −1.9 = + ⇒ 0.801 = + 1.278 7 R1 5 R1 ⇒ R1 = KCL at node a: 1.9 = 3.983 ≈ 4 Ω 1.278 − 0.801 3.71 3.71 − 5.61 3.71 − 12 −8.29 + + = 0 ⇒ 1.855 + ( −0.475 ) + =0 2 4 R2 R2 ⇒ R2 = 8.29 = 6.007 ≈ 6 Ω 1.855 − 0.475 (checked using LNAP 8/16/02) P 3.2-10 The circuit shown in Figure P 3.2-10 consists of five voltage sources and four current sources. Express the power supplied by each source in terms of the voltage source voltages and the current source currents. Figure P 3.2-10 Solution: The subscripts suggest a numbering of the sources. Apply KVL to get v1 = v 2 + v 5 + v 9 − v 6 i1 and v1 do not adhere to the passive convention, so p 1 = i1 v 1 = i1 ( v 2 + v 5 + v 9 − v 6 ) is the power supplied by source 1. Next, apply KCL to get i 2 = − ( i1 + i 4 ) i 2 and v 2 do not adhere to the passive convention, so p 2 = i 2 v 2 = − ( i1 + i 4 ) v 2 is the power supplied by source 2. Next, apply KVL to get v3 = v6 − (v5 + v9 ) i 3 and v 3 adhere to the passive convention, so ( p 3 = −i 3 v 3 = −i 3 v 6 − ( v 5 + v 9 ) is the power supplied by source 3. Next, apply KVL to get v 4 = v 2 + v5 + v8 i 4 and v 4 do not adhere to the passive convention, so p 4 = i4 v 4 = i4 (v 2 + v5 + v8 ) is the power supplied by source 4. Next, apply KCL to get ) ( ) i 5 = i 3 − i 2 = i 3 − − ( i1 + i 4 ) = i1 + i 3 + i 4 i 5 and v 5 adhere to the passive convention, so p 5 = −i 5 v 5 = − ( i1 + i 3 + i 4 ) v 5 is the power supplied by source 5. Next, apply KCL to get i 6 = i 7 − ( i1 + i 3 ) i 6 and v 6 adhere to the passive convention, so ( ) p 6 = −i 6 v 6 = − i 7 − ( i1 + i 3 ) v 6 is the power supplied by source 6. Next, apply KVL to get v 7 = −v 6 i 7 and v 7 adhere to the passive convention, so p 7 = −i 7 v 7 = −i 7 ( −v 6 ) = i 7 v 6 is the power supplied by source 7. Next, apply KCL to get i 8 = −i 4 i 8 and v 8 do not adhere to the passive convention, so p 8 = i 8 v 8 = ( −i 4 ) v 8 = −i 4 v 8 is the power supplied by source 8. Finally, apply KCL to get i 9 = i1 + i 3 i 9 and v 9 adhere to the passive convention, so p 9 = −i 9 v 9 = − ( i1 + i 3 ) v 9 is the power supplied by source 9. (Check: 9 ∑p n =1 n = 0 .) P 3.2-11 Determine the power received by each of the resistors in the circuit shown in Figure P 3.2-11. Figure P 3.2-11 Solution The subscripts suggest a numbering of the circuit elements. Apply KCL to get i 2 + 0.2 + 0.3 = 0 ⇒ i 2 = −0.5 A The power received by the 6 Ω resistor is p 2 = 6 i 2 2 = 6 ( −0.5 ) = 1.5 W 2 Next, apply KCL to get i 5 = 0.2 + 0.3 + 0.5 = 1.0 A The power received by the 8 Ω resistor is p 5 = 8 i 5 2 = 8 (1) = 8 W 2 Next, apply KVL to get v 7 = 15 V The power received by the 20 Ω resistor is p7 = v 72 20 = 15 2 = 11.25 W 20 is the power supplied by source 7. Finally, apply KCL to get i 9 = 0.2 + 0.5 = 0.7 A The power received by the 5 Ω resistor is p 9 = 5 i 9 2 = 5 ( 0.7 ) = 2.45 W 2 P 3.2-12 Determine the voltage and current of each of the circuit elements in the circuit shown in Figure P 3.2-12. Figure P 3.2-12 Solution: We can label the circuit as follows: The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get i 4 + 0.25 + 0.75 = 0 ⇒ i 4 = −1.0 A Next, apply KCL at node d to get i 3 = i 4 + 0.25 = −1.0 + 0.25 = −0.75 A Next, apply KVL to the loop consisting of the voltage source and the 60 Ω resistor to get v 2 − 15 = 0 ⇒ v 2 = 15 V Apply Ohm’s law to each of the resistors to get i2 = and v2 60 = 15 = 0.25 A , 60 v 3 = 10 i 3 = 10 ( −0.75 ) = −7.5 V v 4 = 20 i 4 = 20 ( −1) = −20 V Next, apply KCL at node c to get i1 + i 2 = i 3 ⇒ i1 = i 3 − i 2 = −0.75 − 0.25 = −1.0 A Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get v 6 − v 4 − v 3 − v 2 = 0 ⇒ v 6 = v 4 + v 3 + v 2 = −20 + (−7.5) + 15 = −12.5 V Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 Ω resistor to get v 5 + v 4 = 0 ⇒ v 5 = −v 4 = − ( −20 ) = 20 V (Checked: LNAPDC 8/28/04) P 3.2-13 Determine the value of the current that is measured by the meter in Figure P 3.2-13. Figure P 3.2-13 Solution: We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Apply KVL to node the left mesh to get 15 i1 + 25 i1 − 20 = 0 ⇒ i1 = 20 = 0.5 A 40 Apply KVL to node the left mesh to get v 2 − 25 i1 = 0 ⇒ v 2 = 25 i1 = 25 ( 0.5 ) = 12.5 V Apply KCL to get i m = i 2 . Finally, apply Ohm’s law to the 50 Ω resistor to get im = i2 = v2 50 = 12.5 = 0.25 A 50 (Checked: LNAPDC 9/1/04) P 3.2-14 Determine the value of the voltage that is measured by the meter in Figure P 3.2-14. Figure P 3.2-14 Solution: We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Ohm’s law to the 48 Ω resistor to get v1 = 48 i1 Apply KCL at the top node of the CCCS to get i1 + 5 i1 = i 2 ⇒ i 2 = 6 i1 Ohm’s law to the 4 Ω resistor to get v m = 4 i 2 = 4 ( 6 i1 ) = 24 i1 Apply KVL to the outside loop to get v1 + v m − 24 = 0 ⇒ 48 i1 + 24 i1 = 24 ⇒ i1 = Finally, 24 1 = A 72 3 ⎛1⎞ v m = 24 i1 = 24 ⎜ ⎟ = 8 V ⎝ 3⎠ (Checked: LNAPDC 9/1/04) P 3.2-15 Determine the value of the voltage that is measured by the meter in Figure P 3.2-15. Figure P 3.2-15 Solution: We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Apply KCL at the top node of the current source to get i1 = i 2 + 0.25 Apply Ohm’s law to the resistors to get v1 = 20 i1 and v 2 = 60 i 2 = 60 ( i1 − 0.25 ) = 60 i1 − 15 Apply KVL to the outside to get v 2 + 80 i1 + v1 = 0 ⇒ Finally, ( 60 i 1 − 15 ) + 80 i1 + 20 i1 = 0 ⇒ i1 = 15 = 0.09375 A 160 v m = 80 i1 = 80 ( 0.09375 ) = 7.5 V (Checked: LNAPDC 9/1/04) P 3.2-16 The voltage source in Figure P 3.2-16 supplies 4.8 W of power. The current source supplies 3.6 W. Determine the values of the resistances, R1 and R2. Figure P 3.2-16 Solution: i= R1 = 3.6 4.8 = 0.3 A and v = = 9.6 V 12 0.5 12 − 9.6 9.6 = 8 Ω and R 2 = = 12 Ω 0.3 0.3 + 0.5 (Checked: LNAPDC 7/27/08) P 3.2-17 Answer: Determine the current i in Figure P 3.3-17. i=4A Figure P 3.3-17 Solution: Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A P 3.2-18 Determine the value of the current im in Figure P 3.2-18a. Hint: Apply KVL to the closed path a-b-d-c-a in Figure P 3.2-18b to determine va. Then apply KCL at node b to find im. Answer: im = 9 A. Figure P 3.2-18 Solution: −18 + 0 − 12 − va = 0 ⇒ va = −30 V and im = 2 va + 3 ⇒ im = 9 A 5 P3.2-19 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-19. Figure P3.2-23 Solution: Apply KCL at the left node: 0.25 + i 2 + 0.22 = 0 ⇒ i 2 = −0.47 A Use the element equation of the dependent source: v 5 = 10 i 2 = 10 ( −0.47 ) = −4.7 V Apply KVL to the right mesh v 5 − v 6 − 12 = 0 ⇒ v 6 = v 5 − 12 = −4.7 − 12 = −16.7 V P3.2-20 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-20. Figure P3.2-20 Solution: Apply KCL at the left node: 0.015 + 0.025 = i 2 ⇒ i 2 = 0.040 A From Ohm’s law v 2 = 25 i 2 = 25(0.04) = 1 V Use the element equation of the dependent source: v 5 = 10 v 2 = 10 (1) = 10 V Apply KVL to the right mesh v 5 + v 6 + 12 = 0 ⇒ v 6 = −v 5 − 12 = −10 − 12 = −22 V P3.2-21 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-21. Figure P3.2-21 Solution: Apply KVL to the left mesh: v 2 − 18 + 12 = 0 ⇒ v 2 = 6 V Use the element equation of the dependent source: i 6 = 0.10 v 2 = 0.10 ( 6 ) = 0.6 A Apply KCL at the right node v5 25 + i 6 = 0.25 ⇒ v 5 = 25 ( 0.25 − i 6 ) = 25 ( 0.25 − 0.6 ) = −8.75 V P3.2-22 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-22. Figure P3.2-22 Solution: Apply KVL to the left mesh: 48 i 2 − 18 + 12 = 0 ⇒ i 2 = 0.125 A Use the element equation of the dependent source: i 6 = 1.5 i 2 = 1.5 ( 0.125 ) = 0.1875 A Apply KCL at the right node v5 40 + i 6 = 0.25 ⇒ v 5 = 40 ( 0.25 − i 6 ) = 40 ( 0.25 − 0.1875 ) = 2.5 V P3.2-23 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-23. Figure P3.2-23 Solution: Apply KVL to the left mesh: 50 i 2 + 18 + 12 = 0 ⇒ i 2 = −30 = −0.6 A 50 Use the element equation of the dependent source: v 5 = 10 i 2 = 10 ( −0.6 ) = −6 V Apply KVL to the right mesh v 5 + v 6 + 12 = 0 ⇒ v 6 = −v 5 − 12 = − ( −6 ) − 12 = −6 V P3.2-24 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-24. Figure P3.2-24 Solution: Apply KCL at the left node: 0.015 = 0.025 + v2 250 ⇒ v 2 = 250 ( −0.01) = −2.5 V Use the element equation of the dependent source: i 6 = 0.5 v 2 = 0.5 ( −2.5 ) = −1.25 A Apply KCL at the right node v5 25 + i 6 = 0.25 ⇒ v 5 = 25 ( 0.25 − i 6 ) = 25 ( 0.25 − ( −1.25 ) ) = 37.5 V P3.2-25 The voltage source in the circuit shown in Figure P3.2-25 supplies 2 W of power. The value of the voltage across the 25 Ω is v5 = 4 V. Determine the values of the resistance R1 and of the gain, G, of the CCVS. Figure P3.2-25 Solution: The voltage source current is calculated from the values of the source voltage and power: 2 is = = 0.1 A 20 Apply KCL at the bottom node to get v2 4 is + G v2 = ⇒ G v2 = − 0.1 = 0.06 A 25 25 Then G v 2 0.06 G= = = 0.015 A/V = 15 mA/V v2 4 Next, use Ohm’s law to determine the value of the resistance R1: R1 = 20 − v 2 20 − 4 = = 100 Ω 4 v2 25 25 P3.2-26 Consider the circuit shown in Figure P3.2-26. Determine the values of (a) The current i a in the 20-Ω resistor. (b) The voltage v b across the 10-Ω resistor. (c) The current i c in the independent voltage source. Figure P3.2-26 Solution: 25 = −1.25 A . 20 (b) From KVL 4 i a − v b + 20 i a = 0 ⇒ v b = 24 i a = 24 ( −1.25 ) = −30 V (a) From Ohm’s law i a = − (c) From KCL i c = i a + vb 10 = −1.25 + −30 = −4.25 A 10 P3.2-27 Consider the circuit shown in Figure 3.2-27. a.) Determine the values of the resistances. b.) Determine the values of the power supplied by each current source. c.) Determine the values of the power received by each resistor. Figure 3.2-27 Solution: Using KCL and Ohm’s law: va = Ra ( I 2 − I1 ) , vb = − Rb ( I1 + I 3 ) and vc = Rc ( I 2 + I 3 ) Using KVL, the power supplied the current sources are: I 2 ( va + vc ) , − I1 ( va + vb ) and I 3 ( −vb + vc ) The power received the resistors are: va ( I 2 − I1 ) , −vb ( I1 + I 3 ) and vc ( I 2 + I 3 ) a.) and b.) and Ra = 4 −3 = 4 Ω , Rb = − =6Ω 3− 2 2 + ( −1.5 ) Rc = 12 =8 Ω 3 + ( −1.5 ) 3 ( 4 + 12 ) = 48 W , −2 ( 4 + ( −3) ) = −2 W −1.5 ( − ( −3) + 12 ) = −22.5 W c.) 4 ( 3 − 2 ) = 4 W , − ( −3) ( 2 + ( −1.5 ) ) = 1.5 W and 12 ( 3 + ( −1.5 ) ) = 18 W LNAPDC (8/30/10): P3.2-28 Consider the circuit shown in Figure 3.2-28. a. Determine the value of the power supplied by each independent source. b. Determine the value of the power received by each resistor. c. Is power conserved? Figure 3.2-28 Solution: Apply KCL twice and KVL to get Apply Ohm’s law 3 times to get Apply KCL twice and KVL to get a. Notice that the voltage and current reference directions for each independent source do not adhere to passive convention. Consequently the power supplied by the 2 mA current source is 15 (2) = 30 mW The power supplied by the 3 mA current source is 9 (3) = 27 mW The power supplied by the 12 V voltage source is 12 (11/3) = 44 mW b. Notice that the voltage and current reference directions for each resistor do adhere to passive convention. Consequently the power received by the 18 kΩ resistor is 12 (2/3) = 8 mW The power received by the 3 kΩ resistor is 15 (5) = 75 mW The power received by the 2 kΩ resistor is 6 (3) = 18 mW c. Power is conserved; the sum of the powers supplied by the independent sources is equal to the sum of the powers received by the resistors. 30 + 27 + 44 = 101 mW = 8 +75 +18 P3.2-29 P3.2-29 The voltage across the capacitor in Figure P3.2-29 is v ( t ) = 24 − 10 e −25 t V for t ≥ 0. Determine the voltage source current i(t) for t > 0. Solution: Notice that i(t) is the current in the 20 Ω resistor. Apply KVL to the left mesh to get −20 i ( t ) + ⎡⎣ 24 − 10 e ⎡ 24 − 10 e −25 t ⎤ − 30 ⎣ ⎦ = −0.3 − 0.5 e −25 t for t > 0. ⎦ − 30 = 0 ⇒ i ( t ) = 20 − 25 t ⎤ P3.2-30 P3.2-30 The current the inductor in Figure P3.2-30 is given by i ( t ) = 8 − 6 e −25 t A for t ≥ 0. Determine the voltage v(t) across the 80 Ω resistor for t > 0. Solution: Notice that i(t) is the current in the 20 Ω resistor. Apply KCL at the top node of the 80 Ω resistor to get v (t ) = 10 − 8 − 6 e −25 t 80 ( ) ( ) ⇒ v ( t ) = 80 ⎡10 − 8 − 6 e −25 t ⎤ = 160 + 480 e −25 t for t ≥ 0 ⎣ ⎦ Section 3-3 Series Resistors and Voltage Division P 3.3-1 Use voltage division to determine the voltages v1, v2, v3, and v4 in the circuit shown in Figure P 3.3-1. Figure P 3.3-1. Solution: 6 6 v = 12 = 12 = 4 V 1 6+3+5+ 4 18 3 5 10 v = 12 = 2 V ; v = 12 = V 2 18 3 18 3 4 8 v = 12 = V 4 18 3 (checked using LNAP 8/16/02) P 3.3-2 Consider the circuits shown in Figure P 3.3-2. (a) Determine the value of the resistance R in Figure P 3.3-2b that makes the circuit in Figure P 3.32b equivalent to the circuit in Figure P 3.3-2a. (b) Determine the current i in Figure P 3.3-2b. Because the circuits are equivalent, the current i in Figure P 3.3-2a is equal to the current i in Figure P 3.3-2b. (c) Determine the power supplied by the voltage source. Solution: (a ) R = 6 + 3 + 2 + 4 = 15 Ω 28 28 (b) i = = = 1.867 A R 15 ( c ) p = 28 ⋅ i =28(1.867)=52.27 W (28 V and i do not adhere to the passive convention.) (checked using LNAP 8/16/02) P 3.3-3 The ideal voltmeter in the circuit shown in Figure P 3.3-3 measures the voltage v. (a) Suppose R2 = 100 Ω. Determine the value of R1. (b) Suppose, instead, R1 = 100 Ω. Determine the value of R2. (c) Suppose, instead, that the voltage source supplies 1.2 W of power. Determine the values of both R1 and R2. Figure P 3.3-3 Solution: i R2 = v = 8 V 12 = i R1 + v = i R1 + 8 ⇒ 4 = i R1 8 8 4 4 ⋅ 50 ; R1 = = = = 25 Ω R 2 50 i 8 4 4 8 8 ⋅ 50 = 100 Ω (b) i = = ; R2 = = 4 R1 50 i 4 8 ( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω i i (a) i= (Checked using LNAP 8/16/02) P 3.3-4 Determine the voltage v in the circuit shown in Figure P 3.3-4. Figure P 3.3-4 Solution: Voltage division 16 v1 = 12 = 8 V 16 + 8 4 v3 = 12 = 4 V 4+8 KVL: v3 − v − v1 = 0 v = −4 V (checked using LNAP 8/16/02) P 3.3-5 The model of a cable and load resistor connected to a source is shown in Figure P 3.3-5. Determine the appropriate cable resistance, R, so that the output voltage, vo, remains between 9 V and 13 V when the source voltage, vs, varies between 20 V and 28 V. The cable resistance can only assume integer values in the range 20 < R < 100 Ω. Figure P 3.3-5 Solution: ⎛v ⎞ ⇒ R = 50 ⎜ s − 1⎟ ⎜v ⎟ ⎝ o ⎠ with v = 20 V and v > 9 V, R < 61.1 Ω ⎫ ⎪ s 0 ⎬ R = 60 Ω with v = 28 V and v < 13 V, R > 57.7 Ω ⎪ s 0 ⎭ ⎛ 100 ⎞ using voltage divider: v = ⎜ v 0 ⎝ 100 + 2 R ⎟⎠ s P 3.3-6 The input to the circuit shown in Figure P 3.3-6 is the voltage of the voltage source, va. The output of this circuit is the voltage measured by the voltmeter, vb. This circuit produces an output that is proportional to the input, that is vb = k va where k is the constant of proportionality. (a) Determine the value of the output, vb, when R = 240 Ω and va = 18 V. (b) Determine the value of the power supplied by the voltage source when R = 240 Ω and va = 18 V. (c) Determine the value of the resistance, R, required to cause the output to be vb = 2 V when the input is va = 18 V. (d) Determine the value of the resistance, R, required to cause vb = 0.2va (that is, the value of the constant of proportionality is k = 102 ). Figure P 3.3-6 Solution: ⎛ 180 ⎞ a.) ⎜ ⎟ 18 = 10.8 V ⎝ 120 + 180 ⎠ 18 ⎛ ⎞ b.) 18 ⎜ ⎟ = 1.08 W ⎝ 120 + 180 ⎠ ⎛ R ⎞ c.) ⎜ ⎟ 18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω ⎝ R + 120 ⎠ R ⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω d.) 0.2 = R + 120 (Checked using LNAP 8/16/02) P 3.3-7 Determine the value of voltage v in the circuit shown in Figure P 3.3-7. Figure P 3.3-7 Solution: All of the elements are connected in series. Replace the series voltage sources with a single equivalent voltage having voltage 12 + 12 – 18 = 6 V. Replace the series 15 Ω, 5 Ω and 20 Ω resistors by a single equivalent resistance of 15 + 5 + 20 = 40 Ω. By voltage division 6 ⎛ 10 ⎞ v=⎜ ⎟ 6 = = 1.2 V 5 ⎝ 10 + 40 ⎠ (Checked: LNAP 2/6/07) P 3.3-8 Determine the power supplied by the dependent source in the circuit shown in Figure P 3.3-8. Figure P 3.3-8 Solution: Use voltage division to get ⎛ 10 ⎞ va = ⎜ ⎟ (120 ) = 20 V ⎝ 10 + 50 ⎠ Then i a = 0.2 ( 20 ) = 4 A The power supplied by the dependent source is given by p = (120 ) i a = 480 W (Checked: LNAP 6/21/04) P 3.3-9 A potentiometer can be used as a transducer to convert the rotational position of a dial to an electrical quantity. Figure P 3.3-9 illustrates this situation. Figure P 3.3-9a shows a potentiometer having resistance Rp connected to a voltage source. The potentiometer has three terminals, one at each end and one connected to a sliding contact called a wiper. A voltmeter measures the voltage between the wiper and one end of the potentiometer. Figure P 3.3-9b shows the circuit after the potentiometer is replaced by a model of the potentiometer that consists of two resistors. The parameter a depends on the angle, θ, of the dial. Here θ a = 360 ° , and θ is given in degrees. Also, in Figure P 3.39b, the voltmeter has been replaced by an open circuit and the voltage measured by the voltmeter, vm, has been labeled. The input to the circuit is the angle θ, and the output is the voltage measured by the meter, vm. (a) Show that the output is proportional to the input. (b) Let Rp = 1 kΩ and vs = 24 V. Express the output as a function of the input. What is the value of the output when θ = 45º? What is the angle when vm = 10 V? Figure P 3.3-9 Solution: (a) Use voltage division to get vm = Therefore aR p (1 − a ) R p + R p v s = av s ⎛ vs ⎞ vm = ⎜ ⎟θ ⎝ 360 ⎠ So the input is proportional to the input. ⎛1⎞ (b) When vs = 24 V then v m = ⎜ ⎟ θ . When θ = 45o then vm = 3 V. When vm = 10 V then θ = 150o . ⎝ 15 ⎠ (Checked: LNAP 6/12/04) P 3.3-10 Determine the value of the voltage measured by the meter in Figure P 3.3-10. Figure P 3.3-10 Solution: Replace the (ideal) voltmeter with the equivalent open circuit. Label the voltage measured by the meter. Label some other element voltages and currents. Apply KVL the left mesh to get 8 i a + 4 i a − 24 = 0 ⇒ i a = 2 A Use voltage division to get vm = 5 5 4 ia = 4 ( 2) = 5 V 5+3 5+3 (Checked using LNAP 9/11/04) P 3.3-11 For the circuit of Figure P 3.3-11, find the voltage v3 and the current i and show that the power delivered to the three resistors is equal to that supplied by the source. Figure P 3.3-11 Solution: ⎛ 3 ⎞ ⎟ ⎝ 3+9 ⎠ From voltage division v3 = 12 ⎜ then i = v3 =1A 3 The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W The power supplied by the source is (12)(1) = 12 W. = 3V P 3.3-12 Consider the voltage divider shown in Figure P 3.3-12 when R1 = 6 Ω. It is desired that the output power absorbed by R1 = 6 Ω be 6 W. Find the voltage vo and the required source vs. Answer: vs = 14 V, vo = 6 V Figure P 3.3-12 Solution: P = 4.5 W and R1 = 8 Ω i2 = P 4.5 = = 0.5625 or i = 0.75 A R1 8 v0 = i R1 =(0.75) (8)=6 V from KVL: − v + i (2 + 4 + 8 + 2) = 0 s ⇒ v = 16 i = 16 ( 0.75 ) = 12 V s Figure P3.3-13 P3.3-13 Consider the voltage divider circuit shown in Figure P3.3-13. The resistor R represents a temperature sensor. The resistance R, in Ω, is related to the temperature T, in °C, by the equation 1 R = 50 + T 2 a) Determine the meter voltage, v m , corresponding to temperatures 0 °C, 75 °C and 100 °C. b) Determine the temperature, T, corresponding to the meter voltages 8 V, 10 V and 15 V. P3.3-13 Using voltage division ⎛ R ⎞ vm = ⎜ ⎟ 20 ⎝ 75 + R ⎠ R= Solving for R yields 75 v m 20 − v m The temperature can be calculated from the resistance using ⎛ 75 v m ⎞ 150 v m T = 2 ( R − 50 ) = 2 ⎜ − 50 ⎟ = − 100 ⎜ 20 − v m ⎟ 20 − v m ⎝ ⎠ ⎛ 50 ⎞ a) At 0 °C the resistance is R = 50 Ω so v m = ⎜ ⎟ 20 = 8 V . At 75 °C the resistance is ⎝ 75 + 50 ⎠ ⎛ 87.5 ⎞ R = 87.5 Ω so v m = ⎜ ⎟ 20 = 10.77 V . At 100 °C the resistance is R = 100 Ω so ⎝ 75 + 87.5 ⎠ ⎛ 100 ⎞ vm = ⎜ ⎟ 20 = 11.43 V . ⎝ 75 + 100 ⎠ 150 ( 8 ) − 100 = 0 °C . When v m = 10 V, the temperature is 20 − 8 150 (10 ) 150 (15 ) T= − 100 = 50 °C . When v m = 15 V, the temperature is T = − 100 = 350 °C . 20 − 10 20 − 15 b) When v m = 8 V, the temperature is T = P3.3-14 Consider the circuit shown in Figure P3.3-14. (a) Determine the value of the resistance R required to cause v o = 17.07 V. (b) Determine the value of the voltage v o when R = 14 Ω . (c) Determine the power supplied by the voltage source when v o = 14.22 V. Figure P3.3-14 Solution: ⎛ 8 ⎞ Voltage division indicates that v o = ⎜ ⎟ 32 . ⎝8+ R ⎠ (8 ) 32 − 8 = 6.997 7 Ω . ⎛ 8 ⎞ (a) When v o = 17.07 V, then 17.07 = ⎜ ⎟ 32 ⇒ R = 17.07 ⎝8+ R ⎠ ⎛ 8 ⎞ (b) When R = 14 Ω then v o = ⎜ ⎟ 32 = 11.6363 V . ⎝ 8 + 14 ⎠ (c) The power supplied by the voltage source is given by 32 vo = 4 v o since vo 8 8 the series elements. When v o = 14.22 V the voltage source supplies 56.88 W. is the current in each of Figure P3.3-15 P3.3-15. Figure P3.3-15 shows four similar but slightly different circuits. Determine the values of the voltages v1 , v 2 , v 3 and v 4 . Solution: Using voltage division: ⎛ 60 ⎞ ⎛ 130 ⎞ v1 = ⎜ ⎟ 26 = 12 V , v 2 = − ⎜ ⎟ 26 = −16.9 V , ⎝ 60 + 70 ⎠ ⎝ 70 + 130 ⎠ ⎛ 30 ⎞ ⎛ 70 ⎞ v3 = − ⎜ ⎟ 26 = −7.8 V and v 4 = − ⎜ ⎟ 26 = 18.2 V ⎝ 70 + 30 ⎠ ⎝ 30 + 70 ⎠ Figure P3.3-16 P3.3-16. Figure P3.3-16 shows four similar but slightly different circuits. Determine the values of the voltages v1 , v 2 , v 3 and v 4 . Solution: Using voltage division: ⎛ 30 ⎞ ⎛ 60 ⎞ v1 = − ⎜ ⎟ 28 = −16.8 V , v 2 = − ⎜ ⎟ 28 = 16.8 V , ⎝ 20 + 30 ⎠ ⎝ 40 + 60 ⎠ ⎛ 30 ⎞ ⎛ 20 ⎞ v3 = ⎜ ⎟ 28 = 12 V and v 4 = − ⎜ ⎟ 28 = −7 V ⎝ 40 + 30 ⎠ ⎝ 60 + 20 ⎠ Figure P3.3-17 P3.3-17. The input to the circuit shown in Figure P3.3-19 is the voltage source voltage v s ( t ) = 12 cos ( 377 t ) mV The output is the voltage vo(t). Determine vo(t). P3.3-17 Using voltage division: v a ( t ) = Using voltage division again: 110 12 cos ( 377 t ) = 11cos ( 377 t ) mV 10 + 110 9900 v o (t ) = 1000 v a ( t ) 100 + 9900 Therefore: v o (t ) = 9900 (1000 )11cos ( 377 t ) = 10890 cos ( 377 t ) mV 100 + 9900 = 10.89 cos ( 377 t ) V Section 3-4 Parallel Resistors and Current Division Use current division to determine the currents i1, i2, i3, and i4 in the circuit shown in Figure P P 3.4-1 3.4-1. Figure P 3.4-1. Solution: i = 1 i = 2 i = 3 i = 4 1 1 1 6 4= 4= A 1 + 1 + 1 +1 1+ 2 + 3 + 6 3 6 3 2 1 1 2 3 4 = A; 1 + 1 + 1 +1 3 6 3 2 1 1 2 4 =1 A 1 + 1 + 1 +1 6 3 2 1 1 4=2 A 1 + 1 + 1 +1 6 3 2 Consider the circuits shown in Figure P 3.4-2. P 3.4-2 Figure P 3.4-2 (a) (b) (c) Determine the value of the resistance R in Figure P 3.4-2b that makes the circuit in Figure P 3.42b equivalent to the circuit in Figure P 3.4-2a. Determine the voltage v in Figure P 3.4-2b. Because the circuits are equivalent, the voltage v in Figure P 3.4-2a is equal to the voltage v in Figure P 3.4-2b. Determine the power supplied by the current source. Solution: (a) (b) (c) 1 1 1 1 1 = + + = ⇒ R = 2Ω R 6 12 4 2 v = 6 ⋅ 2 = 12 V p = 6 ⋅12 = 72 W The ideal voltmeter in the circuit shown in Figure P 3.4-3 measures the voltage v. P 3.4-3 Figure P 3.4-3 (a) (b) (c) Suppose R2 = 12 Ω. Determine the value of R1 and of the current i. Suppose, instead, R1 = 12 Ω. Determine the value of R2 and of the current i. Instead, choose R1 and R2 to minimize the power absorbed by any one resistor. Solution: i= 8 8 or R1 = R1 i 8 = R 2 (2 − i ) ⇒ i = 2 − 8 8 or R 2 = 2−i R2 8 2 8 = A ; R1 = = 12 Ω 2 6 3 3 8 4 8 ( b ) i = = A ; R 2 = 4 = 12 Ω 6 3 2− 3 1 ( c ) R1 = R 2 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A. 2 R1 R 2 1 = 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω 2 ⋅ R1 + R 2 2 (a) i = 2− P 3.4-4 Determine the current i in the circuit shown in Figure P 3.4-4. Figure P 3.4-4 Solution: Current Division: 8 ( −6 ) = −2 A 16 + 8 8 i2 = ( −6 ) = −3 A 8+8 i1 = KCL: i = i 1 − i 2 = −2 − ( −3 ) = 1 A P 3.4-5 Consider the circuit shown in Figure P 3.4-5 when 4 Ω ≤ R1 ≤ 6 Ω and R2 = 10 Ω. Select the source is so that vo remains between 9 V and 13 V Figure P 3.4-5 Solution: ⎛ ⎞ R 1 ⎟ i and current division: i = ⎜ 2 ⎜R + R ⎟ s ⎝ 1 2⎠ Ohm's Law: v = i R yields o 2 2 ⎛ v ⎞⎛ R + R ⎞ 2⎟ i = ⎜ o ⎟⎜ 1 s ⎜ R ⎟⎜ R ⎟ ⎝ 2 ⎠⎝ 1 ⎠ plugging in R = 4Ω, v > 9 V gives i > 3.15 A o s 1 and R = 6Ω, v < 13 V gives i < 3.47 A o s 1 So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V. s o P 3.4-6 Figure P 3.4-6 shows a transistor amplifier. The values of R1 and R2 are to be selected. Resistances R1 and R2 are used to bias the transistor, that is, to create useful operating conditions. In this problem, we want to select R1 and R2 so that vb = 5 V. We expect the value of ib to be approximately 10 μA. When i1 ≤ 10ib, it is customary to treat ib as negligible, that is, to assume ib = 0. In that case R1 and R2 comprise a voltage divider. (a) (b) Figure P 3.4-6 Select values for R1 and R2 so that vb = 5 V and the total power absorbed by R1 and R2 is no more than 5 mW. An inferior transistor could cause ib to be larger than expected. Using the values of R1 and R2 from part (a), determine the value of vb that would result from ib = 15 μA. Solution: (a) To insure that ib is negligible we require 15 i1 = ≥ 10 (10 × 10−6 ) = 10−3 R1 + R 2 R1 + R 2 ≤ 150 kΩ so To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require 152 R1 + R 2 ≥ 45 kΩ ≤ 5 × 10−3 ⇒ R1 + R 2 Next to cause vb = 5 V we require 5 = vb = R2 R1 + R 2 15 ⇒ R1 = 2 R 2 For example, R1 = 40 kΩ, R 2 = 80 kΩ, satisfy all three requirements. (b) KVL gives KCL gives Therefore Finally (80 ×10 ) i 3 i1 = 1 + v b − 15 = 0 vb + 15 × 10−6 40 ×10 ⎛ v ⎞ (80 ×103 ) ⎜ 40 ×b103 + 15 ×10−6 ⎟ + v b = 15 ⎝ ⎠ 13.8 3v b + 1.2 = 15 ⇒ vb = = 4.6 V 3 3 P 3.4-7 Determine the value of the current i in the circuit shown in Figure P 3.4-7. Figure P 3.4-7 Solution: All of the elements of this circuit are connected in parallel. Replace the parallel current sources by a single equivalent 2 – 0.5 + 1.5 = 3 A current source. Replace the parallel 12 Ω and 6 Ω resistors by a 12 × 6 single = 4 Ω resistor. 12 + 6 By current division 12 ⎛ 4 ⎞ i = −⎜ ⎟ 3 = − = −1.714 A 7 ⎝ 3+ 4 ⎠ (checked: LNAP 62/6/07) P 3.4-8 Determine the value of the voltage v in Figure P 3.4-8. Figure P 3.4-8 Solution: Each of the resistors is connected between nodes a and b. The resistors are connected in parallel and the circuit can be redrawn like this: 40 & 20 & 40 = 10 Ω Then v = 10 ( 0.003) = 0.03 = 30 mV so (checked: LNAP 6/21/04) P 3.4-9 Determine the power supplied by the dependent source in Figure P 3.4-9. Figure P 3.4-9 Solution: Use current division to get ia = − so 75 30 × 10−3 ) = −22.5 mA ( 25 + 75 v b = 50 ( −22.5 × 10−3 ) = −1.125 V The power supplied by the dependent source is given by p = − ( 30 × 10−3 ) ( −1.125 ) = 33.75 mW (checked: LNAP 6/12/04) . P 3.4-10 Determine the values of the resistances R1 and R2 for the circuit shown in Figure P 3.4-10. Figure P 3.4-10 Solution: Using voltage division R1 × 24 8= 40 R 2 R1 + R 2 + 40 ⇒ Using KVL R1 ( R 2 + 40 ) 1 = 3 R1 R 2 + 40 ( R1 + R 2 ) ⇒ R1 R 2 + 40 ( R1 + R 2 ) = 3R1 R 2 + 120 R1 24 = 8 + R 2 (1.6 ) Then R1 = ⇒ ⇒ R1 = 40 R 2 2R 2 + 80 R 2 = 10 Ω 40 (10 ) =4Ω 2 (10 ) + 80 P 3.4-11 Determine the values of the resistances R1 and R2 for the circuit shown in Figure P 3.4-11. Figure P 3.4-11 Solution: Using KCL .024 = 0.0192 + 0.384 R2 ⇒ R2 = 0.384 = 80 Ω 0.0048 Using current division R1 0.384 = × 0.024 R2 R1 + ( R 2 + 80 ) ⇒ 16 = R1 R 2 R1 + R 2 + 80 = 80 R1 R1 + 160 ⇒ R1 = 40 Ω Determine the value of the current measured by the meter in Figure P 3.4-12. P 3.4-12 Figure P 3.4-12 Solution: Replace the (ideal) ammeter with the equivalent short circuit. Label the current measured by the meter. Apply KCL at the left node of the VCCS to get 1.2 = va 10 + 0.2 v a = 0.3 v a ⇒ va = 1.2 =4V 0.3 Use current division to get im = 30 30 0.2 v a = 0.2 ( 4 ) = 0.6 A 30 + 10 30 + 10 (checked using LNAP 9/11/04) P3.4-13 Consider the combination of resistors shown in Figure P3.4-13. Let R p denote the equivalent resistance. (a) Suppose 20 Ω ≤ R ≤ 320 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 40 Ω. Determine the Figure P3.4-13 value of R. Solution: (a) First, when R = 20 Ω then R p = 80 || 20 = R p = 80 || 320 = 80 ( 20 ) = 16 Ω . Next, when R = 320 Ω then 80 + 20 80 ( 320 ) = 64 Ω . Consequently 80 + 320 16 Ω ≤ R p ≤ 64 Ω. (b) When R = 0 then R p = 80 || 0 = (c) When R = ∞ then R p = 80 ( 0 ) = 0 Ω. 80 + 0 80 ( ∞ ) 80 = = 80 Ω . 80 + ∞ 80 + 1 ∞ (d) When R p = 40 Ω then 40 = 80 || R = 80 R 80 + R ⇒ 80 + R = 2 R ⇒ R = 80 Ω P3.4-14 Consider the combination of resistors shown in Figure P3.4-14. Let R p denote the equivalent resistance. (a) Suppose 40 Ω ≤ R ≤ 400 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 80 Ω. Determine the value of R. Figure P3.4-14 Solution: 160 ( 80 ) 160 = = 53.33 Ω . Next, when R 160 + 80 3 160 ( 440 ) 16 ( 44 ) = = 117.33 Ω . = 400 Ω then R p = 160 || ( 40 + 400 ) = 160 || 440 = 160 + 440 6 (a) First, when R = 40 Ω then R p = 160 || ( 40 + 40 ) = 160 || 80 = Consequently 53.33 Ω ≤ R p ≤ 117.33 Ω. (b) When R = 0 then R p = 160 || ( 0 + 40 ) = 160 || 40 = (c) When R = ∞ then R p = 160 || ( ∞ + 40 ) = 160 || ∞ = (d) When R p = 80 Ω then 80 = 160 || ( R + 40 ) = 160 ( 40 ) 16 ( 4 ) = = 32 Ω . 160 + 40 2 1 = 160 Ω . 1 1 + 160 ∞ 160 ( R + 40 ) 160 ⇒ R + 200 = ( R + 40 ) = 2 R + 80 ⇒ R = 120 Ω . 160 + R + 40 80 P3.4-15 Consider the combination of resistors shown in Figure P3.4-15. Let R p denote the equivalent resistance. (a) Suppose 50 Ω ≤ R ≤ 800 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 150 Ω. Determine the value of R. Figure P3.4-15 Solution: 200 ( 50 ) 200 = 50 + = 90 Ω . Next, when R = 200 + 50 5 200 ( 800 ) 800 800 Ω then R p = 50 + ( 200 || 800 ) = 50 + = 50 + = 210 Ω . 200 + 800 5 (a) First, when R = 50 Ω then R p = 50 + ( 200 || 50 ) = 50 + Consequently 90 Ω ≤ R p ≤ 210 Ω. (b) When R = 0 then R p = 50 + ( 200 || 0 ) = 50 + 200 ( 0 ) = 50 Ω . 200 + 0 (c) When R = ∞ then R p = 50 + ( 200 || ∞ ) = 50 + 200 ( ∞ ) 200 = 50 + = 250 Ω . 200 200 + ∞ +1 ∞ (d) When R p = 150 Ω then 150 = 50 + ( 200 || R ) = 50 + 200 R 200 + R ⇒ 100 ( 200 + R ) = 200 R ⇒ R = 200 Ω . P3.4-16 The input to the circuit shown in Figure P3.4-16 is the source current, i s . The output is the current measured by the meter, i o . A current divider connects the source to the meter. Given the following observations: A. The input i s = 5 A causes the output to be i o = 2 A . B. When i s = 2 A the source supplies 48 W. Determine the values of the resistances R1 and R 2 . Figure P3.4-16 Solution: ⎛ R1 ⎞ R1 2 From current division, i o = ⎜ so i s . When i s = 5 A and i o = 2 A then = ⎟ ⎜ R1 + R 2 ⎟ 5 R1 + R 2 ⎝ ⎠ 2 ( R1 + R 2 ) = 5 R1 or 2 R 2 = 3 R1 . ⎡⎛ R1 R 2 ⎞ ⎤ The power supplied by the source is given by i s ⎢⎜ ⎟ i s ⎥ . When i s = 2 A the source supplies 48 ⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦ ⎡⎛ R1 R 2 ⎞ ⎤ R1 R 2 . W, so 48 = 2 ⎢⎜ 2 ⎥ ⇒ 12 = ⎟ R1 + R 2 ⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦ ⎛3 ⎞ 3 R1 ⎜ R1 ⎟ R1 3 5 2 ⎠ ⎝ 2 Combining these results gives 12 = = = R1 ⇒ R1 = (12 ) = 20 Ω and 5 5 3 ⎛3 ⎞ R1 + ⎜ R1 ⎟ 2 ⎝2 ⎠ 3 R1 = 30 Ω . 2 Figure P3.4-17 P3.4-17. Figure P3.4-17 shows four similar but slightly different circuits. Determine the values of the currents i1 , i 2 , i 3 and i 4 . Solution: Using current division: ⎛ 75 ⎞ ⎛ 45 ⎞ i1 = − ⎜ ⎟ 320 = −240 mA , i 2 = ⎜ ⎟ 300 = 100 mA , ⎝ 75 + 25 ⎠ ⎝ 45 + 90 ⎠ ⎛ 15 ⎞ ⎛ 120 ⎞ i3 = − ⎜ ⎟ 500 = −100 mA and i 4 = ⎜ ⎟ 250 = 200 mA ⎝ 15 + 60 ⎠ ⎝ 120 + 30 ⎠ Figure P3.4-18 P3.4-18. Figure P3.4-18 shows four similar but slightly different circuits. Determine the values of the currents i1 , i 2 , i 3 and i 4 . Solution: Using current division: ⎛ 30 ⎞ ⎛ 60 ⎞ i1 = ⎜ ⎟ 240 = 80 mA , i 2 = − ⎜ ⎟ 240 = −144 mA , ⎝ 30 + 60 ⎠ ⎝ 60 + 40 ⎠ ⎛ 20 ⎞ ⎛ 60 ⎞ i3 = ⎜ ⎟ 240 = 60 mA and i 4 = − ⎜ ⎟ 240 = −192 mA ⎝ 60 + 20 ⎠ ⎝ 60 + 15 ⎠ Figure P3.4-19 P3.4-19. The input to the circuit shown in Figure P3.4-19 is the current source current Is. The output is the current io. The output of this circuit is proportion to the input, that is io = k I s Determine the value of the constant of proportionality, k. Solution: Replace six resistors at the right of the circuit by an equivalent resistance to get Using current division: ⎛ 5 ⎞ ⎜ 2R ⎟ 2 i1 = ⎜ Is = Is ⎟ 5 7 ⎜ R+ R⎟ 2 ⎠ ⎝ Return to the original circuit Using current division: io = Therefore R 1 1⎛2 ⎞ 1 i1 = i 1 = ⎜ I s ⎟ = I s R+R 2 2⎝7 ⎠ 7 1 k= 7 Figure P3.4-20 P3.4-20. The input to the circuit shown in Figure P3.4-20 is the voltage source voltage Vs. The output is the voltage vo. The output of this circuit is proportion to the input, that is v o = k Vs Determine the value of the constant of proportionality, k. Solution: Replace six resistors at the right of the circuit by an equivalent resistance to get Using voltage division: 5 ⎞ ⎛ ⎞ ⎛ 5 ⎜ R || 2 R ⎟ ⎜ 7 R ⎟ 5 v1 = ⎜ Vs ⎜ Vs = Vs ⎟ ⎟ 5 5 19 ⎜ 2 R + R || R ⎟ ⎜ 2 R + R ⎟ 2 ⎠ ⎝ 7 ⎠ ⎝ Return to the original circuit Using current division: vo = Therefore R 2 1 1⎛ 5 ⎞ 1 v1 = v1 = ⎜ I s ⎟ = V s R 5 5 ⎝ 19 ⎠ 19 2R+ 2 1 k= 19 Section 3-5 Series Voltage Sources and Parallel Current Sources P 3.5-1 Determine the power supplied by each source in the circuit shown in Figure P 3.5-1. . Figure P 3.5-1 Solution: The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source. After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get i1 + 1.75 = i 2 Apply KVL to the outside loop to get 5 + 2 i 2 + 2 i1 = 0 so and 5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = − 8.5 = −2.125 A 4 i 2 = −2.125 + 1.75 = −0.375 A The power supplied by each sources is: Source 8-V voltage source Power delivered −8 i1 = 17 W 3-V voltage source 3 i1 = −6.375 W 3-A current source 3 × 2 i 2 = −2.25 W 1.25-A current source −1.25 × 2 i 2 = 0.9375 W (Checked using LNAP, 9/14/04) 1 P 3.5-2 Determine the power supplied by each source in the circuit shown in Figure P 3.5-2. Figure P 3.5-2 Solution: 20 × 5 = 4 Ω. 20 + 5 The 7-Ω resistor is connected in parallel with a short circuit, a 0-Ω resistor. The equivalent 0× 7 resistance is = 0 Ω , a short circuit. 0+7 The 20-Ω and 5-Ω resistors are connected in parallel. The equivalent resistance is The voltage sources are connected in series and can be replaced by a single equivalent voltage source. After doing so, and labeling the resistor currents, we have the circuit shown. The parallel current sources can be replaced by an equivalent current source. Apply KVL to get −5 + v1 − 4 ( 3.5 ) = 0 ⇒ v1 = 19 V The power supplied by each sources is: Source 8-V voltage source Power delivered −2 ( 3.5 ) = −7 W 3-V voltage source −3 ( 3.5 ) = −10.5 W 3-A current source 0.5-A current source 3 ×19 = 57 W 0.5 ×19 = 9.5 W (Checked using LNAP, 9/15/04) 2 P 3.5-3 Determine the power received by each resistor in the circuit shown in Figure P 3.5-3. Figure P 3.5-3 Solution: The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source. After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get i1 + 1.75 = i 2 Apply KVL to the outside loop to get 5 + 2 i 2 + i1 = 0 so 5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = − 8.5 = −2.125 A 4 and i 2 = −2.125 + 1.75 = −0.375 A The power supplied by each sources is: Source 8-V voltage source Power delivered −8 i1 = 17 W 3-V voltage source 3 i1 = −6.375 W 3-A current source 3 × 2 i 2 = −2.25 W 1.25-A current source −1.25 × 2 i 2 = 0.9375 W (Checked using LNAP, 9/14/04) 3 Section 3-6 Circuit Analysis P 3.6-1 The circuit shown in Figure P 3.6-1a has been divided into two parts. In Figure P 3.61b, the right-hand part has been replaced with an equivalent circuit. The left-hand part of the circuit has not been changed. (a) Determine the value of the resistance R in Figure P 3.6-1b that makes the circuit in Figure P 3.6-1b equivalent to the circuit in Figure P 3.6-1a. (b) Find the current i and the voltage v shown in Figure P 3.6-1b. Because of the equivalence, the current i and the voltage v shown in Figure P 3.6-1a are equal to the current i and the voltage v shown in Figure P 3.6-1b. (c) Find the current i2 shown in Figure P 3.61a using current division. Figure P 3.6-1 Solution: (a) (b) (c) 48 ⋅ 24 = 32 Ω 48 + 24 32 ⋅ 32 v = 32 + 32 24 = 16 V ; 32 ⋅ 32 8+ 32 + 32 16 1 i= = A 32 2 48 1 1 i2 = ⋅ = A 48 + 24 2 3 R = 16 + P 3.6-2 The circuit shown in Figure P 3.6-2a has been divided into three parts. In Figure P 3.6-2b, the rightmost part has been replaced with an equivalent circuit. The rest of the circuit has not been changed. The circuit is simplified further in Figure 3.6-2c. Now the middle and rightmost parts have been replaced by a single equivalent resistance. The leftmost part of the circuit is still unchanged. Figure P 3.6-2 (a) (b) (c) (d) (e) Determine the value of the resistance R1 in Figure P 3.6-2b that makes the circuit in Figure P 3.62b equivalent to the circuit in Figure P 3.6-2a. Determine the value of the resistance R2 in Figure P 3.6-2c that makes the circuit in Figure P 3.62c equivalent to the circuit in Figure P 3.6-2b. Find the current i1 and the voltage v1 shown in Figure P 3.6-2c. Because of the equivalence, the current i1 and the voltage v1 shown in Figure P 3.6-2b are equal to the current i1 and the voltage v1 shown in Figure P 3.6-2c. Hint: 24 = 6(i1–2) + i1R2 Find the current i2 and the voltage v2 shown in Figure P 3.6-2b. Because of the equivalence, the current i2 and the voltage v2 shown in Figure P 3.6-2a are equal to the current i2 and the voltage v2 shown in Figure P 3.6-2b. Hint: Use current division to calculate i2 from i1. Determine the power absorbed by the 3-Ω resistance shown at the right of Figure P 3.6-2a. Solution: 3⋅ 6 =6Ω 3+ 6 1 1 1 = + + ⇒ R p = 2.4 Ω then 12 6 6 (a ) R1 = 4 + (b) 1 Rp R 2 = 8 + R p = 10.4 Ω (c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0 ⇒ − 24 + 6 (i1 − 2) + 10.4i1 = 0 ⇒ i1 = 36 =2.195 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.83 V 16.4 1 6 ( d ) i2 = ( 2.195 ) = 0.878 A, 1 1 1 + + 6 6 12 v2 = ( 0.878 ) (6) = 5.3 V (e) i3 = 6 i2 = 0.585 A ⇒ 3+ 6 P = 3 i32 = 1.03 W P 3.6-3 Find i using appropriate circuit reductions and the current divider principle for the circuit of Figure P 3.6-3. Figure P 3.6-3 Solution: Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with a 2 Ω resistor by the equivalent 1 Ω resistor This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors: i1 = 1+1 (1.5 ) = 0.75 A 2 + (1 + 1) P 3.6-4 (a) Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b equivalent to the circuit in Figure P 3.6-4a. (b) Analyze the circuit in Figure P 3.6-4b to determine the values of the currents ia and ib (c) Because the circuits are equivalent, the currents ia and ib shown in Figure P 3.6-4b are equal to the currents ia and ib shown in Figure P 3.6-4a. Use this fact to determine values of the voltage v1 and current i2 shown in Figure P 3.6-4a. Figure P 3.6-4 Solution: (a) 1 1 1 1 = + + ⇒ R2 = 4 Ω R2 24 12 8 and R1 = (10 + 8) ⋅ 9 = 6Ω 10 + 8 + 9 b g (b) First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next, apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A . (c) 1 8 i2 = 2.25 = 1125 . A 1 1 1 + + 24 8 12 b gLM b10 +98g + 9 3OP = −10 V Q N and v1 = − 10 P 3.6-5 The voltmeter in the circuit shown in Figure P 3.6-5 shows that the voltage across the 30-Ω resistor is 6 volts. Determine the value of the resistance R1. Hint: Use the voltage division twice. Answer: R1 = 40 Ω Figure P 3.6-5 P3.6-5 30 v1 = 6 ⇒ v1 = 8 V 10 + 30 R2 12 = 8 ⇒ R2 = 20 Ω R2 + 10 20 = b g R1 10 + 30 R1 + 10 + 30 b g ⇒ R1 = 40 Ω Alternate values that can be used to change the numbers in this problem: meter reading, V 6 4 4 4.8 Right-most resistor, Ω 30 30 20 20 R1, Ω 40 10 15 30 P 3.6-6 Determine the voltages va and vc and the currents ib and id for the circuit shown in Figure P 3.6-6. Answer: va = –2 V, vc = 6 V, ib = –16 mA, and id = 2 mA Figure P 3.6-6 Solution: P 3.6-7 Determine the value of the resistance R in Figure P 3.6-7. Answer: R = 28 kΩ Figure P 3.6-7 Solution: 1× 10−3 = 24 12 ×103 + R p ⇒ R p = 12 × 103 = 12 kΩ ( 21×10 ) R = ( 21×10 ) + R 3 12 × 10 = R p 3 3 ⇒ R = 28 kΩ P 3.6-8 Most of us are familiar with the effects of a mild electric shock. The effects of a severe shock can be devastating and often fatal. Shock results when current is passed through the body. A person can be modeled as a network of resistances. Consider the model circuit shown in Figure P 3.6-8. Determine the voltage developed across the heart and the current Figure P 3.6-8 flowing through the heart of the person when he or she firmly grasps one end of a voltage source whose other end is connected to the floor. The heart is represented by Rh. The floor has resistance to current flow equal to Rf, and the person is standing barefoot on the floor. This type of accident might occur at a swimming pool or boat dock. The upper-body resistance Ru and lower-body resistance RL vary from person to person. Solution: ⎛ ⎞ 130 500 Voltage division ⇒ v = 50 ⎜ 15.963 V ⎜ 130 500 + 200 + 20 ⎟⎟ = ⎝ ⎠ ⎛ 100 ⎞ ⎛ 10 ⎞ ∴v = v ⎜ ⎟ = (15.963) ⎜ ⎟ = 12.279 V h ⎝ 100 + 30 ⎠ ⎝ 13 ⎠ v ∴ i = h = .12279 A h 100 P 3.6-9 Determine the value of the current i in Figure 3.6-9. Answer: i = 0.5 mA Figure 3.6-9 Solution: P 3.6-10 Determine the values of ia, ib, and vc in Figure P 3.6-10. Figure P 3.6-10 Solution: Req = 15 ( 20 + 10 ) = 10 Ω 15 + ( 20 + 10 ) ia = − 60 ⎛ 30 ⎞ ⎛ 60 ⎞ ⎛ 20 ⎞ = −6 A, ib = ⎜ ⎟⎟ = 4 A, vc = ⎜ ⎟ ⎜⎜ ⎟ ( −60 ) = −40 V Req ⎝ 30 + 15 ⎠ ⎝ Req ⎠ ⎝ 20 + 10 ⎠ P 3.6-11 Find i and Req a–b if vab = 40 V in the circuit of Figure P 3.6-11. Answer: Req a–b = 8 Ω, i = 5/6 A Figure P 3.6-11 Solution: a) Req = 24 12 = b) (24)(12) =8Ω 24 + 12 from voltage division: 100 5 ⎛ 20 ⎞ 100 v = 40 ⎜ = V∴ i = 3 = A ⎟ x x 20 3 3 ⎝ 20 + 4 ⎠ from current division: i = i 5 ⎛ 8 ⎞ A = x ⎜⎝ 8 + 8 ⎟⎠ 6 P 3.6-12 The ohmmeter in Figure P 3.6-12 measures the equivalent resistance, Req, of the resistor circuit. The value of the equivalent resistance, Req, depends on the value of the resistance R. Figure P 3.6-12 (a) Determine the value of the equivalent resistance, Req, when R = 18 Ω. (b) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 18 Ω. Solution: 9 + 10 + 17 = 36 Ω a.) b.) 36 R = 12 ⇒ 24 R = (12 )( 36 ) ⇒ R = 18 Ω 36+R 36 ( 9 ) = 7.2 Ω 36+9 P 3.6-13 Find the Req at terminals a–b in Figure P 3.6-13. Also determine i, i1, and i2. Answer: Req = 8 Ω, i = 5 A, i1 = 5/3 A, i2 = 5/2 A Figure P 3.6-13 Solution: Req = 2 + 1 + Using current division ⎛ 6 ⎞ i1 = i ⎜ ⎟ = ⎝ 6 + 12 ⎠ ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω ( 5) ( 13 ) =5 3 A so i= 40 40 = =5 A 8 Req ( ) ⎛ 2 ⎞ 5 1 and i2 = i ⎜ ⎟ = ( 5) 2 = 2 A ⎝ 2+2⎠ P 3.6-14 All of the resistances in the circuit shown in Figure P 3.6-14 are multiples of R. Determine the value of R. Figure P 3.6-14 Solution: 4 6 R + R = 2R 5 5 = R + ( 2R & 2R ) = 2R ( R & 4 R ) + ( 2 R & 3R ) = ( R + 2R & ( R + ( 2R & 2R )) ) So the circuit is equivalent to Then 12 = 0.1( R + ( 2 R & 2 R ) ) = 0.1( 2 R ) ⇒ R = 60 Ω (checked: ELAB 5/31/04) P 3.6-15 The circuit shown in Figure P 3.6-15 contains seven resistors, each having resistance R. The input to this circuit is the voltage source voltage, vs. The circuit has two outputs, va and vb. Express each output as a function of the input. Figure P 3.6-15 Solution: The circuit can be redrawn as va = R & ( R + ( R & 2R )) 2R + R & ( R + ( R & 2R )) vc = vs = 5 vs 21 R & 2R 2 2 vs = va = vs R + ( R & 2R ) 5 21 vb = R 1 1 vc = vc = vs R+R 2 21 (Checked using LNAP 5/23/04) P 3.6-16 The circuit shown in Figure P 3.6-16 contains three 10-Ω, 1/4-W resistors. (Quarter-watt resistors can dissipate 1/4 W safely.) Determine the range of voltage source voltages, vs, such that none of the resistors absorbs more than 1/4 W of power. Figure P 3.6-16 Solution: vo = (10 & 10 ) v = 5 v = v s s s 10 + (10 & 10 ) 15 3 vR + vo − vs = 0 iR = vR 10 ⇒ = vR = 2 vs 3 2 vs 30 2 4 1 ⎛ 2 ⎞ P = ⎜ v s ⎟ (10 ) = v s 2 ≤ 90 4 ⎝ 30 ⎠ ⇒ vs ≤ 90 3 10 = = 2.37 V 16 4 (checked: LNAP 5/31/04) P 3.6-17 The four resistors shown in Figure P 3.6-17 represent strain gauges. Strain gauges are transducers that measure the strain that results when a resistor is stretched or compressed. Strain gauges are used to measure force, displacement, or pressure. The four strain gauges in Figure P 3.6-17 each have a nominal (unstrained) resistance of 120 Ω and can each absorb 0.2 mW safely. Determine the range of voltage source voltages, vs, such that no strain gauge absorbs more than 0.2 mW of power. Figure P 3.6-17 Solution: vs so the current in each strain gauge is vs . The power 2 400 v s2 v s ⎛ v s ⎞ v s2 −3 dissipated by each resistor is given by so we require 0.5 × 10 ≤ or = ⎜ ⎟ 800 2 ⎝ 400 ⎠ 800 v s ≤ 0.4 = 0.6325 V . The voltage across each strain gauge is (checked: LNAP 6/9/04) P 3.6-18 The circuit shown in Figure P 3.6-18b has been obtained from the circuit shown in Figure P 3.6-18a by replacing series and parallel combinations of resistances by equivalent resistances. (a) (b) (c) Determine the values of the resistances R1, R2, and R3 in Figure P 3.6-18b so that the circuit shown in Figure P 3.6-18b is equivalent to the circuit shown in Figure P 3.6-18a. Determine the values of v1, v2, and i in Figure P 3.618b. Because the circuits are equivalent, the values of v1, v2, and i in Figure P 3.6-18a are equal to the values of v1, v2, and i in Figure P 3.6-18b. Determine the values of v4, i5, i6, and v7 in Figure P 3.6-18a. Solution: (a) R1 = 10 & ( 30 + 10 ) = 8 Ω , R 2 = 4 + (18 & 9 ) = 10 Ω and R3 = 6 & ( 6 + 6 ) = 4 Ω (b) i = 1 A , v1 = 8 V and v 2 = 4 V (c) 10 9 1 8 = −2 V , i 5 = − 1= − A, 10 + 30 9 + 18 3 4 1 ⎛ 1⎞ v 7 = −18 ⎜ − ⎟ = +6 V and i 6 = = A 12 3 ⎝ 3⎠ v4 = − (checked: LNAP 6/6/04) Figure P 3.6-18 P 3.6-19 Determine the values of v1, v2, i3, v4, v5, and i6 in Figure P 3.6-19. Figure P 3.6-19 Solution: Replace series and parallel combinations of resistances by equivalent resistances. Then KVL gives ( 20 + 4 + 8 + 16 ) i = 48 ⇒ i = 0.5 A v a = 20 i = 10 V , v b = 16 i = 8 V and v c = 8 i = 4 V Compare the original circuit to the equivalent circuit to get ⎛ 10 || (10 + 30) ⎞ ⎛ 8 ⎞ v1 = − ⎜ ⎟va = −⎜ ⎟10 = −4 V ⎝ 12 + 8 ⎠ ⎝ 12 + 10 || (10 + 30) ⎠ v 2 = −v c = −4 V ⎛ 20 ⎞ ⎛1⎞ i3 = − ⎜ ⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.1 A ⎝ 20 + 80 ⎠ ⎝5⎠ ⎛ 30 ⎞ ⎛1⎞ v4 = −⎜ ⎟ v1 = − ⎜ ⎟ ( −4 ) = 1 V ⎝ 10 + 30 ⎠ ⎝ 4⎠ 4 ⎛ ⎞ ⎛1⎞ v5 = ⎜ ⎟ v c = ⎜ ⎟ ( 4) = 1 V ⎝ 5+6+6⎠ ⎝ 4⎠ ⎛ ⎞ 16 ⎛1⎞ i 6 = − ⎜⎜ ⎟⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.25 A ⎝ 2⎠ ⎝ 16 + ( 4 + 6 + 6 ) ⎠ (checked: LNAP 6/10/04) P 3.6-20 Determine the values of i, v, and Req by the circuit model shown in Figure P 3.6-20, given that vab = 18 V. Figure P 3.6-20 Solution: Replace parallel resistors by equivalent resistors: 6 || 30 = 5 Ω and 72 || 9 = 8 Ω A short circuit in parallel with a resistor is equivalent to a short circuit. R eq = 36 || ( 8 + 10 ) = 12 Ω Using voltage division when vab = 18 V: v= 8 4 v ab = (18 ) = 8 V 8 + 10 9 i= v =1 A 8 (checked: LNAP 6/21/04) P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that Req = 9 Ω. Answer: R = 15 Ω Figure P 3.6-22 Solution: Replace parallel resistors by an equivalent resistor: 8 || 24 = 6 Ω A short circuit in parallel with a resistor is equivalent to a short circuit. Replace series resistors by an equivalent resistor: 4+6 = 10 Ω Now 9 = R eq = 5 + (12 || R ||10 ) so 60 R× 11 ⇒ R = 15 Ω 4= 60 R+ 11 (checked: LNAP 6/21/04) P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that Req = 50 Ω. Figure P 3.6-22 Solution: R eq = ( R || ( R + R ) || R ) || ( R || ( R + R) || R ) R || ( R + R) || R = 2 R || R eq = R 2 2 R || R = 5 5 5 R 2 = R 2 5 ⇒ R = 5 R eq = 200 Ω (checked: LNAP 6/21/04) P 3.6-23 Determine the values of r, the gain of the CCVS, and g, the gain of the VCCS, for the circuit shown in Figure P 3.6-23. Figure P 3.6-23 Solution: ia = 9.74 = 1.2175 A 8 ⎛ 9.74 ⎞ 9.74 − 6.09 = r i a = r ⎜ ⎟ ⎝ 8 ⎠ ⇒ V ⎛ 9.74-6.09 ⎞ r =⎜ ⎟8 = 3 A ⎝ 9.74 ⎠ v b = 12 − 9.74 = 2.26 V gv b + 6.09 9.74 2.26 + − =0 8 8 8 g= gv b vb = ⇒ gv b = −1.696 A −6.696 = −0.75 2.26 (checked: LNAP 6/21/04) P 3.6-24 The input to the circuit in Figure P 3.6-24 is the voltage of the voltage source, vs. The output is the voltage measured by the meter, vo. Show that the output of this circuit is proportional to the input. Determine the value of the constant of proportionality. Figure P 3.6-24 Solution: va = 20 & 20 1 vs = vs 20 + ( 20 & 20 ) 3 3 1 ⎛ 12 ⎞ vo = ⎜ ⎟ (10v a ) = × 10 × v s = 2v s 5 3 ⎝ 12 + 8 ⎠ V . So vo is proportional to vs and the constant of proportionality is 2 V P 3.6-25 The input to the circuit in Figure P 3.6-25 is the voltage of the voltage source, vs. The output is the current measured by the meter, io. Show that the output of this circuit is proportional to the input. Determine the value of the constant of proportionality. Figure P 3.6-25 Solution: vs ⎛ 40 ⎞ ⎛ 4 ⎞ ⎛ vs ⎞ 4 ia = ⎜ = ⎟ ⎜ ⎟ ⎜ ⎟ = vs ⎝ 40 + 10 ⎠ 2 + ( 40 & 10 ) ⎝ 5 ⎠ ⎝ 10 ⎠ 50 100 ⎛ 4 ⎞ 8 ⎛ 40 ⎞ io = − ⎜ ⎟ ( 50i a ) = − ⎜ ⎟ vs = − vs 3 ⎝ 50 ⎠ 3 ⎝ 20 + 40 ⎠ The output is proportional to the input and the constant of proportionality is − 8 A . 3 V P 3.6-26 Determine the voltage measured by the voltmeter in the circuit shown in Figure P 3.6-26. Figure P 3.6-26 Solution: Replace the voltmeter by the equivalent open circuit and label the voltage measured by the meter as vm. The 10-Ω resistor at the right of the circuit is in series with the open circuit that replaced the voltmeter so it’s current is zero as shown. Ohm’s law indicates that the voltage across that 10-Ω resistor is also zero. Applying KVL to the mesh consisting of the dependent voltage source, 10-Ω resistor and open circuit shows that vm = 8 ia The 10-Ω resistor and 40-Ω resistor are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor with a resistance equal to 40 ×10 =8 Ω 40 + 10 Figure a shows part of the circuit. In Figure b, an equivalent resistor has replaced the parallel resistors. Now the 4-Ω resistor and 8-Ω resistor are connected in series. The series combination of these resistors is equivalent to a single resistor with a resistance equal to 4 + 8 = 12 Ω . In Figure c, an equivalent resistor has replaced the series resistors. Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works toward the original circuit. Consider Figure a. Using Ohm’s law, we see that the current in the 12-Ω resistor is 2 A. The current in the voltage source is also 2 A. Replacing series resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the voltage source must also be 2 A in Figure b. The currents in resistors in Figure b are equal to the current in the voltage source. Next, Ohm’s law is used to calculate the resistor voltages as shown in Figure b. Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the 4-Ω resistor in Figure c must be equal to the current in the 4Ω resistor in Figure b. Using current division in Figure c are yields ⎛ 40 ⎞ ia = ⎜ ⎟ 2 = 1.6 A ⎝ 40 + 10 ⎠ Finally, v m = 8 i a = 8 ×1.6 = 12.8 V P 3.6-27 Determine the current measured by the ammeter in the circuit shown in Figure P 3.6-27. Figure P 3.6-27 Solution: Replace the ammeter by the equivalent short circuit and label the current measured by the meter as im. The 10-Ω resistor at the right of the circuit is in parallel with the short circuit that replaced the ammeter so it’s voltage is zero as shown. Ohm’s law indicates that the current in that 10-Ω resistor is also zero. Applying KCL at the top node of that 10-Ω resistor shows that i m = 0.8 v a Figure a shows part of the circuit. The 2-Ω resistor and 4-Ω resistor are connected in series. The series combination of these resistors is equivalent to a single 6-Ω resistor. In Figure b, an equivalent resistor has replaced the series resistors. Now the 3-Ω resistor and 6-Ω resistor are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor with a resistance equal to 3× 6 =2Ω 3+ 6 In Figure c, an equivalent resistor has replaced the parallel resistors. Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works toward the original circuit. Consider Figure a. Using Ohm’s law, we see that the voltage across the 2-Ω resistor is 6 V. The voltage across the current source is also 6 V. Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the voltage across the current source must also be 6 V in Figure b. The voltage across each resistor in Figure b is equal to the voltage across the current source. Replacing series resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the voltage across the 3-Ω resistor in Figure c must be equal to the voltage across the 3-Ω resistor in Figure b. Using voltage division in Figure c yields ⎛ 4 ⎞ va = ⎜ ⎟6=4 V ⎝ 2+4⎠ Finally, i m = 0.8 v a = 0.8 × 4 = 3.2 V P 3.6-28 Determine the value of the resistance R that causes the voltage measured by the voltmeter in the circuit shown in Figure P 3.6-28 to be 6 V. Figure P 3.6-28 Solution: Use current division in the top part of the circuit to get ⎛ 40 ⎞ ia = ⎜ ⎟ ( −3) = −2.4 A ⎝ 40 + 10 ⎠ Next, denote the voltage measured by the voltmeter as vm and use voltage division in the bottom part of the circuit to get ⎛ R ⎞ ⎛ −5 R ⎞ vm = ⎜ ⎟ ( −5 i a ) = ⎜ ⎟ ia ⎝ 18 + R ⎠ ⎝ 18 + R ⎠ Combining these equations gives: 12 R ⎛ −5 R ⎞ vm = ⎜ ⎟ ( −2.4 ) = 18 + R ⎝ 18 + R ⎠ When vm = 6 V, 6= 12 R 18 + R ⇒ R= 6 × 18 = 18 Ω 12 − 6 P 3.6-29 The input to the circuit shown in Figure P 3.6-29 is the voltage of the voltage source, vs. The output is the current measured by the meter, im. (a) (b) (c) Figure P 3.6-29 Suppose vs = 15 V. Determine the value of the resistance R that causes the value of the current measured by the meter to be im = 5 A. Suppose vs = 15 V and R = 24 Ω. Determine the current measured by the ammeter. Suppose R = 24 Ω. Determine the value of the input voltage, vs, that causes the value of the current measured by the meter to be im = 3 A. Soluton: Use voltage division in the top part of the circuit to get 2 ⎛ 12 ⎞ va = ⎜ ⎟ ( −v s ) = − v s 5 ⎝ 12 + 18 ⎠ Next, use current division in the bottom part of the circuit to get 80 ⎞ ⎛ 16 ⎞ ⎛ im = − ⎜ ⎟ (5 v a ) = ⎜ − ⎟ va ⎝ 16 + R ⎠ ⎝ 16 + R ⎠ Combining these equations gives: 80 ⎞ ⎛ 2 ⎞ ⎛ 32 ⎞ ⎛ im = ⎜ − ⎟ ⎜ − vs ⎟ = ⎜ ⎟ vs ⎝ 16 + R ⎠ ⎝ 5 ⎠ ⎝ 16 + R ⎠ a. When vs = 15 V and im = 12 A 288 ⎛ 32 ⎞ 12 = ⎜ = 24 Ω ⎟ 15 ⇒ 192 + 12 R = 480 ⇒ R = 12 ⎝ 16 + R ⎠ b. When vs = 15 V and R = 80 Ω ⎛ 32 ⎞ im = ⎜ ⎟ 15 = 5 A ⎝ 16 + 80 ⎠ c. When im = 3 A and R = 24 Ω 4 ⎛ 32 ⎞ 3=⎜ ⎟ vs = vs 5 ⎝ 16 + 24 ⎠ ⇒ vs = 15 = 3.75 V 4 P 3.6-30 The ohmmeter in Figure P 3.6-30 measures the equivalent resistance of the resistor circuit connected to the meter probes. Figure P 3.6-31 (a) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 12 Ω. (b) Determine the value of the equivalent resistance when R = 14 Ω. Solution: R eq = ( ( R + 4 ) || 20 ) + 2 = (a) (b) 12 = ( R + 4 ) × 20 + 2 = 20 R + 80 + 2 R + 24 ( R + 4 ) + 20 20 R + 80 20 R + 80 + 2 ⇒ 10 = ⇒ R + 24 = 2 R + 8 ⇒ R = 16 Ω R + 24 R + 24 R eq = 20 (14 ) + 80 + 2 = 11.5 Ω 14 + 24 (Checked: LNAPDC 9/28/04) P 3.6-31 The voltmeter in Figure P 3.6-31 measures the voltage across the current source. Figure P 3.6-31 (a) (b) Determine the value of the voltage measured by the meter. Determine the power supplied by each circuit element. Solution: Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by the meter. Label the element voltages and currents as shown in (b). Using units of V, A, Ω and W: Using units of V, mA, kΩ and mW: a.) Determine the value of the voltage measured by the meter. a.) Determine the value of the voltage measured by the meter. Kirchhoff’s laws give Kirchhoff’s laws give 12 + v R = v m and −i R = −i s = 2 mA 12 + v R = v m and −i R = −i s = 2 ×10−3 A Ohm’s law gives Ohm’s law gives ( ) v R = −25 i R v R = − 25 ×103 i R Then Then ( ) ( )( ) v R = − 25 ×103 i R = − 25 ×103 −2 ×10−3 = 50 V v m = 12 + v R = 12 + 50 = 62 V v R = −25 i R = −25 ( −2 ) = 50 V v m = 12 + v R = 12 + 50 = 62 V b.) Determine the power supplied by each element. voltage source current source resistor ( ) ( 12 i s = −12 −2 × 10−3 ( ) = 50 ( −2 × 10 ) 62 2 ×10−3 = 124 ×10−3 W vR iR −3 P 3.6-32 0 voltage source ( ) 12 i s = −12 ( −2 ) = −24 mW = −24 ×10−3 W = −100 × 10 −3 W total ) b.) Determine the power supplied by each element. current source 62 ( 2 ) = 124 mW resistor v R i R = 50 ( − 2 ) = −100 mW total 0 Determine the resistance measured by the ohmmeter in Figure P 3.6-32. Figure P 3.6-32 Solution: 12 + P 3.6-33 40 × 10 + 4 = 12 Ω 40 + 10 Determine the resistance measured by the ohmmeter in Figure P 3.6-33. Figure P 3.6-33 Solution: ( 60 + 60 + 60 ) × 60 = 45 Ω ( 60 + 60 + 60 ) + 60 P3.6-34 Consider the circuit shown in Figure P3.6-34. Given the values of the following currents and voltages: i1 = 0.625 A , v 2 = −25 V , i 3 = −1.25 A and v 4 = −18.75 V Determine the values of R1 , R 2 , R 3 and R 4 . Figure P3.6-34 Solution: 50 + v 2 − v1 = 0 ⇒ v1 = 50 + ( −25 ) = 25 V From KVL R1 = From Ohm’s law v1 i1 = 25 = 40 Ω 0.625 From KCL v ⎞ 25 ⎞ ⎛ ⎛ i1 + i 5 + i 2 = 0 ⇒ i 2 = − i1 + i 5 = − ⎜ 0.625 + 1 ⎟ = − ⎜ 0.625 + ⎟ = −1.25 A 40 ⎠ 40 ⎠ ⎝ ⎝ v2 −25 R2 = = = 20 Ω From Ohm’s law i 2 −1.25 ( ) From KCL v2 v −25 = i 6 + i 3 ⇒ i 6 = −i 3 + 2 = − ( −1.25 ) + = −5 A 4 4 4 From KVL v 3 + v 4 − 5 i 6 = 0 ⇒ v 3 = −v 4 + 5 i 6 = − ( −18.75 ) + 5 ( −5 ) = −6.25 V From Ohm’s law R3 = v3 i3 = v 4 −18.75 −6.25 = = 15 Ω = 5 Ω and R 4 = −1.25 −1.25 i3 P3.6-35 Consider the circuits shown in Figure P3.6-35. The equivalent circuit on the right is obtained from the original circuit on the left by replacing series and parallel combinations of resistors by equivalent resistors. The value of the current in the equivalent circuit is is = 0.8 A. Determine the values of R1, R2, R5, v2 and i3. Figure P3.6-35 Solution: R1 ||18 = 6 ⇒ 18 R1 18 + R1 = 6 ⇒ 3 R1 = 18 + R1 ⇒ R1 = 9 Ω R 2 + 10 = 28 ⇒ R 2 = 18 Ω 40 = ( 6 + 28 + R 5 ) i s ⇒ 40 = 34 + R 5 ⇒ R 5 = 50 − 34 = 16 Ω v 2 = −10 i s = −10 ( 0.8 ) = −8 V and i 3 = − 32 0.8 is = − = −0.4 A 32 + 32 2 0.8 P3.6-36 Consider the circuit shown in Figure P3.6-36. Given 2 1 3 v 2 = v s , i 3 = i1 and v 4 = v 2 . 3 5 8 Determine the values of R1 , R 2 and R 4 . Figure P3.6-36 2 1 3 Hint: Interpret v 2 = v s , i 3 = i1 and v 4 = v 2 as current and voltage division. 3 5 8 Solution: From voltage division v 4 = so R4 50 + R 4 50 + R 4 = From current division i 3 = so R4 v2 3 150 ⇒ 8 R 4 = 3 ( 50 + R 4 ) ⇒ R 4 = = 30 Ω . 8 8−3 R2 R 2 + ( 50 + R 4 ) R2 R 2 + 80 = i1 = R2 R 2 + 80 i1 1 ⇒ 5 R 2 = R 2 + 80 ⇒ R 2 = 20 Ω . 5 Notice that R 2 || ( 50 + R 4 ) = 20 || ( 50 + 30 ) = 20 || 80 = 16 Ω . From voltage division v1 = so R 2 || ( 50 + R 4 ) ( R1 + R 2 || ( 50 + R 4 ) ) vs = 16 vs R1 + 16 16 2 48 − 32 = ⇒ 48 = 2 ( R1 + 16 ) ⇒ R1 = =8 Ω. R1 + 16 3 2 P3.6-37 Consider the circuit shown in Figure P3.6-37. Given 2 2 4 i 2 = i s , v 3 = v1 and i 4 = i 2 . 5 3 5 Determine the values of R1 , R 2 and R 4 . 2 2 4 Hint: Interpret i 2 = i s , v 3 = v1 and i 4 = i 2 as current and voltage division. 5 3 5 Figure P3.6-37 Solution: From current division i 4 = so 80 4 400 − 320 = ⇒ 400 = 4 ( 80 + R 4 ) ⇒ R 4 = = 20 Ω . 80 + R 4 5 4 From voltage division v 3 = so 80 i2 80 + R 4 80 || R 4 R 2 + ( 80 || R 4 ) v1 = 80 || 20 16 v1 = v1 R 2 + ( 80 || 20 ) R 2 + 16 16 2 48 − 32 = ⇒ 48 = 2 ( R 2 + 16 ) ⇒ R 2 = =8 Ω. R 2 + 16 3 2 Notice that R 2 + ( 80 || R 4 ) = 8 + ( 80 || 20 ) = 8 + 16 = 24 Ω . From current division i1 = so R1 R1 + 24 = ( R1 R1 + R 2 + ( 80 || R 4 ) ) is = R1 R1 + 24 is 2 48 ⇒ 5 R1 = 2 ( R1 + 24 ) ⇒ R1 = = 16 Ω 5 3 P3.6-38 Consider the circuit shown in Figure P3.6-38. Figure P3.6-38 1 (a) Suppose i 3 = i1 . What is the value of the resistance R? 3 (b) Suppose instead v 2 = 4.8 V . What is the value of the equivalent resistance of the parallel resistors? (c) Suppose instead R = 20 Ω. What is the value of the current in the 40 Ω resistor? 1 Hint: Interpret i 3 = i1 as current division. 3 Solution: (a) From current division i 3 = 20 20 1 i1 so = ⇒ 60 = 20 + R ⇒ R = 40 Ω . 20 + R 20 + R 3 (b) From voltage division v 2 = so (c) 4.8 = Rp 40 + R p Rp 40 + R p 24 ⇒ i1 = 24 4.8 ⎡ 40 + R p ⎤⎦ = R p 24 ⎣ ⇒ Rp = 24 24 24 = = = 0.48 A 40 + ( 20 || 20 ) 40 + 10 50 ( 0.2 ) 40 = 10 Ω . 1 − 0.2 P3.6-39 Consider the circuit shown in Figure P3.6-39. Figure P3.6-39 1 (a) Suppose v 3 = v1 . What is the value of the resistance R? 4 (b) Suppose i 2 = 1.2 A . What is the value of the resistance R? (c) Suppose R = 70 Ω . What is the voltage across the 20 Ω resistor? (d) Suppose R = 30 Ω . What is the value of the current in this 30 Ω resistor? 1 Hint: Interpret v 3 = v1 as voltage division. 4 Solution: (a) From voltage division v 3 = (b) 1.2 = 10 10 1 v1 so = ⇒ 40 = 10 + R ⇒ R = 30 Ω . 10 + R 10 + R 4 20 ( 2.4 ) 20 20 2.4 = 2.4 ⇒ R + 30 = = 40 ⇒ R = 10 Ω 20 + ( R + 10 ) R + 30 1.2 (c) 20 || ( 70 + 10 ) = (d) i2 = 20 ( 80 ) = 16 Ω so v1 = (16 ) 2.4 = 38.4 V 20 + 80 20 20 20 2.4 = 2.4 = 2.4 = 0.8 A 20 + ( R + 10 ) 20 + ( 30 + 10 ) 60 P3.6-40 Consider the circuit shown in Figure P3.6-40. Given that the voltage of the dependent voltage source is v a = 8 V , determine the values of R1 and v o . Figure P3.6-40 Solution: First, vo = − 20 8 = −3.2 V 20 + 30 Next, ⎛ ⎞ ⎜ ⎟ ⎞ 8 40 40 ⎛ 10 40 ⎜ 10 400 400 ⎟= ic = = ib = = ⎜⎜ ⎟⎟ = 40 R1 ⎟ 12 40 + R1 + 40 R1 480 + 52 R1 20 40 + R1 40 + R1 ⎝ 12 + 40 || R1 ⎠ 40 + R1 ⎜ ⎜⎜ 12 + ⎟ 40 + R1 ⎟⎠ ⎝ then 400 ( 20 ) 8 400 1000 − 480 = ⇒ 480 + 52 R1 = = 1000 ⇒ = 10 Ω 20 480 + 52 R1 8 52 ( ) P3.6-41 Consider the circuit shown in Figure P3.6-41. Given that the current of the dependent current source is i a = 2 A , determine the values of R1 and i o . Figure P3.6-41 Solution: First, io = − 15 2 = −0.5 A 15 + 45 Next, (( 2 25 25 vb = 2 10 || 25 + R1 = vc = 0.2 25 + R1 25 + R1 ( ( then 2 500 = 0.2 35 + R1 ) ⎛ 10 25 + R1 ⎜ 1⎜ ⎝ 10 + 25 + R1 ) ) = 2550+ R ⇒ 35 + R1 = 50 ⇒ R1 = 15 Ω ) ⎞ 500 ⎟= ⎟ 35 + R1 ⎠ P3.6-42 Determine the values of i a , i b , i 2 , and v1 in the circuit shown in Figure P3.6-42. Figure P3.6-42 Solution: Use equivalent resistances to reduce the circuit to From KCL i b = 4 i a + i a ⇒ i b = −3 i a . From KVL 12 i a + 10 i b − 6 = 0 ⇒ 12 i a + 10 ( −3 i a ) = 6 So i a = − 1 A , v a = −4 A and i b = 1 A . 3 Returning our attention to the original circuit, notice that i a and i b were not changed when the circuit was reduced. Now v1 = ( 5 || 20 ) i a = ( 4 )( −0.333) = −1.333 V and i 2 = 12 i = 0.333 A . 12 + 24 b P3.6-43 The Ohmmeter in Figure P3.6-43 measures R eq , the equivalent resistance of the part of the circuit to the left of the terminals. (a) Suppose R eq = 12 Ω. Determine the value of the resistance R. (b) Suppose instead that R = 14 Ω. Determine the value of the equivalent resistance R eq . Figure P3.6-43 Solution: R eq = 2 + ( 20 || ( 4 + R ) ) . (a) When R eq = 12 Ω then 12 = 2 + ( 20 || ( 4 + R ) ) ⇒ 10 = 20 ( 4 + R ) ⇒ 24 + R = 2 ( 4 + R ) ⇒ R = 16 Ω 20 + ( 4 + R ) (b) When R = 14 Ω then R eq = 2 + ( 20 || ( 4 + 14 ) ) = 2 + ( 20 ||18 ) = 2 + 20 (18 ) = 11.4736 Ω 20 + 18 P3.6-43. Determine the values of the resistance R and current i a in the circuit shown in Figure P3.6-43. Figure P3.6-43 Solution. Replace the series resistors by an equivalent resistor. Then use KVL to write 80 i a + 8 − 24 = 0 ⇒ i a = 24 − 8 = 0.2 A 80 Use KCL to write 24 − 8 8 8 = + ⇒ R 200 80 8 16 8 8 = − = 0.16 ⇒ R = = 50 Ω R 80 200 0.16 Figure P3.6-44 P3.6-44 The input to the circuit shown in Figure P3.6-44 is the voltage of the voltage source, 32 V. The output is the current in the 10 Ω resistor, io. Determine the values of the resistance, R1, and of the gain of the dependent source, G, that cause both the value of voltage across the 12 Ω to be v a = 10.38 V and the value of the output current to be i o = 0.4151 A. Solution: Using voltage division 10.38 = v a = 12 ( 32 ) 12 = 36.9942 ≈ 37 Ω ⇒ R1 = 25 Ω ( 32 ) ⇒ R1 + 12 = R1 + 12 10.38 Using current division 0.4151 = i o = 40 0.4151 A G v a = ( 0.8 ) G (10.38 ) ⇒ G = = 0.05 40 + 10 V ( 0.8)10.38 P3.6-45 The equivalent circuit in Figure 3.6-45 is obtained from the original circuit by replacing series and parallel combinations of resistors by equivalent resistors. The values of the currents in the equivalent circuit are ia = 3.5 A and ib = −1.5 A. Determine the values of the voltages v1 and v2 in the original circuit. Figure P3.6-45 Solution: Label the currents in the equivalent circuit that correspond to the give currents in the equivalent circuit: Use current division: Using Ohm’s law: v1 = −35 i a = −35 ( 3.5 ) = −122.5 V and v 2 = −50 ( −0.75 ) = 37.5 V P3.6-46 Figure 3.6-46 shows three separate, similar circuits. In each a 12 V source is connected to a subcircuit consisting of three resistors. Determine the values of the voltage source currents i1, i2 and i3. Conclude that while the voltage source voltage is 12 V in each circuit, the voltage source current depends on the subcircuit connected to the voltage source. Figure P3.6-46 Solution: Replace the resistor subcircuit by an equivalent resistor in each circuit: Using Ohm’s law: i1 = 12 12 12 = 0.75 mA , i 2 = = 3 mA and i 3 = = 4 mA 16 kΩ 4 kΩ 3 kΩ Figure P3.6-47 P3.6-47 Determine the values of the voltages, v1 and v2, and of the current, i3, in the circuit shown in Figure P3.6-47. Solution: Replace series and parallel combinations of resistors by equivalent resistors to get ( 4 + ( 80 || 20 ) = 4 + 16 = 20 Ω and 40|| ( 80 + 80 ) = 40 ||160 = 32 Ω. ) Next, apply KVL to the left mesh to get 50 + 30 i a − 20 i a = 0 ⇒ i a = ib = Ohm’s law gives 30 i a 32 = 50 = −5 A and 30 i a = −150 V 20 − 10 −150 = −4.6875 A 32 Label ib on the original circuit Finally and v1 = ( 80 || 20 ) i a = 16 ( −5 ) = −80 V , v 2 = i3 = 1 ( 30 i a ) = −75 V 2 80 + 80 4 i b = ( −4.6875 ) = −3.75 A 40 + ( 80 + 80 ) 5 Section 3-7 Analyzing Resistive Circuits using MATLAB P3.7-1 Determine the power supplied by each of the sources, independent and dependent, in the circuit shown in Figure P3.7-1. Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the equations representing the circuit. Figure 3.7-1 Solution: We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and some element voltages: Notice that the 8 Ω resistor, the 10 Ω resistor and the two independent current sources are all connected in parallel. Consequently, the element voltages of theses elements can be labeled so that they are equal. Similarly, the 4 Ω resistor and the dependent current source are connected in parallel so their voltages can be labeled so as to be equal. Using Ohm’s Law we see that the current directed downward in the 8 Ω resistor is downward in the 10 Ω resistor is v1 v1 8 , current directed , and the current directed from left to right in the 2 Ω resistor is 10 Applying Kirchhoff’s Current Law (KCL) at node a gives 5= v1 8 + 2.5 + v1 10 + v2 2 ⇒ 0.225 v1 + 0.5 v 2 = 2.5 Using Ohm’s Law we see that the current directed downward in the 4 Ω resistor is v2 2 . (1) v3 4 . Applying Kirchhoff’s Current Law (KCL) at node a gives 1 v2 + 1.5 v1 = v3 ⇒ 1.5 v1 + 0.5 v 2 − 0.25 v 3 = 0 (2) 2 4 Applying Kirchhoff’s Voltage Law (KVL) to the mesh consisting of the 10 Ω resistor, the 2 Ω resistor and the dependent source to get v 2 + v 3 − v1 = 0 (3) Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 , v 2 and v 3 . We can write these equations in matrix form as 0 ⎤ ⎡ v1 ⎤ ⎡ 2.5⎤ ⎡ 0.225 0.5 ⎢ 1.5 0.5 −0.25⎥ ⎢ v ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −1 1 1 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ We can solve this matrix equation using MATLAB: Hence v1 = −4.1096 V, v 2 = 6.8493 V and v 3 = −10.9589 V 2 The power supplied by the 5 A current source is 5 v1 = 5 ( −4.1096 ) = −20.548 W . The power supplied by the 2.5 A current source is −2.5 v1 = −2.5 ( −4.1096 ) = 10.247 W . The power supplied by the dependent current source is (1.5 v1 ) v 3 = 1.5 ( −4.1096 )( −10.9589 ) = 67.555 W . Observation: Changing the order of the 8 Ω resistor, the 10 Ω resistor and the two independent current sources only changes the order of the terms in the KCL equation at node a. We know that addition is commutative, so change the order of the terms will not affect the values of the voltages v1 , v 2 and v 3 . For example, if the positions of the 2.5 A current source and 8 Ω resistor are swithched: The KCL equation at node a is 5 = 2.5 + v1 8 + v2 + 2.5 + v2 + v1 10 2 ⇒ 0.225 v1 + 0.5 v 2 = 2.5 Similarly, when the circuit is drawn as The KCL equation at node a is 5= v1 10 + v1 8 2 ⇒ 0.225 v1 + 0.5 v 2 = 2.5 The changes do not affect the values of the voltages v1 , v 2 and v 3 . 3 P3.7-2 Determine the power supplied by each of the sources, independent and dependent, in the circuit shown in Figure P3.7-2. Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the equations representing the circuit. Figure 3.7-2 Solution: We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and some element currents: Notice that two 4 Ω resistors, an 8 Ω resistor and the two independent voltage sources are all connected in series. Consequently, the element currents of theses elements can be labeled so that they are equal. Similarly, a 4 Ω resistor and the dependent voltage source are connected in series so their currents can be labeled so as to be equal. The current in each resistor has been labeled so we can use Ohm’s Law to calculate resistor voltages from the resistor currents and the resistances. Apply Kirchhoff’s Voltage Law (KVL) to the left mesh to get 6 + 8 i1 + 8 i 2 + 4 i1 − 15 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9 (1) Apply Kirchhoff’s Voltage Law (KVL) to the right mesh to get 4 i 3 + 5 i1 − 8 i 2 = 0 (2) Applying Kirchhoff’s Current Law (KCL) at node a to get i 1 = i 2 + i 3 ⇒ − i1 + i 2 + i 3 = 0 (3) 4 Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 , v 2 and v 3 . We can write these equations in matrix form as ⎡16 8 0 ⎤ ⎡ i1 ⎤ ⎡9 ⎤ ⎢ 5 −8 4 ⎥ ⎢i ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −1 1 1 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣0 ⎥⎦ We can solve this matrix equation using MATLAB: Hence i1 = 0.4091 A, i 2 = 0.3068 A and i 3 = 0.1023 A The power supplied by the 15 V voltage source is 15 i1 = 15 ( 0.4091) = 6.1365 W . The power supplied by the 6 V voltage source is −6 i1 = −6 ( 0.4091) = −2.4546 W . The power supplied by the dependent voltage source is − ( 5 i 2 ) i 3 = −5 ( 0.3068 )( 0.1023) = 0.1569 W . Observation: Changing the order of the two 4 Ω resistors, an 8 Ω resistor and the two independent voltage sources in the left mesh changes the order of the terms in the KVL equation for that mesh. We 5 know that addition is commutative, so change the order of the terms will not affect the values of the currents i1 , i 2 and i 3 . For example, when the circuit is drawn as The KVL equation for the left mesh is 6 − 15 + 8 i 2 + 4 i1 + 8 i1 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9 When the circuit is drawn as The KVL equation for the left mesh is 4 i1 + 4 i1 + 8 i 2 + 6 − 15 + 8 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9 These changes do not affect the values of the currents i1 , i 2 and i 3 . 6 Figure P3.7-3 P3.7-3. Determine the power supplied by each of the independent sources in the circuit shown in Figure P3.7-3. P3.7-3 Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit Variables: Apply KCL at the top left node: Apply KVL to the left mesh: In matrix form: Solving using MATLAB: v1 4 + v1 4 + 2 + i2 = 0 8 i 2 − 12 + 12 i 2 − v1 = 0 ⎡1 1 ⎤ 1 ⎥ ⎡v1 ⎤ ⎡ −2 ⎤ ⎢4 + 6 = ⎢ ⎥ ⎢⎣ i 2 ⎥⎦ ⎢⎣12 ⎥⎦ ⎣ −1 8 + 12 ⎦ ⎡ v1 ⎤ ⎡ −5.5714 ⎤ ⎢i ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ 0.3214 ⎦ The current source supplies −2 v1 = −2 ( −5.5714 ) = 11.1428 W The voltage source supplies 12 i 2 = 12 ( 0.3214 ) = 3.8568 W 7 Figure P3.7-4 P3.7-4. Determine the power supplied by each of the sources in the circuit shown in Figure P3.7-4. P3.7-4 Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit Variables: Apply KVL to the loop consisting of 30 Ω, 40 Ω and 50 Ω resistors: Apply KVL to the right mesh: v c = 40 i d + 12 v c Apply KCL at the top node of the current source: va 50 Apply KCL at the top node of a 40 Ω resistor: i b = In matrix form: Solving using MATLAB: The current source supplies The VCVS supplies 30 i b + v c − v a = 0 + i b = 2.4 vc 40 + id 1 0 ⎤ ⎡v a ⎤ ⎡ 0 ⎤ ⎡ −1 30 ⎢ 0 ⎥ ⎢i ⎥ ⎢ 0 ⎥ 0 11 40 ⎢ ⎥⎢ b ⎥ = ⎢ ⎥ ⎢ 0.02 1 0 0 ⎥ ⎢ v c ⎥ ⎢ 2.4 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ −1 0.025 1 ⎦ ⎢⎣ i d ⎥⎦ ⎣ 0 ⎦ ⎣ 0 ⎡ v a ⎤ ⎡ 41.0526 ⎤ ⎢i ⎥ ⎢ ⎥ ⎢ b ⎥ = ⎢ 1.5789 ⎥ ⎢ v ⎥ ⎢ −6.3158⎥ ⎢ c⎥ ⎢ ⎥ ⎢⎣ i d ⎥⎦ ⎣ 1.7368 ⎦ 2.4 v a = 2.4 ( 41.0526 ) = 98.5262 W − (12 v c ) i d = −12 ( −6.3158 )(1.7368 ) = 131.6314 W 8 Section 3-8 How Can We Check … P 3.8-1 A computer analysis program, used for the circuit of Figure P 3.8-1, provides the following branch currents and voltages: i1 = –0.833 A, i2 = –0.333 A, i3 = –1.167 A, and v = –2.0 V. Are these answers correct? Hint: Verify that KCL is satisfied at the center node and that KVL is satisfied around the outside loop consisting of the two 6-Ω resistors and the voltage source. Solution: Figure P 3.8-1 KCL at node a: i = i + i 3 1 2 − 1.167 = −0.833 + ( −0.333) − 1.167 = − 1.166 OK KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resistors, and the voltage source: 6i + 3i + v + 12 = 0 3 2 yields v = −4.0 V not v = −2.0 V The answers are not correct. The circuit of Figure P 3.8-2 was assigned as P 3.8-2 a homework problem. The answer in the back of the textbook says the current, i, is 1.25 A. Verify this answer using current division. Figure P 3.8-2 Solution: Apply current division to get: 1 ⎛ ⎞ ⎜ ⎟ ⎛1⎞ 20 i=⎜ 5 = ⎜ ⎟ 5 = 1.25 A √ 1 1 1 ⎟ ⎝4⎠ ⎜ ⎟ + + ⎝ 20 20 5 + 5 ⎠ The answer is correct. P 3.8-3 The circuit of Figure P 3.8-3 was built in the lab and vo was measured to be 6.25 V. Verify this measurement using the voltage divider principle. Figure P 3.8-3 Solution: Apply voltage division to get: 320 ⎛ ⎞ ⎛ 320 ⎞ v=⎜ ⎟ 24 = ⎜ ⎟ 24 = 6.4 V ≠ 6.25 V ⎝ 650 + 320 + 230 ⎠ ⎝ 1200 ⎠ The measurement is not correct. P 3.8-4 The circuit of Figure P 3.8-4 represents an auto’s electrical system. A report states that iH = 9 A, iB = –9 A, and iA = 19.1 A. Verify that this result is correct. Hint: Verify that KCL is satisfied at each node and that KVL is satisfied around each loop. Figure P 3.8-4 Solution: KVL bottom loop: KVL right loop: KCL at left node: − 14 + 0.1i + 1.2i = 0 A H − 12 + 0.05i + 1.2i = 0 B H i +i =i A B H Solving the three above equations yields: i = 16.8 A i = 10.3 A i = −6.49 A and A H B These are not the given values. Consequently, the report is incorrect. P 3.8-5 Computer analysis of the circuit in Figure P 3.8-5 shows that ia = –0.5 mA and ib = –2 mA. Was the computer analysis done correctly? Hint: Verify that the KVL equations for all three meshes are satisfied when ia = –0.5 mA and ib = –2 mA. Figure P 3.8-5 Solution: 1 ⎛ ⎞ Top mesh: 0 = 4 i a + 4 i a + 2 ⎜ i a + − i b ⎟ = 10 ( −0.5 ) + 1 − 2 ( −2 ) 2 ⎝ ⎠ Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V The KVL equations are satisfied so the analysis is correct. P 3.8-6 Computer analysis of the circuit in Figure P 3.8-6. shows that ia = 0.5 mA and ib = 4.5 mA. Was the computer analysis done correctly? Hint: First, verify that the KCL equations for all five nodes are satisfied when ia = 0.5 mA and ib = 4.5 mA. Next, verify that the KVL equation for the lower left mesh (a-e-d-a) is satisfied. (The KVL equations for the other meshes aren’t useful because each involves an unknown voltage.) Solution: Apply KCL at nodes b and c to label the circuit as Apply KCL to get i c + 4 = 4.5 ⇒ i c = 0.5 A i a + i c = 1 ⇒ 0.5 + 0.5 = 1 √ 6 = 1 + i a + i b = 1 + 0.5 + 4.5 √ (node d) (node a) (node e) Apply KVL to get to mesh (a-e-d-a) −3 ( 0.5 ) + 2 ( 4.5 ) − 5 ( 0.5 ) = −1.5 + 9 − 2.5 ≠ 0 The given currents do not satisfy these five Kirchhoff’s laws equations and therefore are not correct. P 3.8-7 Verify that the element currents and voltages shown in Figure P 3.8-7 satisfy Kirchhoff’s laws: (a) Verify that the given currents satisfy the KCL equations corresponding to nodes a, b, and c. (b) Verify that the given voltages satisfy the KVL equations corresponding to loops a-b-d-c-a and a-b-c-d-a. Figure P 3.8-7 Solution: (a) 7 + ( −3) = 4 4 + ( −2 ) = 2 − 5 = − 2 + ( −3 ) (b) −1 − ( −6 ) + ( −8 ) + 3 = 0 −1 − 2 − ( −8 ) − 5 = 0 (node a ) (node b) (node c) (loop a - b - d - c - a ) (loop a - b - c - d - a ) The given currents and voltages satisfy these five Kirchhoff’s laws equations P 3.8-8 Figure P 3.8-8 shows a circuit and some corresponding data. The tabulated data provides values of the current, i, and voltage, v, corresponding to several values of the resistance R2. (a) Use the data in rows 1 and 2 of the table to find the values of vs and R1. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the missing entries in the table. Figure P 3.8-8 Solution: i= (a) R1 + R 2 2.4 = from row 1 1.2 = from row 2 so vs vs R1 vs R1 + 10 2.4 R1 = v s = 1.2 ( R1 + 10 ) then (b) ⇒ R1 = 10 Ω v s = 2.4 (10 ) = 24 V i= 24 R 2 24 and v = 10 + R 2 10 + R 2 24 480 = 0.8 A and v = = 16 V . 30 30 720 When R2 = 30 Ω then v = = 18 V . 40 24 When R2 = 40 Ω the i = = 0.48 A . 50 When R2 = 20 Ω then i = 24 = 0.6 A . 40 960 When R2 = 40 W then v = = 19.2 V . 50 (c) When R2 = 30 Ω then i = (checked: LNAP 6/21/04) P 3.8-9 Figure P 3.8-9 shows a circuit and some corresponding data. The tabulated data provide values of the current, i, and voltage, v, corresponding to several values of the resistance R2. (a) Use the data in rows 1 and 2 of the table to find the values of is and R1. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the missing entries in the table. Figure P 3.8-9 Solution: i= (a) R1 R1 + R 2 is From row 1 R1 4 = is 3 R1 + 10 ⇒ 4R1 + 40 = 3R1i s From row 2 R1 6 = is 7 R1 + 20 ⇒ 6R1 + 120 = 7 R1i s ⇒ 28R1 + 280 = 18 R1 + 360 ⇒ is = 3 A So 4 R1 + 40 3R1 = is = When R2 = 40 Ω then i = 7 R1 4 8 = is 3 8 + 10 Then (b) 6 R1 + 120 i= ⇒ R1 = 8 Ω 24 R 2 8 24 and v = R 2 i = ( 3) = 8 + R2 8 + R2 8 + R2 24 960 = 0.5 A and v = = 20 V . These are the values in the table so tabulated 48 48 data is consistent. (c) When R2 = 80 Ω then i = 24 ( 80 ) 240 24 3 = A and v = = V. 88 11 88 11 (checked: LNAP 6/21/04) Design Problems DP 3-1 The circuit shown in Figure DP 3.1 uses a potentiometer to produce a variable voltage. The voltage vm varies as a knob connected to the wiper of the potentiometer is turned. Specify the resistances R1 and R2 so that the following three requirements are satisfied: 1. The voltage vm varies from 8 V to 12 V as the wiper moves from one end of the potentiometer to the other end of the potentiometer. 2. The voltage source supplies less than 0.5 W of power. 3. Each of R1, R2, and RP dissipates less than 0.25 W. Figure DP 3.1 Solution: Using voltage division: vm = R 2 + aR p R1 + (1 − a ) R p + R 2 + aR p vm = 8 V when a = 0 ⇒ vm = 12 V when a = 1 ⇒ 24 = R 2 + aR p R1 + R 2 + R p R2 R1 + R 2 + R p R2 + R p R1 + R 2 + R p = 1 3 = 1 2 24 The specification on the power of the voltage source indicates 242 1 ≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω R1 + R 2 + R p 2 Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives 3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives R1 = 6000 Ω and R 2 = 4000 Ω . With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate 24 mW, 16 mW and 8 mW respectively. Therefore the design is complete. DP 3-2 The resistance RL in Figure DP 3.2 is the equivalent resistance of a pressure transducer. This resistance is specified to be 200 Ω ± 5 percent. That is, 190 Ω ≤ RL ≤ 210 Ω. The voltage source is a 12 V ± 1 percent source capable of supplying 5 W. Design this circuit, using 5 percent, 1/8-watt resistors for R1 and R2, so that the voltage across RL is vo = 4V ± 10% (A 5 percent, 1/8-watt 100-Ω resistor has a resistance between 95 and 105 Ω and can safely dissipate 1/8-W continuously.) Figure DP 3.2 Solution: Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage 200 division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2 R 2 + 200 82 1 = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the series is 400 8 combination of two 200 Ω resistors. The power required to be dissipated by each of these resistors is 42 1 = 0.08 W < W . 200 8 Now let’s check the voltage: 190 210 11.88 < v < 12.12 0 190 + 420 210 + 380 3.700 < v0 < 4.314 4 − 7.5% < v0 < 4 + 7.85% Hence, vo = 4 V ± 8% and the design is complete. DP 3-3 A phonograph pickup, stereo amplifier, and speaker are shown in Figure DP 3.3a and redrawn as a circuit model as shown in Figure DP 3.3b. Determine the resistance R so that the voltage v across the speaker is 16 V. Determine the power delivered to the speaker. Figure DP 3.3 Solution: Vab ≅ 200 mV 10 10 120 Vab = (120) (0.2) 10 + R 10 + R 240 ⇒ R=5Ω let v = 16 = 10 + R 162 = 25.6W ∴P= 10 v= DP 3-4 A Christmas tree light set is required that will operate from a 6-V battery on a tree in a city park. The heavy-duty battery can provide 9A for the four-hour period of operation each night. Design a parallel set of lights (select the maximum number of lights) when the resistance of each bulb is 12 Ω. Solution: N 1 N ⎛1⎞ i = G v = v where G = ∑ = N⎜ ⎟ T T R ⎝ R⎠ n = 1 Rn ∴N= iR ( 9 )(12 ) = = 18 bulbs 6 v DP 3-5 The input to the circuit shown in Figure DP 3.5 is the voltage source voltage, vs. The output is the voltage vo. The output is related to the input by vo = R2 vs = gvs R1 + R2 The output of the voltage divider is proportional to the input. The constant of proportionality, g, is called the gain of the voltage divider and is given by R2 R1 + R2 The power supplied by the voltage source is ⎛ vs ⎞ vs2 vs2 p = vsis = vs ⎜ = ⎟= ⎝ R1 + R2 ⎠ R1 + R2 Rin where Rin = R1 + R2 is called the input resistance of the voltage divider. g= (a) (b) Figure DP 3.5 Design a voltage divider to have a gain, g = 0.65. Design a voltage divider to have a gain, g = 0.65, and an input resistance, Rin = 2500 Ω. Solution: R2 ⇒ g R1 = (1 − g ) R 2 R1 + R 2 Thus either resistance can be determined from the other resistance and the gain of the voltage divider. Also R2 R2 g= = ⇒ R 2 = g R in R1 + R 2 R in Notice that Consequently g= g R1 = (1 − g ) R 2 = (1 − g ) g R in ⇒ R1 = (1 − g ) R in (a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2 that will cause g = 0.65. Let’s pick a convenient value for R1, say R1 = 100 Ω Then (b) and g R1 = (1 − g ) R 2 ⇒ R2 = g R1 1− g = 0.65 ×100 = 186 Ω 1 − 0.65 R 2 = g R in = 0.65 × 2500 = 1625 Ω R1 = (1 − g ) R in = (1 − 0.65 ) 2500 = 875 Ω DP 3-6 The input to the circuit shown in Figure DP 3.6 is the current source current, is. The output is the current io. The output is related to the input by R1 io = is = gis R1 + R2 The output of the current divider is proportional to the input. The constant of proportionality, g, is called the gain of the current divider and is given by R1 g= R1 + R2 The power supplied by the current source is ⎡ ⎛ R R ⎞⎤ RR p = vsis = ⎢is ⎜ 1 2 ⎟ ⎥ is = 1 2 is2 = Rin is2 R1 + R2 ⎣ ⎝ R1 + R2 ⎠ ⎦ RR where Rin = 1 2 R1 + R2 is called the input resistance of the current divider. (a) (b) Figure DP 3.6 Design a current divider to have a gain, g = 0.65. Design a current divider to have a gain, g = 0.65, and an input resistance, Rin = 10000 Ω. Solution: Notice that g= R1 ⇒ g R 2 = (1 − g ) R1 R1 + R 2 Thus either resistance can be determined from the other resistance and the gain of the current divider. Also R1 R in R in = ⇒ R2 = g= R1 + R 2 R 2 g Consequently (1 − g ) R1 = g R 2 = R in ⇒ R1 = R in (1 − g ) Thus specified values of g and Rin uniquely determine the required values of R1 and R2. (a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2 that will cause g = 0.65. Let’s pick a convenient value for R1, say R1 = 100 Ω Then g R 2 = (1 − g ) R1 ⇒ R 2 = R2 = (b) and R1 = R in R in g (1 − g ) = = (1 − g ) R1 (1 − 0.65 ) ×100 = = 54 Ω g 10000 = 15385 Ω 0.65 10000 = 28571 Ω (1 − 0.65 ) 0.65 DP 3-7 Design the circuit shown in Figure DP 3-7 to have an output vo = 8.5 V when the input is vs = 12 V. The circuit should require no more than 1 mW from the voltage source. Figure DP 3.7 Solution: vo g= The required gain is 8.5 = 0.7083 12 = vs 0.7083 R1 = (1 − 0.7083) R 2 Consequently, ⇒ R 2 = 2.428 R1 It is also required that 122 0.001 ≥ R1 + R 2 ⇒ R1 + R 2 ≥ 144000 ⇒ 3.428 R1 ≥ 144000 ⇒ R1 ≥ 42007 Ω R1 = 45 kΩ and R 2 = 109.26 kΩ For example, DP 3-8 Design the circuit shown in Figure DP 3.8 to have an output io = 1.8 mA when the input is is = 5 mA. The circuit should require no more than 1 mW from the current source. Figure DP 3.8 Solution: g= The required gain is io is = 1.8 = 0.36 5 0.36 R 2 = (1 − 0.36 ) R1 ⇒ R 2 = 1.778 R1 Consequently, It is also required that 0.001 ≥ 0.005 2 R1 R 2 R1 + R 2 ⇒ R1 + R 2 R1 R 2 ⇒ R1 ≤ For example, = 2.778 R1 1.778 R12 ≥ 0.005 2 = 0.025 0.001 2.778 = 62.5 Ω 1.778 ( 0.025 ) R1 = 60 Ω and R 2 = 106.7 Ω DP 3.9 A thermistor is a temperature dependent resistor. The thermistor resistance, RT, is related to the temperature by the equation β 1 T −1 T o ) RT = RT e ( where T has units of °K and R is in Ohms. R0 is resistance at temperature T0 and the parameter β is in °K. For example, suppose that a particular thermistor has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. At T = 70 °C = 343 °K the resistance of this thermistor will be 3330 ( 1 342 −1 293 ) R T = 620 e = 121.68 Ω In Figure DP 3-9 this particular thermistor in used in a voltage divider circuit. Specify the value of the resistor R that will cause the voltage vT across the thermistor to be 4 V when the temperature is 100 °C. Solution At T = 373°K R T = 620 e 3330 ( 1 372 −1 293 ) = 54.17 Ω Using voltage division 4 = vT = 54.17 54.17 ( 40 ) ⇒ R + 54.17 = ( 40 ) 4 R + 54.17 R = 541.7 − 54.17 = 487.54 Ω Figure DP 3-9 DP3-10 The circuit shown in Figure DP 3-10 contains a thermistor that has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. (See problem DP 3-9.) Design this circuit (that is, specify the values of R and Vs) so that the thermistor voltage is vT = 4 V when T = 100 °C and vT = 20 V when T = 0 °C. Figure DP 3-10 Solution 3330 ( 1 273 −1 293 ) = 1425.6 Ω 3330 ( 1 373 −1 293 ) = 54.17 Ω At T = 0°C = 273°K R T = 620 e At T = 100°C = 373°K R T = 620 e Using voltage division 54.17 54.17 V s ) ⇒ R + 54.17 = ( (Vs ) 4 R + 54.17 1425.6 1425.6 20 = v T = V s ) ⇒ R + 1425.6 = ( (Vs ) 20 R + 1425.6 4 = vT = In matrix form: Solving gives ⎡ 54.17 ⎢ 4 ⎢ ⎢1425.6 ⎢⎣ 20 ⎤ −1⎥ ⎡V s ⎤ ⎡ 54.17 ⎤ ⎥⎢ ⎥ = ⎢ R ⎦ ⎣1425.6 ⎥⎦ ⎣ ⎥ −1 ⎥⎦ V s = 23.7528 V and R = 267.5029 Ω DP3-11 The circuit shown in Figure DP 3-11 is designed help orange grower protect their crops against frost by sounding an alarm when the temperature falls below freezing. It contains a thermistor that has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. (See problem DP 3-9.) The alarm will sound when the voltage at the − input of the comparator is less than the voltage at the + input. Using voltage division twice, we see that the alarm sounds whenever R2 RT + R2 < R4 R3 + R 4 Figure DP 3-11 Determine values of R2, R3 and R4 that cause the alarm to sound whenever the temperature is below freezing. Solution: The solution is not unique. For example, pick R3 = 30 kΩ and R4 = 10 kΩ. The alarm turns on and off R2 10 = = 0.25 when R T + R 2 30 + 10 At freezing the temperature is T = 0°C = 273°K and the thermistor resistance is R T = 620 e 3330 ( 1 273 −1 293 ) = 1425.6 Ω At freezing, we have R2 1425.6 + R 2 = 0.25 ⇒ 0.75 R 2 = 356.4 R 2 = 475.2 Ω Let’s check. At −2 °C R T = 620 e 3330 ( 1 271−1 293 ) = 1560 Ω 475.2 = 0.2335 < 0.25 1560 + 475.2 so the alarm is on. At 2 °C R T = 620 e 3330 ( 1 275 −1 293 ) = 1305 Ω 475.2 = 0.2669 > 0.25 1305 + 475.2 so the alarm is off. Chapter 4 Exercises Exercise 4.2-1 Determine the node voltages, va and vb, for the circuit of Figure E 4.2-1. Answer: va = 3 V and vb = 11 V Figure E 4.2-1 Solution: KCL at a: v −v a a b + + 3 = 0 ⇒ 5 v − 3 v = −18 a b 3 2 KCL at b: v −v b a − 3 −1 = 0 ⇒ v − v = 8 b a 2 v Solving these equations gives: va = 3 V and vb = 11 V Exercise 4.2-2 Determine the node voltages, va and vb, for the circuit of Figure E 4.2-2. Answer: va = –4/3 V and vb = 4 V Figure E 4.2-2 Solution: v −v a a b + + 3 = 0 ⇒ 3 v − 2 v = −12 a b 4 2 v v −v b a b − −4= 0 ⇒ − 3 v + 5 v = 24 a b 3 2 v KCL at a: KCL at b: Solving: va = −4/3 V and vb = 4 V Exercise 4.3-1 Find the node voltages for the circuit of Figure E 4.3-1. Hint: Write a KCL equation for the supernode corresponding to the 10-V voltage source. Answer: 2+ vb + 10 vb + = 5 ⇒ vb = 30 V and va = 40 V 20 30 Figure E 4.3-1 Solution: Apply KCL to the supernode to get v + 10 v 2+ b + b =5 20 30 Solving: v = 30 V and v = v + 10 = 40 V b a b Exercise 4.3-2 Find the voltages va and vb for the circuit of Figure E 4.3-2. Answer: (vb + 8) − (−12) vb + = 3 ⇒ vb = 8 V and va = 16 V 10 40 Figure E 4.3-2 Solution: ( vb + 8) − ( −12) + vb = 3 10 40 ⇒ v = 8 V and v = 16 V b a Exercise 4.4-1 Find the node voltage vb for the circuit shown in Figure E 4.4-2. Hint: Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb = 4ia and solve for vb. 6 v v Answer: − + b − b = 0 ⇒ vb = 4.5 V 8 4 12 Figure E 4.4-2 Solution: Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb = 4 ia and solve for vb . 6 vb + =i 8 12 a ⎛ 9 + vb ⎞ ⇒ v = 4i = 4 ⎜ ⎟ ⇒ v = 4.5 V b a b ⎜ 12 ⎟ ⎝ ⎠ Exercise 4.4-2 Find the node voltages for the circuit shown in Figure E 4.4-2. Hint: The controlling voltage of the dependent source is a node voltage, so it is already expressed as a function of the node voltages. Apply KCL at node a. Answer: va − 6 va − 4va + = 0 ⇒ va = −2 V 20 15 Figure E 4.4-2 Solution: The controlling voltage of the dependent source is a node voltage so it is already expressed as a function of the node voltages. Apply KCL at node a. v −6 v −4v a a = 0 ⇒ v = −2 V + a a 20 15 Exercise 4.5-1 Determine the value of the voltage measured by the voltmeter in Figure E 4.5-1. Answer: –1 V Figure E 4.5-1 Solution: Mesh equations: −12 + 6 i + 3 ⎛⎜ i − i ⎞⎟ − 8 = 0 ⇒ 9 i − 3 i = 20 1 1 2 ⎝ 1 2⎠ 8 − 3 ⎛⎜ i − i ⎞⎟ + 6 i = 0 ⇒ − 3 i + 9 i = −8 2 1 2 ⎝ 1 2⎠ Solving these equations gives: 13 1 i = A and i = − A 1 6 2 6 The voltage measured by the meter is 6 i2 = −1 V. Exercise 4.6-1 Determine the value of the voltage measured by the voltmeter in Figure E 4.6-1. Hint: Write and solve a single mesh equation to determine the current in the 3-Ω resistor. Answer: –4 V Figure E 4.6-1 Solution: −12 ⎛ 3⎞ 9 + 3 i + 2 i + 4 ⎜ i + ⎟ = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i = A 9 ⎝ 4⎠ The voltmeter measures 3 i = −4 V Mesh equation: Exercise 4.6-2 Determine the value of the current measured by the ammeter in Figure E 4.6-2. Hint: Write and solve a single mesh equation. Answer: –3.67 A Figure E 4.6-2 Solution: Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒ ( 3 + 6 ) i = −15 − 6 ( 3) ⇒ i= −33 2 = −3 A 9 3 Section 4-2 Node Voltage Analysis of Circuits with Current Sources P 4.2-1 The node voltages in the circuit of Figure P 4.2-1 are v1 = –4 V and v2 = 2 V. Determine i, the current of the current source. Answer: i = 1.5 A Figure P 4.2-1 Solution: v −v −4 − 4 − 2 1 1 2 0= + +i = + + i = −1.5 + i ⇒ i = 1.5 A 8 6 8 6 (checked using LNAP 8/13/02) v KCL at node 1: P 4.2-2 Determine the node voltages for the circuit of Figure P 4.2-2. Answer: v1 = 2 V, v2 = 30 V, and v3 = 24 V Figure P 4.2-2 Solution: KCL at node 1: KCL at node 2: KCL at node 3: v −v v 1 2 1 + + 1 = 0 ⇒ 5 v − v = −20 1 2 20 5 v −v v −v 1 2 2 3 +2= ⇒ − v + 3 v − 2 v = 40 1 2 3 20 10 v −v v 2 3 3 +1 = ⇒ − 3 v + 5 v = 30 2 3 10 15 Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V. (checked using LNAP 8/13/02) Figure P4.2-3 P4.2-3 The encircled numbers in the circuit shown Figure P4.2-3 are node numbers. Determine the values of the corresponding node voltages, v1 and v2. Solution: First, express the resistor currents in terms of the node voltages: Apply KCL at node 1 to get v1 15 + 0.025 + v1 − v 2 40 =0 Multiply both sides by 40 and simplify to get Apply KCL at node 2 to get 11 v 1 − v 2 = −1 3 v1 − v 2 v 2 0.025 + = 40 25 Multiply both sides by 40 and simplify to get −v1 + Solving, we get 13 v2 = 1 5 v1 = −0.1875 V and v 2 = 0.3125 V P 4.2-4 Consider the circuit shown in Figure P 4.2-4. Find values of the resistances R1 and R2 that cause the voltages v1 and v2 to be v1 = 1 V and v2 = 2 V. Figure P 4.2-4 Solution: −.003 + Write the node equations: − When v1 = 1 V, v2 = 2 V v1 v1 − v2 + =0 R1 500 v1 − v2 v2 + − .005 = 0 500 R2 1 1 −1 + = 0 ⇒ R1 = = 200 Ω 1 R1 500 .003 + 500 2 −1 2 − + − .005 = 0 ⇒ R2 = = 667 Ω 1 500 R2 .005 − 500 −.003 + (checked using LNAP 8/13/02) P 4.2-5 Find the voltage v for the circuit shown in Figure P 4.2-5. Answer: v = 21.7 mV Figure P 4.2-5 Solution: Write node equations: Solving gives: Finally: v1 v − v 2 v1 − v3 + 1 + =0 500 125 250 v − v3 v − v2 − 1 − .001 + 2 =0 125 250 v − v3 v1 − v3 v3 − 2 − + =0 250 250 500 v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V v = v1 − v3 = 0.022 V (checked using LNAP 8/13/02) P 4.2-6 Simplify the circuit shown in Figure P 4.2-6 by replacing series and parallel resistors with equivalent resistors; then analyze the simplified circuit by writing and solving node equations. (a) Determine the power supplied by each current source. (b) Determine the power received by the 12-Ω resistor. Figure P 4.2-6 Solution: Replacing series and parallel resistors with equivalent resistors we get 12 Ω + ( 40 Ω & 10 Ω ) = 20 Ω 60 Ω & 120 Ω = 40 Ω The node equations are 3 × 10−3 = 2 × 10−3 + v2 − v3 v1 − v 2 20 v1 − v 2 + = 20 v1 − v 3 v1 − v 3 20 v 2 − v3 10 v3 ⇒ 0.06 = 2v1 − ( v 2 − v 3 ) ⇒ 0.04 = −v1 + 3v 2 − 2v 3 ⇒ 0 = − ( 2v1 + 4v 2 ) + 7v 3 ⎡ 2 −1 −1⎤ ⎡ v1 ⎤ ⎡.06 ⎤ ⎢ −1 3 −2 ⎥ ⎢v ⎥ = ⎢.04 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −2 −4 +7 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ v1 ⎤ ⎡0.244 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢v 2 ⎥ = ⎢ 0.228⎥ ⎢ v 3 ⎥ ⎢⎣ 0.200 ⎥⎦ ⎣ ⎦ 10 + 20 = 40 Solving, e.g. using MATLAB, gives ⇒ (a) The power supplied by the 3 mA current source is ( 3 ×10−3 ) ( 0.244 ) = 0.732 mW. The power supplied by the 2 mA source is ( 2 ×10−3 ) ( 0.228 ) = 0.456 mW. (b) The current in the 12 Ω resistor is equal to the current i = power received by the 12 Ω resistor is ( 0.8 ×10 v1 − v 2 20 ) (12 ) = 7.68 ×10 −3 2 −b = 0.244 − 0.228 = 0.8 mA so the 20 = 7.68 μ W. (checked: LNAP and MATLAB 5/31/04) P 4.2-7 The node voltages in the circuit shown in Figure P 4.2-7 are va = 7 V and vb = 10 V. Determine values of the current source current, is, and the resistance, R. Figure P 4.2-7 Solution Apply KCL at node a to get 2= va R + va 4 + va − vb 2 = 7 7 7 − 10 7 1 + + = + ⇒ R=4Ω R 4 R 4 2 Apply KCL at node b to get is + va − vb 2 = vb 8 + vb 8 = is + 7 − 10 10 10 = + ⇒ is = 4 A 2 8 8 (checked: LNAP 6/21/04) Figure P4.2-8 P4.2-8 The encircled numbers in the circuit shown Figure P4.2-8 are node numbers. The corresponding node voltages are v1 and v2.The node equation representing this circuit is ⎡ 0.225 −0.125⎤ ⎡ v1 ⎤ ⎡ −3⎤ ⎢ −0.125 0.125 ⎥ ⎢v ⎥ = ⎢ 2 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ a) Determine the values of R and Is in Figure P4.2-8 b) Determine the value of the power supplied by the 3 A current source. Solution: a) The node equations representing the circuit are v1 R + v1 − v 2 8 + 3 = 0 and v1 − v 2 8 + Is = 0 In matrix form, we have 1 ⎡ ⎢ 0.125 + R ⎢ ⎣ 0.125 ⎤ −0.125⎥ ⎡ v1 ⎤ ⎡ −3 ⎤ = ⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ − I s ⎥⎦ −0.125⎦ After multiplying the 2nd row by -1, we have 1 ⎡ ⎢ 0.125 + R ⎢ ⎣ −0.125 ⎤ −0.125⎥ ⎡ v1 ⎤ ⎡ −3⎤ = ⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ I s ⎥⎦ 0.125 ⎦ Comparing to the given node equation, we see that 1 = 0.1 ⇒ R = 10 Ω and I s = 2 A R b) Solving the given node equations, we get v1 = −10 V and v 2 = 6 V The power supplied by the 3 A current source is given by −v1 ( 3) = − ( −10 )( 3) = 30 W Section 4-3 Node Voltage Analysis of Circuits with Current and Voltage Sources P 4.3-1 The voltmeter in Figure P 4.3-1 measures vc, the node voltage at node c. Determine the value of vc. Figure P 4.3-1 Solution: Express the voltage of the voltage source in terms of its node voltages: 0 − va = 6 ⇒ va = −6 V KCL at node b: va − vb v −v +2= b c 6 10 KCL at node c: Finally: ⇒ −6 − vb v −v +2= b c 6 10 vb − vc vc = 10 8 ⇒ −1− ⇒ 4 vb − 4 vc = 5 vc ⎛9 ⎞ 30 = 8 ⎜ vc ⎟ − 3 vc ⎝4 ⎠ vb v −v +2= b c 6 10 ⇒ vb = ⇒ 30 = 8 vb − 3 vc 9 vc 4 ⇒ vc = 2 V (checked using LNAP 8/13/02) P 4.3-2 The voltages va, vb, vc, and vd in Figure P 4.3-2 are the node voltages corresponding to nodes a, b, c, and d. The current i is the current in a short circuit connected between nodes b and c. Determine the values of va, vb, vc, and vd and of i. Answer: va = –12 V, vb = vc = 4 V, vd = –4 V, i = 2 mA Figure P 4.3-2 Solution: Express the branch voltage of each voltage source in terms of its node voltages to get: va = −12 V, vb = vc = vd + 8 KCL at node b: vb − va = 0.002 + i ⇒ 4000 vb − ( −12 ) = 0.002 + i ⇒ vb + 12 = 8 + 4000 i 4000 KCL at the supernode corresponding to the 8 V source: v 0.001 = d + i ⇒ 4 = vd + 4000 i 4000 so vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd Consequently vb = vc = vd + 8 = 4 V and i = ⇒ vd = −4 V 4 − vd = 2 mA 4000 (checked using LNAP 8/13/02) Figure P4.3-3 P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-3. Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages. Identify the supernode corresponding to the 24 V source Apply KCL to the supernode to get 12 − ( v a − 24 ) 10 + 0.6 = v a − 24 40 + va 40 ⇒ 196 = 6 v a ⇒ v a = 32 V The 12 V source supplies ⎛ 12 − ( v a − 24 ) ⎞ ⎛ 12 − ( 32 − 24 ) ⎞ ⎟ = 12 ⎜ 12 ⎜ ⎟ = 4.8 W ⎜ ⎟ 10 10 ⎝ ⎠ ⎝ ⎠ The 24 V source supplies va ⎞ ⎛ 32 ⎞ ⎛ 24 ⎜ −0.6 + ⎟ = 24 ⎜ −0.6 + ⎟ = 4.8 W 40 ⎠ 40 ⎠ ⎝ ⎝ The current source supplies 0.6 v a = 0.6 ( 32 ) = 19.2 W Figure P4.3-4 P4.3-4. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.3-13. Solution: First, express the resistor currents in terms of the node voltages: Apply KCL to the supernode to get −15 − v1 20 + v 2 − ( v1 + 15 ) Apply KCL at node 2 to get In matrix form: Solving using MATLAB: 50 v1 − v 2 25 = = v2 40 v1 − v 2 25 + + v1 + 5 10 v 2 − ( v1 + 15 ) 50 ⇒ 0.21 v1 − 0.06 v 2 = −1.55 ⇒ − 0.06 v1 + 0.085 v 2 = 0.3 ⎡ 0.21 −0.06 ⎤ ⎡ v1 ⎤ ⎡ −1.55⎤ ⎢ −0.06 0.085 ⎥ ⎢v ⎥ = ⎢ 0.3 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ v1 = −7.9825 V and v 2 = −2.1053 V P 4.3-5 The voltages va, vb, and vc in Figure P 4.3-5 are the node voltages corresponding to nodes a, b, and c. The values of these voltages are: va = 12 V, vb = 9.882 V, and vc = 5.294 V Determine the power supplied by the voltage source. Figure P 4.3-5 Solution: The power supplied by the voltage source is ⎛v −v v −v ⎞ ⎛ 12 − 9.882 12 − 5.294 ⎞ va ( i1 + i 2 ) = va ⎜ a b + a c ⎟ = 12 ⎜ + ⎟ 6 ⎠ 4 6 ⎝ ⎠ ⎝ 4 = 12(0.5295 + 1.118) = 12(1.648) = 19.76 W (checked using LNAP 8/13/02) P 4.3-6 The voltmeter in the circuit of Figure P 4.3-6 measures a node voltage. The value of that node voltage depends on the value of the resistance R. (a) Determine the value of the resistance R that will cause the voltage measured by the voltmeter to be 4 V. (b) Determine the voltage measured by the voltmeter when R = 1.2 kΩ = 1200 Ω. Answer: (a) 6 kΩ (b) 2 V Figure P 4.3-6 Solution: Label the voltage measured by the meter. Notice that this is a node voltage. Write a node equation at the node at which the node voltage is measured. ⎛ 12 − v m ⎞ v m v −8 −⎜ + 0.002 + m =0 ⎟+ 3000 ⎝ 6000 ⎠ R That is 6000 ⎞ 6000 ⎛ ⎜3 + ⎟ v m = 16 ⇒ R = 16 R ⎠ ⎝ −3 vm (a) The voltage measured by the meter will be 4 volts when R = 6 kΩ. (b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ. P 4.3-7 Determine the values of the node voltages, v1 and v2, in Figure P 4.3-7. Determine the values of the currents ia and ib. Figure P 4.3-7 Solution: Apply KCL at nodes 1 and 2 to get 10 − v1 1000 10 − v 2 v1 = + v1 − v 2 3000 5000 v1 − v 3 v3 + = 4000 5000 2000 ⇒ 23v1 − 3v 2 = 150 ⇒ -4v1 + 19v 3 = 50 Solving, e.g. using MATLAB, gives ⎡ 23 −3⎤ ⎡ v1 ⎤ ⎡150 ⎤ ⎢ −4 19 ⎥ ⎢ v ⎥ = ⎢ 50 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ ib = Then v1 − v 2 5000 = ⇒ v1 = 7.06 V and v1 = 4.12 V 7.06 − 4.12 = 0.588 mA 5000 Apply KCL at the top node to get ia = v 1 − 10 1000 + v 2 − 10 4000 = 7.06 − 10 4.12 − 10 + = −4.41 mA 1000 4000 (checked: LNAP 5/31/04) P 4.3-8 The circuit shown in Figure P 4.3-8 has two inputs, v1 and v2, and one output, vo. The output is related to the input by the equation vo = av1 + bv2 where a and b are constants that depend on R1, R2 and R3. (a) (b) Determine the values of the coefficients a and b when R1 = 10 Ω, R2 = 40 Ω and R3 = 8 Ω. Determine the values of the coefficients a and b when R1 = R2 and R3 = R1 || R2. Figure P 4.3-8 Solution: vo R3 + v o − v1 R1 + vo − v2 R2 =0 ⇒ vo = v1 v2 + R R R R 1+ 1 + 1 1 + 2 + 2 R 2 R3 R1 R 3 (a) When R 1 = 10 Ω, R 2 = 40 Ω and R 3 = 8 Ω vo = v1 v2 + = 0.4v1 + 0.1v 2 1 5 1+ 4 + 5 1+ + 4 4 So a = 0.4 and b = 0.1. (b) When R 1 = R 2 and R 3 = R1 & R 2 = R1 / 2 vo = v1 + v2 1+1+ 2 1+1+ 2 = 0.25v1 + 0.25v 2 So a = 0.25 and b = 0.25. (checked: LNAP 5/31/04) P 4.3-9 Determine the values of the node voltages of the circuit shown in Figure P 4.3-9. Figure P 4.3-9 Solution: Express the voltage source voltages as functions of the node voltages to get v 2 − v1 = 5 and v 4 = 15 Apply KCL to the supernode corresponding to the 5 V source to get 1.25 = v1 − v 3 8 + v 2 − 15 20 =0 ⇒ 80 = 5v1 + 2v 2 − 5v 3 Apply KCL at node 3 to get v1 − v 3 8 = v3 40 + v 3 − 15 12 ⇒ − 15v1 + 28v 3 = 150 Solving, e.g. using MATLAB, gives ⎡ −1 1 0 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤ ⎢ 5 2 −5⎥ ⎢v ⎥ = ⎢ 80 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −15 0 28 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣150 ⎥⎦ ⇒ ⎡ v1 ⎤ ⎡ 22.4 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢v 2 ⎥ = ⎢ 27.4 ⎥ ⎢ v 3 ⎥ ⎢⎣17.4 ⎥⎦ ⎣ ⎦ So the node voltages are: v1 = 22.4 V, v 2 = 27.4 V, v 3 = 17.4 V, and v 4 = 15 (checked: LNAP 6/9/04) P 4.3-10 Figure P 4.3-10 shows a measurement made in the laboratory. Your lab partner forgot to record the values of R1, R2, and R3. He thinks that the two resistors were 10-kΩ resistors and the other was a 5-kΩ resistor. Is this possible? Which resistor is the 5-kΩ resistor? Figure P 4.3-10 Solution: Write a node equation to get ⎛ 12 − 4.5 ⎞ 4.5 4.5 − 6 7.5 4.5 1.5 −⎜ + + =0 ⇒ − + − =0 ⎜ R1 ⎟⎟ R 3 R R R R 2 1 3 2 ⎝ ⎠ Notice that 7.5 4.5 is either 0.75 mA or 1.5 mA depending on whether R1 is 10 kΩ or 5 kΩ. Similarly, R1 R3 is either 0.45 mA or 0.9 mA and 1.5 is either 0.15 mA or 0.3 mA. Suppose R1 and R2 are 10 kΩ R2 resistors and R3 is a 5 kΩ resistor. Then − 7.5 4.5 1.5 + − = −0.75 + 0.9 − 0.15 = 0 R1 R 3 R 2 It is possible that two of the resistors are 10 kΩ and the third is 5 kΩ. R3 is the 5 kΩ resistor. (checked: LNAP 6/9/04) Figure P4.3-11 P4.3-11. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-11. Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages. Identify the supernode corresponding to the 10 V source Apply KCL to the supernode to get 15 − ( v a + 10 ) 15 − v a v a + 10 v a + = + 4 6 8 3 ⇒ 60 = 21 v a ⇒ v a = 2.857 V The 15 V source supplies ⎛ 15 − ( v a + 10 ) 15 − v a + 15 ⎜ ⎜ 4 6 ⎝ ⎞ 15 − 12.857 15 − 2.857 ⎞ ⎟ = 15 ⎛⎜ + ⎟ = 15 ( 2.56 ) = 38.4 W ⎟ 4 6 ⎝ ⎠ ⎠ ⎛ 15 − v a v a ⎞ ⎛ 15 − 2.857 2.857 ⎞ The 10 V source supplies 10 ⎜ + ⎟ = 10 ⎜ + ⎟ = 10 (1.071) = 10.71 W 6 3 6 3 ⎝ ⎠ ⎝ ⎠ P 4.3-12 Determine the values of the node voltages of the circuit shown in Figure P 4.3-12. Figure P 4.3-12 Solution: Express the voltage source voltages in terms of the node voltages: v 2 − v1 = 8 and v 3 − v1 = 12 Apply KVL to the supernode to get v2 10 so + v1 4 + v3 5 =0 ⇒ 2v 2 + 5v1 + 4v 3 = 0 2 ( 8 + v1 ) + 5v1 + 4 (12 + v1 ) = 0 ⇒ v1 = − 64 V 11 The node voltages are v1 = −5.818 V v 2 = 2.182 V v 3 = 6.182 V (checked: LNAP 6/21/04) Figure P4.3-13 P4.3-13. Determine the values node voltages, v1 and v2, in the circuit shown in Figure P4.3-13. Solution: Apply KCL at node 1 to get v1 50 + v1 − v 2 65 v1 − 60 + 80 1 1 ⎞ 60 ⎛ 1 ⎛ 1 ⎞ = 0 ⇒ ⎜ + + ⎟ v1 − ⎜ ⎟ v 2 = 80 ⎝ 50 65 80 ⎠ ⎝ 65 ⎠ Apply KCL at node 2 to get 0.1 = v 2 − v1 65 + v 2 − 60 75 1 ⎞ ⎛ 1 ⎞ ⎛ 1 = ⇒ − ⎜ ⎟ v1 + ⎜ + ⎟ v 2 = 0.1 ⎝ 65 ⎠ ⎝ 65 75 ⎠ In matrix form 1 1 ⎡1 ⎢ 50 + 65 + 80 ⎢ 1 ⎢ − ⎢⎣ 65 1 ⎤ ⎡ 60 ⎤ 65 ⎥ ⎡ v1 ⎤ ⎢ ⎥ ⎥ ⎢ ⎥ = 80 1 1 ⎥ ⎣v 2 ⎦ ⎢ ⎥ + ⎣0.1⎦ 65 75 ⎥⎦ Solving, we get v1 = 13.2356 V and v2 = 22.3456 V − Figure P4.3-14 P4.3-14. The voltage source in the circuit shown in Figure P4.3-14 supplies 83.802 W. The current source supplies 17.572 W. Determine the values of the node voltages v1 and v2. Solution: From the power supplied by the current source we calculate 17.572 = v 2 ( 0.25 ) ⇒ v 2 = i3 = Using Ohm’s law 17.572 = 70.288 V 0.25 70.288 − 80 = −0.4856 A 20 From the power supplied by the voltage source we calculate 83.802 = 80 i 6 ⇒ i6 = 83.802 = 1.0475 V 80 i1 = − ( i 3 + i 6 ) = − ( −0.4856 + 1.0475 ) = −0.5619 A Using Ohm’s law In summary −0.5619 = v1 − 80 50 ⇒ v1 = 80 + 50 ( −0.5619 ) = 51.905 V v1 = 51.905 V and v2 = 70.288 V Section 4-4 Node Voltage Analysis with Dependent Sources P 4.4-1 The voltages va, vb, and vc in Figure P 4.4-1 are the node voltages corresponding to nodes a, b, and c. The values of these voltages are: va = 8.667 V, vb = 2 V, and vc = 10 V Determine the value of A, the gain of the dependent source. Figure P 4.4-1 Solution: Express the resistor currents in terms of the node voltages: va − vc = 8.667 − 10 = −1.333 A and 1 v −v 2 − 10 = −4 A i 2= b c = 2 2 i 1= Apply KCL at node c: i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333) ⇒ A= −5.333 =4 −1.333 (checked using LNAP 8/13/02) P 4.4-2 Find ib for the circuit shown in Figure P 4.4-2. Answer: ib = –12 mA Figure P 4.4-2 Solution: Write and solve a node equation: va − 6 v v − 4va + a + a = 0 ⇒ va = 12 V 1000 2000 3000 ib = va − 4va = −12 mA 3000 (checked using LNAP 8/13/02) P 4.4-3 Determine the node voltage vb for the circuit of Figure P 4.4-3. Answer: vb = 1.5 V Figure P 4.4-3 Solution: First express the controlling current in terms of the node voltages: 2 − vb i = a 4000 Write and solve a node equation: − 2 − vb v ⎛ 2 − vb ⎞ + b − 5⎜ ⎟ = 0 ⇒ vb = 1.5 V 4000 2000 ⎝ 4000 ⎠ (checked using LNAP 8/14/02) P 4.4-4 The circled numbers in Figure P 4.4-4 are node numbers. The node voltages of this circuit are v1 = 10 V, v2 = 14 V, and v3 = 12 V. (a) Determine the value of the current ib. (b) Determine the value of r, the gain of the CCVS. Answers: (a) –2 A (b) 4 V/A Figure P 4.4-4 Solution: Apply KCL to the supernode of the CCVS to get 12 − 10 14 − 10 1 + − + i b = 0 ⇒ i b = −2 A 4 2 2 Next 10 − 12 1⎫ =− ⎪ V −2 =4 4 2⎬ ⇒ r = 1 A − r i a = 12 − 14 ⎪⎭ 2 ia = (checked using LNAP 8/14/02) P 4.4-5 Determine the value of the current ix in the circuit of Figure P 4.4-5. Answer: ix = 2.4 A Figure P 4.4-5 Solution: First, express the controlling current of the CCVS in v2 terms of the node voltages: i x = 2 Next, express the controlled voltage in terms of the node voltages: v2 24 12 − v 2 = 3 i x = 3 V ⇒ v2 = 2 5 so ix = 12/5 A = 2.4 A. (checked using ELab 9/5/02) Figure P4.4-6 P4.4-6 The encircled numbers in the circuit shown Figure P4.4-6 are node numbers. Determine the value of the power supplied by the CCVS . P4.4-6 First, express the contolling current of the CCVS in terms of the node voltages: ia = v2 20 ⎛ v2 ⎞ v1 = 12 V and v 3 = 40 i a = 40 ⎜ ⎟ = 2 v 2 ⎝ 20 ⎠ Next, express the resistor currents in terms of the node voltages: Notice that Apply KCL at node 2 to get Then The CCVS supplies 12 − v 2 + v2 + v2 − 2v2 = 0 ⇒ v 2 = 16 V 5 20 10 v 2 16 ia = = = 0.8 A and v 3 = 40 i a = 40 ( 0.8 ) = 32 V 20 20 ⎛ v3 − v2 ⎞ ⎛ 32 − 16 ⎞ v3 ⎜ ⎟ = 32 ⎜ ⎟ = 51.2 W ⎝ 10 ⎠ ⎝ 10 ⎠ Figure P4.4-7 P4.4-7 The encircled numbers in the circuit shown Figure 4.4-27 are node numbers. The corresponding node voltages are: v1 = 9.74 V and v2 = 6.09 V Determine the values of the gains of the dependent sources, r and g. Solution: Using Ohm’s law, i b = Then v1 8 = 9.74 = 1.2175 A . Using KVL, the voltage across the CCVS is 8 r i b = v1 − v 2 = 9.74 − 6.09 = 3.65 V r= r ib ib = 3.65 = 2.9979 V/A 1.2175 Using KVL, v b = 12 − v1 = 12 − 9.74 = 2.26 V . Apply KCL to the supernode corresponding to the CCVS to get 12 − v1 v1 v 2 12 − 9.74 9.74 6.09 = + + g vb ⇒ = + + g v b ⇒ g v b = −1.6963 A 8 8 8 8 8 8 g v b −1.6963 g= = = −0.7506 A/V Then vb 2.26 P 4.4-8 Determine the value of the power supplied by the dependent source in Figure P 4.4-8. Figure P 4.4-8 Solution: Label the node voltages. First, v2 = 10 V due to the independent voltage source. Next, express the controlling current of the dependent source in terms of the node voltages: ia = v3 − v 2 16 = v 3 − 10 16 Now the controlled voltage of the dependent source can be expressed as ⎛ v 3 − 10 ⎞ v1 − v 3 = 8 i a = 8 ⎜ ⎟ ⎝ 16 ⎠ ⇒ v1 = 3 v3 − 5 2 Apply KCL to the supernode corresponding to the dependent source to get v1 − v 2 + 4 v1 12 + v3 − v 2 16 + v3 8 =0 Multiplying by 48 and using v2 = 10 V gives 16v1 + 9v 3 = 150 Substituting the earlier expression for v1 ⎛3 ⎞ 16 ⎜ v 3 − 5 ⎟ + 9v 3 = 150 ⎝2 ⎠ ⇒ v 3 = 6.970 V Then v1 = 5.455 V and ia = -0.1894 A. Applying KCL at node 2 gives v1 12 So = ib + 10 − v1 4 ⇒ 12 i b = −3 + 4 v1 = −30 + 4 ( 5.455 ) i b = −0.6817 A. Finally, the power supplied by the dependent source is p = ( 8 i a ) i b = 8 ( −0.1894 ) ( −0.6817 ) = 1.033 W (checked: LNAP 5/24/04) P 4.4-9 The node voltages in the circuit shown in Figure P 4.4-9 are v1 = 4 V, v2 = 0 V, and v3 = –6 V Determine the values of the resistance, R, and of the gain, b, of the CCCS. Figure P 4.4-9 Solution: Apply KCL at node 2: i a + bi a = i b = but ia = v3 − v 2 20 v 2 − v1 40 = = −6 − ( 0 ) = −0.3 A 20 0−4 = −0.1 40 so (1 + b )( −0.1) = ( −0.3) ⇒ b=2 A A Next apply KCL to the supernode corresponding to the voltage source. v1 10 + 2 ia + v3 R =0 ⇒ 4 −6 + 2 ( −0.1) + =0 10 R ⇒ R= 6 = 30 Ω .2 (checked: LNAP 6/9/04) P 4.4-10 The value of the node voltage at node b in the circuit shown in Figure P 4.4-10 is vb = 18 V. (a) Determine the value of A, the gain of the dependent source. (b) Determine the power supplied by the dependent source. Figure P 4.4-10 Solution: (a) Express the controlling voltage of the dependent source in terms of the node voltages: va = 9 − vb Apply KCL at node b to get 9 − vb 100 = A (9 − v b ) + vb 200 ⇒ A= 18 − 3v b 200 ( 9 − v b ) = 0.02 (b) The power supplied by the dependent source is − ( Av a ) v b = − ( 0.02 ( 9 − 18 ) ) (18 ) = 3.24 W (checked: LNAP 6/06/04) P 4.4-11 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.4-11. Figure P 4.4-11 Solution: This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single supernode. That single supernode separates the two voltage sources and their nodes from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get v x − 20 v x + +4=0 ⇒ v x = 10 V . 2 10 The power supplied by the dependent source is ( 0.1 v ) ( −30 ) = −30 W . x (checked: LNAP 6/6/04) P 4.4-12 Determine values of the node voltages, v1, v2, v3, v4, and v5, in the circuit shown in Figure P 4.4-12. Figure P 4.4-12. Solution: Express the voltages of the independent voltage sources in terms of the node voltages v1 − v 2 = 16 and v 4 − v 5 = 8 Express the controlling current of the dependent source in terms of the node voltages ix = v3 6 Express the controlled voltage of the dependent source in terms of the node voltages ⎛ v3 ⎞ v 2 − v 4 = 4i x = 4 ⎜ ⎟ ⎝6⎠ ⇒ − 6v 2 + 4v 3 + 6v 4 = 0 Apply KCL to the supernode to get v1 − v 3 2 + v 4 − v3 3 + v5 8 =1 ⇒ 12v1 − 20v 3 + 8v 4 + 3v 5 = 24 Apply KCL at node 3 to get v 3 − v1 2 + v3 6 + v3 − v 4 3 =0 ⇒ − 3v1 + 6v 2 − 2v 4 = 0 Solving, e.g. using MATLAB, gives 0 0 ⎤ ⎡ v1 ⎤ ⎡16 ⎤ ⎡ 1 −1 0 ⎢ ⎥ ⎢0 0 0 1 −1⎥⎥ ⎢v 2 ⎥ ⎢⎢ 8 ⎥⎥ ⎢ ⎢ 0 −6 4 6 0 ⎥ ⎢v 3 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 12 0 20 8 3 − ⎢ ⎥ ⎢v 4 ⎥ ⎢ 24 ⎥ ⎢⎣ −3 0 6 −2 0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⇒ ⎡ v1 ⎤ ⎡ 24 ⎤ ⎢v ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ 8 ⎥ ⎢ v 3 ⎥ = ⎢12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥ ⎢ v 5 ⎥ ⎢⎣ −8⎥⎦ ⎣ ⎦ (checked: LNAP 6/13/04) P 4.4-13 Determine values of the node voltages, v1, v2, v3, v4, and v5, in the circuit shown in Figure P 4.4-13. Figure P 4.4-13 Solution: Express the voltage source voltages in terms of the node voltages: v1 − v 2 = 8 and v 4 − v 3 = 16 Express the controlling current of the dependent source in terms of the node voltages: ix = v 2 − v3 10 + v1 − v 3 5 = 0.2v1 + 0.1v 2 − 0.3v 3 Express the controlled voltage of the dependent source in terms of the node voltages: v 5 = 4i x = 0.8v1 = 0.4v 2 − 1.2v 3 ⇒ 0.8v1 + 0.4v 2 − 1.2v 3 − v 5 = 0 Apply KVL to the supernodes v1 − v 5 + 2 v2 − v4 v4 − v2 4 4 + v3 8 + + v 2 − v3 10 v3 − v 2 10 + + v1 − v 3 5 v 3 − v1 5 =0 =2 ⇒ ⇒ 14v1 + 7v 2 − 6v 3 − 5v 4 − 10v 5 = 0 − 8v1 − 14v 2 + 17v 3 + 10v 4 = 80 Solving, e.g. using MATLAB, gives 0 0 0 ⎤ ⎡ v1 ⎤ ⎡ 8 ⎤ −1 ⎡1 ⎢ ⎥ ⎢0 0 1 0 ⎥⎥ ⎢v 2 ⎥ ⎢⎢16 ⎥⎥ −1 ⎢ ⎢ 0.8 0.4 −1.2 0 −1 ⎥ ⎢ v 3 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 7 −6 −5 −10 ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥ ⎢ 14 ⎢⎣ −8 −14 17 10 0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣80 ⎥⎦ ⇒ ⎡ v1 ⎤ ⎡11.32 ⎤ ⎢v ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ 3.32 ⎥ ⎢ v 3 ⎥ = ⎢ 2.11 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢v 4 ⎥ ⎢18.11⎥ ⎢ v 5 ⎥ ⎢⎣ 7.85 ⎥⎦ ⎣ ⎦ (checked: LNAP 6/13/04) P 4.4-14 The voltages v1, v2, v3, and v4 are the node voltages corresponding to nodes 1, 2, 3, and 4 in Figure P 4.4-14. Determine the values of these node voltages. Figure P 4.4-14 Solution: Express the controlling voltage and current of the dependent sources in terms of the node voltages: v3 − v4 v a = v 4 and i b = R2 Express the voltage source voltages in terms of the node voltages: v1 = V s and v 2 − v 3 = Av a = Av 4 Apply KCL to the supernode corresponding to the dependent voltage source v 2 − v1 R1 + v3 − v4 R2 = Is ⇒ − R 2 v1 + R 2 v 2 + R1 v 3 − R1 v 4 = R1 R 2 I s Apply KCL at node 4: B v3 − v4 R2 + v3 − v4 R2 = v4 R3 ⇒ ⎛ ( B + 1) v 3 − ⎜⎜ B + 1 + ⎝ R2 ⎞ ⎟ v4 = 0 R 3 ⎟⎠ Organizing these equations into matrix form: ⎡ 1 ⎢ 0 ⎢ ⎢− R2 ⎢ ⎢ ⎢ 0 ⎢⎣ With the given values: 0 1 R2 0 ⎤ ⎥ ⎡ v1 ⎤ ⎡ V s ⎤ ⎥ ⎢v ⎥ ⎢ ⎥ ⎥ ⎢ 2⎥ = ⎢ 0 ⎥ ⎥ ⎢v 3 ⎥ ⎢ R R I ⎥ ⎛ R 2 ⎞⎥ ⎢ ⎥ ⎢ 1 2 s ⎥ B +1 − ⎜ B +1+ ⎟⎟ ⎥ ⎣⎢v 4 ⎦⎥ ⎣ 0 ⎦ ⎜ R 3 ⎝ ⎠ ⎥⎦ 0 −1 R1 0 −A − R1 ⎡ v1 ⎤ ⎡ 25 ⎤ 0 0 0 ⎤ ⎡ v1 ⎤ ⎡ 25 ⎤ ⎡ 1 ⎢v ⎥ ⎢ ⎢v ⎥ ⎢ ⎢ 0 ⎥ ⎥ −5 ⎥ ⎢ 2 ⎥ ⎢ 0 ⎥ 1 −1 44.4 ⎥⎥ 2⎥ ⎢ ⎢ ⎢ = ⇒ = ⎢ v 3 ⎥ ⎢ 8.4 ⎥ ⎢ −20 20 10 −10 ⎥ ⎢ v 3 ⎥ ⎢ 400 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎥ 0 4 −4.667 ⎦ ⎢⎣v 4 ⎥⎦ ⎣ 0 ⎦ ⎢⎣ v 4 ⎥⎦ ⎣ 7.2 ⎦ ⎣ 0 (Checked using LNAP 9/29/04) P 4.4-15 The voltages v1, v2, v3, and v4 in Figure P 4.4-15 are the node voltages corresponding to nodes 1, 2, 3, and 4. The values of these voltages are v1 = 10 V, v2 = 75 V, v3 = –15 V, and v4 = 22.5 V Determine the values of the gains of the dependent sources, A and B, and of the resistance R1. Figure P 4.4-15 Solution: Express the controlling voltage and current of the dependent sources in terms of the node voltages: v a = v 4 = 22.5 V ib = and v3 − v4 R2 = −15 − 22.5 = −0.75 50 Express the dependent voltage source voltage in terms of the node voltages: v 2 − v 3 = Av a = Av 4 A= so v 2 − v3 v4 = 75 − ( −15 ) = 4 V/V 22.5 Apply KCL to the supernode corresponding to the dependent voltage source v 2 − v1 R1 + v3 − v4 R2 = Is ⇒ 75 − 10 −15 − 22.5 + = 2.5 ⇒ R1 = 20 Ω R1 50 Apply KCL at node 4: v3 − v 4 R2 = v4 R3 +B v3 − v 4 R2 ⇒ −15 − 22.5 22.5 −15 − 22.5 = +B ⇒ B = 2.5 A/A 50 20 50 (Checked using LNAP 9/29/04) P 4.4-16 The voltages v1, v2, and v3 in Figure P 4.4-16 are the node voltages corresponding to nodes 1, 2, and 3. The values of these voltages are v1 = 12 V, v2 = 21 V, and v3 = –3 V (a) Determine the values of the resistances R1 and R2. (b) Determine the power supplied by each source. Figure P 4.4-16 Solution: v 2 − v1 R1 = (b) The power supplied by the voltage source is 12 ( 0.5 + 1.25 − 2 ) = −3 W . The power supplied by 2 − 0.5 = v2 21 − 12 −3 = 6 Ω and R 2 = = =4Ω 1.5 1.25 − 2 −0.75 (a) the 1.25-A current source is 1.25 ( −3 − 12 ) = −18.75 W . The power supplied by the 0.5-A current source is −0.5 ( 21) = −10.5 W . The power supplied by the 2-A current source is 2 ( 21 − ( −3) ) = 48 W . P 4.4-17 The voltages v1, v2, and v3 in Figure P 4.4-17 are the node voltages corresponding to nodes 1, 2, and 3. The values of these voltages are v1 = 12 V, v2 = 9.6 V, and v3= –1.33 V (a) Determine the values of the resistances R1 and R2. (b) Determine the power supplied by each source. Figure P 4.4-18 Solution: 12 − ( −1.33) = 1.666 A 8 9.6 i2 = = 2.4 A 4 v 2 − v1 9.6 − 12 = = 6 Ω and R1 = 2 − i1 2 − 2.4 i1 = and (a) R2 = v3 i1 − 2 = −1.33 = 3.98 4 Ω 1.666 − 2 (b) The power supplied by the voltage source is 12 ( 2.4 + 1.66 − 2 ) = 24.7 W . The power supplied by the current source is 2 ( 9.6 − ( −1.33) ) = 21.9 W . (Checked using LNAP 10/2/04) P4.4-18 The voltages v 2 , v 3 and v 4 for the circuit shown in Figure P4.4-18 are: v 2 = 16 V, v 3 = 8 V and v 4 = 6 V Determine the values of the following: (a) The gain, A, of the VCVS (b) The resistance R 5 (c) The currents i b and i c (d) The power received by resistor R 4 Figure P4.4-18 Solution: Given the node voltages v 2 = 16 V, v 3 = 8 V and v 4 = 6 V A= Av a va = ⎛ v3 − v4 ⎞ R5 ⎜ ⎟ = v4 ⎝ 15 ⎠ ib = 16 − 8 V =4 8−6 V ⇒ R5 = 15 ( 6 ) = 45 Ω , 8−6 40 − 24 40 − 16 16 = 2 A and i c = − = 0.6667 A 12 12 12 p4 = va2 15 = 22 = 0.2667 W 15 P4.4-19 Determine the values of the node voltages v1 and v 2 for the circuit shown in Figure P4.4-19. Figure P4.4-19 P4.4-19 The node equations are 28 − v1 4 and + 3 v1 = v1 − v 2 5 v1 − v 2 5 = 3 v1 + v2 6 ⇒ 5 ( 28 − v1 ) + 20 ( 3 v1 ) = 4 ( v1 − v 2 ) ⇒ 140 = −51 v1 − 4 v 2 + 4 v 3 = 3 v1 + v2 6 + 4 ( v1 − v 2 ) ⇒ 0 = 204 v1 − 109 v 2 Using MATLAB to solve these equations: Consequently v1 = −2.3937 V and v 2 = −4.4800 V P4.4-20 The encircled numbers in Figure P4.4-20 are node numbers. Determine the values of v1 , v 2 and v 3 , the node voltages corresponding to nodes 1, 2 and 3. Figure P4.4-20 Solution: Apply KCL to the supernode corresponding to the horizontal voltage source to get v1 10 = va 2 = v3 − v 2 2 = v 3 − ( v1 + 10 ) 2 ( ⇒ v1 = 5 v 3 − ( v1 + 10 ) ) ⇒ 50 = −6 v1 + 5 v 3 Looking at the dependent source we notice that ( v 3 = 5 v a = 5 ( v 3 − v 2 ) = 5 v 3 − ( v1 + 10 ) ) Using MATLAB to solve these equations: Consequently Then v1 = −50 V and v 3 = −50 V v 2 = v1 + 10 = −40 V ⇒ 50 = −5 v1 + 4 v 3 P4.4-21 Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-21. Figure P4.4-21 Solution: Represent the controlling current of the dependent source in terms of the node voltages: i a = Represent the controlled voltage of the dependent source in terms of the node voltages: 4 i a = v1 − v 2 ⎛ v3 ⎞ ⇒ = 4 ⎜ ⎟ = v1 − v 2 ⎝ 2⎠ ⇒ 0 = v1 − v 2 − 2 v 3 Apply KCL to the supernode corresponding to the dependent voltage source: 12 − v1 2 = v2 2 + v 2 − v3 2 ⇒ 12 − v1 = v 2 + v 2 − v 3 ⇒ 12 = v1 + 2 v 2 − v 3 Apply KCL to top node of the current source: v 2 − v3 2 +1 = v3 2 ⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3 Solving these equations using MATLAB gives v1 = 8.2857 V , v 2 = 3.1459 V and v 3 = 2.5714 V v3 2 P4.4-22 Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-22. Figure P4.4-22 Solution: The node equations are: 12 − v1 2 4 ia = v2 2 ⎛ v3 ⎞ = 4 i a = 4 ⎜ ⎟ ⇒ 12 − v1 = 4 v 3 ⇒ 12 = v1 + 4 v 3 ⎝ 2⎠ + v 2 − v3 v 2 − v3 2 ⎛ v3 ⎞ v 2 v 2 − v3 ⇒ 4⎜ ⎟ = + 2 ⎝ 2⎠ 2 2 +1 = v3 2 ⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3 Solving these equations using MATLAB gives v1 = 28 V , v 2 = −10 V and v 3 = −4 V ⇒ 0 = 2 v 2 − 5 v3 Section 4-5 Mesh Current Analysis with Independent Voltage Sources P 4.5-1 Determine the mesh currents, i1, i2, and i3, for the circuit shown in Figure P 4.5-1. Answers: i1 = 3 A, i2 = 2 A, and i3 = 4 A Figure P 4.5-1 Solution: The mesh equations are 2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0 15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0 −6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0 or 14 i1 − 3 i 2 − 9 i 3 = 0 −3 i1 + 9 i 2 − 6 i 3 = −15 −9 i1 − 6 i 2 + 15 i 3 = 21 so i1 = 3 A, i2 = 2 A and i3 = 4 A. (checked using LNAP 8/14/02) P 4.5-2 The values of the mesh currents in the circuit shown in Figure P 4.5-2 are i1 = 2 A, i2 = 3 A, and i3 = 4 A. Determine the values of the resistance R and of the voltages v1 and v2 of the voltage sources. Answers: R = 12 Ω, v1 = –4 V, and v2 = –28 V Figure P 4.5-2 Solution: The mesh equations are: Top mesh: so Bottom, right mesh: so Bottom left mesh so 4 (2 − 3) + R(2) + 10 (2 − 4) = 0 R = 12 Ω. 8 (4 − 3) + 10 (4 − 2) + v 2 = 0 v2 = −28 V. −v1 + 4 (3 − 2) + 8 (3 − 4) = 0 v1 = −4 V. (checked using LNAP 8/14/02) P 4.5-3 The currents i1 and i2 in Figure P 4.5-3 are the mesh currents. Determine the value of the resistance R required to cause va = –6 V. Answer: R = 4 Ω Figure P 4.5-3 Solution: Ohm’s Law: i 2 = −6 = −0.75 A 8 KVL for loop 1: R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0 KVL for loop 2 + (−6) − 3 − 4 ( i1 − i 2 ) = 0 ⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0 ⇒ i1 = −3 A R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω (checked using LNAP 8/14/02) P 4.5-4 Determine the mesh currents, ia and ib, in the circuit shown in Figure P 4.5-4. Figure P 4.5-4 Solution: KVL loop 1: KVL loop 2: Solving these equations: 25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0 450 ia −100 ib = −2 −100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0 −100 ia + 500 ib = − 4 ia = − 6.5 mA , ib = − 9.3 mA (checked using LNAP 8/14/02) Find the current i for the circuit of Figure P 4.5-5. P 4.5-5 Hint: A short circuit can be treated as a 0-V voltage source. Figure P 4.5-5 Solution: Mesh Equations: mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0 Solving: 5 i = i2 ⇒ i = − = −0.294 A 17 (checked using LNAP 8/14/02) P 4.5-6 Simplify the circuit shown in Figure P 4.5-6 by replacing series and parallel resistors by equivalent resistors. Next, analyze the simplified circuit by writing and solving mesh equations. (a) Determine the power supplied by each source. (b) Determine the power absorbed by the 30-Ω resistor. Figure P 4.5-6 Solution: Replace series and parallel resistors with equivalent resistors: 60 Ω & 300 Ω = 50 Ω , 40 Ω + 60 Ω = 100 Ω and 100 Ω + 30 Ω + ( 80 Ω & 560 Ω ) = 200 Ω so the simplified circuit is The mesh equations are 200 i1 + 50 ( i1 − i 2 ) − 12 = 0 100 i 2 + 8 − 50 ( i1 − i 2 ) = 0 or ⎡ 250 −50 ⎤ ⎡ i1 ⎤ ⎡12 ⎤ ⎢ −50 150 ⎥ ⎢i ⎥ = ⎢ −8⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ ⇒ ⎡ i1 ⎤ ⎡ 0.04 ⎤ ⎢i ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ −0.04 ⎦ The power supplied by the 12 V source is 12 i1 = 12 ( 0.04 ) = 0.48 W . The power supplied by the 8 V source is −8i 2 = −8 ( −0.04 ) = 0.32 W . The power absorbed by the 30 Ω resistor is i12 ( 30 ) = ( 0.04 ) ( 30 ) = 0.048 W . 2 (checked: LNAP 5/31/04) Section 4-6 Mesh Current Analysis with Voltage and Current Sources P 4.6-1 Find ib for the circuit shown in Figure P 4.6-1. Answer: ib = 0.6 A Figure P 4.6-1 Solution: 1 A 2 mesh 2: 75 i2 + 10 + 25 i2 = 0 mesh 1: i1 = ⇒ i2 = − 0.1 A ib = i1 − i2 = 0.6 A (checked using LNAP 8/14/02) P 4.6-2 Find vc for the circuit shown in Figure P 4.6-2. Answer: vc = 15 V Figure P 4.6-2 Solution: Mesh currents: mesh a: ia = − 0.25 A mesh b: ib = − 0.4 A Ohm’s Law: vc = 100(ia − ib ) = 100(0.15) =15 V (checked using LNAP 8/14/02) P 4.6-3 Find v2 for the circuit shown in Figure P 4.6-3. Answer: v2 = 2 V Figure P 4.6-3 Solution: Express the current source current as a function of the mesh currents: i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5 Apply KVL to the supermesh: 30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10 50 i2 − 15 = − 10 ⇒ i2 = 5 = .1 A 50 i1 =−.4 A and v2 = 20 i2 = 2 V (checked using LNAP 8/14/02) Find vc for the circuit shown in P 4.6-4 Figure P 4.6-4. Figure P 4.6-4 Solution: Express the current source current in terms of the mesh currents: ib = ia − 0.02 Apply KVL to the supermesh: 250 ia + 100 (ia − 0.02) + 9 = 0 ∴ ia = − .02 A = − 20 mA vc = 100(ia − 0.02) = −4 V (checked using LNAP 8/14/02) P 4.6-5 Determine the value of the voltage measured by the voltmeter in Figure P 4.6-5. Answer: 8 V Figure P 4.6-5 Solution: Label the mesh currents: Express the current source current in terms of the mesh currents: i 3 − i 1 = 2 ⇒ i1 = i 3 − 2 Supermesh: Lower, left mesh: 6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8 −12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3 Eliminating i1 and i2 from the supermesh equation: 6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24 ⎛ 24 ⎞ The voltage measured by the meter is: 3 i 3 = 3 ⎜ ⎟ = 8 V ⎝ 9 ⎠ (checked using LNAP 8/14/02) Determine the value of the current measured by the ammeter in Figure P 4.6-6. P 4.6-6 Hint: Write and solve a single mesh equation. Answer: –5/6 A Figure P 4.6-6 Solution: Mesh equation for right mesh: 4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = − 10 5 A=− A 12 6 (checked using LNAP 8/14/02) Figure P4.6-7 P4.6-7 The mesh currents are labeled in the circuit shown Figure 4.6-7. The value of these mesh currents are: i1 = −1.1014 A, i 2 = 0.8986 A and i 3 = −0.2899 A a.) Determine the values of the resistances R1 and R 3 . b.) Detemine the value of the current sourc current. c.) Determine the value of the power supplied by the 12 V voltage source. Solution: Label the resistor currents and the current source currrents in terms of the mesh currents: a.) Apply KVL to the supermesh corresponding to the current source to get R1 i1 + 12 + 24 ( i 2 − i 3 ) − 24 = 0 ⇒ R1 = 12 − 24 ( i 2 − i 3 ) i1 = 12 − 24 ( 0.8986 − ( −0.2899 ) ) −1.1014 = 15 Ω Apply KVL to the rightmost mesh to get R 3 i 3 + 32 − 24 ( i 2 − i 3 ) = 0 ⇒ R 3 = b.) −32 + 24 ( i 2 − i 3 ) i3 = −32 + 24 ( 0.8986 − ( −0.2899 ) ) −0.2899 = 12 Ω I s = i 2 − i1 = 0.8986 − ( −1.1014 ) = 2 A c.) Noticing that 12 V and i 2 adhere to the passive convention, the power supplied by the 12 V voltage source is −12 i 2 = −12 ( 0.8986 ) = −10.783 W . P 4.6-8 Determine values of the mesh currents, i1, i2, and i3, in the circuit shown in Figure P 4.6-8. Figure P 4.6-8 Solution: Use units of V, mA and kΩ. Express the currents to the supermesh to get i1 − i 3 = 2 Apply KVL to the supermesh to get 4 ( i 3 − i 3 ) + (1) i 3 − 3 + (1) ( i1 − i 2 ) = 0 ⇒ i1 − 5 i 2 + 5 i 3 = 3 Apply KVL to mesh 2 to get 2i 2 + 4 ( i 2 − i 3 ) + (1) ( i 2 − i1 ) = 0 ( −1) i1 + 7i 2 − 4i 3 = 0 ⇒ Solving, e.g. using MATLAB, gives ⎡ 1 0 −1⎤ ⎡ i1 ⎤ ⎡ 2 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 1 −5 5 ⎥ ⎢i 2 ⎥ = ⎢ 3 ⎥ ⎢⎣ −1 7 −4 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⇒ ⎡ i 1 ⎤ ⎡ 3⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢1⎥ ⎢ i 3 ⎥ ⎢⎣1⎥⎦ ⎣ ⎦ (checked: LNAP 6/21/04) Figure P4.6-9 P4.6-9 The mesh currents are labeled in the circuit shown Figure 4.6-9. Determine the value of the mesh currents i1 and i 2 . Solution: Determine the value of the mesh currents i1 and i 2 . i 2 = i1 + 4 A , 8 i1 + 12 i1 + 5 i 2 = 0 ⇒ 8 i1 + 12 i1 + 5 ( 4 + i1 ) = 0 ⇒ i1 = −0.8 A i 2 = i1 + 4 = −0.8 + 4 = 3.2 A and P 4.6-10 The mesh currents in the circuit shown in Figure P 4.6-10 are i1 = –2.2213 A, i2 = 0.7787 A, and i3 = 0.0770 A (a) Determine the values of the resistances R1 and R3. (b) Determine the value of the power supplied by the current source. Figure P 4.6-10 Solution: (a) 50 ( i 3 − i 2 ) + R 3i 3 + 32 = 0 ⇒ 50 ( 0.0770 − 0.7787 ) + R 3 ( 0.0770 ) + 32 = 0 ⇒ R 3 = 40 Ω i1 R1 + 20i 2 + 50 ( i 2 − i 3 ) − 24 = 0 ⇒ R1 ( −2.2213) + 20 ( 0.7787 ) + 50 ( 0.7787 − 0.0770 ) = 24 ⇒ (b) R1 = 12 Ω I s = i 2 − i1 = 0.7787 − ( −2.2213) = 3 A The power supplied by the current source is p = I s ( 24 − R1 i1 ) = 3 ( 24 − 12 ( −2.2213) ) = 152 W (checked: LNAP 6/19/04) P 4.6-11 Determine the value of the voltage measured by the voltmeter in Figure P 4.6-11. Hint: Apply KVL to a supermesh to determine the current in the 2-Ω resistor. Answer: 4/3 V Figure P 4.6-11 Solution: 3 3 = i1 − i 2 ⇒ i1 = + i 2 . 4 4 ⎛3 ⎞ Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4 ⎜ + i 2 ⎟ + 5 i 2 = 9 ⇒ 9 i 2 = 6 ⎝4 ⎠ 2 4 so i 2 = A and the voltmeter reading is 2 i 2 = V 3 3 Express the current source current in terms of the mesh currents: P 4.6-12 Hint: Determine the value of the current measured by the ammeter in Figure P 4.6-12. Apply KVL to a supermesh. Answer: –0.333 A Figure P 4.6-12 Solution: Express the current source current in terms of the mesh currents: 3 = i1 − i 2 ⇒ i1 = 3 + i 2 . Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3 Finally, i 2 = − 1 A is the current measured by the ammeter. 3 Figure P4.6-13 P4.6-13 Determine the values of the mesh currents i1, i2 and i3 and the output voltage vo in the circuit shown Figure 4.6-13. Solution: Notice that the current source are each in a single mesh. Consequently, i1 = 2.4 A and i2 = 1.2 A. Label the resistor currents in terms of the mesh currents: Apply KVL to mesh 3 to get 12 i 3 + 15 − 16 (1.2 − i 3 ) − 18 ( 2.4 − i 3 ) = 0 ⇒ 46 i 3 = 47.4 ⇒ i 3 = 1.0304 A Apply KVL to the rightmost mesh to get v o − 15 − 12 i 3 = 0 ⇒ v o = 15 + 12 (1.0304 ) = 27.3648 V Figure P4.6-14 P4.6-14 Determine the values of the power supplied by the sources in the circuit shown Figure P4.6-14. Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents: Apply KVL to the middle mesh: 15 i + 25 ( i + 3) + 10 ( i − 5 ) = 0 ⇒ i = − ⎛ 1 ⎞ The 5 A current source supplies 5 (10 )( i − 5 ) = 5 (10 ) ⎜ − − 5 ⎟ = 275 W ⎝ 2 ⎠ ⎛ 1 ⎞ The 3 A current source supplies 3 ( 25 )( i + 3) = 3 ( 25 ) ⎜ − + 3 ⎟ = 187.5 W ⎝ 2 ⎠ 1 A 2 Figure P4.6-15 P4.6-15 Determine the values of the resistance R and of the power supplied by the 6 A current source in the circuit shown Figure P4.6-15. Solution: First, label the mesh currents: Notice that i1 = −2.5 A, i 2 = 1 A and 6 = i 3 + 2.5 ⇒ i 3 = 3.5 A Next, express the resistor currents in terms of the mesh currents: Apply KVL to the bottom, left mesh: Apply KVL to the right mesh The 6 A current source supplies 4 ( 3.5 ) − 10 ( 2.5 ) + R (1) = 0 ⇒ R = 11 Ω 3.5 ( 5 ) + 2.5 (10 ) − v = 0 ⇒ v = 42.5 V 6 v = 6 ( 42.5 ) = 255 W Section 4-7 Mesh Current Analysis with Dependent Sources P 4.7-1 Find v2 for the circuit shown in Figure P 4.7-1. Answer: v2 = 10 V Figure P 4.7-1 Solution: Express the controlling voltage of the dependent source as a function of the mesh current v2 = 50 i1 Apply KVL to the right mesh: −100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A v2 = 50 i1 = 10 V (checked using LNAP 8/14/02) P4.7-2 Determine the values of the power supplied by the voltage source and by the CCCS in the circuit shown Figure P4.7-2 Figure P4.7-2 Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents: Apply KVL to the left mesh: 4000 i a + 2000 ( 6 i a ) − 2 = 0 ⇒ i a = The 2 A voltage source supplies The CCCS supplies 2 i a = 2 ( 0.125 × 10−3 ) = 0.25 mW ( 5 i a ) ⎡⎣( 2000 ) ( 6 i a )⎤⎦ = ( 60 ×10 )( 0.125 ×10 ) 3 −3 2 1 = 0.125 mA 8 = 0.9375 × 10−3 = 0.9375 mW P 4.7-3 Find vo for the circuit shown in Figure P 4.7-3. Answer: vo = 2.5 V Figure P 4.7-3 Solution: Express the controlling current of the dependent source as a function of the mesh current: ib = .06 − ia Apply KVL to the right mesh: −100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒ ia = 10 mA vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V Finally: (checked using LNAP 8/14/02) P 4.7-4 Determine the mesh current ia for the circuit shown in Figure P 4.7-4. Answer: ia = –24 mA Figure P 4.7-4 Solution: Express the controlling voltage of the dependent source as a function of the mesh current: vb = 100 (.006 − ia ) Apply KVL to the right mesh: −100 (.006 − ia ) + 3 [100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA (checked using LNAP 8/14/02) P 4.7-5 Although scientists continue to debate exactly why and how it works, the process of utilizing electricity to aid in the repair and growth of bones—which has been used mainly with fractures—may soon be extended to an array of other problems, ranging from osteoporosis and osteoarthritis to spinal fusions and skin ulcers. An electric current is applied to bone fractures that have not healed in the normal period of time. The process seeks to imitate natural electrical forces within the body. It takes only a small amount of electric stimulation to accelerate bone recovery. The direct current method uses an electrode that is implanted at the bone. This method has a success rate approaching 80 percent. The implant is shown in Figure P 4.7-5a and the circuit model is shown in Figure P 4.7-5b. Find the energy delivered to the cathode during a 24-hour period. The cathode is represented by the dependent voltage source and the 100-kΩ resistor. Figure P 4.7-5 Solution: Apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3 Apply KVL to right mesh : 5 ×103 i1 + 100 ×103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2 Solving (1) & ( 2 ) simultaneously ⇒ i1 = 6 3 mA, i2 = mA 55 220 (1) ( 2) Power delivered to cathode = ( 5 i1 ) ( i2 ) + 100 ( i2 )2 ( 55)( 3 220) + 100 ( 3 220) = 0.026 mW ( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J 2 = 5 6 ∴ Energy in 24 hr. = Figure P4.7-6 P4.7-6 Determine the value of the power supplied by the VCCS in the circuit shown Figure P4.7-6. Solution: First, label the mesh currents. Next, express the controlling voltage of the VCCS in terms of the mesh currents: v a = 20 i 2 i1 = 2 A and i 3 = va = 10 i 2 2 Next, express the resistor currents in terms of the mesh currents: Notice that Apply KVL to the middle mesh: Consequently The VCCS supplies 20 i 2 + 2 ( i 2 − 10 i 2 ) + 8 ( i 2 − 2 ) = 0 ⇒ i 2 = 1.6 A v a = 20 i 2 = 20 (1.6 ) = 32 V and i 3 = va 2 = 32 = 16 A 2 ⎡ 2 ( i 3 − i 2 ) ⎤ = 32 ( 2 )(16 − 1.6 ) = 460.8 W ⎦ 2 2 ⎣ va P 4.7-7 The currents i1, i2 and i3 are the mesh currents of the circuit shown in Figure P 4.7-7. Determine the values of i1, i2, and i3. Figure P 4.7-7 Solution: Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh currents v a = 5 ( i 2 − i 3 ) and i b = −i 2 Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh currents 20 i b = −20 i 2 and 3 v a = 15 ( i 2 − i 3 ) Apply KVL to the meshes −15 ( i 2 − i 3 ) + ( −20 i 2 ) + 10 i1 = 0 − ( −20 i 2 ) + 5 ( i 2 − i 3 ) + 20 i 2 = 0 10 − 5 ( i 2 − i 3 ) + 15 ( i 2 − i 3 ) = 0 These equations can be written in matrix form ⎡10 −35 15 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤ ⎢ 0 45 −5 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ 0 10 −10 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦ Solving, e.g. using MATLAB, gives i1 = −1.25 A, i 2 = +0.125 A, and i 3 = +1.125 A (checked: MATLAB & LNAP 5/19/04) P 4.7-8 Determine the value of the power supplied by the dependent source in Figure P 4.7-8. Figure P 4.7-8 Solution: Label the mesh currents: Express ia, the controlling current of the CCCS, in terms of the mesh currents i a = i 3 − i1 Express 2 ia, the controlled current of the CCCS, in terms of the mesh currents: i 1 − i 2 = 2 i a = 2 ( i 3 − i 1 ) ⇒ 3 i1 − i 2 − 2 i 3 = 0 Apply KVL to the supermesh corresponding to the CCCS: 80 ( i1 − i 3 ) + 40 ( i 2 − i 3 ) + 60 i 2 + 20 i1 = 0 ⇒ 100i1 + 100i 2 − 120i 3 = 0 Apply KVL to mesh 3 10 + 40 ( i 3 − i 2 ) + 80 ( i 3 − i1 ) = 0 ⇒ -80 i1 − 40 i 2 + 120 i 3 = −10 These three equations can be written in matrix form −1 −2 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤ ⎡ 3 ⎢100 100 −120 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −80 −40 120 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦ Solving, e.g. using MATLAB, gives i1 = −0.2 A, i 2 = −0.1 A and i 3 = −0.25 A Apply KVL to mesh 2 to get v b + 40 ( i 2 − i 3 ) + 60 i 2 = 0 ⇒ v b = −40 ( −0.1 − ( −0.25 ) ) − 60 ( −0.1) = 0 V So the power supplied by the dependent source is p = v b ( 2i a ) = 0 W . (checked: LNAP 6/7/04) P 4.7-9 Determine the value of the resistance R in the circuit shown in Figure P 4.7-9. Figure P 4.7-9 Solution: Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get 1 mA 6 Apply KVL to the supermesh corresponding to the dependent source to get i b + 0.5 × 10−3 = 4 i b ⇒ ib = ( ) −5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0 ( ) ⎛1 ⎞ −5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25 ⎝6 ⎠ 125 6 R= = 41.67 kΩ 0.5 ×10−3 P 4.7-10 The circuit shown in Figure P 4.7-10 is the small signal model of an amplifier. The input to the amplifier is the voltage source voltage, vs. The output of the amplifier is the voltage vo. (a) The ratio of the output to the input, vo/vs, is called the gain of the amplifier. Determine the gain of the amplifier. (b) The ratio of the current of the input source to the input voltage, ib/vs, is called the input resistance of the amplifier. Determine the input resistance. (checked: LNAP 6/21/04) Figure P 4.7-10 Solution: The controlling and controlled currents of the CCCS, i b and 40 i b, are the mesh currents. Apply KVL to the left mesh to get 1000 i b + 2000 i b + 300 ( i b + 40i b ) − v s = 0 ⇒ 15300i b = v s The output is given by (a) The gain is (b) The input resistance is v o = −3000 ( 40 i b ) = −120000 i b vo vs =− 120000 = −7.84 V/V 15300 vs ib = 15300 Ω (checked: LNAP 5/24/04) P 4.7-11 Determine the values of the mesh currents of the circuit shown in Figure P 4.7-11. Figure P 4.7-11 Solution: Label the mesh currents. Express ix in terms of the mesh currents: i x = i1 Express 4ix in terms of the mesh currents: 4i x = i 3 Express the current source current in terms of the mesh currents to get: 0.5 = i1 − i 2 ⇒ i 2 = i x − 0.5 Apply KVL to supermesh corresponding to the current source to get 5i1 + 20 ( i1 − i 3 ) + 10 ( i 2 − i 3 ) + 25i 2 = 0 Substituting gives 5i x + 20 ( −3i x ) + 10 ( i x − 0.5 − 4i x ) + 25 ( i x − 0.5 ) = 0 ⇒ ix = − 35 = −0.29167 120 So the mesh currents are i1 = i x = −0.29167 A i 2 = i x − 0.5 = −0.79167 A i 3 = 4i x = −1.1667 A (checked: LNAP 6/21/04) P 4.7-12 The currents i1, i2, and i3 are the mesh currents corresponding to meshes 1, 2, and 3 in Figure P 4.7-12. Determine the values of these mesh currents. Figure P 4.7-12 Solution: Express the controlling voltage and current of the dependent sources in terms of the mesh currents: v a = R 3 ( i1 − i 2 ) and i b = i 3 − i 2 Express the current source currents in terms of the mesh currents: i 2 = − I s and i1 − i 3 = B i b = B ( i 3 − i 2 ) Consequently i1 − ( B + 1) i 3 = B I s Apply KVL to the supermesh corresponding to the dependent current source R1 i 3 + A R 3 ( i1 − i 2 ) + R 2 ( i 3 − i 2 ) + R 3 ( i1 − i 2 ) − V s = 0 or ( A + 1) R 3 i1 − ( R 2 + ( A + 1) R3 ) i 2 + ( R1 + R 2 ) i 3 = Vs Organizing these equations into matrix form: ⎡ 0 ⎢ 1 ⎢ ⎢ ⎢⎣( A + 1) R 3 1 0 − ( R 2 + ( A + 1) R 3 ) 0 ⎤ ⎡ i1 ⎤ ⎡ − I s ⎤ ⎥⎢ ⎥ ⎢ ⎥ − ( B + 1) ⎥ ⎢i 2 ⎥ = ⎢ B I s ⎥ ⎥ R1 + R 2 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ V s ⎥⎦ With the given values: ⎡ i1 ⎤ ⎡ −0.8276 ⎤ 1 0 ⎤ ⎡ i1 ⎤ ⎡ −2 ⎤ ⎡0 ⎢ ⎥ ⎢1 ⎥ i = ⎢ 6 ⎥ ⇒ ⎢i ⎥ = ⎢ −2 ⎥ A − 0 4 ⎢ 2⎥ ⎢ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎥ ⎢ i 3 ⎥ ⎢⎣ −1.7069 ⎥⎦ ⎢⎣ 60 −80 50 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 25 ⎥⎦ ⎣ ⎦ (Checked using LNAP 9/29/04) P 4.7-13 The currents i1, i2, and i3 are the mesh currents corresponding to meshes 1, 2, and 3 in Figure P 4.7-13. The values of these currents are i1 = –1.375 A, i2 = –2.5 A, and i3 = –3.25 A Determine the values of the gains of the dependent source, A and B. Figure P 4.7-13 Solution: Express the controlling voltage and current of the dependent sources in terms of the mesh currents: v a = 20 ( i1 − i 2 ) = 20 ( −1.375 − ( −2.5 ) ) = 22.5 and i b = i 3 − i 2 = −3.25 − ( −2.5 ) = −0.75 A Express the current source currents in terms of the mesh currents: i 2 = −2.5 A and i 3 − i1 = B i b ⇒ − 1.375 − ( −2.5 ) = B ( −0.75 ) ⇒ B = 2.5 A/A Apply KVL to the supermesh corresponding to the dependent current source 0 = 20 i 3 + Av a + 50 i b + v a − 10 = 20 ( −3.25 ) + A ( 22.5 ) + 50 ( −0.75 ) + 22.5 − 10 ⇒ A = 4 V/V (Checked using LNAP 9/29/04) P 4.7-14 Determine the current i in the circuit shown in Figure P 4.7-14. Answer: i = 3 A Figure P 4.7-14 Solution: Label the node voltages as shown. The controlling currents of the CCCS is expressed as i = The node equations are and 12 = va 28 + va − vb 4 va − vb 4 + va 14 = + va 28 . va 14 vb 8 Solving the node equations gives v a = 84 V and v b = 72 V . Then i = va 28 = 84 =3A . 28 (checked using LNAP 6/16/05) P4.7-15 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-15. Figure P4.7-15 Solution: Expressing the dependent source currents in terms of the mesh currents we get: i1 = 4 i a = 4 ( i 2 + 1) ⇒ 4 = i1 − 4 i 2 Apply KVL to mesh 2 to get 2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2 Solving these equations using MATLAB we get i1 = −8 A and i2 = −3 A P4.7-16 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-16 . Figure P4.7-16 Apply KVL to mesh 1 to get 2 i1 + 4 i a + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 2 i1 + 4 ( i 2 + 1) + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 8 = 4 i1 + 2 i 2 Apply KVL to mesh 2 to get 2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2 Solving these equations using MATLAB we get i1 = 1.8571 A and i2 = 0.2857 A Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a) Apply KVL to meshes 1 and 2: 32i1 − v s + 96 ( i1 − i s ) = 0 v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is = + vs + vo 120 va − vo 32 + vo 30 = = vs + vo 120 vs 120 + + vo 30 vo 24 Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04) P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages. v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50 = va 10 ⇒ v a = 20 V and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W (b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 ) Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1 ⇒ i1 = 2i 2 Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0 ⇒ ⇒ i 2 − i1 = 2 So i 2 − 2i 2 = 2 i 2 = −2 A ⇒ i 1 = −4 A Then v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04) Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a) Apply KVL to meshes 1 and 2: 32i1 − v s + 96 ( i1 − i s ) = 0 v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is = + vs + vo 120 va − vo 32 + vo 30 = = vs + vo 120 vs 120 + + vo 30 vo 24 Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04) P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages. v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50 = va 10 ⇒ v a = 20 V and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W (b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 ) Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1 ⇒ i1 = 2i 2 Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0 ⇒ ⇒ i 2 − i1 = 2 So i 2 − 2i 2 = 2 i 2 = −2 A ⇒ i 1 = −4 A Then v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04) Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a) Apply KVL to meshes 1 and 2: 32i1 − v s + 96 ( i1 − i s ) = 0 v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is = + vs + vo 120 va − vo 32 + vo 30 = = vs + vo 120 vs 120 + + vo 30 vo 24 Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04) P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages. v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50 = va 10 ⇒ v a = 20 V and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W (b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 ) Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1 ⇒ i1 = 2i 2 Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0 ⇒ ⇒ i 2 − i1 = 2 So i 2 − 2i 2 = 2 i 2 = −2 A ⇒ i 1 = −4 A Then v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04) Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a) Apply KVL to meshes 1 and 2: 32i1 − v s + 96 ( i1 − i s ) = 0 v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is = + vs + vo 120 va − vo 32 + vo 30 = = vs + vo 120 vs 120 + + vo 30 vo 24 Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04) P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages. v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50 = va 10 ⇒ v a = 20 V and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W (b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 ) Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1 ⇒ i1 = 2i 2 Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0 ⇒ ⇒ i 2 − i1 = 2 So i 2 − 2i 2 = 2 i 2 = −2 A ⇒ i 1 = −4 A Then v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04) Section 4.9 Circuit Analysis Using MATLAB Figure P4.9-1 P4.9-1. The encircled numbers in the circuit shown Figure P4.9-1 are node numbers. Determine the values of the corresponding node voltages, v1, v2 and v3. Solution: First, express the resistor currents in terms of the node voltages: Apply KCL at node 1 to get 5 = Apply KCL at node 2 to get Apply KCL at node 3 to get In matrix form: Solving using MATLAB: v1 − v 2 5 v1 − v 2 5 v2 − v3 4 = + + v2 10 v1 − v 3 2 + v2 − v3 v1 − v 3 2 4 ⇒ 0.7 v1 − 0.2 v 2 − 0.5 v 3 = 5 ⇒ − 0.2 v1 + 0.55 v 2 − 0.25 v 3 = 0 = 3 ⇒ − 0.5 v1 − 0.25 v 2 + 0.75 v 3 = −3 ⎡ 0.7 −0.2 −0.5 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤ ⎢ −0.2 0.55 −0.25⎥ ⎢v ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ −0.5 −0.25 0.75 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ −3⎥⎦ v1 = 28.1818 V, v 2 = 20 V and v 3 = 21.4545 Figure P4.9-2 P4.9-2. Determine the values of the node voltages, v1 and v2, in the circuit shown Figure P4.9-2. Solution: First, express the resistor currents in terms of the node voltages: Apply KCL at node 1 to get Apply KCL at node 2 to get In matrix form: Solving using MATLAB: 15 − v1 25 v1 − v 2 50 = = v1 − v 2 50 v2 10 + + v1 + 8 20 v2 + 8 40 ⇒ 0.11v1 − 0.02 v 2 = 0.2 ⇒ − 0.02 v1 + 0.145 v 2 = −0.2 ⎡ 0.11 −0.02 ⎤ ⎡ v1 ⎤ ⎡ 0.2 ⎤ ⎢ −0.02 0.145 ⎥ ⎢v ⎥ = ⎢ −0.2 ⎥ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦ v1 = 1.6077 V and v 2 = −1.1576 V Figure P4.9-3 P4.9-3. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.9-3. Solution: First, express the resistor currents in terms of the node voltages: Apply KCL at node 1 to get Apply KCL at node 2 to get Apply KCL at node 3 to get In matrix form: Solving using MATLAB: −15 − v1 25 v1 − v 2 20 v1 − v 2 50 = + = v1 − v 2 20 v2 25 + v1 − v 3 50 v2 − v3 v 2 − v3 40 + 40 = ⇒ − 0.05 v1 + 0.115 v 2 − 0.025 v 3 = 0 v 3 − 10 10 ⇒ 0.11v1 − 0.05 v 2 − 0.02 v 3 = −0.6 ⇒ − 0.02 v1 − 0.025 v 2 + 0.145 v 3 = 1 ⎡ 0.11 −0.05 −0.02 ⎤ ⎡ v1 ⎤ ⎡ −0.6 ⎤ ⎢ −0.05 0.115 −0.025⎥ ⎢v ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ −0.02 −0.025 0.145 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 1 ⎥⎦ v1 = 1.6077 V and v 2 = −1.1576 V P4.9-4 Determine the node voltages, v1 and v2, for the circuit shown in Figure P4.9-4. Figure P.4.9-4 Solution: Emphasize and label the nodes: Notice the 24 V source connected between node 3 and the reference node. Consequently v 3 = 24 V Apply KCL at node 1 to get v1 8 v1 + v1 − v 2 25 +2=0 is the current directed downward in the 8 Ω resistor and v1 − v 2 is the current 8 25 directed from left to right in the 25 Ω resistor. We will simplify this equation by doing two things: In this equation 1. Multiplying each side by 8 × 25 = 200 to eliminate fractions. 2. Move the terms that don’t involve node voltages to the right side of the equation. The result is 33 v1 − 8 v 2 = −400 Next, apply KCL at node 2 to get v2 9 + v 2 − 24 14 = v1 − v 2 25 In this equation v2 9 is the current directed downward in the 9 Ω resistor, v1 − v 2 v 2 − 24 14 is the current directed is the current directed from left to right in the 25 Ω 25 resistor. We will simplify this equation by doing two things: from left to right in the 14 Ω resistor and 1. Multiplying each side by 8 × 25 × 14 = 2800 to eliminate fractions. 2. Move the terms that involve node voltages to the left side of the equation and move the terms that don’t involve node voltages to the right side of the equation. The result is − ( 9 ×14 ) v1 + ( 9 × 14 + 25 × 14 + 25 × 9 ) v 2 = 24 × 25 × 9 ⇒ − 126 v1 + 701v 2 = 5400 The simultaneous equations can be written in matrix form 33 v1 − 8 v 2 = −400 −8 ⎤ ⎡ v1 ⎤ ⎡ −400 ⎤ ⎡ 33 ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −126 v1 + 701 v 2 = 5400 ⎣ −126 701⎦ ⎣ v 2 ⎦ ⎣ 5400 ⎦ We can use MATLAB to solve the matrix equation: Then ⎡ v1 ⎤ ⎡ −10.7209 ⎤ ⎢v ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ 5.7763 ⎦ That is, the node voltages are v1 = −10.7209 V and v 2 = 5.7763 V . P4.9-5 Determine the mesh currents, i1 and i2, for the circuit shown in Figure P4.9-5. Figure P4.9.5 Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh currents: Notice that the 2 A source on the outside of the circuit is in mesh 3 and that the currents 2 A and i 3 have the same direction. Consequently i3 = 2 A Apply KVL to mesh 1 to get 25 ( i1 − i 3 ) + 9 ( i1 − i 2 ) + 8 i1 = 0 In this equation 25 ( i1 − i 3 ) is the voltage across the 25 Ω resistor (+ on the left), 9 ( i1 − i 2 ) is the voltage across the 9 Ω resistor (+ on top) and 8i1 is the voltage across the 8 Ω resistor (+ on bottom). Substituting i 3 = 2 A and doing a little algebra gives 42 i1 − 9 i 2 = 50 Next, apply KVL to mesh 2 to get 14 ( i 2 − i 3 ) + 24 − 9 ( i1 − i 2 ) = 0 In this equation 14 ( i 2 − i 3 ) is the voltage across the 14 Ω resistor (+ on the left), 24 is the voltage source voltage and 9 ( i1 − i 2 ) is the voltage across the 9 Ω resistor (+ on top). Substituting i 3 = 2 A and doing a little algebra gives −9 i1 + 23 i 2 = −24 + 14 ( 2 ) = 4 The simultaneous equations can be written in matrix form 42 i1 − 9 i 2 = 50 ⎡ 42 −9 ⎤ ⎡ i1 ⎤ ⎡50 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −9 i1 + 23 i 2 = 4 ⎣ −9 23 ⎦ ⎣i 2 ⎦ ⎣ 4 ⎦ We can use MATLAB to solve the matrix equation: Then ⎡ i1 ⎤ ⎡1.3401 ⎤ ⎢ ⎥=⎢ ⎥ ⎣i 2 ⎦ ⎣0.6983⎦ That is, the mesh currents are i1 = 1.3401 A and i 2 = 0.6983 A . P4.9-6 Represent the circuit shown in Figure P4.9-6 by the matrix equation ⎡ a 11 ⎢ ⎣ a 21 a 12 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤ = a 22 ⎦⎥ ⎣⎢ v 2 ⎦⎥ ⎢⎣ −228⎥⎦ Determine the values of the coefficients a11 , a12 , a 21 and a 22 . Figure P4.9-6 Solution: Emphasize and label the nodes: Noticing the 10 V source connected between node 3 and the reference node, we determine that node voltage at node 3 is v 3 = −10 V Apply KCL at node 1 to get In this equation v1 v1 10 + 0.4 + v1 − v 2 10 =0 is the current directed downward in the vertical 10 Ω resistor and v1 − v 2 is the 10 10 current directed from left to right in the horizontal 10 Ω resistor. We will simplify this equation by doing two things: 1. Multiplying each side by 10 to eliminate fractions. 2. Move the terms that don’t involve node voltages to the right side of the equation. 2 v1 − v 2 = −4 The result is Next, apply KCL at node 2 to get v2 19 In this equation v2 19 + v 2 − ( −10 ) 22 = v1 − v 2 10 + 0.4 is the current directed downward in the 19 Ω resistor, directed from left to right in the 22 Ω resistor and v 2 − ( −10 ) v1 − v 2 22 is the current is the current directed from left to right in the 10 horizontal 10 Ω resistor. We will simplify this equation by doing two things: 1. Multiplying each side by 19 × 22 × 10 = 4180 to eliminate fractions. 2. Move the terms that involve node voltages to the left side of the equation and move the terms that don’t involve node voltages to the right side of the equation. The result is − (19 × 22 ) v1 + (19 × 10 + 22 ×10 + 19 × 22 ) v 2 = −10 × 10 × 19 + 0.4 × 19 × 22 × 10 ⇒ − 418 v1 + 828 v 2 = −228 Comparing our equations to the given equations, we see that we need to multiply both sides of our first equation by 10. Then 20 v1 − 10 v 2 = −40 ⎡ 20 −10 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −418 v1 + 828 v 2 = −228 ⎣ −418 828 ⎦ ⎣v 2 ⎦ ⎣ −228⎦ Comparing coefficients gives a 11 = 20 , a 12 = −10 , a 21 = −418 and a 22 = 828 . P4.9-7 Represent the circuit shown in Figure P4.9-7 by the matrix equation ⎡ a 11 ⎢ ⎣ a 21 a12 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤ = a 22 ⎦⎥ ⎣⎢i 2 ⎦⎥ ⎢⎣10 ⎥⎦ Determine the values of the coefficients a11 , a12 , a 21 and a 22 . Figure P4.9-7 Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh currents: Notice that the 0.4 A source on the inside of the circuit is in both mesh 1 and mesh 3. Mesh current i1 is directed in the same way as current source current but the mesh current i 3 is directed opposite to the current source current. Consequently i1 − i 3 = 0.4 A The current source is in both mesh 1 and mesh 3 so we apply KVL to the supermesh corresponding to the current source (i.e. the perimeter of meshes 1 and 3). The result is 10 i 3 + 19 ( i1 − i 2 ) + 10 i1 = 0 In this equation 10i 3 is the voltage across the horizontal 10 Ω resistor (+ on the left), 19 ( i1 − i 2 ) is the voltage across the 19 Ω resistor (+ on top) and 10i1 is the voltage across the vertical 10 Ω resistor (+ on bottom). Substituting i 3 = i1 − 0.4 and doing a little algebra gives 39 i1 − 19 i 2 = 4 Next, apply KVL to mesh 2 to get 22 i 2 − 10 − 19 ( i1 − i 2 ) = 0 In this equation 22i 2 is the voltage across the 22 Ω resistor (+ on the left), 10 is the voltage source voltage and 19 ( i1 − i 2 ) is the voltage across the 19 Ω resistor (+ on top). Doing a little algebra gives −19 i1 + 41i 2 = 10 To summarize, the circuit is represented by the simultaneous equations: 39 i1 − 19 i 2 = 4 ⎡ 39 −19 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −19 i1 + 41i 2 = 10 ⎣ −19 41 ⎦ ⎣i 2 ⎦ ⎣10 ⎦ Comparing these equations to the given equations shows a 11 = 39 , a 12 = −19 , a 21 = −19 and a 22 = 41 . P4.9-8 Determine the values of the power supplied by each of the sources for the circuit shown in Figure P4.9-8. Figure P.4.9-8 Solution: First, label the mesh currents and then label the element currents: Notice the 2.4 A source in both mesh 2 and mesh 3. We have i 3 − i 2 = 2.4 A Apply KVL to mesh 1 to get 40 i1 − 5 ( i 3 − i1 ) − 5 ( i 2 − i1 ) = 0 ⇒ 50 i1 − 5 i 2 − 5 i 3 = 0 Identify the supermesh corresponding to the 2.4 A current source: Apply KVL to the supermesh to get 5 ( i 2 − i1 ) + 5 ( i 3 − i1 ) + 24 + 40 i 2 = 0 ⇒ − 10 i1 + 45 i 2 + 5 i 3 = −24 Writing the mesh equations in matrix form gives ⎡ 0 −1 1 ⎤ ⎡ i1 ⎤ ⎡ 2.4 ⎤ ⎢ 50 −5 −5⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −10 45 5 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ −24 ⎥⎦ Solving using MATLAB: That is, the mesh currents are i1 = 0.1 A, i 2 = −0.7 A and i 3 = 1.7 A. The 24 V source supplies −24 i 3 = ( −24 )(1.7 ) = −40.8 W The power supplied by the current source depends on vs, the voltage across the current source. Apply KVL to mesh 3 to get 5 ( i 3 − i1 ) + 24 − v s = 0 ⇒ v s = 5 (1.7 − 0.1) + 24 = 32 V The current source supplies 2.4 vs = 2.4 ( 32 ) = 76.8 W Figure P4.9-9 P4.9-9 The mesh currents are labeled in the circuit shown Figure 4.9-9. Determine the value of the mesh currents i1 and i 2 . Solution: Determine the value of the mesh currents i1 and i 2 . Replace series resistors with an equivalent resistor and series voltage sources with and equivalent voltage source to get Apply KVL to mesh 1 16 i1 + 8 ( i1 − i 2 ) − 9 = 0 ⇒ 24 i1 − 8 i 2 = 9 Apply KVL to mesh 2 4 i 2 + 5 i1 − 8 ( i1 − i 2 ) = 0 ⇒ − 3 i1 + 12 i 2 = 0 In matrix form Solving using MATLAB ⎡ 24 −8⎤ ⎡ i1 ⎤ ⎡9 ⎤ ⎢ −3 12 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ So the mesh currents are i1 = 0.4091 A and i 2 = 0.1023 A Figure P4.9-10 P4.9-10 The encircled numbers in the circuit shown Figure P4.9-10 are node numbers. Determine the values of the corresponding node voltages, v1 and v2. Solution: Determine the value of the node voltages, v1 and v2. Replace parallel resistors with an equivalent resistor and parallel sources with and equivalent current source to get Apply KCL at node 1 Apply KCL at node 2 In matrix form Solving using MATLAB 2.5 = v1 4.44 v1 − v 2 2 1 ⎡ 1 ⎢ 4.44 + 2 ⎢ ⎢ 1 + 1.5 ⎣⎢ 2 + v1 − v 2 2 + 1.5 v1 = =0 v2 4 1 ⎤ 2 ⎥ ⎡ v1 ⎤ ⎡ 2.5⎤ ⎥⎢ ⎥ = 1 1 ⎥ ⎣v 2 ⎦ ⎢⎣ 0 ⎥⎦ − − 2 4 ⎦⎥ − So the node voltages are v1 = −4.1111 V and v 2 = −10.9630 V Section 4.10 How Can We Check … ? P 4.10-1 Computer analysis of the circuit shown in Figure P 4.10-1 indicates that the node voltages are va = 5.2 V, vb = –4.8 V, and vc = 3.0 V. Is this analysis correct? Hint: Use the node voltages to calculate all the element currents. Check to see that KCL is satisfied at each node. Figure P 4.10-1 Solution: Apply KCL at node b: vb − va v −v 1 − + b c = 0 4 2 5 −4.8 − 5.2 1 − 4.8 − 3.0 − + ≠0 4 2 5 The given voltages do not satisfy the KCL equation at node b. They are not correct. P 4.10-2 An old lab report asserts that the node voltages of the circuit of Figure P 4.10-2 are va = 4 V, vb = 20 V, and vc = 12 V. Are these correct? Solution: Apply KCL at node a: v ⎛v −v ⎞ −⎜ b a ⎟ − 2 + a = 0 2 ⎝ 4 ⎠ 4 ⎛ 20 − 4 ⎞ −⎜ = −4≠ 0 ⎟−2+ 2 ⎝ 4 ⎠ The given voltages do not satisfy the KCL equation at node a. They are not correct. P 4.10-3 Your lab partner forgot to record the values of R1, R2, and R3. He thinks that two of the resistors in Figure P 4.10-3 had values of 10 kΩ and that the other had a value of 5 kΩ. Is this possible? Which resistor is the 5-kΩ resistor? Figure P 4.10-3 Solution: ⎛ 12 − 7.5 ⎞ 7.5 7.5 − 6 −⎜ + =0 ⎟+ R2 ⎝ R1 ⎠ R3 Writing a node equation: 4.5 7.5 1.5 + + =0 R1 R3 R2 There are only three cases to consider. Suppose R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then − So − 4.5 7.5 1.5 −0.9 + 0.75 + 0.15 + + = = 0 R1 R3 R2 1000 This choice of resistance values corresponds to branch currents that satisfy KCL. Therefore, it is indeed possible that two of the resistances are 10 kΩ and the other resistance is 5 kΩ. The 5 kΩ is R1. P 4.10-4 Computer analysis of the circuit shown in Figure P 4.10-4 indicates that the mesh currents are i1 = 2 A, i2 = 4 A, and i3 = 3 A. Verify that this analysis is correct. Hint: Use the mesh currents to calculate the element voltages. Verify that KVL is satisfied for each mesh. Figure P 4.10-4 Solution: Applying KVL to each mesh: Top mesh: Bottom right mesh 10 (2 − 4) + 12(2) + 4 (2 − 3) = 0 8 (3 − 4) + 4 (3 − 2) + 4 = 0 Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source was entered incorrectly.) KVL is not satified for the bottom, left mesh so the computer analysis is not correct. PSpice Problems SP 4-1 Use PSpice to determine the node voltages of the circuit shown in Figure SP 4-1 Figure SP 4-1 Solution: The PSpice schematic after running a “Bias Point” simulation: SP 4-2 Use PSpice to determine the mesh currents of the circuit shown in Figure SP 4-2. Figure SP 4-2 Solution: The PSpice schematic after running a “Bias Point” simulation: From the PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E+00 -2.250E+00 V_V3 -7.500E-01 The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current in the voltage source labeled V3). SP 4-3 The voltages va, vb, vc, and vd in Figure SP 4-3 are the node voltages corresponding to nodes a, b, c and d. The current i is the current in a short circuit connected between nodes b and c. Use PSpice to determine the values of va, vb, vc, and vd and of i. Solution: The PSpice schematic after running a “Bias Point” simulation: The PSpice output file: **** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2 V_V4 0 N01588 12Vdc R_R4 N01588 N01565 4k V_V5 N01542 N01565 0Vdc R_R5 0 N01516 4k V_V6 N01542 N01516 8Vdc I_I1 0 N01565 DC 2mAdc I_I2 0 N01542 DC 1mAdc VOLTAGE SOURCE CURRENTS NAME CURRENT V_V4 V_V5 V_V6 -4.000E-03 2.000E-03 -1.000E-03 From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA. SP 4-4 Determine the current, i, shown in Figure SP 4-4. Answer: i = 0.56 A Figure SP 4-4 Solution: The PSpice schematic after running a “Bias Point” simulation: The PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V7 V_V8 -5.613E-01 -6.008E-01 The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of the circuit. However this current is directed from right to left in the 2 Ω resistor while the current i is directed from left to right. Consequently, i = +5.613 A. Design Problems DP 4-1 An electronic instrument incorporates a 15-V power supply. A digital display is added that requires a 5-V power supply. Unfortunately, the project is over budget and you are instructed to use the existing power supply. Using a voltage divider, as shown in Figure DP 4-1, you are able to obtain 5 V. The specification sheet for the digital display shows that the display will operate properly over a supply voltage range of 4.8 V to 5.4 V. Furthermore, the display will draw 300 mA (I) when the display is active and 100 mA when quiescent (no activity). (a) Select values of R1 and R2 so that the display will be supplied with 4.8 V to 5.4 V under all conditions of current I. (b) Calculate the maximum power dissipated by each resistor, R1 and R2, and the maximum current drawn from the 15-V supply. (c) Is the use of the voltage divider a good engineering solution? If not, why? What problems might arise? Figure DP 4-1 Solution: Model the circuit as: (a) We need to keep v2 across R2 in the range 4.8 ≤ v2 ≤ 5.4 display is active ⎧0.3 A For I = ⎨ ⎩ 0.1 A display is not active KCL at a: v2 − 15 v2 + +I =0 R1 R2 Assumed that maximum I results in minimum v2 and visa-versa. Then ⎧4.8 V v2 = ⎨ ⎩5.4 V when I = 0.3 A when I = 0.1 A Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances 4.8 − 15 4.8 + + 0.3 = 0 R1 R2 5.4 − 15 5.4 + + 0.1 = 0 R1 R2 ⇒ R1 = 7.89 Ω, R2 = 4.83 Ω (b) 15 − 4.8 = 1.292 A ⇒ PR = (1.292)2 (7.89) = 13.17 W 1max 1max 7.89 5.4 )2 ( 5.4 IR = = 1.118 A ⇒ PR = = 6.03 W 2max 2 max 4.83 4.83 maximum supply current = I R = 1.292 A IR = 1max (c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop below 4.8V or rise above 5.4V. The power dissipated in the resistors is excessive. Most of the power from the supply is dissipated in the resistors, not the display. DP 4-2 For the circuit shown in Figure DP 4-2, it is desired to set the voltage at node a equal to 0 V in order to control an electric motor. Select voltages v1 and v2 in order to achieve va = 0 V when v1 and v2 are less than 20 V and greater than zero and R = 2 Ω. Figure DP 4-2 Solution: Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL to the supernode corresponding to the 8 V source: va − v1 R + va vb vb − ( −v 2 ) + + = 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0 R R R ⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0 ⇒ 4 va − v1 + v 2 + 16 = 0 ⇒ va = Next set va = 0 to get 0= v1 − v 2 For example, v1 = 18 V and v2 = 2 V. 4 v1 − v 2 4 − 4 ⇒ v1 − v 2 = 16 V −4 DP 4-3 A wiring circuit for a special lamp in a home is shown in Figure DP 4-3. The lamp has a resistance of 2 Ω, and the designer selects R = 100 Ω. The lamp will light when I ≥ 50 mA but will Figure DP 4-3 burn out when I > 75 mA. (a) Determine the current in the lamp and determine if it will light for R = 100 Ω. (b) Select R so that the lamp will light but will not burn out if R changes by ± 10 percent because of temperature changes in the home. Solution: (a) Apply KCL to left mesh: −5 + 50 i1 + 300 (i1 − I ) = 0 Apply KCL to right mesh: ( R + 2) I + 300 ( I − i1 ) = 0 150 1570 + 35 R We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA ⇒ l amp so the lamp will not light. Solving for I: I= (b) From the equation for I, we see that decreasing R increases I: try R = 50 Ω ⇒ I = 45 mA (won't light) try R = 25Ω ⇒ I = 61 mA ⇒ will light Now check R±10% to see if the lamp will light and not burn out: −10% → 22.5Ω → I = 63.63 mA ⎫ lamp will ⎬ +10% → 27.5Ω → I = 59.23 mA ⎭ stay on DP 4-4 In order to control a device using the circuit shown in Figure DP 4-4, it is necessary that vab = 10 V. Select the resistors when it is required that all resistors be greater than 1 Ω and R3 + R4 = 20 Ω. Figure DP 4-4 Solution: R = R1 || R 2 || ( R 3 + R 4 ) Equivalent resistance: R ( 25 ) 10 + R We require vab = 10 V. Apply the voltage division principle in the left circuit to get: Voltage division in the equivalent circuit: v1 = 10 = ( ) R1 R 2 ( R3 + R4 ) R4 R4 v1 = × × 25 R3 + R4 R3 + R4 10 + R1 R 2 ( R3 + R4 ) ( ) This equation does not have a unique solution. Here’s one solution: choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω then 10 = (12.5 20 ) × 25 ⇒ R = 18.4Ω R4 × 4 20 10 + (12.5 20 ) and R3 + R4 = 20 ⇒ R3 = 1.6 Ω DP 4-5 The current i shown in the circuit of Figure DP 4-5 is used to measure the stress between two sides of an earth fault line. Voltage v1 is obtained from one side of the fault, and v2 is obtained from the other side of the fault. Select the resistances R1, R2, and R3 so that the magnitude of the current i will remain in the range between 0.5 mA and 2 mA when v1 and v2 may each vary independently between +1 V and +2 V (1 V ≤ vn ≤ 2 V). Solution: Apply KCL to the left mesh: (R 1 + R 3 ) i1 − R 3 i2 − v1 = 0 Apply KCL to the left mesh: (R − R 3 i1 + 2 + R 3 ) i2 + v2 = 0 Solving for the mesh currents using Cramer’s rule: − R3 ⎡v1 ⎤ ⎡( R1 + R 3 ) ⎢ ⎥ ⎢−R ⎢⎣ −v2 ( R 2 + R 3 ) ⎥⎦ 3 ⎣ and i2 = i1 = Δ Δ 2 where Δ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3 v1 ⎤ − v2 ⎦⎥ Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then Δ = 3 MΩ. The mesh currents will be given by i = [ 2v1 − v2 ] 1000 and i 2 = [ −2v2 + v1 ] 1000 3 × 10 3 ×10 Now check the extreme values of the source voltages: 1 6 6 if v1 = v2 = 1 V ⇒ i = 2 mA 3 if v1 = v2 = 2 V ⇒ i = 4 mA 3 ⇒ i = i1 − i2 = okay okay v1 + v2 3000 Chapter 5 Circuit Theorems Exercises R Exercise 5.2-1 Determine values of R and is so that the circuits shown in Figures E 5.2-1a,b are equivalent to each other due to a source transformation. + – Answer: R = 10 Ω and is = 1.2 A 12 V 10 Ω is (a) (b) Figures E 5.2-1 Exercise 5.2-2 Determine values of R and is so that the circuits shown in Figures E 5.2-2a,b are equivalent to each other due to a source transformation. R – + Hint: Notice that the polarity of the voltage source in Figure E 5.2-2a is not the same as in Figure E 5.2-1a. 12 V 10 Ω is (b) (a) Figures E 5.2-2 Answer: R = 10 Ω and is = –1.2 A 8Ω Exercise 5.2-3 Determine values of R and vs so that the circuits shown in Figures E 5.2-3a,b are equivalent to each other due to a source transformation. + – Answer: R = 8 Ω and vs = 24 V vs R 3A (a) (b) Figure E 5.2-3 Exercise 5.2-4 Determine values of R and vs so that the circuits shown in Figures E 5.2-4a,b are equivalent to each other due to a source transformation. Hint: Notice that the reference direction of the current source in Figure E 5.2-4b is not the same as in Figure E 5.2-3b. Answer: R = 8 Ω and vs = –24 V 8Ω + – vs R 3A (b) (a) Figure E 5.2-4 3Ω Exercise 5.4-1 Determine values of Rt and voc that cause the circuit shown in Figure E 5.4-1b to be the Thévenin equivalent circuit of the circuit in Figure E 5.4-1a. + – 6Ω + – 6Ω 3V Answer: Rt = 8 Ω and voc = 2 V b (b) Figure E 5.2-1 a voc b (a) Solution: Rt a 6Ω Exercise 5.4-2 Determine values of Rt and voc that cause the circuit shown in Figure E 5.4-2b to be the Thévenin equivalent circuit of the circuit in Figure E 5.4-2a. 12 V 3Ω + – + – ia Rt a + – 2ia voc b Answer: Rt = 3 Ω and voc = –6 V (a) Solution: ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = −6 =3Ω −2 b (b) Figure E 5.2-2 ⇒ i a = −3 A ⇒ i sc = 2 ( − 3 ) = −2 A 3 a 3Ω Exercise 5.5-1 Determine values of Rt and isc that cause the circuit shown in Figure E 5.5-1b to be the Norton equivalent circuit of the circuit in Figure E 5.5-1a. Answer: Rt = 8 Ω and isc = 0.25 A + – 6Ω a 6Ω 3V isc Rt b (a) b (b) Figure E 5.5-1 Solution: a 3Ω Exercise 5.6-1 Find the maximum power that can be delivered to RL for the circuit of Figure E 5.6-1 using a Thévenin equivalent circuit. Answer: 9 W when RL = 4 Ω 18 V + – 2Ω 6Ω Figure E 5.6-1 Solution: voc = 6 (18) = 12 V 6+3 Rt = 2 + ( 3)( 6 ) = 4 Ω 3+ 6 For maximum power, we require R L = Rt = 4 Ω Then 2 pmax = voc 122 = =9 W 4 Rt 4 ( 4 ) RL Section 5-2: Source Transformations P 5.2-1 The circuit shown in Figure P 5.2-1a has been divided into two parts. The circuit shown in Figure P 5.2-1b was obtained by simplifying the part to the right of the terminals using source transformations. The part of the circuit to the left of the terminals was not changed. 4Ω i 2V 2Ω – + + 9V + – v ia 4Ω 0.5 A – (a) (a) Determine the values of Rt and vt in Figure P 5.2-1b. (b) + Determine the values of the + v vt current i and the voltage v in 9 V +– – Figure P 5.2-1b. The circuit in ia – Figure P 5.2-1b is equivalent to the circuit in Figure P 5.2-1a. (b) Consequently, the current i and Figure P 5.2-1 the voltage v in Figure P 5.2-1a have the same values as do the current i and the voltage v in Figure P 5.2-1b. (c) 4Ω Determine the value of the current ia in Figure P 5.2-1a. Solution: (a) i Rt 2Ω ∴ Rt = 2 Ω (b) −9 − 4i − 2i + (−0.5) = 0 −9 + (−0.5) i = = −1.58 A 4+2 v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V (c) ia = i = − 1.58 A vt = − 0.5 V (checked using LNAP 8/15/02) 8Ω P 5.2-2 Consider the circuit of Figure P 5.2-2. Find ia by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find ia. 3Ω 10 V + – ia 6Ω 2A 4Ω Figure P 5.2-2 Solution: Finally, apply KVL: −10 + 3 ia + 4 ia − 16 =0 3 ∴ ia = 2.19 A (checked using LNAP 8/15/02) 3A P 5.2-3 Find vo using source transformations if i = 5/2 A in the circuit shown in Figure P 5.2-3. Hint: Reduce the circuit to a single mesh that contains the voltage source labeled vo. 6Ω 8V 3Ω 10 Ω +– 2A 16 Ω Answer: vo = 28 V 12 Ω 20 Ω v0 7Ω i +– Figure P 5.2-3 Solution: Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left: Finally, apply KVL to loop − 6 + i (9 + 19) − 36 − vo = 0 i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02) P 5.2-4 Determine the value of the current ia in the circuit shown in Figure P 5.2-4. 6 kΩ 10 V 4 kΩ 4 kΩ +– ia + – 12 V 4 kΩ 3 kΩ – + 6V Figure P 5.2-4 Solution: − 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0 ∴ ia = 375 μ A (checked using LNAP 8/15/02) 4A P 5.2-5 Use source transformations to find the current ia in the circuit shown in Figure P 5.2-5. 6Ω 6V – + Answer: ia = 1 A 12 V + – 1A ia 3Ω Figure P 5.2-5. Solution: −12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A (checked using LNAP 8/15/02) P 5.2-6 Use source transformations to find the value of the voltage va in Figure P 5.2-6. Answer: va = 7 V 8V 100 Ω + – + – 10 V + va – 100 Ω 100 Ω 30 mA Figure P 5.2-6 Solution: A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit: Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: Finally, va = 50 (100 ) 100 ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02) P5.2-7 Figure P5.2-7 The equivalent circuit in Figure P5.2-7 is obtained from the original circuit using source transformations and equivalent resistances. (The lower case letters a and b identify the nodes of the capacitor in both the original and equivalent circuits.) Determine the values of Ra, Va, Rb and I b in the equivalent circuit. Solution Performing a source transformation at each end of the circuit yields Thenx where V a = 2.2 (10 ) − 32 = −10 V , R a = 18 + 10 = 28 Ω , R b = 18 || 9 = 6 Ω and I b = 2.5 + 36 18 = 4.5 A P 5.2-8 The circuit shown in Figure P 5.2-8 contains an unspecified resistance R. (a) Determine the value of the current i when R = 4 Ω. (b) Determine the value of the voltage v when R = 8 Ω. (c) Determine the value of R that will cause i = 1 A. (d) Determine the value of R that will cause v = 16 V. i R + 24 Ω + – 12 V v – 18 Ω 2A 24 Ω Figure P 5.2-8 Solution: Replace series and parallel resistors by an equivalent resistor. 18 & (12 + 24 ) = 12 Ω Do a source transformation, then replace series voltage sources by an equivalent voltage source. 12 Ω Do two more source transformations Now current division gives 24 ⎛ 8 ⎞ i=⎜ ⎟3 = 8+ R ⎝8+ R ⎠ Then Ohm’s Law gives 24 R v = Ri = 8+ R (a ) i= 24 =2A 8+ 4 (b) v = 24 ( 8 ) = 12 V 8+8 (c) 1 = 24 8+ R (d) 16 = 24 R 8+ R ⇒ ⇒ R = 16 Ω R = 16 Ω (checked: LNAP 6/9/04) 15 Ω P 5.2-9 Determine the value of the power supplied by the current source in the circuit shown in Figure P 5.2-9. + 24 V – 2A 25 Ω 24 Ω – 32 V + Figure P 5.2-9 Solution: Use source transformations and equivalent resistances to reduce the circuit as follows The power supplied by the current source is given by p = ⎡⎣ 23.1 + 2 (10.3125 ) ⎤⎦ 2 = 87.45 W 12 Ω Section 5-3 Superposition P5.3-1 The inputs to the circuit shown in Figure P5.3-1 are the voltage source voltages v1 and v2.. The output of the circuit is the voltage vo. The output is related to the inputs by vo = a v1 + b v2 where a and b are constants. Determine the values of a and b. Figure P5.3-1 Solution: Let vo1 = a v1 be the output when v2 = 0. In this case, the right voltage source acts like a short circuit so we have the circuit show to the right. Then v o1 = 20 || 5 4 1 1 v1 = v1 = v1 ⇒ a = 20 + ( 20 || 5 ) 20 + 4 6 6 Let vo2 = b v2 be the output when v1 = 0. In this case, the left voltage source acts like a short circuit so we have the circuit show to the right. Then v o2 = 20 || 20 10 2 v2 = v2 = v2 5 + ( 20 || 20 ) 5 + 10 3 ⇒ b= 2 3 P5.3-2 A particular linear circuit has two inputs, v1 and v2, and one output, vo. Three measurements are made. The first measurement shows that the output is vo = 4 V when the inputs are v1 = 2 V and v2 = 0. The second measurement shows that the output is vo = 10 V when the inputs are v1 = 0 and v2 = −2.5 V. In the third measurement the inputs are v1 = 3 V and v2 = 3 V. What is the value of the output in the third measurement? Solution: The output of a linear circuit is a linear combination of the inputs: v o = a1v1 + a 2 v 2 From the first two measurements we have: ⎛ a1 ⎞ ⎛ 2 ⎞ 0 ⎞ ⎛ a1 ⎞ ⎛ 4 ⎞ ⎛2 ⎜ ⎟=⎜ ⎟⎜ a ⎟ ⇒ ⎜ a ⎟ = ⎜ ⎟ ⎝10 ⎠ ⎝ 0 −2.5 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ −4 ⎠ Now the output of the third measurement can be determine to be v o = a 1 ( 3) + a 2 ( 3) = ( 2 )( 3) + ( −4 )( 3) = −6 V P5.3-3 The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is related to the inputs by the equation io = a is + b vs Given the following two facts: The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V. and The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V. Determine the values of the constants a and b and the values of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω. Figure P5.3-3 Solution: From the 1st fact: 0.45 = a ( 0.25 ) + b (15 ) From the 2nd fact: 0.30 = a ( 0.50 ) + b ( 0 ) ⇒ a = 0.30 = 0.60 0.50 0.45 = ( 0.60 )( 0.25 ) + b (15 ) ⇒ b = Substituting gives Next, consider the circuit: a i s = i o1 = i o so 0.60 = R1 + R 2 b v s = i o2 = i o and so R1 0.02 = 1 R1 + R 2 v s =0 ⎛ R1 ⎞ i =⎜ ⎜ R1 + R 2 ⎟⎟ s ⎝ ⎠ ⇒ 2 R1 = 3 R 2 i s =0 = vs R1 + R 2 ⇒ R1 + R 2 = 1 = 50 Ω 0.02 Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω. 0.45 − ( 0.60 )( 0.25 ) = 0.02 15 P 5.3-4 Use superposition to find the value of 10 Ω the voltage v in Figure P 5.3-4. 9A 20 Ω + v 6A 15 Ω – Figure P 5.3-4 Solution: Consider 6 A source only (open 9 A source) Use current division: v1 ⎡ 15 ⎤ = 6 ⎢ ⇒ v1 = 40 V 20 ⎣15 + 30 ⎥⎦ Consider 9 A source only (open 6 A source) Use current division: v2 ⎡ 10 ⎤ = 9 ⎢ ⇒ v2 = 40 V 20 ⎣10 + 35 ⎥⎦ ∴ v = v1 + v2 = 40 + 40 = 80 V (checked using LNAP 8/15/02) Figure P5.3-5 P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5. Solution; We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t). Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate 40||10 = 8 Ω. Next, using voltage division we calculate 8 v1 ( t ) = (12 ) = 2 V 8 + 40 Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first determine 40||40 = 20 Ω and then calculate 20 (12 cos ( 5t ) ) = 8cos ( 5t ) V 10 + 20 v ( t ) = v1 ( t ) + v 2 ( t ) = 2 + 8cos ( 5 t ) V v 2 (t ) = Using superposition 15 mA P 5.3-6 Use superposition to find the value of the current i in Figure P 5.3-6. 4 kΩ Answer: i = 3.5 mA 15 V +– 2 kΩ 30 mA 12 kΩ 6 kΩ i Figure P5.3-6 Solution: Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source. ⎛ 2 ⎞ ⎛ 6 ⎞ ia = 30 ⎜ ⎟ = 6 mA ⇒ i1 = ia ⎜ ⎟ = 2 mA ⎝ 2+8⎠ ⎝ 6 + 12 ⎠ Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source. ⎛ 4 ⎞ ib = 15 ⎜ ⎟ = 6 mA ⇒ ⎝ 4+6⎠ ⎛ 6 ⎞ i2 = ib ⎜ ⎟ = 2 mA ⎝ 6 + 12 ⎠ Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source. ⎛ 6 || 6 ⎞ ⎛ 3 ⎞ i3 = − 2.5 ⎜⎜ ⎟⎟ = − 10 ⎜ ⎟ = −0.5 mA ⎝ 3 + 12 ⎠ ⎝ ( 6 || 6 ) + 12 ⎠ Finally, i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA (checked using LNAP 8/15/02) Figure P5.3-7 P5.3-7 Determine v(t), the voltage across the 40 Ω resistor in the circuit in Figure P5.3-7. Solution: We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t). Using voltage division we calculate v1 ( t ) = − 40 (12 + 15cos (8 t ) ) = −9.6 − 12 cos (8t ) 10 + 40 Using equivalent resistance we first determine 10||40 = 8 Ω and then calculate v 2 ( t ) = 8 (1 + sin ( 5t ) ) = 8 + 8sin ( 5t ) Using superposition v ( t ) = v1 ( t ) + v 2 ( t ) = −1.6 + 8sin ( 5 t ) − 12 cos ( 8 t ) V ix P 5.3-8 Use superposition to find the value of the current ix in Figure P 5.3-8. Answer: i = 3.5 mA + – 6Ω 3Ω 8V 2A + – 3ix Figure P5.3-8 Solution: Consider 8 V source only (open the 2 A source) Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh: 6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0 8 2 = A 12 3 Let i2 be the part of ix due to the 2 A current source. i1 = Consider 2 A source only (short the 8 V source) Apply KVL to the supermesh: 6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 i2 = Finally, i x = i1 + i 2 = 2 1 1 − = A 3 2 6 −6 1 =− A 12 2 P 5.3-9 The input to the circuit shown in Figure P 5.3-9 is the voltage source voltage, vs. The output is the voltage vo. The current source current, ia, is used to adjust the relationship between the input and output. Design the circuit so that input and output are related by the equation vo = 2vs + 9. Hint: Determine the required values of A and ia. A ix 6Ω + – ix + – vs 12 Ω 12 Ω ia + vo − Figure P 5.3-9 Soluton: ix = vs − va R1 va − vo = A ix = A va = vs − va R1 R1 v o + A v s R1 + A Apply KCL to the supernode corresponding to the CCVS to get va − vs R1 R1 + R 2 R1 R 2 + va R2 va − + ia + vs R1 vo R3 + ia + =0 vo R3 =0 R1 + R 2 ⎛ R 1 v o + A v s ⎞ v s vo + ia + =0 ⎜ ⎟− R1 R 2 ⎜⎝ R1 + A ⎟⎠ R1 R3 ( ) ⎛ R +R ⎛ R1 + R 2 A 1 ⎞⎟ 1 ⎞ 1 2 ⎜ + vo + ⎜ − ⎟ vs + ia = 0 ⎜ R 2 R1 + A R 3 ⎟ ⎜ R1 R 2 R1 + A R1 ⎟ ⎝ ⎠ ⎝ ⎠ ( ( ) ) ( R 2 R 3 ( R1 + A ) R 3 R1 + R 2 + R 2 R 1 + A )v ( o+ ) A − R2 ( R 2 R1 + A ) vs + ia = 0 vo = ( R 2 R 3 ( R1 + A ) ) vs − ia R 3 ( R1 + R 2 ) + R 2 ( R 1 + A ) R 3 ( R 1 + R 2 ) + R 2 ( R1 + A ) R3 R 2 − A When R1 = 6 Ω, R 2 = 12 Ω and R 3 = 12 Ω vo = 12 ( 6 + A ) 12 − A vs − ia 24 + A 24 + A Comparing this equation to v o = 2 v s + 9 , we requires 12 − A =2 ⇔ 24 + A A = −12 V A Then 2 v s + 9 = v o = 2v s + 6i a so we require 9 = 6ia ⇒ i a = 1.5 A (checked: LNAP 6/22/04) P 5.3-10 The circuit shown in Figure P 5.3-10 has three inputs: v1, v2, and i3. The output of the circuit is vo. The output is related to the inputs by vo = av1 + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c. v2 + 8Ω – + + – v1 40 Ω 10 Ω vo i3 – Figure P 5.3-10 Solution: vo1 = vo2 = − 40 ||10 1 1 v1 = v1 ⇒ a = 8 + 40 ||10 2 2 10 3 v1 = − v 2 8 || 40 + 10 5 ⇒ b=− 3 5 vo3 = ( 8 ||10 || 40 ) i 3 = 4 i 3 ⇒ c = 4 (checked: LNAP 6/22/04) P 5.3-11 Determine the voltage vo(t) for the circuit shown in Figure P 5.3-11. – + 12 cos 2t V 4 ix 10 Ω + 40 Ω + – 2 V 10 Ω ix 5Ω vo(t) – Figure P 5.3-11 Solution: Using superposition: v x = 10 i x and v x − 12 cos 2t 40 + vx 10 + vx 10 = 4ix so 10 i x − 12 cos 2t 40 = 2ix ⇒ ix = − 12 cos 2t 70 Finally, v o1 = −5 4 i x = 3.429 cos 2t V ( ) v x = 10 i x and vx 40 + vx − 2 10 + vx 10 = 4ix so −0.2 = 1.75 i x Finally, ⇒ i x = −0.11429 A ( ) v o1 = −5 4 i x = 2.286 V v o = v o1 + v o2 = 3.429 cos 2t + 2.286 V (checked: LNAP 6/22/04) P 5.3-12 Determine the value of the voltage vo in the circuit shown in Figure P 5.3-12. 96 Ω 32 Ω 20 V + 0.3 A – 120 Ω 30 Ω + vo – Figure P 5.3-12 Solution: Using superposition: v o1 = 24 ( 0.3) = 7.2 V v o2 = − 30 20 = −4 V 120 + 30 v o = v o1 + v o 2 = 3.2 V (checked: LNAP 5/24/04) P 5.3-13 Determine the value of the voltage vo in the circuit shown in Figure P 5.3-13. – a vo + b 2Ω i1 R 4Ω 8Ω 8Ω i2 Figure P 5.3-13 Solution: Using superposition ⎛ R || 4 ⎞ ⎛ ⎞ 4 v o = −2 ⎜⎜ ⎟⎟ i1 + 2 ⎜⎜ ⎟⎟ i 2 6 + R || 4 2 + R || 4 + 4 ( ) ( ) ⎝ ⎠ ⎝ ⎠ Comparing to v o = −0.5 i1 + 4 , we require ⎛ R || 4 ⎞ −2 ⎜⎜ ⎟⎟ = −0.5 ⇒ 4 ( R || 4 ) = 6 + ( R || 4 ) ⇒ R || 4 = 2 ⇒ R = 4 Ω ⎝ 6 + ( R || 4 ) ⎠ and ⎛ ⎞ ⎛ ⎞ 4 4 2 ⎜⎜ ⎟⎟ i 2 = 4 ⇒ 2 ⎜⎜ ⎟⎟ i 2 = 4 ⇒ i 2 = 4 A ⎝ 2 + ( R || 4 ) + 4 ⎠ ⎝ 2 + ( 4 || 4 ) + 4 ⎠ (checked LNAP 6/12/04) P 5.3-14 Determine values of the current, ia, and the resistance, R, for the circuit shown in Figure P 5.3-14. + 8V – ia 5 kΩ 20 kΩ 7 mA 4 kΩ Figure P 5.3-14 Solution: Use units of mA, kΩ and V. 4 + (5||20) = 8 kΩ (a) Using superposition 8 ⎛ 8 ⎞ 2=⎜ ⇒ 2 ( R + 8 ) = 48 ⇒ R = 16 kΩ ⎟7− R +8 ⎝ R +8⎠ (b) Using superposition again 8 ⎤ 4⎛ 2 1⎞ ⎛ 5 ⎞ ⎡⎛ 16 ⎞ ia = ⎜ = ⎜ × 7 + ⎟ = 4 mA ⎟ ⎢⎜ ⎟7+ ⎥ 8 + 16 ⎦ 5 ⎝ 3 3⎠ ⎝ 5 + 20 ⎠ ⎣⎝ 8 + 16 ⎠ R 2 mA 20 Ω P 5.3-15 The circuit shown in Figure P 5.3-15 has three inputs: v1, i2, and v3. The output of the circuit is the v1 current io. The output of the circuit is related to the inputs by + – i2 io 12 Ω i1 = avo + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c. 40 Ω v3 – + 10 Ω Figure P 5.3-15 Solution: ⎞ ⎞ ⎛ v1 10 ⎞ ⎛ 10 ⎞ ⎛ 20 ⎛ + − io = ⎜ − ⎜ ⎟ i2 ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ 10 + 40 ⎠ ⎜⎝ 20 + 12 + ( 40 & 10 ) ⎟⎠ ⎝ 10 + 40 ⎠ ⎜⎝ 20 + ⎡⎣12 + ( 40 & 10 ) ⎤⎦ ⎟⎠ ⎞ ⎛ ⎞⎛ v3 20 + 12 + ⎜⎜ − ⎟ ⎟⎟ ⎜⎜ ⎟ ⎝ 40 + ( 20 + 12 ) ⎠ ⎝ 10 + ⎡⎣ 40 & ( 20 + 12 ) ⎤⎦ ⎠ ⎛ 1 ⎞ ⎛ 1⎞ ⎛ 1 ⎞ io = ⎜ − ⎟ v1 + ⎜ − ⎟ i 2 + ⎜ − ⎟ v3 ⎝ 200 ⎠ ⎝ 10 ⎠ ⎝ 62.5 ⎠ So a = −0.05, b = −0.1 and c = −0.016 (checked: LNAP 6/19/04) 25 V P 5.3-16 Using the superposition principle, find the value of the current measured by the ammeter in Figure P 5.3-16a. 5A Hint: Figure P 5.3-16b shows the circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter, im. Answer: 25 3 im = − 5 = 5−3 = 2 A 3+ 2 2+3 Ammeter –+ 3Ω 2Ω (a) 25 V –+ 3Ω 5A 2Ω (b) Figure P 5.3-16 Solution: im = 25 3 − ( 5) = 5 − 3 = 2 A 3+ 2 2+3 im Section 5-4: Thèvenin’s Theorem P 5.4-1 Determine values of Rt and voc that cause the circuit shown in Figure P 5.4-1b to be the Thévenin equivalent circuit of the circuit in Figure P 5.4-1a. Hint: Use source transformations and equivalent resistances to reduce the circuit in Figure P 5.4-1a until it is the circuit in Figure P 5.4-1b. Answer: Rt = 5 Ω and voc = 2 V 3Ω + – 12 V 3Ω 6Ω Rt a + – 3A voc b (a) a b (b) Figure P 5.4-1 Solution: (checked using LNAP 8/15/02) P 5.4-2 The circuit shown in Figure P 5.4-2b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-2a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = –12 V and Rt = 16 Ω 10 Ω 15 V 8Ω 40 Ω + – (a) Rt + – voc (b) Figure P 5.4-2 Solution: The circuit from Figure P5.4-2a can be reduced to its Thevenin equivalent circuit in four steps: (a) (b) (c) (d) Comparing (d) to Figure P5.4-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V. P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = 2 V and Rt = 4 Ω 12 V Rt –+ 1A 6Ω 6Ω + – voc 6Ω (a) (b) Figure P 5.4-3 Solution: The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps: (a) (c) (b) (d) (e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02) 12 Ω P 5.4-4 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-4. 6Ω + – 10 Ω a 3Ω 18 V b Figure P 5.4-4 Find Rt: Rt = Write mesh equations to find voc: 12 (10 + 2 ) =6Ω 12 + (10 + 2 ) Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i1 = 18 36 i1 = 18 ⇒ i1 = i2 = Finally, 1 A 2 14 ⎛ 1 ⎞ 7 ⎜ ⎟= A 3 ⎝2⎠ 3 ⎛7⎞ ⎛1⎞ voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V ⎝ 3⎠ ⎝ 2⎠ (checked using LNAP 8/15/02) 0.75va P 5.4-5 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-5. 8Ω Answer: voc = –2 V and Rt = –8/3 Ω + – 6V a – va + 4Ω b Figure P 5.4-4 Solution: Find voc: Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a: ⎛ 6 − voc ⎞ voc ⎛ 3 ⎞ −⎜ + ⎜ − voc ⎟ = 0 ⎟+ ⎝ 8 ⎠ 4 ⎝ 4 ⎠ −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces Apply KCL at node a: va = 0 ⎛ 6−0⎞ 0 ⎛ 3 ⎞ −⎜ ⎟ + + ⎜ − 0 ⎟ + i sc = 0 ⎝ 8 ⎠ 4 ⎝ 4 ⎠ i sc = Rt = 6 3 = A 8 4 voc −2 8 = =− Ω 3 i sc 3 4 (checked using LNAP 8/15/02) 3Ω 3Ω P 5.4-6 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-6. 2va + – 6Ω + va – a 3A b Figure P 5.4-6 Solution: Find voc: 2 va − va va = + 3 + 0 ⇒ va = 18 V 3 6 The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Apply KCL at the top, middle node: Find isc: 2 va − va va v = + 3 + a ⇒ va = −18 V 3 6 3 v −18 Apply Ohm’s law to the right-hand 3 Ω resistor : = −6 V i sc = a = 3 3 v 18 R t = oc = = −3 Ω Finally: i sc −6 Apply KCL at the top, middle node: (checked using LNAP 8/15/02) P5.4-7 The equivalent circuit in Figure P5.4-7 is obtained by replacing part of the original circuit by its Thevenin equivalent circuit. The values of the parameters of the Thevenin equivalent circuit are v oc = 15 V and R t = 60 Ω Determine the following: a.) The values of V s and R a . (Three resistors in the original circuit have equal resistance, Ra.) b.) The value of R b required to cause i = 0.2 A. c.) The value of R b required to cause v = 12 V. Figure P5.4-7 Solution: a.) From We see that v oc = b.) i = c.) v = v oc Rt + Rb Rb Rt + Rb Vs 2 and R t = R a . With the given values of voc and Rt we calculate 5 5 V 2 15 = s ⇒ Vs = 75 V and 60 = R a ⇒ R a = 150 Ω . 5 5 ⇒ 0.2 = v oc 15 60 + R b ⇒ 12 = ⇒ R b = 15 Ω 15 R b 60 + R b ⇒ R b = 240 Ω P 5.4-8 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-8. The voltage, v, was measured. The resistance was changed, and the voltage was measured again. The results are shown in the table. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ. i + Circuit v – R R v 2 kΩ 4 kΩ 6V 2V Figure P 5.4-8 Solution: From the given data: 2000 ⎫ voc ⎪ R t + 2000 ⎪ ⎧ voc = 1.2 V ⎬ ⇒ ⎨ 4000 ⎩ R t = −1600 Ω voc ⎪ 2= R t + 4000 ⎪⎭ 6= When R = 8000 Ω, v= R voc Rt + R v= 8000 (1.2 ) = 1.5 V −1600 + 8000 P 5.4-9 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-9. The current, i, was measured. The resistance was changed, and the current was measured again. The results are shown in the table. (a) (b) Hint: i R i 2 kΩ 4 kΩ 4 mA 3 mA + Circuit Specify the value of R required to cause i = 2 mA. Given that R > 0, determine the maximum possible value of the current i. v R – Figure P 5.4-9 Use the data in the table to represent the circuit by a Thévenin equivalent. Solution: From the given data: voc ⎫ R t + 2000 ⎪⎪ ⎧ voc = 24 V ⎬ ⇒ ⎨ voc ⎩ R t = 4000 Ω ⎪ 0.003 = R t + 4000 ⎪⎭ 0.004 = i= voc Rt + R (a) When i = 0.002 A: 24 ⇒ R = 8000 Ω 0.002 = 4000 + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA ⇒ i ≤ 6 mA 4000 P 5.4-10 For the circuit of Figure P 5.4-10, specify the resistance R that will cause current ib to be 2 mA. The current ia has units of amps. Hint: Find the Thévenin equivalent circuit of the circuit connected to R. 2000ia 6 kΩ + – 12 V + – 1 kΩ ia ib R Figure P 5.4-10 Solution: −12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A voc = 1000 ia = 4 V 3 ia = 0 due to the short circuit −12 + 6000 isc = 0 ⇒ isc = 2 mA 4 voc Rt = = 3 = 667 Ω isc .002 4 3 ib = 667 + R ib = 0.002 A requires 4 3 − 667 = 0 R = 0.002 (checked using LNAP 8/15/02) 4i a P 5.4-11 For the circuit of Figure P 5.4-11, specify the value of the resistance RL that will cause current iL to be –2 A. + – 10 A 2Ω iL RL i Answer: RL = 12 Ω b Figure P 5.4-11 Solution: 10 = i + 0 ⇒ i = 10 A voc + 4 i − 2 i = 0 ⇒ voc = −2 i = −20 V i + i sc = 10 ⇒ i = 10 − i sc 4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A Rt = −2 = iL = voc −20 = = −2 Ω isc 10 −20 ⇒ RL = 12 Ω RL − 2 (checked using LNAP 8/15/02) P 5.4-12 The circuit shown in Figure P 5.4-12 contains an adjustable resistor. The resistance R can be set to any value in the range 0 ≤ R ≤ 100 kΩ. (a) Determine the maximum value of the current ia that can be obtained by adjusting R. Determine the corresponding value of R. (b) Determine the maximum value of the voltage va that can be obtained by adjusting R. Determine the corresponding value of R. (c) Determine the maximum value of the power supplied to the adjustable resistor that can be obtained by adjusting R. Determine the corresponding value of R. ia + + – 12 kΩ R 12 V va − 2 mA 18 kΩ 24 kΩ Figure P 5.4-12 Solution: Replace the part of the circuit that is connected to the variable resistor by its Thevenin equivalent circuit: 18 kΩ || (12 kΩ + 24 kΩ ) = 18 kΩ || 36 kΩ = 12 kΩ ia = R 36 36 and v a = R + 12000 R + 12000 2 36 ⎛ ⎞ p = ia va = ⎜ ⎟ R ⎝ R + 12000 ⎠ 36 = 3 mA when R = 0 Ω (a short circuit). 0 + 12000 105 36 = 32.14 V when R is as large as possible, i.e. R = 100 kΩ. (b) v a = 5 10 + 12000 (c) Maximum power is delivered to the adjustable resistor when R = R t = 12 kΩ . Then (a) i a = 2 36 ⎛ ⎞ p = ia va = ⎜ ⎟ 12000 = 0.027 = 27 mW ⎝ 12000 + 12000 ⎠ (checked: LNAP 6/22/04) P 5.4-13 The circuit shown in Figure P 5.4-13 consists of two parts, the source (to the left of the terminals) and the load. The load consists of a single adjustable resistor having resistance 0 ≤ RL ≤ 20 Ω. The resistance R is fixed, but unspecified. When RL = 4 Ω, the load current is measured to be io = 0.375 A. When RL = 8 Ω, the value of the load current is io = 0.300 A. (a) Determine the value of the load current when RL = 10 Ω. (b) Determine the value of R. 48 Ω io 24 V + – R source RL load Figure P 5.4-13 Solution: Replace the source by it’s Thevenin equivalent circuit to get io = v oc R t +R L Using the given formation v oc ⎫ ⎪ R t + 4⎪ ⎬ v oc ⎪ 0.300 = R t + 8 ⎪⎭ 0.375 = So Rt = ⇒ 0.375 ( R t + 4 ) = 0.300 ( R t + 8 ) ( 0.300 ) 8 − ( 0.375) 4 = 12 Ω and v 0.075 6 (a) When R L = 10 Ω, i o = = 0.2727 A. 12 + 10 (b) 12 Ω = R t = 48 11R ⇒ R = 16 Ω . oc = 0.3 (12 + 8 ) = 6 V (checked: LNAP 5/24/04) P 5.4-14 The circuit shown in Figure P 5.4-14 contains an unspecified resistance, R. Determine the value of R in each of the following two ways. (a) (b) Write and solve mesh equations. Replace the part of the circuit connected to the resistor R by a Thévenin equivalent circuit. Analyze the resulting circuit. 40 Ω 20 Ω R + – 40 V 0.25 A 20 Ω 10 Ω Figure P 5.4-14 Solution: (a) i 3 − i 2 = 0.25 A Apply KVL to mesh 1 to get 20 ( i1 − i 2 ) + 20 ( i1 − i 3 ) − 40 = 0 Apply KVL to the supermesh corresponding to the unspecified resistance to get 40i 2 + 10i 3 − 20 ( i1 − i 3 ) − 20 ( i1 − i 2 ) = 0 Solving, for example using MATLAB, gives −1 1 ⎤ ⎡ i1 ⎤ ⎡0.25⎤ ⎡ 0 ⎢ 40 −20 −20 ⎥ ⎢i ⎥ = ⎢ 40 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −40 60 30 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⇒ Apply KVL to mesh 2 to get 40i 2 + R ( i 2 − i 3 ) − 20 ( i1 − i 2 ) = 0 ⇒ R= ⎡ i1 ⎤ ⎡1.875 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢0.750 ⎥ ⎢ i 3 ⎥ ⎢⎣1.000 ⎥⎦ ⎣ ⎦ 20 ( i1 − i 2 ) − 40i 2 i2 − i3 = 30 Ω (b) ⎛ 20 ⎞ ⎛ 40 ⎞ v oc = ⎜ ⎟ 40 − ⎜ ⎟ 40 = −12 V ⎝ 20 + 20 ⎠ ⎝ 10 + 40 ⎠ R t = 18 Ω 0.25 = 12 18 + R ⇒ R = 30 Ω (checked: LNAP 5/25/04) a 96 Ω io 32 Ω 20 V + P 5.4-15 Consider the circuit shown in Figure P 5.4-15. Replace the part of the circuit to the left of terminals a–b by its Thévenin equivalent circuit. Determine the value of the current io. 32 Ω – + 120 Ω 30 Ω vo – b Figure P 5.4-15 Solution: Find the Thevenin equivalent circuit for the part of the circuit to the left of the terminals a-b. Using voltage division twice v oc = 32 30 20 − 20 = 5 − 4 = 1 V 32 + 96 120 + 30 R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit gives io = 1 = 0.0125 A = 12.5 mA 48 + 32 (checked: LNAP 5/24/04) P 5.4-16 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 5.4-16a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 5.4-16b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. The voltmeter measures vmi, the ideal value of vm. Determine the value of vmi. (b) Express the measurement error that occurs when Rm = 1000 Ω as a percentage of vmi. (c) 10 Ω + – 25 V 50 Ω vm – (a) 200 Ω 10 Ω + – 25 V 50 Ω vmi – (b) 200 Ω 10 Ω + + – 25 V 50 Ω Rm vm – (c) Figure P 5.4-16 Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi. Solution: Replace the circuit by its Thevenin equivalent circuit: (a) Voltmeter + + As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi. When Rm < ∞, the voltmeter is not ideal and vm > vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a) 200 Ω ⎛ Rm ⎞ vm = ⎜ 5 ⎜ R m + 50 ⎟⎟ ⎝ ⎠ v mi = lim v m = 5 V R m →∞ (b) When R m = 1000 Ω, v m = 4.763 V so % error = 5 − 4.762 × 100 = 4.76% 5 (c) ⎛ Rm ⎞ 5−⎜ 5 ⎜ R m + 50 ⎟⎟ ⎝ ⎠ 0.02 ≥ 5 ⇒ Rm R m + 50 ≥ 0.98 ⇒ R m ≥ 2450 Ω (checked: LNAP 6/16/04) P5.4-17 Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.4-17, consider these two observations: Observation 1: When R = 2 Ω then v R = 4 V and i R = 2 A. Observation 1: When R = 6 Ω then v R = 6 V and i R = 1 A. Determine the following a) The maximum value of i R and the value of R that causes i R to be maximal. b) The maximum value of v R and the value of R that causes v R to be maximal. c) The maximum value of p R = i R v R and the value of R that causes p R to be maximal. Figure P5.4-17 Solution: We can replace the part of the circuit to the left of the terminals by its Thevenin equivalent circuit: Using voltage division v R = iR = v oc R + Rt R v oc and using Ohm’s law R + Rt . v oc R v oc = will be maximum when Rt R + Rt 1+ R R = ∞. The maximum value of v R will be v oc . Similarly, By inspection , v R = iR = v oc R + Rt will be maximum when R = 0. The maximum value of i R will be v oc Rt = i sc . The maximum power transfer theorem tells use that p R = i R v R will be maximum when R = R t . ⎛ v oc Then p R = i R v R = ⎜ ⎜ R + Rt ⎝ ⎞⎛ R ⎞ ⎛ v oc v oc ⎟ = R ⎜ ⎟⎜ ⎟⎜ R + R t ⎟ ⎜ R + Rt ⎠⎝ ⎠ ⎝ 2 ⎞ ⎟⎟ . ⎠ Let’s substitute the given data into the equation i R = When R = 2 Ω we get 2 = 1= v oc 6 + Rt v oc 2 + Rt v oc R + Rt . ⇒ 4 + 2 R t = v oc . When R = 6 Ω we get ⇒ 6 + R t = v oc . So 6 + R t = 4 + 2 R t ⇒ R t = 2 Ω and v oc = 4 + 2 R t = 8 V . Also i sc = v oc Rt = 8 = 4 A. 2 P5.4-18 Consider the circuit shown in Figure P5.4-18. Determine a) The value of v R that occurs when R = 9 Ω. b) The value of R that causes v R = 5.4 V. c) The value of R that causes i R = 300 mA. Figure P5.4-18 Solution: Reduce this circuit using source transformations and equivalent resistance: 9 ⎛ R ⎞ so the questions can be easily answered: Now v R = ⎜ ⎟ 9 and i R = R + 18 ⎝ R + 18 ⎠ a) When R = 9 Ω then v R = 3 V. b) When R = 27 Ω then v R = 5.4 V. c) When R = 12 Ω then i R = 300 mA. P5.4-19 The circuit shown in Figure P5.4-19a can be reduced to the circuit shown in Figure P5.4-19b using source transformations and equivalent resistances. Determine the values of the source voltage v oc and the (a) resistance R. (b) Figure P5.4-19 Solution 46 = R t = R + ( 42 || 84 ) = R + 28 ⇒ R = 18 Ω v oc = 84 (18) = 12 V 42 + 84 P5.4-20 The equivalent circuit in Figure P5.4-20 is obtained by replacing part of the original circuit by its Thevenin equivalent circuit. The values of the parameters of the Thevenin equivalent circuit are v oc = 15 V and R t = 60 Ω Determine the following: a.) The values of V s and R a . (Three resistors in the original circuit have equal resistance, Ra.) b.) The value of R b required to cause i = 0.2 A. c.) The value of R b required to cause v = 5 V. Figure P5.4-20 Solution a.) From We see that v oc = b.) i = c.) v = v oc Rt + Rb Rb Rt + Rb Vs 3 and R t = R a . With the given values of voc and Rt we calculate 2 2 V 3 15 = s ⇒ Vs = 30 V and 60 = R a ⇒ R a = 40 Ω . 2 2 ⇒ 0.2 = v oc 15 60 + R b ⇒ 5= 15 R b 60 + R b ⇒ R b = 15 Ω ⇒ R b = 30 Ω Section 5-5: Norton’s Theorem P5.5-1 The part of the circuit shown in Figure P5.3-1a to the left of the terminals can be reduced to its Norton equivalent circuit using source transformations and equivalent resistance. The resulting Norton equivalent circuit, shown in Figure P5.3-1b, will be characterized by the parameters: i sc = 0.5 A and R t = 20 Ω a) Determine the values of v s and R1 . b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum values of the voltage, v, and of the power, p = vi. Answers: v s = 37.5 V, R1 = 25 Ω, max v = 10 V and max p = 1.25 W (a) (b) Figure P5.5-1 Solution: Two source transformations reduce the circuit as follows: Recognizing the parameters of the Norton equivalent circuit gives: 0.5 = i sc = 12.5 + v s 100 ⇒ v s = 37.5 V and 20 = R t = 100 || R1 = ( ) Next, the voltage across resistor R 2 is given by v = i sc R t || R 2 = 100 R1 100 + R1 R t R 2 i sc Rt + R2 = ⇒ R1 = 25 Ω R t i sc so this Rt +1 R2 voltage is maximum when R 2 = ∞ and max v = R t i sc =10 V. The power p = vi will be maximum when R 2 = R t = 20 Ω . Then v = R t i sc 2 = 20 ( 0.5 ) 2 =5 V, i= v 5 = = 0.25 A and R 2 20 p = v i = 5 ( 0.25 ) = 1.25 W. P 5.5-2 Two black boxes are shown in Figure P 5.5-2. Box A contains the Thévenin equivalent of some linear circuit, and box B contains the Norton equivalent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one shorting wire? Box A 1Ω 1V Box B a + – a 1A b 1Ω b Figure P 5.5-2 Solution: When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. (a) (b) Figure P5.5-3 P5.5-3 The circuit shown in Figure P5.5-3a can be reduced to the circuit shown in Figure P5.53b using source transformations and equivalent resistances. Determine the values of the source current i sc and the resistance R. Solution: 48 = R t = R || ( 80 + 160 ) = i sc = 240 R ⇒ R = 60 Ω R + 240 80 ( 4.8 ) = 1.6 A 80 + 160 P 5.5-4 Find the Norton equivalent circuit for the circuit shown in Figure P 5.5-4. 3Ω 5Ω a 4A 8Ω 5A b Figure P 5.5-4 Solution: P 5.5-5 The circuit shown in Figure P 5.5-5b is the Norton equivalent circuit of the circuit shown in Figure P 5.5-5a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. Answer: isc = 1.13 A and Rt = 7.57 Ω 3Ω 5Ω – + + – 10 V 2ia 6Ω isc Rt ia (a) (b) Figure P 5.5-5 Solution: To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − isc Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 Apply KVL to mesh 2 to get 5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 = 11 i2 6 Substituting into equation 1 gives ⎛ 11 ⎞ 7 ⎜ i 2 ⎟ − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A ⎝6 ⎠ Figure (a) Calculating the short circuit current, isc, using mesh equations. (1) To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v Rt = T iT In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 + i T Apply KVL to mesh 1 to get 3 i 1 − 2 ( i 1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i 1 − 4 i 2 = 0 ⇒ i 1 = 4 i2 7 (2) Apply KVL to mesh 2 to get 5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT Substituting for i1 using equation 2 gives ⎛4 ⎞ −6 ⎜ i 2 ⎟ + 11 i 2 = −vT ⎝7 ⎠ Finally, Rt = ⇒ 7.57 i 2 = −vT vT −vT −vT = = = 7.57 Ω iT i2 −iT Figure (b) Calculating the Thevenin resistance, R t = vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 − 0 = i1 Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0 ⇒ i1 = 10 = 1.43 A 7 Apply KVL to mesh 2 to get 5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V Figure (c) Calculating the open circuit voltage, voc, using mesh equations. As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc (checked using LNAP 8/16/02) 3Ω P 5.5-6 The circuit shown in Figure P 5.5-6b is the Norton equivalent circuit of the circuit shown in Figure P 5.5-6a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. 6Ω + – + 24 V va 1.33va isc Rt – Answer: isc = –24 A and Rt = –3 Ω (a) (b) Figure P 5.5-6 Solution: To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get v1 − v 2 3 = v 2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 = 3 v a ⇒ v a = −16 V Apply KCL at node 3 to get v 2 − v3 6 + 4 v 2 = isc 3 ⇒ 9 v a = isc 6 ⇒ isc = 9 ( −16 ) = −24 A 6 Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v R th = T iT Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get v1 − v 2 3 = v2 − v3 6 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ vT = 3 v a Apply KCL at node 3 to get v2 − v3 6 + 4 v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0 3 ⇒ 9 v a − vT + 6 iT = 0 ⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT Finally, Rt = vT = −3 Ω iT vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. Figure (b) Calculating the Thevenin resistance, R th = In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get v1 − v 2 = v 2 − v3 3 Apply KCL at node 3 to get 6 v2 − v3 ⇒ 2 v1 + v 3 = 3 v 2 ⇒ − 48 + v oc = 3 v a 4 v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc 6 3 Combining these equations gives + 3 ( −48 + voc ) = 9 v a = voc ⇒ voc = 72 V Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that R th I sc = ( −3)( −24 ) = 72 = Voc (checked using LNAP 8/16/02) P 5.5-7 Determine the value of the resistance R in the circuit shown in Figure P 5.5-7 by each of the following methods: (a) (b) 5 kΩ 10 kΩ ib 25 V +– Replace the part of the circuit to the left of terminals a–b by its the Norton equivalent circuit. Use current division to determine the value of R. 4 ib a R 0.5 mA b Figure P 5.5-7 Analyze the circuit shown Figure P 5.5-6 using mesh equations. Solve the mesh equations to determine the value of R. Solution: (a) Replace the part of the circuit that is connected to the left of terminals a-b by its Norton equivalent circuit: Apply KCL at the top node of the dependent source to see that i b = 0 A . Then ( ) v oc = 25 + 5000 i b = 25 V Apply KVL to the supermesh corresponding to the dependent source to get ( ) −5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA Apply KCL to get i sc = 3 i b = 3 mA Then Rt = v oc i sc = 8.33 kΩ Current division gives 0.5 = 8333 3 ⇒ R = 41.67 kΩ R + 8333 (b) Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get 1 i b + 0.5 × 10−3 = 4 i b ⇒ i b = mA 6 Apply KVL to the supermesh corresponding to the dependent source to get ( ) −5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0 ( ) ⎛1 ⎞ −5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25 ⎝6 ⎠ 125 6 R= = 41.67 kΩ 0.5 ×10−3 P5.5-8 Find the Norton equivalent circuit of this circuit: Solution Simplify the circuit using a source transformation: Identify the open circuit voltage and short circuit current. Apply KVL to the mesh to get: (10 + 2 + 3) i x − 15 = 0 ⇒ ix = 1 A Then v oc = 3 i x = 3 V Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 and i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 = 4 i sc 3 so ⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠ The Thevenin resistance is Rt = 3 =3Ω 1 Finally, the Norton equivalent circuit is (checked: LNAP 6/21/04) P5.5-9 Find the Norton equivalent circuit of this circuit: Solution Identify the open circuit voltage and short circuit current. ⎛1⎞ v1 = ⎜ ⎟ 3 = 1 V ⎝ 3⎠ Then v oc = v1 − 4 ( 2.5 v 1 ) = −9 V ⎛1 ⎞ v1 = 3 ⎜ − i sc ⎟ = 1 − 3 i sc ⎝3 ⎠ 4 ( 2.5 v1 + i sc ) + 5 i sc − v1 = 0 ⇒ 9 v1 + 9 i sc = 0 9 (1 − 3 i sc ) + 9 i sc = 0 ⇒ i sc = The Thevenin resistance is Rt = 1 A 2 −9 = −18 Ω 0.5 Finally, the Norton equivalent circuit is (checked: LNAP 6/21/04) P 5.5-10 An ideal ammeter is modeled as a short circuit. A more realistic model of an ammeter is a small resistance. Figure P 5.5-10a shows a circuit with an ammeter that measures the current im. In Figure P 5.5-10b the ammeter is replaced by the model of an ideal ammeter, a short circuit. The ammeter measures imi, the ideal value of im. As Rm → 0, the ammeter becomes an ideal ammeter and im → imi. When Rm > 0, the ammeter is not ideal and im < imi. The difference between im and imi is a measurement error caused by the fact that the ammeter is not ideal. (a) Determine the value of imi. (b) Express the measurement error that occurs when Rm = 20 Ω as a percentage of imi. (c) im 4 kΩ 3 mA Ammeter 4 kΩ 2 kΩ (a) imi 4 kΩ 3 mA 4 kΩ 2 kΩ (b) im 4 kΩ 3 mA Determine the maximum value of Rm required to ensure that the measurement error is smaller than 2 percent of imi. 4 kΩ 2 kΩ (c) Figure P 5.5-10 Solution: Replace the circuit by its Norton equivalent circuit: ⎛ 1600 ⎞ im = ⎜ 1.5 × 10−3 ) ⎜ 1600 + R m ⎟⎟ ( ⎝ ⎠ (a) Rm i mi = lim R m →0 i m = 1.5 mA (b) When Rm = 20 Ω then i m = 1.48 mA so % error = 1.5 − 1.48 × 100 = 1.23% 1.5 (c) ⎛ 1600 ⎞ 0.015 − ⎜ 0.015 ) ⎜ 1600 + R m ⎟⎟ ( ⎝ ⎠ 0.02 ≥ 0.015 ⇒ 1600 ≥ 0.98 ⇒ R m ≤ 32.65 Ω 1600 + R m (checked: LNAP 6/18/04) 3Ω 6Ω P 5.5-11 Determine values of Rt and isc that cause the circuit shown in Figure P 5.5-11b to be the Norton equivalent circuit of the circuit in Figure P 5.5-11a. + – + – 12 V ia a a isc 2ia Rt b Answer: Rt = 3 Ω and isc = –2 A b (a) (b) Figure P 5.5-11 Solution: ia = 2 i a − 12 ⇒ i a = −3 A 6 voc = 2 i a = −6 V 12 + 6 i a = 2 i a 3 i sc = 2 i a Rt = −6 =3Ω −2 ⇒ i a = −3 A ⇒ i sc = 2 ( − 3 ) = −2 A 3 12 Ω P 5.5-12 Use Norton’s theorem to formulate a general expression for the current i in terms of the variable resistance R shown in Figure P 5.5-12. 8 Ω a 30 V + – i R b Answer: i = 20/(8 + R) A Figure P 5.5-12 Solution: 12 × 24 12 × 24 = = 8Ω 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 Rt = i= 20 8+ R 16 Ω Section 5-6: Maximum Power Transfer P 5.6-1 The circuit model for a photovoltaic cell is given in Figure P 5.6-1 (Edelson, 1992). The current is is proportional to the solar insolation (kW/m2). (a) (b) Find the load resistance, RL, for maximum power transfer. Find the maximum power transferred when is = 1 A. 8Ω ia R2 + + – vs + – 4 ia v R – i Figure P 5.6-1 Solution: (a) The value of the current in R 2 is 0 A so v oc = 4 i a . Then KVL gives 8 i a + 4 i a − V s = 0 ⇒ V s = 12 i a = 3 ( 4 i a ) = 3 ( v oc ) = 24 V Next, KVL gives 8 i a + 4 i a − 24 = 0 ⇒ i a = 2 A and 4 i a = R 2 i sc ⇒ 4 ( 2) = R2 ( 2) ⇒ R2 = 4 Ω (b) The power delivered to the resistor to the right of the terminals is maximized by setting R equal to the Thevenin resistance of the part of the circuit to the left of the terminals: R = Rt = Then p max = v oc i sc v oc 2 4 Rt = = 8 =4Ω 2 82 =4W 4 ( 4) 1Ω P 5.6-2 For the circuit in Figure P 5.6-2, (a) find R such that maximum power is dissipated in R and (b) calculate is the value of maximum power. 100 Ω Answer: R = 60 Ω and Pmax = 54 mW Figure P 5.6-2 Solution: a) For maximum power transfer, set RL equal to the Thevenin resistance: R L = R t = 100 + 1 = 101 Ω b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit: The voltage across RL is vL = Then pmax 101 (100 ) = 50 V 101 + 101 2 vL 502 = = = 24.75 W R L 101 RL 150 Ω P 5.6-3 For the circuit in Figure P 5.6-3, prove that for Rs variable and RL fixed, the power dissipated in RL is maximum when Rs = 0. 6V + – 100 Ω R + – 2V Figure P 5.6-3 Solution: Reduce the circuit using source transformations: Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R Rs P 5.6-4 Find the maximum power to the load RL if the maximum power transfer condition is met for the circuit of Figure P 5.6-4. vs + – RL Answer: max pL = 0.75 W source network load Figure P 5.6-4 Solution: ⎡ RL ⎤ v L = vS ⎢ ⎥ ⎣⎢ R S + R L ⎦⎥ ∴ pL = v L2 RL = v S2 R L ( RS + R L )2 By inspection, pL is max when you reduce RS to get the smallest denominator. ∴ set RS = 0 8Ω P 5.6-5 Determine the maximum power that can be absorbed by a resistor, R, connected to terminals a–b of the circuit shown in Figure P 5.6-5. Specify the required value of R. 20 Ω a 10 Ω 20 A 120 Ω 50 Ω b Figure P 5.6-5 Solution: The required value of R is R = Rt = 8 + ( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 ) 30 ⎡ 170 voc = ⎢ ( 20 )⎤⎥ 10 − ⎡⎢ ( 20 )⎤⎥ 50 ⎣170 + 30 ⎦ ⎣170 + 30 ⎦ 170(20)(10) − 30(20)(50) 4000 = = = 20 V 200 200 The maximum power is given by 2 voc 202 pmax = = =2W 4 R t 4 ( 50 ) P 5.6-6 Figure P 5.6-6 shows a source Ro connected to a load through an amplifier. The load can safely receive + up to 15 W of power. Consider three + + Ava v 500 mV 100 kΩ RL a cases: – – – (a) A = 20 V/V and Ro = 10 Ω. Determine the value of RL that source amplifier load maximizes the power delivered Figure P 5.6-6 to the load and the corresponding maximum load power. (b) A = 20 V/V and RL = 8 Ω. Determine the value of Ro that maximizes the power delivered to the load and the corresponding maximum load power. (c) Ro = 10 Ω and RL = 8 Ω. Determine the value of A that maximizes the power delivered to the load and the corresponding maximum load power. Solution: iL = A vs Ro +RL PL = i L R L = 2 A 2v s 2 R L (R o +RL) 2 (a) R t =R o so R L =R o = 10 Ω maximizes the power delivered to the load. The corresponding load power is 2 ⎛1⎞ 20 ⎜ ⎟ 10 ⎝2⎠ PL = = 2.5 W . 2 (10 + 10 ) 2 (b) Ro = 0 maximizes PL (The numerator of PL does not depend on Ro so PL can be maximized by making the denominator as small as possible.) The corresponding load power is 2 PL = A 2v s 2 R L R L2 = A 2v s 2 RL ⎛1⎞ 20 ⎜ ⎟ ⎝ 2 ⎠ = 12.5 W. = 8 2 (c) PL is proportional to A2 so the load power continues to increase as A increases. The load can safely receive 15 W. This limit corresponds to 2 ⎛1⎞ A ⎜ ⎟ 8 ⎝2⎠ 15 = 2 (18) 2 ⇒ A = 36 15 = 49.3 V. 8 (checked: LNAP 6/9/04) P 5.6-7 The circuit in Figure P 5.6-7 contains a variable resistance, R, implemented using a potentiometer. The resistance of the variable resistor varies over the range 0 ≤ R ≤ 1000 Ω. The variable resistor can safely receive 1/4 W power. Determine the maximum power received by the variable resistor. Is the circuit safe? 180 Ω + – 10 V R 120 Ω 150 Ω 470 Ω + – 20 V Figure P 5.6-7 Solution: Replace the part of the circuit connected to the variable resistor by its Thevenin equivalent circuit. First, replace the left part of the circuit by its Thevenin equivalent: ⎛ 150 ⎞ v oc1 = ⎜ ⎟10 = 4.545 V ⎝ 150 + 180 ⎠ R t1 = 180 & 150 = 81.8 Ω Next, replace the right part of the circuit by its Thevenin equivalent: ⎛ 470 ⎞ v oc2 = ⎜ ⎟ 20 = 15.932 V ⎝ 470 + 120 ⎠ R t2 = 120 & 470 = 95.6 Ω Now, combine the two partial Thevenin equivalents: v oc = v oc1 − v oc2 = −10.387 V and R t = R t1 + R t2 = 177.4 Ω So The power received by the adjustable resistor will be maximum when R = Rt = 177.4 Ω. The maximum power received by the adjustable ( −11.387 ) = 0.183 W . p= 4 (177.4 Ω ) 2 resistor will be (checked LNAPDC 7/24/04) Rt P 5.6-8 For the circuit of Figure P 5.6-8, find the power delivered to the load when RL is fixed and Rt may be varied between 1 Ω and 5 Ω. Select Rt so that maximum power is delivered to RL. 10 V Answer: 13.9 W + – RL = 5 Ω Figure P 5.6-8 Solution: ⎛ 10 ⎞ ⎡ R L ⎤ 100 R L p=iv=⎜ 10 ) ⎥ = ( ⎟ ⎢ 2 ⎜ Rt + R L ⎟ ⎢ Rt + R L ⎝ ⎠⎣ ⎦⎥ ( R t + R L ) The power increases as Rt decreases so choose Rt = 1 Ω. Then pmax = i v = 100 ( 5 ) (1 + 5) 2 = 13.9 W 5 P 5.6-9 A resistive circuit was connected to a variable resistor, and the power delivered to the resistor was measured as shown in Figure P 5.6-9. Determine the Thévenin equivalent circuit. Power (W) 0 Answer: Rt = 20 Ω and voc = 20 V 10 20 30 R (ohms) Figure P 5.6-9 Solution: From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: Rt = 20 Ω and 2 pmax v = oc 4 Rt ⇒ voc = pmax 4 Rt = 5 ( 4 ) 20 = 20 V 40 Figure P5.6-10 P5.6-10 The part circuit shown in Figure P5.6-10a to left of the terminals can be reduced to its Norton equivalent circuit using source transformations and equivalent resistance. The resulting Norton equivalent circuit, shown in Figure P5.6-10b, will be characterized by the parameters: i sc = 1.5 A and R t = 80 Ω (a) Determine the values of i s and R1 . (b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum value of p = vi, the power delivered to R2. Solution: Two source transformations reduce the circuit as follows: (a) Recognizing the parameters of the Norton equivalent circuit gives: 1.5 = i sc = i s + 0.25 ⇒ i s = 1.25 A and 80 = R t = 100 || R1 = 100 R1 100 + R1 ⇒ R1 = 400 Ω (b) The maximum value of the power delivered to R2 occurs when R 2 = R t = 80 Ω . Then 2 1 ⎛1 ⎞ i = i sc = 0.75 A and p = ⎜ i sc ⎟ R t = ( 0.6252 ) 80 = 45 W 2 ⎝2 ⎠ P5.6-11. Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.6-12, determine (a) maximum value of i a , (b) the maximum value of v a , and (c) the maximum value of pa = iava . Figure P5.6-11 Solution: Replace the parallel combination of resistor R and the 8 Ω resistor by an equivalent resistance. Using voltage division va = R eq 4 + R eq (12 ) = 1 (12 ) 4 +1 R eq Consequently, the maximum value of va corresponds to the is obtained by maximizing Req. The maximum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the maximum value of Req is 8 Ω and the maximum value of va is v a max = Using Ohm’s law 1 4 +1 8 (12 ) = 8 V ia = 12 4 + R eq Consequently, the maximum value of ia corresponds to the is obtained by minimizing Req. The minimum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the minimum value of Req is 0 Ω and the maximum value of ia is i a max = 12 =3A 4+0 The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req = Rt. In this case, Rt = 4 Ω. We require Req = 4 Ω which is accomplished by making R = 8 Ω, an acceptable value since 0 ≤ 8 ≤ ∞. Then 2 2 ⎛ 12 ⎞ ⎛ 12 ⎞ ⎜ ⎟ ⎜ ⎟ 2 2 pa = ⎝ ⎠ = ⎝ ⎠ = 9 W R eq 4 P5.6-12. Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.6-12, determine value of R that maximizes the power p a = i a v a and the corresponding maximum value of p a . Figure P5.6-12 Solution: Replace the combination of resistor R and the 20 Ω and 2 Ω resistors by an equivalent resistance. The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req = Rt. In this case, Rt = 8 Ω. We require 8 = R eq = 20 ( R + 2 ) 20 R + 40 = 20 + ( R + 2 ) R + 22 8 ( R + 22 ) = 20 R + 40 ⇒ R = 8 ( 22 ) − 40 = 11.333 Ω 20 − 8 This isn’t a standard resistance value but it is an acceptable value for this problem since 0 ≤ 11.333 ≤ ∞. Then 2 2 ⎛6⎞ ⎛6⎞ ⎜ ⎟ ⎜ ⎟ 2 2 p a = ⎝ ⎠ = ⎝ ⎠ = 1.125 W 8 R eq Section 5.8 Using PSpice to Determine the Thevenin Equivalent Circuit P5.8-1 a) Here are the results of simulating the circuit in PSpice. The numbers shown in white on a black background are the node voltages. b) Add a resistor across the terminals of Circuit A. Then v oc = v R i sc = vR R when R≈∞ when R≈0 1 Here are the PSpice simulation results: v oc = 10 V i sc = Rt = 1× 10−6 =1 A 1× 10−6 v oc i sc = 10 = 10 Ω 1 c) Here is the result of simulation the circuit after replacing Circuit A by its Thevenin equivalent: d) The node voltages of Circuit B are the same before and after replacing Circuit A by its Thevenin equivalent circuit. 2 Section 5-9 How Can We Check…? P 5.9-1 For the circuit of Figure P 5.9-1, the current i has been measured for three different values of R and is listed in the table. Are the data consistent? 1 kΩ ix R(Ω) i(mA) 5000ix 5000 500 0 16.5 43.8 97.2 + – R i 4 kΩ + – 10 V 4 kΩ Figure P 5.9-1 Solution: Use the data in the first two lines of the table to determine voc and Rt: voc ⎫ R t + 0 ⎪⎪ ⎧voc = 39.9 V ⎬ ⇒ ⎨ voc ⎪ ⎩ R t = 410 Ω 0.0438 = R t + 500 ⎪⎭ 0.0972 = Now check the third line of the table. When R= 5000 Ω: v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table. The data is not consistent. i P 5.9-2 R v – + Your lab partner built the circuit shown in Figure P 5.9-2 and measured the 6 kΩ current i and voltage v corresponding to several values of the resistance R. The results 12 V + – 2 mA 18 kΩ are shown in the table in Figure P 5.9-2. Your lab partner says that RL = 8000 Ω is required to cause i = 1 mA. Do you agree? Justify your answer. 24 kΩ R i v open 10 kΩ short 0 mA 0.857 mA 3 mA 12 V 8.57 V 0V Figure P 5.9-2 Solution: Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table) isc = 3 mA so Rt = (line 3 of the table) voc = 4 kΩ isc Next, check line 2 of the table. When R = 10 kΩ: v 12 i = oc = = 0.857 mA 3 R t + R 10 (10 ) + 5 (103 ) To cause i = 1 mA requires which agrees with the data in the table. v 12 0.001 = i = oc = ⇒ R = 8000 Ω R t + R 10 (103 ) + R I agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 12 kΩ P 5.9-3 In preparation for lab, your lab 3R partner determined the Thévenin equivalent of the circuit connected to RL in 2R Figure P 5.9-3. She says that the Thévenin resistance is Rt = voltage is voc = 60 11 6 R 11 and the open-circuit 30 V 20 V V. In lab, you built the i R + – + – Load 10 V + – circuit using R = 110 Ω and RL = 40 Ω and measured that i = 54.5 mA. Is this Figure P 5.9-3 measurement consistent with the prelab calculations? Justify your answers. Solution: 1 1 1 1 11 = + + = R t R 2 R 3R 6 R ⇒ Rt = 6R 11 and ⎛ 23 ⎞ ⎛ 34 ⎞ ⎛ 65 ⎞ 180 voc = ⎜ ⎟ 30 + ⎜ ⎟ 20 + ⎜ ⎟ 10 = 11 ⎝ 3+ 2 3⎠ ⎝ 2+3 4⎠ ⎝ 1+ 6 5 ⎠ so the prelab calculation isn’t correct. But then 180 180 voc 11 i= = = 11 = 163 mA ≠ 54.5 mA 6 Rt + R (110 ) + 40 60 + 40 11 so the measurement does not agree with the corrected prelab calculation. RL P 5.9-4 Your lab partner claims that the current i in Figure P 5.9-4 will be no greater than 12.0 mA, regardless of the value of the resistance R. Do you agree? i 500 Ω R 12 V + – 3 kΩ 6 kΩ 1500 Ω Figure P 5.9-4 Solution: 6000 & 3000 & ( 500 + 1500 ) = 2000 & 2000 = 1000 Ω i= 12 12 ≤ = 12 mA R + 1000 1000 How about that?! Your lab partner is right. (checked using LNAP 6/21/05) P 5.9-5 Figure P 5.9-5 shows a circuit and some corresponding data. Two resistances, R1 and R, and the current source current are unspecified. The tabulated data provide values of the current, i, and voltage, v, corresponding to several values of the resistance R. (a) Consider replacing the part of the circuit connected to the resistor R by a Thévenin equivalent circuit. Use the data in rows 2 and 3 of the table to find the values of Rt and voc, the Thévenin resistance and the open-circuit voltage. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the blanks in the table. (d) Determine the values of R1 and is. 12 V + – 24 Ω is i + v R – R, Ω i, A v, V 0 10 20 40 80 3 1.333 0.857 0.5 ? 0 13.33 17.14 ? 21.82 18 Ω (b) R1 12 Ω (a) Figure P5.9-5 Solution: (a) KVL gives v oc = ( R t + R ) i from row 2 from row 3 So (R t v oc = ( R t + 10 ) (1.333) v oc = ( R t + 20 ) ( 0.857 ) + 10 ) (1.333) = ( R t + 20 ) ( 0.857 ) 28 ( R t + 10 ) = 18 ( R t + 20 ) Solving gives 10 R t = 360 − 280 = 80 and ⇒ Rt = 8 Ω v oc = ( 8 + 10 )(1.333) = 24 V (b) i= v oc Rt + R = 24 R 24 R and v = v oc = 8+ R R + Rt R+8 When R = 0, i = 3 A, and v = 0 V. 1 When R = 40 Ω, i = A . 2 24 ( 80 ) 240 = = 21.82 . When R = 80 Ω, v = 88 11 These are the values given in the tabulated data so the data is consistent. 24 ( 40 ) (c) When R = 40 Ω, v = = 20 V . 48 24 = 0.2727 A . When R = 80 Ω, i = 88 (d) First 8 = R t = 24 & 18 & ( R1 + 12 ) ⇒ R1 = 24 Ω the, using superposition, 24 = v oc = ( 24 24 + 18 & ( R1 + 12 ) ) ( ) 12 + 24 & 18 ( R1 + 12 ) i s = 8 + 8i s ⇒ is = 2 A (checked using LNAP 6/21/05) PSpice Problems SP 5-1 The circuit in Figure SP 5.1 has three inputs: v1, v2, and i3. The circuit has one output, vo. The equation vo = av1 + bv2 + ci3 expresses the output as a function of the inputs. The coefficients a, b, and c are real constants. (a) Use PSpice, and the principle of superposition, to determine the values of a, b, and c. (b) Suppose v1 = 10 V, v2 = 8 V, and we want the output to be vo = 7 V. What is the required value of i3? Hint: The output is given by vo = a when v1 = 1 V, v2 = 0 V, and i3 = 0 A. Answer: (a) vo = 0.3333v1 + 0.3333v2 + 33.33i3, 100 Ω (b) i3 = 30 mA v2 +– + – v1 + vo – 100 Ω 100 Ω i3 Figure SP 5.1 Solution: a = 0.3333 b = 0.3333 c =33.33 V/A (a) (b) vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3 7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3 18 3 = 3 = 30 mA ⇒ i3 = 100 100 3 7− SP 5-2 The pair of terminals a–b partitions the circuit in Figure SP 5.2 into two parts. Denote the node voltages at nodes 1 and 2 as v1 and v2. Use PSpice to demonstrate that performing a source transformation on the part of the circuit to the left of the terminal does not change anything to the right of the terminals. In particular, show that the current, io, and the node voltages, v1 and v2, have the same values after the source transformation as before the source transformation. 100 Ω a 1 8V 2 +– 10 V + – 100 Ω io 100 Ω b Figure SP 5.2 Solution: Before the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E-02 -4.000E-02 After the source transformation: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V2 -4.000E-02 30 mA 0.75va SP 5-3 Use PSpice to find the Thévenin equivalent circuit for the circuit shown in Figure SP 5.3. 8Ω Answer: voc = –2 V and Rt = –8/3 Ω + – Solution: 6V a – va + 4Ω b Figure SP 5.3 voc = −2 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V3 V_V4 -7.500E-01 7.500E-01 isc = 0.75 A Rt = −2.66 Ω SP 5-4 The circuit shown in Figure SP 5-4b is the Norton equivalent circuit of the circuit shown in Figure SP 5-4a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. Answer: isc = 1.13 V and Rt = 7.57 Ω 5Ω 3Ω – + + – 10 V 2ia 6Ω isc Rt ia (a) (b) Figure SP 5-4 Solution: voc = 8.571 V VOLTAGE SOURCE CURRENTS NAME CURRENT V_V5 -2.075E+00 V_V6 1.132E+00 X_H1.VH_H1 9.434E-01 isc = 1.132 A Rt = 7.571 Ω Design Problems R1 i + DP 5-1 The circuit shown in Figure DP 51a has four unspecified circuit parameters: vs, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-1b describes a relationship between the current i and the voltage v. Specify values of vs, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-1a to satisfy the relationship described by the graph in Figure DP 5-1b. First Hint: The equation representing the straight line in Figure DP 5-1b is v = –Rti + voc That is, the slope of the line is equal to –1 times the Thévenin resistance and the “vintercept” is equal to the open-circuit voltage. Second Hint: There is more than one correct answer to this problem. Try setting R1 = R2. R3 + – R2 vs v – (a) v, V 6 4 2 –6 –4 –2 2 4 6 8 i, mA –2 –4 –6 –8 (b) Figure DP 5-1 Solution: The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0−5 open circuit voltage. Therefore: R t = − = 625 Ω and voc = 5 V. 0.008 − 0 Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then 5= R2 1 vs = vs R1 + R 2 2 ⇒ vs = 10 V and R1 R2 = R3 + 500 ⇒ R3 = 125 Ω R1 + R2 Now vs, R1, R2 and R3 have all been specified so the design is complete. 625 = R 3 + DP 5-2 The circuit shown in Figure DP 5.2a has four unspecified circuit parameters: is, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-2b describes a relationship between the current i and the voltage v. Specify values of is, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-2a to satisfy the relationship described by the graph in Figure DP 5-2b. First Hint: Calculate the open-circuit voltage, voc, and the Thévenin resistance, Rt, of the part of the circuit to the left of the terminals in Figure DP 5-2a. Second Hint: The equation representing the straight line in Figure DP 5-2b is v = –Rti + voc That is, the slope of the line is equal to –1 times the Thévenin resistance and the “v-intercept” is equal to the open-circuit voltage. Third Hint: There is more than one correct answer to this problem. Try setting both R3 and R1 + R2 equal to twice the slope of the graph in Figure DP 5-2b. v, V 6 4 R2 i 2 + is R1 R3 v –6 – –4 –2 2 4 6 8 i, mA –2 (a) –4 –6 –8 (b) Figure DP 5-2 Solution: The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0 − ( −3 ) open circuit voltage. Therefore: R t = − = 500 Ω and voc = −3 V. −0.006 − 0 From the circuit we calculate R 3 ( R1 + R 2 ) R1 R 3 is and voc = − Rt = R1 + R 2 + R 3 R1 + R 2 + R 3 so 500 Ω = R 3 ( R1 + R 2 ) R1 + R 2 + R 3 and −3 V = − R1 R 3 R1 + R 2 + R 3 is Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and −3 = − 1000 R1 is = − R1 i s ⇒ 6 = R1 i s 2000 2 This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω = 400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete. R1 i R3 + DP 5-3 The circuit shown in Figure DP 5-3a has four unspecified circuit parameters: vs, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-3b describes a relationship between the current i and the voltage v. Is it possible to specify values of vs, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-1a to satisfy the relationship described by the graph in Figure DP 5-3b? Justify your answer. + – R2 vs v – (a) v, V 6 4 2 –6 –4 –2 2 4 6 8 i, mA –2 –4 –6 –8 (b) Figure DP 5-3 Solution: The slope of the graph is positive so the Thevenin resistance is negative. This would require R1 R 2 R3 + < 0 , which is not possible since R1, R2 and R3 will all be non-negative. R1 + R 2 Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b. DP 5-4 The circuit shown in Figure DP 5-4a has four unspecified circuit parameters: vs, R1, R2, and d, where d is the gain of the CCCS. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-4b describes a relationship between the current i and the voltage v. Specify values of vs, R1, R2, and d that cause the current i and the voltage v in Figure DP 5-4a to satisfy the relationship described by the graph in Figure DP 5-4b. + – ia i R1 + That is, the slope of the line is equal to –1 times the Thévenin resistance and the “v-intercept” is equal to the open-circuit voltage. Second Hint: There is more than one correct answer to this problem. Try setting R1 = R2. v – (a) v, V 6 4 First Hint: The equation representing the straight line in Figure DP 5-4b is v = –Rti + voc R2 dia vs 2 –6 –4 –2 2 4 6 8 i, mA –2 –4 –6 –8 (b) Figure DP 5-4 Solution: The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open −5 − 0 = −625 Ω and voc = −5 V. circuit voltage. Therefore: R t = − 0 − 0.008 The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by R 2 ( d + 1) voc = vs R1 + ( d + 1) R 2 , ( d + 1) v isc = s R1 and Rt = R1 R 2 R1 + ( d + 1) R 2 Let R1 = R2 = 1 kΩ. Then −625 Ω = R t = and −5 = ( d + 1) vs d +2 1000 1000 ⇒ d= − 2 = −3.6 A/A −625 d +2 ⇒ vs = −3.6 + 2 ( − 5) = −3.077 V −3.6 + 1 Now vs, R1, R2 and d have all been specified so the design is complete. Chapter 6 The Operational Amplifier Exercises Exercise 6.6-1 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to be related to v1 and v2 by the equation v3 = (4)v1 – ( 15 )v2. Answer: R1 = 10 kΩ and R2 = 2.5 kΩ Figure E 6.6-1 Solution: ⎞ R2 R1 ⎞ ⎛ 10 × 103 ⎞ ⎛ ⎛ 10 ×103 ⎞ ⎛ v3 = ⎜ − v + + + v 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 3 3 3 3 ⎟ 1 ⎝ 10 × 10 ⎠ ⎜⎝ R 2 + 10 × 10 ⎟⎠ ⎝ 10 × 10 ⎠ ⎝ 10 × 10 ⎠ ⎛ ⎞ R2 R1 ⎞ ⎛ = −⎜ v2 + 2 ⎜1 + v ⎟ 3 3 ⎟ 1 ⎜ ⎟ ⎝ 10 × 10 ⎠ ⎝ R 2 + 10 × 10 ⎠ ⎛1⎞ We require v3 = ( 4 ) v1 − ⎜ ⎟ v2 , so ⎝5⎠ R1 ⎞ ⎛ 4 = 2 ⎜1 + ⇒ R1 = 10 × 103 = 10 kΩ 3 ⎟ 10 × 10 ⎝ ⎠ and R2 1 = 5 R 2 + 10 × 103 ⇒ R 2 + 10 × 103 = 5 R 2 ⇒ R 2 = 2.5 kΩ Exercise 6.6-2 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to be related to v1 and v2 by the equation v3 = (6)v1 – ( 54 )v2. Answer: R1 = 20 kΩ and R2 = 40 kΩ Figure E 6.6-1 Solution: As in Ex 6.7-1 ⎛ ⎞ R2 R1 ⎞ ⎛ v3 = − ⎜ v + 2 1 + v ⎟ ⎜ 2 3 ⎟ 1 ⎜ R 2 + 10 × 103 ⎟ ⎝ 10 × 10 ⎠ ⎝ ⎠ ⎛ 4⎞ We require v3 = ( 6 ) v1 − ⎜ ⎟ v2 , so ⎝5⎠ R1 ⎞ ⎛ 6 = 2 ⎜1 + ⇒ R1 = 20 ×103 = 20 kΩ 3 ⎟ ⎝ 10 ×10 ⎠ and R2 4 = 5 R 2 + 10 × 103 ⇒ 4 R 2 + 40 × 103 = 5 R 2 ⇒ R 2 = 40 kΩ Exercise 6.7-1 The input offset voltage of a typical μA741 operational amplifier is 1 mV and the bias current is 80 nA. Suppose the operational amplifier in Figure 6.72a is a typical μA741. Show that the output offset voltage of the inverting amplifier will be at most 10 mV. Figure 6.7-2a Solution: Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1 For a μ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA so output offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mV Exercise 6.7-2 Suppose the 10-kΩ resistor in Figure 6.72a is changed to 2 kΩ and the 50-kΩ resistor is changed to 10 kΩ. (These changes will not change the gain of the inverting amplifier. It will still be – 5.) Show that the maximum output offset voltage is reduced to 35 mV. (Use ib = 500 nA and vos = 5 mV to calculate the maximum output offset voltage that could be caused by the μA741 amplifier.) Figure 6.7-2a Solution: vo = − ⎛ R ⎞ R2 vin + ⎜1 + 2 ⎟ vos + R2ib1 R1 ⎝ R1 ⎠ When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and ( ) ( output offset voltage ≤ 6 5 × 10−3 + 10 × 103 ib1 ≤ 500 nA then ) ( 500.10 ) ≤ 35×10 −9 −3 = 35 mV Exercise 6.7-3 Suppose the μA741 operational amplifier in Figure 6.7-2a is replaced with a typical OPA101AM operational amplifier. Show that the output offset voltage of the inverting amplifier will be at most 0.6 mV. Figure 6.7-2a Solution: Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1 For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so output offset voltage ≤ 6 ⎡⎣0.1×10−3 ⎤⎦ + ( 50 ×103 ) ⎡⎣0.012 ×10−9 ⎤⎦ ≤ 0.6 ×10−3 + 0.6 ×10−6 − 0.6 ×10−3 = 0.6 mV Rf Exercise 6.7-4 (a) (b) Ra Determine the voltage ratio vo/vs for the op amp circuit shown in Figure E 6.7-4 5 Calculate vo/vs for a practical op amp with A = 10 , Ro = 100 Ω, and Ri = 500 kΩ. The circuit resistors are Rs = 10 kΩ, Rf = 50 kΩ, and Ra = 25 kΩ. – + + – vs Rs Figure E 6.7-4 Solution: Writing node equations v− − vs v− − vo v− + + =0 Ra Rb Ri + Rs ⎛ ⎞ R i v ⎟ vo − ⎜ − A − ⎜ ⎟ R +R i s ⎠ + vo − v− = 0 ⎝ R0 Rb Av = R0 ( Ri + Rs ) + ARi R f vo = vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi Ra For the given values, Av = −2.00006 V/V. vo – Answer: (b) vo/vs = – 2 After some algebra + Section 6-3: The Ideal Operational Amplifier 20 kΩ P 6.3-1 Determine the value of voltage measured by the voltmeter in Figure P 6.3-1. – Voltmeter + Answer: – 4 V 20 kΩ 50 kΩ – + 4V Figure P 6.3-1 Solution: (checked using LNAP 8/16/02) 3 kΩ P 6.3-2 4 kΩ Find vo and io for the circuit of Figure P 6.3-2. 12 V io + – 2 kΩ – + + 1 kΩ R vo – Figure P 6.3-2 Solution: Apply KVL to loop 1: − 12 + 3000 i1 + 0 + 2000 i1 = 0 12 = 2.4 mA 5000 The currents into the inputs of an ideal op amp are zero so io = i1 = 2.4 mA ⇒ i1 = i2 = − i1 = − 2.4 mA va = i2 (1000 ) + 0 = −2.4 V Apply Ohm’s law to the 4 kΩ resistor vo = va − io ( 4000 ) = −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V (checked using LNAP 8/16/02) 4 kΩ 8 kΩ P 6.3-3 Find vo and io for the circuit of Figure P 6.3-3. Answer: vo = – 30 V and io = 3.5 mA io – 12 V + – + 2V – + + 20 kΩ vo – Figure P 6.3-3 Solution: The voltages at the input nodes of an ideal op amp are equal so va = −2 V . Apply KCL at node a: vo − ( −2 ) 12 − ( −2 ) + = 0 ⇒ vo = −30 V 8000 4000 Apply Ohm’s law to the 8 kΩ resistor io = −2 − vo = 3.5 mA 8000 (checked using LNAP 8/16/02) 10 kΩ P 6.3-4 Find v and i for the circuit of Figure P 6.3-4. – 0.1 mA + v – + + – 5V 20 kΩ i Figure P 6.3-4 Solution: The voltages at the input nodes of an ideal op amp are equal so v = 5 V . Apply KCL at the inverting input node of the op amp: ⎛ v −5 ⎞ −3 −⎜ a ⎟ − 0.1× 10 − 0 = 0 ⇒ va = 4 V ⎝ 10000 ⎠ Apply Ohm’s law to the 20 kΩ resistor va 1 i = = mA 20000 5 (checked using LNAP 8/16/02) 3 kΩ 4 kΩ P 6.3-5 Find vo and io for the circuit of Figure P 6.3-5. Answer: vo = – 15 V and io = 7.5 mA – io + 12 V + – 2 mA 6 kΩ + vo – Figure P 6.3-5 Solution: The voltages at the input nodes of an ideal op amp are equal, so va = 0 V . Apply KCL at node a: ⎛ v − 0 ⎞ ⎛ 12 − 0 ⎞ −3 −⎜ o ⎟−⎜ ⎟ − 2 ⋅10 = 0 3000 4000 ⎝ ⎠ ⎝ ⎠ ⇒ vo = − 15 V Apply KCL at the output node of the op amp: v v io + o + o = 0 ⇒ io = 7.5 mA 6000 3000 (checked using LNAP 8/16/02) 6 kΩ P 6.3-6 Determine the value of voltage measured by the voltmeter in Figure P 6.3-6. Answer: 7.5 V + Voltmeter – 8 kΩ + – 2.5 V 6 kΩ 4 kΩ Figure P 6.3-6 Soluton: The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal so va = 2.5 V . Apply Ohm’s law to the 4 kΩ resistor: v 2.5 ia = a = = 0.625 mA 4000 4000 Apply KCL at node a: ib = ia = 0.625 mA Apply KVL: vo = 8000 ib + 4000 ia = (12 × 103 )( 0.625 × 10−3 ) = 7.5 V (checked using LNAP 8/16/02) P 6.3-7 Find vo and io for the circuit of Figure P 6.3-7. R1 R2 R4 io – + – R3 – + vs + + R5 vo – Figure P 6.3-7 Solution: Label the circuit to account for the properties of the ideal op amps: Apply KCL at the inverting node of the left op amp to get: R2 ⎛ vs − 0 ⎞ ⎛ va − 0 ⎞ −⎜ vs ⎟⎟ + 0 = 0 ⇒ v a = − ⎟ − ⎜⎜ R1 ⎝ R1 ⎠ ⎝ R 2 ⎠ Apply KCL at the output node of the left op amp to get: io = 0 − va R2 + 0 − va R3 =− R 2 + R3 R 2 R3 ⎛ R 2 + R3 ⎞ va = ⎜ v ⎜ R1 R 3 ⎟⎟ s ⎝ ⎠ Apply KCL at the inverting node of the right op amp to get: ⎛ vo − 0 ⎞ ⎛ va − 0 ⎞ R4 R2 R4 −⎜ −⎜ + 0 = 0 ⇒ vo = − va = vs ⎟ ⎟ ⎜ R 4 ⎟ ⎜ R3 ⎟ R3 R1 R 3 ⎝ ⎠ ⎝ ⎠ 6 kΩ + P 6.3-8 Determine the current io for the circuit shown in Figure P 6.3-8. – 8 kΩ Answer: io = 2.5 mA io 6 kΩ + – 4 kΩ 2V 4 kΩ 6 kΩ 8 kΩ – + + – 6 kΩ 5.8 V Figure P 6.3-8 Solution: The node voltages have been labeled using: 1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal. 2. KCL 3. Ohm’s law then and v0 = 11.8 − 1.8 = 10 V 10 io = = 2.5 mA 4000 (checked using LNAP 8/16/02) P 6.3-9 Determine the current io for the circuit shown in Figure P 6.3-9. Answer: io = 2.5 mA 4 kΩ – + 18 V 4 kΩ – a + + b 8 kΩ 8 kΩ vo – Figure P 6.3-9 Solution: Apply KCL at node a: v a − ( −18 ) 4000 + va 8000 + 0 = 0 ⇒ v a = −12 V The node voltages at the input nodes of ideal op amps are equal, so v b = v a . Using voltage division: vo = 8000 v = −8 V 4000 + 8000 b (check using LNAP 8/16/02) R1 P 6.3-10 Determine the current io for the circuit shown in Figure P 6.3-10. is – Answer: io = 2.5 mA + + R2 R3 vo – Figure P 6.3-10 Solution: Label the circuit as shown. The current in resistor R 3 is i s . Consequently: v a = i s R3 Apply KCL at the top node of R 2 to get ⎛ R3 ⎞ va i= + i s = ⎜1 + i ⎜ R 2 ⎟⎟ s R2 ⎝ ⎠ Using Ohm’s law gives vo − va R1 ⎛ R3 ⎞ = i = ⎜1 + i ⎜ R 2 ⎟⎟ s ⎝ ⎠ ⎛ R1 R 3 ⎞ ⇒ v o = ⎜ R1 + R 3 + ⎟ is ⎜ R 2 ⎟⎠ ⎝ We require R1 + R 3 + R1 R 3 R2 = 20 e.g. R1 = 5 kΩ and R 2 = R 3 = 10 kΩ . (checked: LNAP 6/2/04) R1 P 6.3-11 Determine the voltage vo for the circuit shown in Figure P 6.3-11. + vo – + – vs R2 Answer: vo = – 8 V R4 R3 Figure P 6.3-11 Solution: Label the circuit as shown. Apply KCL at the top node of R 2 to get vs − va R1 = ⎛ R2 ⎞ + 0 ⇒ va = ⎜ ⎟⎟ v s ⎜ R2 ⎝ R1 + R 2 ⎠ va Apply KCL at the inverting node of the op amp to get R 2 ( R3 + R 4 ) ⎛ R3 + R 4 ⎞ ⎛ R3 + R 4 ⎞ ⎛ R 2 ⎞ vo − va va = + 0 ⇒ vo = ⎜ va = ⎜ vs = v ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ R3 R4 ( R1 + R 2 ) R 4 s ⎝ R4 ⎠ ⎝ R 4 ⎠ ⎝ R1 + R 2 ⎠ We require R 2 ( R3 + R 4 ) (R 1 + R2 ) R4 =5 e.g. R1 = R 2 = 10 kΩ , R 3 = 90 kΩ and R 4 = 10 kΩ . (checked: LNAP 6/2/04) 5 kΩ P 6.3-12 The circuit shown in Figure P 6.3-12 has one input, is, and one output, vo. Show that the output is proportional to the input. Design the circuit so that the gain is vo is vs + – 20 kΩ – V = 20 . mA + 1.5 V + – + 10 kΩ vo – Figure P 6.3-12 Solution: The node voltage at both input nodes of the op amp is 1.5 V. Apply KCL at the inverting input node of the op amp to get v s − 1.5 5000 + v o − 1.5 20000 = 0 ⇒ 4 ( v s − 1.5 ) + v o − 1.5 = 0 ⇒ v o = −4 ( v s − 1.5 ) + 1.5 = 0 ⇒ v o = −4 v s + 7.5 Comparing this equation to v o = m v s + b , we determine that m = −4 V/V and b = 75. V. R P 6.3-13 The circuit shown in Figure P 6.3-13 has one input, vs, and one output, vo. Show that the output is proportional to the input. Design the circuit so that vo = 5 vs. 0.2 mA 20 kΩ ioa – + 25 kΩ 1.5 V + – 10 kΩ + vo – Figure P 6.3-13 Solution: The output of this circuit is v o = 3.5 V. (a) The current in the 25 kΩ resistor is 0 A because this current is also the input current of an ideal op amp. Consequently, the voltage at the input nodes of the op amp is 1.5 V. Apply KCL at the inverting input of the op amp to get 1.5 3.5 − 1.5 = 0.2 + = 0.3 R 20 so R= 1.5 = 5 kΩ 0.3 (b) The voltage source current is 0A so the voltage source supplies 0 W of power. The voltage across the current source is equal to the node voltage at the inverting input of the op amp, 1.5 V. Notice that this voltage and the given current source current do not adhere to the passive convention so (0.2)(1.5) = 0.3 mW is the power supplied by the current source. (c) Apply KCL at the output node of the op amp to get i oa = 3.5 3.5 − 1.5 + = 0.45 A 10 20 The op amp supplies p oa = i oa × v o = ( 0.45 )( 3.5 ) = 1.575 W P 6-3.14 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in Figure P 6-3.14. 25 kΩ 5 kΩ a 20 kΩ c – + b 15 kΩ – d + 2.5 mA 5 kΩ 5 kΩ 10 kΩ Figure P 6.3-14 Solution: Label the circuit as shown, noticing that some resistor currents are zero because these currents are also input currents of op amps: Now v a = −2.5 ( 5 ) = −12.5 V , v b = v a = −12.5 V, v c = 0 V Apply KCL to get vb − 0 20 = 0+ 0 − vd 15 ⇒ vd = − 15 3 v b = − ( −12.5 ) = 9.375 V 20 4 P 6-3.15 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in Figure P 6-3.15. 10 kΩ a 40 kΩ 20 kΩ – b 10 kΩ c + – + 5V 40 kΩ – d + 10 kΩ 60 kΩ 10 kΩ 40 kΩ Figure P 6.3-15 Solution: Label the circuit as shown, noticing that some resistor currents are zero because these currents are also input currents of op amps: vb ⎛ 20 ⎞ = 5 V and v d = v c = 5 V. Now v a = 0 V, v b = − ⎜ ⎟ ( −5 ) = 10 V, v c = 2 ⎝ 10 ⎠ Figure P6.3-16 P6.3-16 . Figure P6.3-16 shows four similar circuits. The outputs of the circuits are the voltages v1, v2, v3 and v4. Determine the values of these four outputs. Solution: 80 25 80 v1 = − ( 2.1) = −6.72 V , v 2 = − ( −2.1) = 0.65625 V , v 3 = − ( −2.1) = 6.72 V and 25 80 25 25 v 4 = − ( 2.1) = −0.65625 V 80 Figure P6.3-17 P6.3-17 . Figure P6.3-17 shows four similar circuits. The outputs of the circuits are the voltages v1, v2, v3 and v4. Determine the values of these four outputs. Solution ⎛ 80 ⎞ ⎛ 25 ⎞ ⎛ 80 ⎞ v1 = ⎜ 1 + ⎟ ( 2.1) = 8.82 V , v 2 = ⎜1 + ⎟ ( −2.1) = −2.75625 V , v 3 = ⎜ 1 + ⎟ ( −2.1) = −8.82 V a ⎝ 25 ⎠ ⎝ 80 ⎠ ⎝ 25 ⎠ ⎛ 25 ⎞ nd v 4 = ⎜1 + ⎟ ( 2.1) = 2.75625 V . ⎝ 80 ⎠ Section 6-4: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers P 6.4-1 Determine the node voltages for the circuit shown in Figure P 6.4-1. a 20 kΩ b + d – Answer: va = 2 V, vb = – 0.25 V, vc = – 5 V, vd = – 2.5 V, and ve = – 0.25 V 9 kΩ 2V + – c 40 kΩ – + 5V e 40 kΩ 1 kΩ Figure P 6.4-1 Solution: KCL at node b: vb − 2 vb v +5 1 + + b = 0 ⇒ vb = − V 20000 40000 40000 4 The node voltages at the input nodes of an ideal op amp are equal so ve = vb = − KCL at node e: ve v −v 10 V + e d = 0 ⇒ vd = 10 ve = − 1000 9000 4 1 V. 4 (checked using LNAP 8/16/02) 6 kΩ P 6.4-2 Find vo and io for the circuit of Figure P 6.4-2. Answer: vo = – 4 V and io = 1.33 mA 6 kΩ 6 kΩ – + – 12 V 6 kΩ io + + 6 kΩ vo – Figure P 6.4-2 Solution: Apply KCL at node a: 0= va − 12 v v −0 + a + a ⇒ va = 4 V 6000 6000 6000 Apply KCL at the inverting input of the op amp: ⎛v −0⎞ ⎛ 0 − vo ⎞ −⎜ a ⎟+0+⎜ ⎟ = 0 ⎝ 6000 ⎠ ⎝ 6000 ⎠ ⇒ vo = −va = −4 V Apply KCL at the output of the op amp: vo ⎛ 0 − vo ⎞ io − ⎜ = 0 ⎟+ ⎝ 6000 ⎠ 6000 v ⇒ io = − o = 1.33 mA 3000 (checked using LNAP 8/16/02) Figure P6.4-3 P6.4-3. Determine the values of the node voltages, va and vo, of the circuit shown in Figure P6.4-3. Solution: Writing node equations: va 2.25 2.25 + = 0 ⇒ v a = − 40 × 10 3 = −4.5 V 3 3 20 ×10 40 ×10 20 × 10 3 and va va va − vo ⎛ 8 8 8⎞ + + = 0 ⇒ v o = ⎜ + + ⎟ v a = 2 ( −4.5 ) = −9 V 3 3 3 40 ×10 10 ×10 8 × 10 ⎝ 40 10 8 ⎠ ( ) + P 6.4-4 The output of the circuit shown in Figure P 6.4-4 is vo. The inputs are v1 and v2. Express the output as a function of the inputs and the resistor resistances. – + – v1 + R1 vo R2 R3 – + + – v2 Figure P 6.4-4 Solution: Ohm’s law: i= v1 − v2 R2 KVL: v0 = ( R1 + R2 + R3 ) i = R1 + R2 + R3 ( v1 − v2 ) R2 – R3 + P 6.4-5 The outputs of the circuit shown in Figure P 6.4-5 are vo and io. The inputs are v1 and v2. Express the outputs as functions of the inputs and the resistor resistances. R5 – + – v1 R1 io – R7 + + vo R2 – – + + – R4 v2 Figure P 6.4-5 Solution: ⎛ R ⎞ v1 −va v1 −v2 R + + 0 = 0 ⇒ va = ⎜1+ 1 ⎟ v1 − 1 v2 R1 R7 R7 ⎝ R7 ⎠ ⎛ R ⎞ v2 − vb v −v R − 1 2 + 0 = 0 ⇒ vb = ⎜ 1+ 2 ⎟ v2 − 2 v1 R2 R7 R7 ⎝ R7 ⎠ R6 ⎛ v −v ⎞ v −0 R6 − ⎜ b c ⎟ + c + 0 = 0 ⇒ vc = vb R4 + R6 ⎝ R4 ⎠ R6 ⎛ v −v ⎞ ⎛ v −v ⎞ −⎜ a c ⎟ + ⎜ c 0 ⎟ + 0 = 0 ⎝ R3 ⎠ ⎝ R5 ⎠ ⇒ v0 = − R5 R va + (1+ 5 )vc R3 R3 ⎡ R R R (R +R ) ⎡R R (R +R ) R ⎤ R ⎤ R v0 = ⎢ 5 1 + 6 3 5 (1+ 2 ) ⎥ v2 − ⎢ 5 (1+ 1 ) + 6 3 5 2 ⎥ v1 R7 ⎦ R7 R3 ( R4 + R6 ) R7 ⎦ ⎣ R3 R7 R3 ( R4 + R6 ) ⎣ R3 v −v i0 = c 0 = " R5 10 kΩ P 6.4-6 Determine the node voltages for the circuit shown in Figure P 6.4-6. Answer: va = – 0.75 V, vb = 0 V, and vc = – 0.9375 V 40 kΩ a – + 12 V 20 kΩ 40 kΩ b 25 kΩ c – 15 kΩ + Figure P 6.4-6 Solution: KCL at node b: KCL at node a: so va vc 5 + = 0 ⇒ vc = − va 3 3 20 × 10 25 × 10 4 5 ⎛ ⎞ va − ⎜ − va ⎟ va − ( −12 ) va va + 0 ⎝ 4 ⎠ = 0 ⇒ v = − 12 V + + + a 3 3 3 40 ×10 40 × 10 20 × 10 10 × 103 13 vc = − 5 15 va = − 4 13 (checked using LNAP 6/21/05) P 6.4-7 Find vo and io for the circuit shown in Figure P 6.4-7. Assume an ideal operational amplifier. 10 kΩ 30 kΩ 10 kΩ 6V +– 30 kΩ 30 kΩ 10 kΩ – io + + 30 kΩ vo – Figure P 6.4-7 Solution: Label the circuit to account for the properties of the ideal op amp and to identify the supernode corresponding to the voltage source. Apply KCL at the inverting input node of the op amp ⎛ ( va + 6 ) −0 ⎞ ⎛ v −0 ⎞ −⎜ a ⎟ = 0 ⎟+0−⎜ ⎝ 10000 ⎠ ⎝ 30000 ⎠ or va = − 1.5 V Apply KCL to the super node corresponding the voltage source: va − 0 va + 6− 0 va − v b ( va + 6 ) −v b + + + = 0 10000 30000 30000 10000 Multiply both sides by 30000 to get 3va + ( va + 6 ) + ( va − vb ) + 3 ⎡⎣( va + 6 )− v b ⎤⎦ = 0 Solving gives vb = 2 va + 6 = 3 V Apply KCL at node b: ⎛ v −v ⎞ ⎛ ( v a + 6 )−v b ⎞ ⎟ = 0 −⎜ a b ⎟−⎜ 10000 30000 ⎝ 30000 ⎠ ⎜ 10000 ⎟ ⎝ ⎠ Multiply both sides by 30000 to get vb + v b −v o 3v b + ( v b − v o ) − ( v a −v b ) − 3 ⎡⎣( v a + 6 )− v b ⎤⎦ = 0 Solving gives v o = 8 v b − 4 va − 18 = 12 V Apply KCL at the output node of the op amp: io + vo 30000 + v o −v b 30000 = 0 ⇒ i o = −0.7 mA 10 kΩ 20 kΩ + P 6.4-8 Find vo and io for the circuit shown in Figure P 6.4-8. Assume an ideal operational amplifier. – vo + – io 20 kΩ 10 kΩ Figure P 6.4-8 Solution: Apply KVL to the bottom mesh: −i0 (10000) − i0 (20000) + 5 = 0 ⇒ i0 = 1 mA 6 The node voltages at the input nodes of an ideal op amp are equal. Consequently va = 10000 i0 = 10 V 6 Apply KCL at node a: va v −v + a 0 = 0 ⇒ 10000 20000 v0 = 3va = 5 V + – 5V P 6.4-9 Determine the node voltages for the circuit shown in Figure P 6.4-9. Answer: va = – 12 V, vb = – 4 V, vc = – 4 V, vd = – 4 V, ve = – 3.2 V, vf = – 4.8 V, and vg = – 3.2 c 10 kΩ d 20 kΩ e 20 kΩ a 40 kΩ 20 kΩ – b – + 12 V – g + + f 40 kΩ 20 kΩ 40 kΩ Figure P 6.4-9 Solution: vb + 12 vb + = 0 ⇒ vb = −4 V 40000 20000 The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V . KCL at node b: The input currents of an ideal op amp are zero, so vd = vc + 0 × 10 4 = −4 V . KCL at node g: vg ⎛ v f − vg ⎞ 2 −⎜ + = 0 ⇒ vg = v f 3 ⎟ 3 3 ⎝ 20 ×10 ⎠ 40 × 10 The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = 2 vd − v f vd − 3 v f vd − ve KCL at node d: 0 = + = + 20 ×103 20 ×103 20 ×103 20 ×103 vd − v f Finally, ve = vg = 2 16 vf = − V. 3 5 2 vf . 3 6 24 ⇒ v f = vd = − V 5 5 R = Ro + ΔR P 6.4-10 The circuit shown in Figure P 6.4-10 includes a simple strain gauge. The resistor R changes its value by ΔR when it is twisted or bent. Derive a relation for the voltage gain vo/vs and show that it is proportional to the fractional change in R, namely ΔR/Ro. Answer: vo = Ro ΔR Ro + R1 Ro R1 – + vs + + R1 Ideal Ro – – Figure P 6.4-10 Solution: By voltage division (or by applying KCL at node a) R0 va = vs R1 + R0 Applying KCL at node b: vb − vs vb − v0 + = 0 R1 R0 +ΔR ⇒ vo R0 +ΔR ( vb −vs )+ vb = v0 R1 The node voltages at the input nodes of an ideal op amp are equal so vb = va . ⎡⎛ R +ΔR ⎞ R0 ⎛ R +ΔR ⎤ R0 ⎞ ΔR ΔR v0 = ⎢⎜ 0 +1⎟ − 0 v s = ⎜ − vs ⎥ vs = − ⎟ R1 ⎦ R1 + R0 R1 + R0 ⎠ R0 ⎠ R1 + R0 ⎝ ⎣⎝ R1 + P 6.4-11 Find vo for the circuit shown in Figure P 6.4-11. Assume an ideal operational amplifier. – io + – 1.5 V 20 kΩ 20 kΩ 8 kΩ 10 kΩ + vo – Figure P 6.4-11 Solution: Node equations: ⎛ R2 ⎞ vs vs − va + = 0 ⇒ va = ⎜1 + ⎟ vs ⎜ R1 R2 R1 ⎟⎠ ⎝ and vs − va va va − vo = + R2 R3 R4 so ⎛ R4 R4 ⎞ R4 vo = ⎜1 + + va − vs ⎟ ⎜ R 2 R3 ⎟ R 2 ⎝ ⎠ ⎛ R4 R4 ⎞ ⎛ R2 ⎞ ⎛ R4 R4 R2 R2 R4 ⎞ R4 vo = ⎜1 + 1+ vs − vs = ⎜ 1 + v + + + + ⎟ ⎜ ⎟ ⎜ R 2 R3 ⎟ ⎜ ⎜ R 3 R1 R1 R1 R 3 ⎟⎟ s R1 ⎟⎠ R2 ⎝ ⎠⎝ ⎝ ⎠ ⎛ ( R1 + R 2 )( R 3 + R 4 ) + R 3 R 4 ⎞ ⎟ vs =⎜ ⎜ ⎟ R1 R 3 ⎝ ⎠ with the given values: ⎛ ( 20 + 20 )(10 + 8 ) + 10 × 8 ⎞ ⎛ 40 × 18 + 80 ⎞ vo = ⎜ ⎟ vs = ⎜ ⎟ vs = ( 4 ) vs 20 × 10 200 ⎝ ⎠ ⎝ ⎠ (checked: LNAP 5/24/04) + P 6.4-12 The circuit shown in Figure P 6.4-12 has one output, vo, and two inputs, v1 and v2. Show that when R1 vo – v1 +– R3 R3 R6 = , the output is proportional to the R4 R5 R4 difference of the inputs, v1 – v2. Specify resistance values to cause vo = 5 (v1 – v2). + R2 v2 – + – R5 R6 Figure P 6.4-12 Solution: Notice that the currents in resistance R1 and R2 are both zero, as shown. Consequently, the voltages at the noninverting inputs of the op amps are v1 and v2, as shown. The voltages at the inverting inputs of the ideal op amps are also v1 and v2, as shown. Apply KCL at the top node of R6 to get va − v2 v2 = R5 R6 ⎛ R5 + R 6 ⎞ ⇒ va = ⎜ v ⎜ R 6 ⎟⎟ 2 ⎝ ⎠ Apply KCL at the top node of R4 to get vo − v1 v1 − va = R3 R4 ⎛ R3 ⎞ ⎛ R3 ⎞ ⇒ vo = ⎜ 1 + v1 − ⎜ ⎟ va ⎟ ⎜ R4 ⎟ ⎜ R4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R5 + R6 ⎞ vo = ⎜ 1 + v1 − ⎜ ⎟ ⎜ v ⎟ ⎜ R4 ⎟ ⎜ R 4 ⎟ ⎜ R 6 ⎟⎟ 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ When R3 R4 = R6 R5 ⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R5 ⎞ ⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R 4 ⎞ vo = ⎜1 + v1 − ⎜ ⎟ ⎜1 + v2 = ⎜1 + v1 − ⎜ ⎟ ⎜ 1 + v ⎟ ⎟ ⎟ ⎜ R4 ⎟ ⎜ R4 ⎟ ⎜ R6 ⎟ ⎜ R4 ⎟ ⎜ R 4 ⎟ ⎜ R 3 ⎟⎟ 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ R3 ⎞ v −v = ⎜1 + ⎜ R 4 ⎟⎟ ( 1 2 ) ⎝ ⎠ so vo is proportional to the difference of the inputs, v1 − v2, as required. R3 Next, to make vo = 5 ( v1 − v2 ) , we choose R3 and R4 so that 5 = 1 + , e.g. R3 = 40 kΩ and R4 = R4 10 kΩ. Then with R3 R4 = R6 R5 we have R1 = 50 kΩ, R2 = 50 kΩ, R3 = 40 kΩ, R4 = 10 kΩ, R5 = 10 kΩ and R6 = 40 kΩ. (checked: LNAP 5/24/04) R1 P 6.4-13 The circuit shown in Figure P 6.4-13 has one output, vo, and one input, vi. Show that the output is proportional to the input. Specify resistance values to cause vo = 20 vi. vi – vo + R2 R3 R4 – + Figure P 6.4-13 Solution: Write a node equation at the inverting input of the bottom op amp: vo R3 + va R4 = 0 ⇒ va = − R4 R3 vo Write a node equation at the inverting input of the top op amp: R4 vo − vi va vi R3 R 2 R3 0= + = + ⇒ vo = vi R1 R 2 R 1 R2 R1 R 4 The output is proportional to the input and the constant of proportionality is vo = 20 vi so R 2 R3 R1 R 4 R 2 R3 R1 R 4 . We require = 20 . For example, R1 = R 4 = 10 kΩ, R 2 = 40 kΩ and R 3 = 50 kΩ . R1 vs R2 – + – + P 6.4-14 The circuit shown in Figure P 6.4-14 has one input, vs, and one output, vo. Show that the output is proportional to the input. Design the circuit so that vo = 20vs. + R4 R5 vo – R3 Figure P 6.4-14 Solution: Represent this circuit by node equations. vo − va R2 + vo − vs R1 vo − va R4 + =0 vo R5 ⇒ =0 ⎛ R4 ⎞ v a = ⎜1 + v ⎜ R 5 ⎟⎟ o ⎝ ⎠ ( R 1 + R 2 ) R 5 − R 1 ( R 4R 5 ) v o ⎛ R1 ⎞ ⎛ R1 ⎞⎛ R 4 ⎞ v s = ⎜1 + vo − ⎜ 1+ v = ⎟ ⎜ R2 ⎟ ⎜ R 2 ⎟⎟ ⎜⎜ R 5 ⎟⎟ o ⎝ ⎠ ⎝ ⎠⎝ ⎠ R 2R 5 20 = Then R 2 R 5 - R 1R 4 So For example R 2 v s = ( R 1 + R 2 ) v o − R 1v a ⇒ R 2R 5 ⇒ = R 2R 5 R 2 R 5 − R 1R 4 vs 19 R 1R 4 = 20 R 2 R 5 R 1 = 19 kΩ, R 4 = 10 kΩ, R 2 = 20 kΩ, R 5 = 10 kΩ, R 3 = 10 kΩ. (checked: LNAP 5/24/04) R P 6.4-15 The circuit shown in Figure P 6.4-15 has one input, vs, and one output, vo. The circuit contains seven resistors having equal resistance, R. Express the gain of the circuit, vo/vs, in terms of the resistance R. R R R R R – + – vs + + R vo – Figure P 6.4-15 Solution: Writing node equations: vs + v1 =0 ⇒ v 1 = −v s R R v1 v1 v1 − v 2 + + =0 ⇒ v 2 = 3v1 = −3v s R R R v 2 − v1 v 2 v 2 − v o + + =0 ⇒ v o = 3v 2 − v1 = −8v s R R R vo = −8 , does not depend on R. The gain of this circuit, vs (checked: LNAP 6/21/04) R3 P 6.4-16 The circuit shown in Figure P 6.4-16 has one input, vs, and one output, vo. Express the gain, vo/vs, in terms of the resistances R1, R2, R3, R4, and R5. Design the circuit so that vo = – 20 vs. R1 R4 R2 – + + + – vs R5 vo – Figure P 6.4-16 Solution: Represent this circuit by node equations. vs − va R1 + vo − va R3 = va R2 and va R2 + vo R4 = 0 ⇒ vo = − R4 R2 va So ⎛ 1 ⎛ R1 R 2 + R1 R 3 + R 2 R 3 ⎞ ⎛ R 2 ⎞ 1 1 ⎞ =⎜ + + va = ⎜ vo ⎟ ⎟ ⎟⎟ ⎜⎜ − ⎜ ⎟ R1 R 3 ⎜⎝ R1 R 2 R 2 ⎟⎠ R1 R 2 R 3 ⎝ ⎠ ⎝ R4 ⎠ ⎛ 1 R1 R 2 + R1 R 3 + R 2 R 3 ⎞ ⎛ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎞ vs = −⎜ + vo = − ⎜ ⎟ ⎟⎟ v o ⎜ R3 ⎟ ⎜ R1 R R R R R R 1 3 4 1 3 4 ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ R3 R 4 vo = − ⎜ v ⎜ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎟⎟ s ⎝ ⎠ R3 R 4 20 = We require R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 vs + vo Try R1 = R 2 = R and R 3 = R 4 = aR a2 20 = 3a + 1 2 a − 60a − 20 = 0 Then So a= e.g. +60 ± 3600 + 4 ( 80 ) = 60.332, − 0.332 2 R1 = R 2 = 10 kΩ and R 3 = R 4 = 603.32 kΩ (checked: LNAP 6/9/04) R3 P 6.4-17 The circuit shown in Figure P 6.4-17 has one input, vs, and one output, vo. Express the gain of the circuit, vo/vs, in terms of the resistances R1, R2, R3, R4, R5, and R6. Design the circuit so that vo = – 20 vs. R2 R1 – + + + – – vs R4 R5 + R6 vo – Figure P 6.4-17 Solution: Represent this circuit by node equations. vs R1 + va R2 + vo R3 =0 and va R4 + va − vo R5 ⎛ R4 ⎞ = 0 ⇒ va = ⎜ v ⎜ R 4 + R 5 ⎟⎟ o ⎝ ⎠ vs So R1 + vo R3 =− va R2 =− R4 R 2 ( R 4 + R5 ) vo ⎛ 1 ⎞ R 2 ( R 4 + R5 ) + R 4 R3 R4 ⎟ vo = − = −⎜ + vo ⎜ R3 R 2 ( R 4 R5 ) ⎟ R1 R R R + R ( ) 2 3 4 5 ⎝ ⎠ vs vo = − We require Try 20 = R 2 R3 ( R 4 + R5 ) R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 ) vs R 2 R3 ( R 4 + R5 ) R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 ) R1 = R 4 = R 5 = R and R 2 = R 3 = aR Then e.g. 20 = 2a 2 R 3 2 = a 3aR 3 3 ⇒ a = 30 R1 = R 4 = R 5 = 10 kΩ and R 2 = R 3 = 300 kΩ (checked: LNAP 6/10/04) R2 R1 P 6.4-18 The circuit shown in Figure P 6.4-18 has one input, vs, and one output, io. Express the gain of the circuit, io/vs, in terms of the resistances R1, R2, R3, and Ro. (This circuit contains a pair of resistors having resistance R1 and another pair having resistance R2.) Design the circuit so that io = 0.02 vs. + – – vs + R1 R2 – + R3 Ro io Figure P 6.4-18 Solution: Label the node voltages as shown. Represent this circuit by node equations. vb − va R2 io + vo − va R2 + = va R1 vo − vb R3 vs − va R1 ⇒ =0 =0 ⇒ ⇒ va = R1 R1 + R 2 vb v b = R 3i o + v o ⎛ R1 + R2 ⎞ vo =⎜ va − ⎟ R1 ⎜⎝ R1 R 2 ⎟⎠ R2 vs So vs ⎛ R1 + R 2 ⎞ ⎛ R1 ⎞ v o R3 =⎜ R 3i o + v o ) − = io ⎟ ⎜ ⎟ ( R1 ⎜⎝ R1 R 2 ⎟⎠ ⎜⎝ R1 + R 2 ⎟⎠ R2 R2 io R2 = v s R1 R 3 We require R2 R1 R 3 = 0.02, e.g. R 2 = 8 kΩ, R1 = R 3 = 20 kΩ . (checked: LNAP 6/21/04) P 6.4-19 The circuit shown in Figure P 6.4-19 has one input, vs, and one output, vo. The circuit contains one unspecified resistance, R. (a) Express the gain of the circuit, vo/vs, in terms of the resistance R. (b) Determine the range of values of the gain that can be obtained by specifying a value for the resistance R. (c) Design the circuit so that vo = – 3 vs. Figure P 6.4-19 Solution: (a) Use units of volts, mA, and kΩ. Apply KCL at the inverting input of the left op amp to get vs 10 vo = (b) (c) We require + va 50 + 50 ⎞ ⎛ = 0 ⇒ v a = − ⎜ 5v s + v o ⎟ R R ⎠ ⎝ vo 4 40 ⎛ 40 ⎞ v a = −4v s − v o ⇒ ⎜ 1 + ⎟ v o = − 4v s 5 R R⎠ ⎝ vo 4 4R =− =− 40 vs R + 40 1+ R vo 0≤R≤∞ ⇒ −4≤ ≤0 vs −3 = − 4R R + 40 ⇒ R = 120 kΩ (checked: LNAP 6/21/04) P 6.4-20 The circuit shown in Figure P 6.4-20 has one input, vs, and one output, vo. The circuit contains one unspecified resistance, R. (a) Express the gain of the circuit, vo/vs, in terms of the resistance R. (b) Determine the range of values of the gain that can be obtained by specifying a value for the resistance R. (c) Design the circuit so that vo = – 5 vs. R 30 kΩ 10 kΩ + – + + – – vs + 20 kΩ 10 kΩ vo – Figure P 6.4-20 Solution: (a) Use units of V, mA and kΩ. Apply KCL at the inverting input of the left op amp to get v1 10 + va 30 + 30 ⎞ ⎛ = 0 ⇒ v a = − ⎜ 3v s + v o ⎟ R R ⎠ ⎝ vo v o = 3v a = −9v s − (c) We require ⎛ 90 ⎞ ⇒ ⎜ 1 + ⎟ v o = − 9v s R⎠ ⎝ 9 9R =− 90 vs R + 90 1+ R vo 0≤R≤∞ ⇒ −9 ≤ ≤0 vs vo (b) 90 vo R −5 = =− −9 R R + 90 ⇒ R = 112.5 kΩ (checked: LNAP 7/8/04) P 6.4-21 The circuit shown in Figure P 6.4-21 has three inputs: v1, v2, and v3. The output of the circuit is vo. The output is related to the inputs by vo = av1 + bv2 + cv3 where a, b, and c are constants. Determine the values of a, b, and c. + – 20 kΩ 20 kΩ v1 – 40 kΩ + 120 kΩ – 20 kΩ 120 kΩ + + – + – v2 20 kΩ + vo – 20 kΩ 20 kΩ 30 kΩ – + v3 + – Figure P 6.4-21 Solution: Label the node voltages and identify some standard op amp circuits: Either by recognizing the inverting amplifier or by writing and solving the node equation corresponding to node a, we obtain ⎛ 20 ⎞ v b = ⎜ − ⎟ v 1 = −v 1 ⎝ 20 ⎠ Either by recognizing the voltage divider and voltage follower or by writing and solving the node equation corresponding to node c, we obtain 1 ⎛ 20 ⎞ vd = ⎜ ⎟ v2 = v2 2 ⎝ 20 + 20 ⎠ Either by recognizing the noninverting amplifier or by writing and solving the node equation corresponding to node f, we obtain ⎛ 20 ⎞ v g = ⎜1 + ⎟ v 3 = 3 v 3 ⎝ 20 ⎠ Either by recognizing the summing amplifier or by writing and solving the node equation corresponding to node e, we obtain ⎡⎛ 120 ⎞ ⎛ 120 ⎞ ⎛ 120 ⎞ ⎤ v o = − ⎢⎜ ⎟ vb + ⎜ ⎟ vd + ⎜ ⎟ v g ⎥ = − ⎡⎣3 v b + v d + 4 v g ⎤⎦ ⎝ 120 ⎠ ⎝ 30 ⎠ ⎦ ⎣⎝ 40 ⎠ Substituting for v b , v d and v g gives ⎡ ⎤ ⎛1 ⎞ v o = − ⎢3 ( −v1 ) + ⎜ v 2 ⎟ + 4 ( 3 v 3 ) ⎥ = 3 v1 − 0.5 v 2 − 12 v 3 ⎝2 ⎠ ⎣ ⎦ so a = 3, b = −0.5 and c = −12 (checked: LNAP 6/21/04) P 6.4-22 The circuit shown in Figure P 6.4-22 has two inputs: v1 and v2. The output of the circuit is vo. The output is related to the inputs by vo = av1 + bv2 where a and b are constants. Determine the values of a and b. 20 kΩ 20 kΩ 40 kΩ – + + – – v1 + + – 20 kΩ 20 kΩ v2 – + vo 20 kΩ + 20 kΩ Figure P 6.4-22 Solution: Label the node voltages as shown. Use units of V, mA and kΩ. v 3 = v1 and v 4 = −v 2 – v5 − v3 40 + v5 − v4 20 =0 ⇒ v5 = 1 1 2 v 3 + 2v 2 ) = v 1 − v 2 ( 3 3 3 so a=− 1 2 and b = − 3 3 (checked: LNAP 6/21/04) P6.4-23 The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is the node voltage v o . The output is related to the input by the equation v o = k v s where k = is called the gain of the circuit. Determine the value of the gain k. FigureP6.4-23 Solution: Label the node voltages as shown. Apply KCL at the inverting input node of the op amp to get vs 30000 + vs − va 80000 = 0 ⇒ 11v s = 3 v a ⇒ va = 11 vs 3 Apply KCL at the right node of the 80 kΩ resistor to get vs − va 80000 + vo − va 20000 = va 20000 ⇒ vs − 9 va + 4 vo = 0 ⎛ 11 ⎞ ⇒ vs − 9 ⎜ vs ⎟ + 4 vo = 0 ⎝3 ⎠ ⇒ v s − 33 v s + 4 v o = 0 ⇒ 4 v o = 32 v s ⇒ vo = 8 vs Comparing this equation to v o = k v s , we determine that k = 8 V/V. vo vs P6.4-24 The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is the node voltage v o . The output is related to the input by the equation v o = m i s + b where m and b are constants. Determine the values of m and b. FigureP6.4-24 Solution: Label the node voltages as shown. Apply KCL at the inverting input node of the op amp to get is = 6 − va 50000 = 0 ⇒ v a = 6 − ( 50 × 10 3 ) i s Apply KCL at the top node of the 25 kΩ resistor to get vo − va va is + = 25000 25000 Solving: ( 25 ×10 3 ) i s − 2 v a + v o = 0 ( 25 ×10 ) i s − 2 ( 6 − ( 50 ×10 ) i s ) + v o = 0 (125 ×10 ) i s − 12 + v o = 0 v o = − (125 × 10 ) i s + 12 3 3 3 3 Comparing this equation to v o = m i s + b , we get m = − (125 ×10 3 ) V/A and b =12 V. P6.4-25 The input to the circuit shown in Figure P6.4-25 is the node voltage v s . The output is the node voltage v o . The output is related to the input by the equation v o = k v s where k = the gain of the circuit. Determine the value of the gain k. FigureP6.4-25 Solution: Label the node voltages as shown. The node equations are vs − va 5000 va − 0 5000 = vo − vb 5000 = va − 0 5000 0 − vb vb 25000 Substituting v a = − va − vo 50000 ⇒ v b = −10 v a 50000 = + ⇒ v o = 11v b vo 110 into the first node equation gives: vo vs is called vs + vo − vo − vo − vo 110 = 110 + 110 5000 5000 50000 vo ⎞ ⎛ ⎛ vo ⎞ vo ⇒ 10 ⎜ v s + − vo ⎟ = 10 ⎜ − ⎟− 110 ⎠ ⎝ ⎝ 110 ⎠ 110 131 ⇒ 10 v s = − vo 110 1100 v s = −8.397 v s ⇒ vo = − 131 Comparing this equation to v o = k v s , we determine that k = −8.397 V/V. P6.4-26 The values of the node voltages v1, v2 and vo in Figure P6.4-26 are v1 = 6.25 V, v2 = 3.75 V and vo = −15 V. Determine the value of the resistances R1, R2 and R3: FigureP6.4-26 Solution: Label the node voltages as shown. Using units of Volts, kΩ and mAmps, the node equations are: 2.5 v1 − 2.5 v1 − v 2 v 2 = = , 20 R2 20 R1 And v2 − 0 R3 + vo − 0 20 =0 Using v1 = 6.25 V gives 2.5 6.25 − 2.5 = ⇒ R1 = 30 kΩ 20 R1 Using v1 = 6.25 V and v 2 = 3.75 V gives 6.25 − 3.75 3.75 = ⇒ R 2 = 30 kΩ 20 R2 Using v 2 = 3.75 V and v o = −15 V gives 3.75 − 0 −15 − 0 + = 0 ⇒ R 3 = 5 kΩ R3 20 P6.4-27 The input to the circuit shown in Figure P6.4-27 is the voltage source voltage, v i . The output is the node voltage, v o .The output is related to the input by the equation v o = k v i where k = vo vi is called the gain of the circuit. Determine the value of the gain k. FigureP6.4-27 Solution: Label the node voltages as shown. The node voltages at the input nodes of an ideal op amp are equal so va = vb Consequently i2 = va − vb =0 R The input currents of an ideal op amp are 0 A, so applying KCL at nodes a and b shows that i1 = 0 and i 3 = 0 . Consequently, the voltages across the corresponding resistor are 0 V. Finally va = v b = vc = vi Write an node equation at the right node of the 24 kΩ resistor: 0= vi 10 + vi − vo 40 ⇒ vo = 5 vi Finally k= vo vi =5 Section 6-5: Design Using Operational Amplifier P 6.5-1 Design the operational amplifier circuit in Figure P 6.5-1 so that vout = r · iin where r = 20 Operational amplifier circuit iin + 20 kΩ vout – V mA Figure P 6.5-1 Solution: Use the current-to-voltage converter, entry (g) in Figure 6.6-1. P 6.5-2 Design the operational amplifier circuit in Figure P 6.5-2 so that iout = g · vin where g =2 iout vin + – Operational amplifier circuit mA V Figure P 6.5-2 Solution: Use the voltage –controlled current source, entry (i) in Figure 6.6-1. 5 kΩ P 6.5-3 Design the operational amplifier circuit in Figure P 6.5-3 so that v1 Operational amplifier circuit + – vout = 5 · v1 + 2 · v2 + 20 kΩ v2 vout – + – Figure P 6.5-3 Solution: Use the noninverting summing amplifier, entry (e) in Figure 6.6-1. P 6.5-4 Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · (v1 – v2). Solution: Use the difference amplifier, entry (f) in Figure 6.6-1. P 6.5-5 Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · v1 – 2 · v2 Solution: Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1. P 6.5-6 The voltage divider shown in Figure P 6.5-6 has a gain of vout −10 kΩ = =2 vin 5 kΩ + (−10 kΩ) 5 kΩ + vin Design an operational amplifier circuit to implement the – 10-kΩ resistor. Solution: Use the negative resistance converter, entry (h) in Figure 6.6-1. + – –10 kΩ vout – Figure P 6.5-6 P 6.5-7 Design the operational amplifier circuit in Figure P 6.5-7 so that iin = 0 and vout = 3 · vin iin + 5V + – 10 kΩ –5 V + – Operational amplifier circuit vin – + 20 kΩ vout – Figure P 6.5-7 Solution: Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the current iin to be zero. P 6.5-8 Design an operational amplifier circuit with output vo = 6 v1 + 2 v2, where v1 and v2 are input voltages. Solution: Summing Amplifier: va = − ( 6 v1 + 2 v2 ) ⎫ ⎬ ⇒ vo = 6 v1 + 2 v2 Inverting Amplifier: vo = −va ⎭ 8 kΩ P 6.5-9 Determine the voltage vo for the circuit shown in Figure P 6.5-9. Hint: Use superposition. Answer: vo = (– 3)(3) + (4)(– 4) + (4)(8) = 7 V 4 kΩ 24 kΩ – + + 3V – 4V – + 2 mA + 10 kΩ vo – Figure P 6.5-9 Solution: Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 V P 6.5-10 For the op amp circuit shown in Figure P 6.5-10, find and list all the possible voltage gains that can be achieved by connecting the resistor terminals to either the input or the output voltage terminals. 6 12 12 – vs + + – Figure P 6.5-10 Solution: R1 6 12 24 6||12 6||24 R2 -vo/vs 12||12||24 0.8 6||12||24 0.286 6||12||12 0.125 12||24 2 12||12 1.25 12||24 6||12||12 6||12||24 12||12||24 6||12 0.5 24 8 12 3.5 6 1.25 R1 R2 -vo/vs 12||12 6||24 0.8 24 + vo – P 6.5-11 The circuit shown in Figure P 6.5-11 is called a Howland current source. It has one input, vin, and one output, iout. Show that when the resistances are chosen so that R2 R3 = R1 R4, the output is related to the input by the equation iout = vin R1 R4 R3 – + R1 vin + – R2 RL iout Figure P 6.5-11 Solution: Label the node voltages as shown. Apply KCL at the inverting input of the op amp to get va R3 + va − vb R3 ⎛ R3 + R 4 ⎞ = 0 ⇒ vb = ⎜ v ⎜ R 3 ⎟⎟ a ⎝ ⎠ Apply KCL at the noninverting input of the op amp to get v a − v in R1 + va − vb R2 + i out = 0 Solving gives ⎛ R1 + R 2 ⎞ v in v b ⎛ R1 + R 2 R 3 + R 4 ⎞ v in va ⎜ − − + i out = 0 ⇒ v a ⎜ − + i out = 0 ⎟ ⎟− ⎜ R1 R 2 ⎟ R1 R 2 ⎜ R1 R 2 R 2 R 3 ⎟⎠ R1 ⎝ ⎠ ⎝ When R 2 R 3 = R1 R 4 the quantity in parenthesis vanishes leaving i out = 1 v in R1 P 6.5-12 The input to the circuit shown in Figure P 6.5-12a is the voltage vs. The output is the voltage vo. The voltage vb is used to adjust the relationship between the input and output. (a) Show that the output of this circuit is related to the input by the equation vo = avs + b where a and b are constants that depend on R1, R2, R3, R4, and vb. Design the circuit so that its input and output have the relationship specified by the graph shown in Figure P 6.5-12b. (b) vo , V 8 + R1 6 – + 4 R4 + – R2 vs 2 vo + – R3 vb R5 –4 –6 − –2 2 4 –2 –4 (a) (b) Figure P 6.5-12 Solution: (a) Label the node voltages as shown. The node equations are vs − va R1 + vb − va R2 = va R3 and va R5 = vo − va R4 ⎛ R5 ⎞ ⇒ va = ⎜ v ⎜ R 4 + R 5 ⎟⎟ o ⎝ ⎠ Solving these equations gives vs ⎛ 1 v b ⎛ R1 R 2 + R 2 R 3 + R1 R 3 R5 ⎞ vb 1 1 ⎞ =⎜ + + =⎜ × va − vo − ⎟ ⎟ R1 ⎝⎜ R1 R 2 R 3 ⎟⎠ R 2 ⎜⎝ R1 R 2 R 3 R 4 + R 5 ⎟⎠ R2 So ⎛ R 2 R3 R 4 + R5 ⎞ R1 R 3 R 4 + R5 vo = ⎜ × vs + × × vb ⎟ ⎜ R1 R 2 + R 2 R 3 + R1 R 3 R 5 ⎟⎠ R1 R 2 + R 2 R 3 + R1 R 3 R5 ⎝ So vs , V ⎛ R 2 R3 R 4 + R5 ⎞ R1 R 3 R 4 + R5 a=⎜ × v s and b = × × vb ⎟ ⎜ R1 R 2 + R 2 R 3 + R1 R 3 ⎟ R R R + R R + R R R 5 1 2 2 3 1 3 5 ⎝ ⎠ (b) The equation of the straight line is 5 vo = vs + 5 4 We require R 2 R3 R1 R 2 + R 2 R 3 + R1 R 3 × R 4 + R5 R5 = 5 4 For example, let R1 = R 2 = R 3 = 10 kΩ, R 4 = 55 kΩ and R 5 = 20 kΩ . Next we require 5= R1 R 3 R1 R 2 + R 2 R 3 + R1 R 3 × R 4 + R5 R5 × vb = 5 vb 4 i.e. vb = 4 V (checked: LNAP 6/20/04) R3 R1 P 6.5-13 The input to the circuit shown in Figure P 6.5-13a is the voltage vs. The output is the voltage vo. The voltage vb is used to adjust the relationship between the input and output. (a) Show that the output of this circuit is related to the input by the equation – + – R2 vs + – + + R4 vb vo – vo = avs + b where a and b are constants that depend on R1, R2, R3, R4, and vb. (b) (a) Design the circuit so that its input and output have the relationship specified by the graph shown in Figure P 6.5-13b. vo , V 8 6 4 2 –6 –4 2 –2 4 –2 –4 (b) P 6.5-13 Solution: (a) Apply KCL at the inverting input of the op amp to get: ⎛ R3 ⎞ vs vb vo R3 + + = 0 ⇒ vo = ⎜ − ⎟ vs − vb ⎜ ⎟ R1 R 2 R 3 R2 ⎝ R1 ⎠ R3 R3 so a=− and b = − vb R1 R2 (b) The equation of the straight line is We require − R3 5 =− 2 R1 Next, we require 5=− e.g. R 3v b R2 5 vo = − vs + 5 2 R1 = 20 kΩ and R 3 = 50 kΩ . e.g. R 2 = R 3 = 50 kΩ and v b = −5 V . vs, V P6.5-14 The input to the circuit shown in Figure P6.5-14 is the voltage source voltage v s . The output is the node voltage v o . the output is related to the input by the equation v o = m v s + b where m and b are constants. (a) Specify values of R 3 and v a that cause the output to be related to the input by the equation v o = 4 v s + 7 . (b) Determine the values of m and b when R 3 = 20 kΩ and v a = 2.5 V . Figure P6.5-14 Solution: Label the node voltages: ⎛ 30 ⎞ Recognizing an inverting amplifier and a noninverting amplifier we write v b = ⎜ − ⎟ v s . ⎝ 10 ⎠ Applying KCL at the inverting input node of the right op amp gives ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 30 ⎞ ⇒ v o = ⎜1 + v a − ⎜ ⎟ v b = ⎜1 + va − ⎜ ⎟ ⎜ − ⎟ vs ⎟ ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎜ R 3 ⎟ ⎝ 10 ⎠ R3 20 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 20 ⎞ ⎛ 60 ⎞ ⇒ v o = ⎜1 + va + ⎜ ⎟ vs ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 60 ⎞ ⎛ 20 ⎞ (a) We require ⎜ ⎟ = 4 and ⎜1 + v = 7 V so R 3 = 15 kΩ and then v a = 3 V. ⎜ R3 ⎟ ⎜ R 3 ⎟⎟ a ⎝ ⎠ ⎝ ⎠ ⎛ 60 ⎞ ⎛ 60 ⎞ ⎛ 20 ⎞ V ⎛ 20 ⎞ and b = ⎜1 + v a = ⎜1 + ⎟ 2.5 = 5 V . (b) m = ⎜ ⎟ = ⎜ ⎟ = 3 ⎟ ⎜ R 3 ⎟ ⎝ 20 ⎠ ⎜ R3 ⎟ V ⎝ 20 ⎠ ⎝ ⎠ ⎝ ⎠ vb − va = va − vo P6.5-15. The circuit shown in Figure 6.5-15 uses a potentiometer to implement a variable resistor having a resistance R that varies over the range 0 ≤ R ≤ 200 kΩ The gain of this circuit is G = vo vs . Varying the resistance R over it’s range causes the value of the gain G to vary over the range vo ≤ G max G min ≤ vs Determine the minimum and maximum values of the gain, G min and G max . Solution: Let R eq = 25 + R || 50 Then, recognizing this circuit as a noninverting amplifier, we get R eq 25 + R || 50 G = 1+ = 1+ 25 25 G min corresponds to R = 0: G min = 1 + 25 + 0 || 50 25 + 0 = 1+ = 2 V/V 25 25 G max corresponds to R = 200 kΩ: G max = 1 + 25 + 200 || 50 25 + 40 = 1+ = 3.6 V/V 25 25 Figure P6.5-15 P6.5-16 The input to the circuit shown in Figure P6.5-16a is the voltage, vs. The output is the voltage vo. The voltage v b is used to adjust the relationship between the input and output. Determine values of R 4 and v b that cause the circuit input and output have the relationship specified by the graph shown in Figure P6.5-21b. Answers: v b = 1.62 V and R 4 = 62.5 kΩ. (a) (b) Figure P6.5-16 Solution: Recognize the voltage divider, voltage follower and noninverting amplifier to write R4 ⎞ R4 ⎞ 2 R4 ⎞ R4 ⎞ ⎛ ⎞⎛ ⎛ ⎛ ⎛ 20 × 10 3 vo = ⎜ 3 3 ⎟⎜ − 3 ⎟ v s + ⎜1 + 3 ⎟ vb = ⎜ − 3 ⎟ v s + ⎜1 + 3 ⎟ vb ⎝ 20 × 10 + 5 × 10 ⎠ ⎝ 30 × 10 ⎠ ⎝ 30 × 10 ⎠ ⎝ 75 × 10 ⎠ ⎝ 30 × 10 ⎠ (Alternately, this equation can be obtained by writing two node equations: one at the noninverting node of the left op amp and the other at the inverting node of the right op amp.) The equation of the straight line is 5 vo = − vs + 5 3 Comparing coefficients gives 5 5 75 ×10 3 − =− ⇒ R4 = × = 62.5 × 10 3 = 62.5 kΩ 3 75 ×10 3 3 2 2 R4 and R4 ⎞ ⎛ ⎛ 62.5 × 10 3 ⎞ 5 = ⎜1 + v = ⎜1 + 3 ⎟ b 3 ⎟ v b = 3.08333 v b 30 10 30 10 × × ⎝ ⎠ ⎝ ⎠ ⇒ vb = 5 = 1.62 V 3.08333 Figure P6.5-17 P6.5-17 Figure P6.5-17 shows three similar circuits. The outputs of the circuits are the voltages v1, v2 and v3. Determine the values of these three outputs. Solution: 80 80 ⎛ 80 ⎞ v1 = ⎜ 1 + ⎟ (1.8 ) = 7.56 V , v 2 = (1.8) = 1.3714 V and v 3 = − (1.8) = −5.76 V 80 + 25 25 ⎝ 25 ⎠ Figure P6.5-18 P6.5-18 . The input to the circuit shown in Figure P6.5-18 is the source voltage vs. The output is the voltage across the 25 kΩ resistor, vo. The output is related to the input by the equation vo = (g) vi where g is the gain of the circuit. Determine the value of g. Solution: Using equivalent resistance we can draw the circuit as shown. 20 = (15 + 45 ) || 30 and 80 = 64 + ( 80 || 20 ) Recognizing this circuit as an inverting amplifier, we write g= vo vs =− 80 V = −4 20 V Section 6-6: Operational Amplifier Circuits and Linear Algebraic Equations P 6.6-1 Design a circuit to implement the equation z = 4w + x − 3y 4 The circuit should have one output, corresponding to z, and three inputs, corresponding to w, x, and y. Solution: P 6.6-2 Design a circuit to implement the equation 0 = 4w + x + 10 – (6y + 2z) The output of the circuit should correspond to z. Solution: Section 6-7: Characteristics of the Practical Operational Amplifier 10 kΩ P 6.7-1 Consider the inverting amplifier shown in Figure P 6.7-1. The operational amplifier is a typical OP-07E (Table 6.7-1). Use the offsets model of the operational amplifier to calculate the output offset voltage. (Recall that the input, vin, is set to zero when calculating the output offset voltage.) 100 kΩ – + – vin + + vo – Answer: 0.45 mV Figure P 6.7-1 Solution: The node equation at node a is: vout − vos vos = + ib1 3 100×10 10×103 Solving for vout: ⎛ 100×103 ⎞ vout = ⎜1+ v + (100 × 103 ) ib1 = 11vos + (100 × 103 ) ib1 3 ⎟ os 10 10 × ⎝ ⎠ ( ) = 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mV + P 6.7-2 Consider the noninverting amplifier shown in Figure P 6.7-2. The operational amplifier is a typical LF351 (Table 6.7-1). Use the offsets model of the operational amplifier to calculate the output offset voltage. (Recall that the input, vin, is set to zero when calculating the output offset voltage.) – + 90 kΩ + – vin vo 10 kΩ – Figure P 6.7-2 Solution: The node equation at node a is: vos v −v + ib1 = 0 os 10000 90000 Solving for vo: ⎛ 90×103 ⎞ vo = ⎜1+ v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1 3 ⎟ os ⎝ 10×10 ⎠ = 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV R1 P 6.7-3 Consider the inverting amplifier shown in Figure P 6.7-3. Use the finite gain model of the operational amplifier (Figure 6.7-1c) to calculate the gain of the inverting amplifier. Show that vo Rin ( Ro − AR2 ) = vin ( R1 + Rin )( Ro − R2 ) + R1 Rin (1 + A) R2 – + – vin + Figure P 6.7-3 Solution: v1 −vin v1 v1 −v0 ⎫ + + = 0⎪ R1 Rin R2 v0 Rin ( R0 − AR2 ) ⎪ = ⎬ ⇒ v0 + Av1 v0 − v1 vin ( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A) + =0 ⎪ ⎪⎭ R0 R2 + vo – R1 P 6.7-4 Consider the inverting amplifier shown in Figure P 6.7-3. Suppose the operational amplifier is ideal, R1 = 5 kΩ, and R2 = 50 kΩ. The gain of the inverting amplifier will be R2 – + – vin + vo = −10 vin Use the results of Problem P 6.7-3P 6.7-3 to find the gain of the inverting amplifier in each of the following cases: (a) + vo – Figure P 6.7-3 The operational amplifier is ideal, but 2 percent resistors are used and R1 = 5.1 kΩ and R2 = 49 kΩ. (b) The operational amplifier is represented using the finite gain model with A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5 kΩ and R2 = 50 kΩ. (c) The operational amplifier is represented using the finite gain model with A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5.1 kΩ and R2 = 49 kΩ. Solution: a) v0 R 49×103 = − 2 = − = −9.6078 vin R1 5.1×103 ( ) b) 2×106 ) 75−( 200,000 )( 50×103 ) ( v0 = = −9.9957 vin (5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000) c) v0 2×106 (75− (200,000)(49×103 )) = = −9.6037 vin (5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000) R1 P 6.7-5 The circuit in Figure P 6.7-5 is called a difference amplifier and is used for instrumentation circuits. The output of a measuring element is represented by the common mode signal vcm and the differential signal (vn + vp). Using an ideal operational amplifier, show that R4 (vn + vp ) R1 R4 R3 = R1 R2 vo = − When R4 vn + – – vp + – + + R2 vo vcm + – R3 – Figure P 6.7-5 Solution: Apply KCL at node b: R3 vb = (vcm − v p ) R2 + R3 Apply KCL at node a: va −v0 va −(vcm + vn ) + = 0 R4 R1 The voltages at the input nodes of an ideal op amp are equal so va = vb . R R +R v0 = − 4 (vcm + vn ) + 4 1 va R1 R1 R v0 = − 4 (vcm + vn ) + R1 ( R4 + R1 ) R3 (vcm − v p ) R1 ( R2 + R3 ) when R4 R1 R4 +1 ( R4 + R1 ) R3 R3 R R R1 then = = × 3 = 4 R3 R2 R1 ( R2 + R3 ) R1 +1 R2 R2 so v0 = − R4 R R (vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p ) R1 R1 R1 Section 6-10 How Can We Check…? 6 kΩ 4 kΩ P 6.10-1 Analysis of the circuit in Figure P 6.10-1 shows that io = – 1 mA and vo = 7 V. Is this analysis correct? io – + – Hint: Is KCL satisfied at the output node of the op amp? + 2V 5V – + + 10 kΩ Figure P 6.10-1 Solution: Apply KCL at the output node of the op amp io = v − ( −5 ) vo + o =0 10000 4000 Try the given values: io =−1 mA and vo = 7 V −1×10−3 ≠ 3.7 × 10−3 = 7 − ( −5 ) 7 + 10000 4000 KCL is not satisfied. These cannot be the correct values of io and vo. vo – P 6.10-2 Your lab partner measured the output voltage of the circuit shown in Figure P 6.10-2 to be vo = 9.6 V. Is this the correct output voltage for this circuit? Hint: Ask your lab partner to check the polarity of the voltage that he or she measured. 4 kΩ 10 kΩ + 2 mA – – + + vo – Figure P 6.10-2 Solution: va = ( 4 ×103 )( 2 ×10−3 ) = 8 V 12 ×103 va = −1.2 ( 8 ) = −9.6 V 10 ×103 So vo = −9.6 V instead of 9.6 V. vo = − 12 kΩ 4 kΩ 2 kΩ + P 6.10-3 Nodal analysis of the circuit shown in Figure P 6.10-3 indicates that vo = – 12 V. Is this analysis correct? Hint: Redraw the circuit to identify an inverting amplifier and a noninverting amplifier. – – + + 2V vo 6 kΩ – + 3V + – 2 kΩ – Figure P 6.10-3 Solution: First, redraw the circuit as: Then using superposition, and recognizing of the inverting and noninverting amplifiers: ⎛ 6 ⎞⎛ 4 ⎞ ⎛ 4⎞ vo = ⎜ − ⎟ ⎜ − ⎟ ( −3) + ⎜ 1 + ⎟ ( 2 ) = −18 + 6 = −12 V ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2⎠ The given answer is correct. P 6.10-4 Computer analysis of the circuit in Figure P 6.10-4 indicates that the node voltages are va = – 5 V, vb = 0 V, vc = 2 V, vd = 5 V, ve = 2 V, vf = 2 V, and vg = 11 V. Is this analysis correct? Justify your answer. Hint: Verify that the resistor currents indicated by these node voltages satisfy KCL at nodes b, c, d, and f. a 10 kΩ – + b 4 kΩ c 6 kΩ 5 kΩ 40 kΩ 5V e 10 kΩ + – 2V 4 kΩ d – f g + 10 kΩ Figure P 6.10-4 Solution: First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op amp are equal.) The given voltages satisfy all the node equations at nodes b, c and d: node b: 0− (−5) 0 0−2 + + =0 10000 40000 4000 node c: 0− 2 2 −5 = +0 4000 6000 node d: 2 −5 5 5−11 = + 6000 5000 4000 Therefore, the analysis is correct. a 20 kΩ b + P 6.10-5 Computer analysis of the d – noninverting summing amplifier shown in Figure P 6.10-5 indicates that the node 9 kΩ voltages are va = 2 V, vb = – 0.25 V, vc = – 5 V, c 40 kΩ + vd = – 2.5 V, and ve = – 0.25 V. 2V – e (a) Is this analysis correct? – (b) Does this analysis verify that the circuit + 5V 1 kΩ 40 kΩ is a noninverting summing amplifier? Justify your answers. 1st Hint: Verify that the resistor currents indicated by these node voltages satisfy KCL Figure P 6.10-5 at nodes b and e. 2nd Hint: Compare to Figure 6.5-1e to see that Ra = 10 kΩ and Rb = 1 kΩ. Determine K1, K2, and K4 from the resistance values. Verify that vd = K4(K1va + K2vc). Solution: The given voltages satisfy the node equations at nodes b and e: node b: −.25− 2 −.25 −.25−( −5 ) + + =0 20000 40000 40000 node e: −2.5−( −0.25 ) −0.25 ≠ +0 9000 1000 Therefore, the analysis is not correct. Notice that −2.5−( +0.25 ) +0.25 = +0 9000 1000 So it appears that ve = +0.25 V instead of ve = −0.25 V. Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2 = 1/ 4 and K4 = 9. The given node voltages satisfy the equation −2.5 = vd = K 4 ( K v + K v ) = 10 ⎛⎜⎝ 12 ( 2 )+ 14 ( −5) ⎞⎟⎠ None-the-less, the analysis is not correct. 1 a 2 c PSpice Problems SP 6-1 The circuit in Figure SP 6-1 has three inputs: vw, vx, and vy. The circuit has one output, vz. The equation vz = avw + bvx + cvy expresses the output as a function of the inputs. The coefficients a, b, and c are real constants. (a) Use PSpice and the principle of superposition to determine the values of a, b, and c. (b) Suppose vw = 2 V, vx = x, vy = y and we want the output to be vz = z. Express z as a function of x and y. Hint: The output is given by vz = a when vw = 1 V, vx = 0 V, and vy = 0 V. Answer: (a) vz = vw + 4 vx – 5 vy (b) z = 4 x – 5 y + 2 20 kΩ 60 kΩ – 20 kΩ + vx + + – 20 kΩ + – 100 kΩ 20 kΩ – 60 kΩ vy – 20 kΩ + vw + – 20 kΩ 20 kΩ Figure SP 6-1 Solution: (a) vz = a vw + b vx + c v y The following three PSpice simulations show 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V vz 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V: 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V: -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V: Therefore v z = vw + 4 v x − 5 v y (b) When vw= 2 V: vz = 4 vx − 5 v y + 2 25 kΩ SP 6-2 The input to the circuit in Figure SP 6-2 is vs, and the output is vo. + – (a) Use superposition to express vo as a function of vs. 80 kΩ – vs + (b) Use the DC Sweep feature of PSpice to plot vo as a function of vs. + vo 2V – + – (c) Verify that the results of parts (a) and (b) agree with each other. Figure SP 6-2 Solution: a) Using superposition and recognizing the inverting and noninverting amplifiers: vo = − 80 ⎛ 80 ⎞ vs + ⎜1 + ⎟ ( −2 ) = −3.2 vs − 8.4 25 ⎝ 25 ⎠ b) Using the DC Sweep feature of PSpice produces the plot show below. Two points have been labeled in anticipation of c). c) Notice that the equation predicts ( −3.2 ) ( −5) − 8.4 = 7.6 and ( −3.2 ) ( 0 ) − 8.4 = −8.4 Both agree with labeled points on the plot. 1 SP 6-3 A circuit with its nodes identified is shown in Figure SP 6-3. Determine v34, v23, v50, and io. 10 kΩ 10 kΩ 30 kΩ 6V 2 3 +– 30 kΩ 10 kΩ 30 kΩ 4 – + 5 io 30 kΩ + vo – Figure SP 6-3 Solution: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2 -3.000E-04 -7.000E-04 v34 = −1.5 − −12 × 10−6 ≅ −1.5 V v23 = 4.5 − ( −1.5 ) = 6 V v50 = 12 − 0 = 12 V io = −7 × 10−4 = −0.7 mA 2 kΩ SP 6-4 Use PSpice to analyze the VCCS shown in Figure SP 6-4. Consider two cases: (a) The operational amplifier is ideal. (b) The operational amplifier is a typical μA741 represented by the offsets and finite gain model. 10 kΩ – + 2 kΩ 10 kΩ iout 20 mV + – 50 kΩ Figure SP 6-4 Solution: V4 is a short circuit used to measure io. The input of the VCCS is the voltage of the left-hand voltage source. (The nominal value of the input is 20 mV.) The output of the VCCS is io. A plot of the output of the VCCS versus the input is shown below. The gain of the VCVS is gain = 50 ×10−6 − ( −50 ×10−6 ) 1 A = × 10−3 V 100 ×10 − ( −100 ×10 ) 2 −3 −3 Design Problems DP 6-1 Design the operational amplifier circuit in Figure DP 6-1 so that iin iout 1 = ⋅ iin 4 iout Operational Amplifier Circuit 5 kΩ Figure DP 6-1 Solution: From Figure 6.6-1g, this circuit is described by vo = R f i in . Since i out = Notice that i oa = i in + i in < (1250 ) i in 5000 = Rf vo 1 i out , we require = = , or Rf = 1250 Ω 5000 4 i in 5000 5 5 i in . To avoid current saturation requires i in < i sat or 4 4 4 i sat . For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation. 5 DP 6-2 Figure DP 6-2a shows a circuit that has one input, vi, and one output, vo. Figure DP 62b shows a graph that specifies a relationship between vo and vi. Design a circuit having input, vi, and output, vo, that have the relationship specified by the graph in Figure DP 6-2b. Hint: A constant input is required. Assume that a 5-V source is available. vo, V 8 6 4 2 vi vo –6 –4 –2 2 4 6 8 –2 –4 –6 –8 (a) (b) Figure DP 6-2 Solution: 3 3 3 ⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 12 ⎞ vo = − vi + 3 = − vi + ⎜1 + ⎟ ⎜ ⎟ 5 = − vi + ⎜ 1 + ⎟ ⎜ ⎟5 4 4 4 ⎝ 4 ⎠ ⎝ 35 ⎠ ⎝ 4 ⎠ ⎝ 12 + 23 ⎠ vi, V DP 6-3 Design a circuit having input, vi, and output, vo, that are related by the equations (a) vo = 12vi + 6, (b) vo = 12vi – 6, (c) vo = – 12vi + 6, and (d) vo = – 12vi – 6. Hint: A constant input is required. Assume that a 5-V source is available. Solution: 1 1 1 ⎛1 ⎞ (a) 12 v i + 6 = 24 ⎜ v i + ( 5 ) ⎟ ⇒ K 4 = 24, K 1 = , and K 2 = . Take Ra = 18 kΩ and 20 2 20 ⎝2 ⎠ Rb = 1 kΩ to get (b) (c) (d) DP 6-4 Design a circuit having three inputs, v1, v2, v3, and two outputs, va, vb, that are related by the equation ⎡ v1 ⎤ ⎡ va ⎤ ⎡12 3 −2 ⎤ ⎢ ⎥ ⎡ 2 ⎤ ⎢ ⎥=⎢ ⎥ ⎢ v2 ⎥ + ⎢ ⎥ ⎣ vb ⎦ ⎣ 8 −6 0 ⎦ ⎢ v ⎥ ⎣ −4 ⎦ ⎣ 3⎦ Hint: A constant input is required. Assume that a 5-V source is available. Solution: x DP 6-5 A microphone has an unloaded voltage vs = 20 mV, as shown in Figure DP 6-5a. An op amp is available as shown in Figure DP 6-5b. It is desired to provide an output voltage of 4 V. Design an inverting circuit and a noninverting circuit and contrast the input resistance at terminals x–y seen by the microphone. Which configuration would you recommend in order to achieve good performance in spite of changes in the microphone resistance Rs? Hint: We plan to connect terminal a to terminal x and terminal b to terminal y, or visa versa. Rs vs +– 10 kΩ Microphone y (a) R2 R1 – a b + + vo – (b) Figure DP 6-5 Solution: We require a gain of 4 = 200 . Using an inverting amplifier: 20 ×10−3 Here we have 200 = − R2 10 ×103 + R1 noninverting amplifier: . For example, let R1 = 0 and R2 = 1 MΩ. Next, using the Here we have 200 = 1 + R2 R1 . For example, let R1 = 1 kΩ and R2 = 199 kΩ. The gain of the inverting amplifier circuit does not depend on the resistance of the microphone. Consequently, the gain does not change when the microphone resistance changes. Chapter 7 Exercises Exercise 7.2-1 Determine the current i(t) for t > 0 for the circuit of Figure E 7.2-1b when vs(t) is the voltage shown in Figure E 7.2-1a. Hint: Determine iC(t) and iR(t) separately, then use KCL. vs(t)(V) 5 i(t) 4 iR(t) iC(t) 3 vs(t) 2 + – 1Ω 1F 1 1 2 3 4 5 6 7 8 9 t (s) (a) (b) Figure E 7.2-1 Answer: ⎧ 2t − 2 2 < t < 4 ⎪ 4<t <8 v(t ) = ⎨7 − t ⎪0 otherwise ⎩ Solution: ⎧2 2<t <4 d ⎪ i C ( t ) = 1 v s ( t ) = ⎨ −1 4 < t < 8 dt ⎪ 0 otherwise ⎩ so and ⎧2 t − 4 2 < t < 4 ⎪ i R (t ) = 1 v s (t ) = ⎨ 8 − t 4<t <8 ⎪ 0 otherwise ⎩ ⎧2 t − 2 2 < t < 4 ⎪ i (t ) = i C ( t ) + i R ( t ) = ⎨ 7 − t 4<t <8 ⎪ 0 otherwise ⎩ Exercise 7.3-1 A 200-μF capacitor has been charged to 100 V. Find the energy stored by the capacitor. Find the capacitor voltage at t = 0+ if v(0–) = 100 V. Answer: w(1) = 1 J and v(0+) = 100 V Solution: Cv 2 1 2 = ( 2×10 −4 ) (100 ) = 1 J 2 2 + − vc ( 0 ) = vc ( 0 ) = 100 V W = Exercise 7.3-2 A constant current i = 2 A flows into a capacitor of 100μF after a switch is closed at t = 0. The voltage of the capacitor was equal to zero at t = 0–. Find the energy stored at (a) t = 1 s and (b) t = 100 s. Answer: w(1) = 20 kJ and w(100) = 200 MJ Solution: (a) W ( t ) = W ( 0 ) + ∫ 0 vi dt t First, W ( 0 ) = 0 since v ( 0 ) = 0 Next, v( t ) = v( 0 ) + ∴ W (t ) = 1 t 4 t 4 i dt 10 = ∫ ∫ 0 2 dt = 2×10 t C 0 ∫ ( 2×10 ) t ( 2 )dt t 4 0 = 2 × 104 t 2 W (1s ) = 2 ×104 J = 20 kJ (b) W (100s ) = 2 × 104 (100 ) 2 = 2 × 108 J = 200 MJ 6 mF Exercise 7.4-1 Find the equivalent capacitance for the circuit of Figure E 7.4-1 Answer: Ceq = 4 mF 12 mF 9 mF 4 mF Ceq Figure E 7.4-1 Solution: 2 mF Exercise 7.4-2 Determine the equivalent capacitance Ceq for the circuit shown in Figure E 7.4-2. 1 3 mF 1 3 1 mF mF Answer: 10/19 mF Ceq 2 mF 1 3 mF Figure E 7.4-2 Solution: C eq1 = 1 1 = , 1 1 1 9 + + 1/ 3 1/ 3 1/ 3 C eq2 = 1 + C eq1 = 10 1 1 1 10 = = = , C eq = mF 1 1 1 1 1 9 19 9 19 + + + + 2 2 C eq2 2 2 10 10 Exercise 7.5-1 Determine the voltage v(t) for t > 0 for the circuit of Figure E 7.5-1b when is(t) is the current shown in Figure E 7.5-1a. is(t)(V) 5 4 3 1H 1Ω + vL(t) – + vR(t) – + v(t) – 2 1 is(t) 1 2 3 4 5 6 7 8 9 t (s) (a) (b) Figure E 7.5-1b Hint: Determine vL(t) and vR(t) separately, then use KVL. ⎧2t − 2 2 < t < 4 ⎪ Answer: v(t ) = ⎨7 − t 4<t <8 ⎪0 otherwise ⎩ Solution: ⎧ 2 2<t <4 d ⎪ v L ( t ) = 1 i s ( t ) = ⎨ −1 4 < t < 8 dt ⎪ 0 otherwise ⎩ so and ⎧2 t − 4 2 < t < 4 ⎪ v R (t ) = 1 is (t ) = ⎨ 8 − t 4<t <8 ⎪ 0 otherwise ⎩ ⎧2 t − 2 2 < t < 4 ⎪ v(t ) = v L ( t ) + v R ( t ) = ⎨ 7 − t 4<t <8 ⎪ 0 otherwise ⎩ Exercise 7.7-1 Find the equivalent inductance of the circuit of Figure E 7.7-1. Answer: Leq = 14 mH 3 mH 42 mH 5 mH 4 mH Figure E 7.7-1 Solution: 3 mH Exercise 7.7-2 Find the equivalent inductance of the circuit of Figure E 7.7-2. 2 mH 20 mH 4 mH Figure E 7.7-2 Ex. 7.7-2 12 mH Section 7-2: Capacitors P 7.2-1 A 15-μF capacitor has a voltage of 5 V across it at t = 0. If a constant current of 25 mA flows through the capacitor, how long will it take for the capacitor to charge up to 150 μC? Answer: t = 3 ms v (t ) = v ( 0) + Solution: In our case, the current is constant so 1 t i (τ ) dτ C ∫0 and q = Cv t ∫ i (τ ) dτ . 0 ∴ Cv ( t ) = Cv ( 0 ) + i t −6 −6 q −Cv( 0 ) 150×10 −(15×10 )( 5 ) ∴ t= = = 3 ms i 25×10−3 P 7.2-2 The voltage, v(t), across a capacitor and current, i(t), in that capacitor adhere to the passive convention. Determine the current, i(t), when the capacitance is C = 0.125 F and the voltage is v(t) = 12 cos(2t + 30°) V. d d Hint: A cos(ωt + θ ) = − A sin (ωt + θ ) ⋅ (ωt + θ ) dt dt = − Aω sin (ωt + θ ) ⎛ π ⎞⎞ ⎛ = Aω cos ⎜ ωt + ⎜ θ + ⎟ ⎟ 2 ⎠⎠ ⎝ ⎝ Answer: i(t) = 3 cos (2t + 120°) A Solution: i (t ) = C d 1d 1 v (t ) = 12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A dt 8 dt 8 P 7.2-3 The voltage, v(t), across a capacitor and current, i(t), in that capacitor adhere to the passive convention. Determine the capacitance when the voltage is v(t) = 12 cos(500t–45°) V and the current is i(t) = 3 cos (500t + 45°) mA. Answer: C = 0.5 μF Solution: ( 3×10 ) cos ( 500t + 45 ) = C dtd −3 so ° C= 12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° ) = C ( 6000 ) cos ( 500t + 45° ) 3×10−3 1 1 = ×10−6 = μ F 3 6×10 2 2 P 7.2-4 Determine v(t) for the circuit shown in Figure P 7.2-4a when the is(t) is as shown in Figure P 7.2-4b and v0(0–) = –1 mV. is (μA) 4 + is 2 pF 0 v – –2 1 2 3 (a) 4 5 6 t (ns) (b) Figure P 7.2-4 Solution: v (t ) = 0 < t < 2 ×10−9 2 ×10−9 < t < 3 × 10−9 1 t 1 i (τ ) dτ + v ( 0 ) = ∫ C 0 2 × 10−12 is ( t ) = 0 ⇒ v ( t ) = 1 2 × 10−12 t t ∫ i (τ ) dτ − 10 ∫ 0 dτ − 10 0 −3 0 −3 = −10−3 is ( t ) = 4 ×10−6 A t 1 4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t −12 ∫2ns ( 2 × 10 In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 × 106 ) ( 3 × 10−9 ) = 10−3 ⇒ v (t ) = 3 × 10−9 < t < 5 ×10−9 is ( t ) = −2 × 10−6 A t 1 −2 ×10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t −12 ∫3ns ( 2 × 10 In particular, v ( 5 ×10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V ⇒ v (t ) = 5 ×10−9 < t is ( t ) = 0 ⇒ v ( t ) = 1 2 ×10−12 ∫ t 5ns 0 dτ − 10−3 = −10−3 V P 7.2-5 The voltage, v(t), and current, i(t), of a 1-F capacitor adhere to the passive convention. Also, v(0) = 0 V and i(0) = 0 A. (a) Determine v(t) when i(t) = x(t), where x(t) is shown in Figure P 7.2-5 and i(t) has units of A. (b) Determine i(t) when v(t) = x(t), where x(t) is shown in Figure P 7.2-5 and v(t) has units of V. Hint: x(t) = 4t – 4 when 1 < t < 2, and x(t) = –4t + 12 when 2 < t < 3. x 5 4 3 2 1 0 0 1 2 3 4 t (s) Figure P 7.2-5 Solution: (b) (a) ⎧ 0 0 < t <1 ⎪ 4 1< t < 2 d ⎪ i (t ) = C v(t ) = ⎨ dt ⎪ −4 2 < t < 3 3<t ⎩⎪ 0 t 1 t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ 0 C 0 For 0 < t < 1, i(t) = 0 A so t v ( t ) = ∫ 0 dτ + 0 = 0 V 0 For 1 < t < 2, i(t) = (4t − 4) A so t t v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 V 1 1 ( ) v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so t t v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) V 2 1 v(3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 V t For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V 0 P 7.2-6 The voltage, v(t), and current, i(t), of a 0.5-F capacitor adhere to the passive convention. Also, v(0) = 0 V and i(0) = 0 A. (a) Determine v(t) when i(t) = x(t), where x(t) is shown in Figure P 7.2-6 and i(t) has units of A. (b) Determine i(t) when v(t) = x(t), where x(t) is shown in Figure P 7.2-6 and v(t) has units of V. Hint: x(t) = 0.2t – 0.4 when 2 < t < 6. x 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 t (s) Figure P 7.2-6 Solution: (a) (b) ⎧ 0 0<t <2 d ⎪ i (t ) = C v(t ) = ⎨0.1 2 < t < 6 dt ⎪ 6<t ⎩0 t 1 t v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ 0 C 0 t For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V 0 For 2 < t < 6, i(t) = 0.2 t − 0.4 V so t t v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 V 1 2 v(6) = 0.2 ( 62 ) − 0.8 ( 6 ) + 0.8 = 3.2 V . For 6 < t, i(t) = 0.8 A so t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V 6 P 7.2-7 The voltage across a 40-μF capacitor is 25 V at t0 = 0. If the current through the capacitor as a function of time is given by i(t) = 6e–6t mA for t < 0, find v(t) for t > 0. Answer: v(t) = 50 – 25e–6t V Solution: v (t ) = v ( 0) + 1 t 4 t i τ d τ = + × 25 2.5 10 6×10−3 ) e −6τ dτ ( ) ( ∫ ∫ 0 0 C t = 25 + 150∫ 0 e −6τ dτ t ⎡ 1 ⎤ = 2 5 + 150 ⎢ − e −6τ ⎥ = 50 − 25e−6t V 6 ⎣ ⎦0 P 7.2-8 Find i for the circuit of Figure P 7.2-8 if v = 5(1 – 2e–2t) V. + v i – 10 μ F 200 kΩ Figure P 7.2-8 Solution: v 1 = (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) μ A 3 200×10 40 dv = 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t μ A iC = C dt i = iR + i C = 200 e−2t + 25 − 50 e −2t iR = ( ) = 25 + 150e −2t μA is (t) P 7.2-9 Determine v(t) for t ≥ 0 for the circuit of Figure P 7.2-9a when is(t) is the current shown in Figure P 7.2-9b and v(0) = 1 V. 0.5 F + v(t) – (a) is (A) 2 4 1 3 2 –2 (b) Figure P 7.2-9 Solution: t v ( t ) = 2 ∫ i ( t ) dt + 1 0 =1 0≤t ≤2 for t = 2 ∫ 2 dt + 1 = 4 ( t − 2 ) + 1 = 4t − 7 2 3 t 2 3 3 4 2 3 for 2≤t ≤3 = 2 ∫ 2 dt + 2∫ −2 dt + 1 = 4 − 4 ( t − 3) + 1 = −4t + 17 = 2 ∫ 2 dt + 2∫ −2 dt + 1 = 1 for t≥4 In summary 0≤t ≤2 ⎧ 1 ⎪ 4t − 7 2 ≤ t ≤ 3 ⎪ v (t ) = ⎨ ⎪−4t + 17 3 ≤ t ≤ 4 4≤t ⎩⎪ 1 for 3≤ t ≤ 4 t (s) + 2Ω is (t) v(t) 0.1 F – P 7.2-10 Determine v(t) for t ≥ 0 for the circuit of Figure P 7.2-10a when v(0) = –4 V and is(t) is the current shown in Figure P 7.2-10b. (a) is (A) 2 t (s) 0.25 0.5 (b) Figure P 7.2-10 Solution: t 1 t 4 10 i t dt = − + ( ) s ∫ 0 i s ( t ) dt C ∫0 v (t ) = v ( 0) + For 0 ≤ t ≤ 0.25 ( i s ( t ) = 8t for 0 ≤ t ≤ 0.25 ) t ⎛τ 2 ⎞ v ( t ) = −4 + 10 ∫ 8τ dτ = −4 + 80 ⎜ ⎟ = −4 + 40t 2 0 ⎝ 2 ⎠0 t ⎛1⎞ For example v ( 0 ) = −4, v ⎜ ⎟ = −3.375, ⎝8⎠ For 0.25 ≤ t ≤ 0.5 v ( t ) = −1.5 + 10∫ t 0.25 ⎛1⎞ v ⎜ ⎟ = −1.5 ⎝4⎠ 2 dτ = −1.5 + 20 ( t − 0.25 ) = 20t − 6.5 For example v ( 0.25 ) = −1.5, v ( 0.5 ) = 3.5 For 0.5 ≤ t In summary t v ( t ) = 3.5 + 10∫ 0 dτ = 3.5 0.5 ⎧−4 + 40t 2 ⎪ v ( t ) = ⎨ 20t − 6.5 ⎪ 3.5 ⎩ 0 ≤ t ≤ 0.25 s 0.25 ≤ t ≤ 0.5 s t ≥ 0.5 s i(t) P 7.2-11 Determine i(t) for t ≥ 0 for the circuit of Figure P 7.2-11a when vs(t) is the voltage shown in Figure P 7.2-11b. vs(t) + 5Ω 5F – (a) vs (V) 20 t (s) 0 0.5 1.0 1.5 (b) Figure P 7.2-11 Solution: Representing vs(t) using equations of the straight line segments gives 0 ⎧ ⎪ 40t − 20 ⎪ vs (t ) = ⎨ ⎪−20t + 40 ⎪⎩ 0 t ≤ 0.5 0.5 ≤ t ≤ 1.0 1.0 ≤ t ≤ 2.0 2.0 ≤ t Use KCL to get 0 ⎧ ⎪ ⎪ 20 + 40t − 20 vs (t ) ⎪ 1 d 5 i (t ) = vs (t ) + =⎨ 2 dt 5 ⎪ −10 + 40 − 20t ⎪ 5 ⎪ 0 ⎩ ⎧ 0 ⎪ 8t − 16 ⎪ i (t ) = ⎨ ⎪−4t − 2 ⎪⎩ 0 t ≤ 0.5 0.5 ≤ t ≤ 1.0 1.0 ≤ t ≤ 2.0 t ≥ 2.0 t ≤ 0.5 0.5 ≤ t ≤ 1.0 1.0 ≤ t ≤ 2.0 t ≥ 2.0 2.0 The capacitor voltage in the circuit shown in Figure P 7.2-12 is given by v(t ) = 12 − 10e −2t V for t ≥ 0 Determine i(t) for t > 0. P 7.2-12 4Ω + v(t) i(t) 2A 1 F 20 6Ω – Figure P 7.2-12 Solution: 1 d v (t ) 20 dt 1 = ( +20e −2t ) 20 = e −2t A for t > 0 Apply KCL to get iC (t ) = i ( t ) = 2 − i C ( t ) = 2 − e −2t A for t > 0 (checked: LNAP 6/25/04) P 7.2-13 The capacitor voltage in the circuit shown in Figure P 7.2-13 is given by v(t ) = 2.4 + 5.6e −5t V for t ≥ 0 Determine i(t) for t > 0. 20 Ω + v(t) 400 Ω 2 mF 100 Ω + – i(t) 12 V – Figure P 7.2-13 Solution: We’ll write and solve a node equation. Label the node voltages as shown. Apply KCL at node a to get v (t ) − v a 20 = + v a − 12 ⇒ 100 400 v a = 2.4 + 4.48e −5t V So i (t ) = Then va va 100 20v ( t ) + 12 25 for t > 0 va = = 24 + 44.8e −5t mA for t > 0 (checked: LNAP 6/25/04) P 7.2-14 The capacitor voltage in the circuit shown in Figure P 7.2-14 is given by v(t ) = 10 − 8e −5t V for t ≥ 0 Determine i(t) for t > 0. i(t) 20 mF + v(t) – 60 Ω 12 Ω + – 12 V Figure P 7.2-14 Solution: Apply KVL to the outside loop to get 12 i ( t ) + 12 − v ( t ) = 0 ⇒ i ( t ) = 12 − 10 − 8e −5t 1 2 = − − e −5t A for t > 0 12 6 3 (checked: LNAP 6/25/04) P 7.2-15 is(t)(A) 4 Determine the voltage v(t) for t > 0 for the circuit of Figure P 7.2-15b when is(t) is 2 the current shown in Figure P 7.2-15a. The capacitor voltage at time t = 0 is v(0) = –12 V. –4 –2 2 4 6 8 –2 (a) + is(t) 1 3F v(t) – (b) Figure P 7.2-15 Solution: v(t ) = 1 t 1 t ( ) i τ d τ + v t = i (τ ) dτ − 12 ( ) s 0 1 ∫0 s C ∫t 0 3 t v(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 0 t v(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 4 t v(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10 In particular, v(4) = 36 V. In particular, v(10) = 0 V. 10 t (s) P7.2-16 The input to the circuit shown in Figure P7.2-16 is the current i ( t ) = 3.75 e −1.2 t A for t > 0 The output is the capacitor voltage v ( t ) = 4 − 1.25 e −1.2 t V for t > 0 Find the value of the capacitance, C. Figure P7.2-16 Solution: The capacitor voltage is related to the capacitor current by v (t ) = 1 t i (τ ) dτ + v ( 0 ) C ∫0 That is: 4 − 1.25 e −1.2 t = 1 t 3.75 −1.2τ e 3.75 e −1.2τ dτ + v ( 0 ) = ∫ C 0 C ( −1.2 ) t 0 + v ( 0) = −3.125 −1.2 t ( e − 1) + v ( 0 ) C Equating the coefficients of e−1.2 t gives 12.5 = 3.125 3.125 ⇒ C= = 0.25 = 250 mF C 12.5 P7.2-17 The input to the circuit shown in Figure P7.2-17 is the current i ( t ) = 3 e −25 t A for t > 0 The initial capacitor voltage is v C ( 0 ) = −2 V . Determine the current source voltage, v ( t ) , for t > 0 . Figure P7.2-17 Solution: Apply KVL to the mesh to get ⎡ 1 t ⎤ v ( t ) = 4 i ( t ) + vC ( t ) = 4 i ( t ) + ⎢ i (τ ) dτ + v ( 0 ) ⎥ ∫ ⎣ 0.05 0 ⎦ That is, v ( t ) = 4 ( 3 e−25 t ) + 1 t −25τ 3 3 e dτ − 2 = 12 e −25 t + e−25 t − 1) − 2 ( ∫ 0 0.05 0.05 ( −25 ) = 12 e −25 t − 2.4 ( e −25 t − 1) − 2 = 9.6 e−25 t + 0.4 V for t > 0 P7.2-18 The input to the circuit shown in Figure P7.2-18 is the current i ( t ) = 3 e −25 t A for t > 0 The output is the voltage v ( t ) = 9.6 e −25 t + 0.4 V for t > 0 The initial capacitor voltage is v C ( 0 ) = −2 V . Determine the values of the capacitance, C, and resistance, R. Figure P7.2-18 Solution: Apply KVL to the mesh to get ⎡1 t ⎤ v ( t ) = R i ( t ) + vC ( t ) = R i ( t ) + ⎢ ∫ i (τ ) dτ + v ( 0 ) ⎥ 0 ⎣C ⎦ That is ⎡1 t ⎤ 9.6 e −25 t + 0.4 = R ( 3 e −25 t ) + ⎢ ∫ 3 e−25τ dτ − 2 ⎥ 0 ⎣C ⎦ = 3 R e −25 t + ⎛ 3 1 ⎞ −25 t 3 −2 e−25 t − 1) − 2 = 3 ⎜ R − ( ⎟e + 25 C ⎠ 25 C C ( −25 ) ⎝ Equating coefficients gives 0.4 = and 3 − 2 ⇒ C = 0.05 = 50 mF 25 C ⎛ ⎞ ⎛ 1 ⎞ 1 9.6 = 3 ⎜ R − ⎟⎟ = 3 ( R − 0.8 ) R = 4 Ω ⎟ = 3 ⎜⎜ R − 25 C ⎠ 25 0.05 ( ) ⎝ ⎝ ⎠ P7.2-19 The input to the circuit shown in Figure P7.2-19 is the voltage v ( t ) = 8 + 5 e −10 t V for t > 0 Determine the current, i ( t ) for t > 0 . Figure P7.2-19 Solution: Apply KCL at either node to get v (t ) 8 + 5 e −10 t d d + 0.05 v ( t ) = + 0.05 ( 8 + 5 e −10 t ) i (t ) = 4 4 dt dt −10 t = 2 + 1.25 e + 0.05 ( 5 )( −10 ) e −10 t = 2 − 1.25 e −10 t A for t > 0 P7.2-20 The input to the circuit shown in Figure P7.2-20 is the voltage: v ( t ) = 3 + 4 e −2 t A for t > 0 The output is the current: i ( t ) = 0.3 − 1.6 e −2 t V for t > 0 Determine the values of the resistance and capacitance Answer: R = 10 Ω and C = 0.25 F. Figure P7.2-20 Solution: Apply KCL at either node to get 3 + 4 e −2 t d +C 3 + 4 e −2 t R dt −2 t 3+ 4e 3 ⎛4 ⎞ = + ( −2 ) 4 C e −2 t = + ⎜ − 8 C ⎟ e −2 t R R ⎝R ⎠ ( 0.3 − 1.6 e −2 t = ) Equating coefficients: 0.3 = 3 R ⇒ R = 10 Ω and −1.6 = 4 − 8 C ⇒ C = 0.25 F 10 P7.2-21 Consider the capacitor shown in Figure P7.2-21. The current and voltage are given by ⎧0.5 0 < t < 0.5 ⎧2 t + 8.6 0 ≤ t ≤ 0.5 ⎪ ⎪ i ( t ) = ⎨ 2 0.5 < t < 1.5 and v ( t ) = ⎨ a t + b 0.5 ≤ t ≤ 1.5 ⎪0 ⎪ c t > 1.5 t ≥ 1.5 ⎩ ⎩ where a, b and c are real constants. (The current is given in Amps, the voltage in Volts and the time in seconds.) Determine the values of a, b and c. Answer: a = 8 V/s, b = 5.6 V and c = 17.6 V Solution: At t = 0.5 s Figure P7.2-21 v ( 0.5 ) = 2 ( 0.5 ) + 8.6 = 9.6 V For 0.5 ≤ t ≤ 1.5 v (t ) = 1 t 2 dτ + 9.6 = 8τ 0.25 ∫ 0.5 At t = 1.5 s t 0.5 + 9.6 = 8 ( t − 0.5 ) + 9.6 = 8 t + 5.6 V v (1.5 ) = 8 (1.5 ) + 5.6 = 17.6 V For t ≥ 1.5 v (t ) = 1 t 0 dτ + 17.6 = 17.6 0.25 ∫1.5 Checks: At t = 1.0 s At t = 0.5 s i (t ) = 1 d 1 d 1 v (t ) = (8 t + 5.6 ) = (8) = 2 A √ 4 dt 4 dt 4 v ( 0.5 ) = 8 ( 0.5 ) + 5.6 = 9.6 V √ P7.2-22At time t =0, the voltage across the capacitor shown in Figure P7.2-22 is v(0) = −20 V. Determine the values of the capacitor voltage at times 1 ms, 3 ms and 7 ms. Figure P7.2-22 Solution: v (t ) = v (0) + 1 t "area under the curve" "area under the curve" i (τ ) dτ = v ( 0 ) + = −20 + ∫ 0 2.5 × 10−6 C C 1 20 × 10−3 1×10−3 10 v ( 0.001) = −20 + 2 = −20 + = −16 V −6 2.5 × 10 2.5 ( )( ) (When calculating the value of v(0.001), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 1 ms = 0.001 s.) 1 40 × 10−3 2 ×10−3 + 40 × 10−3 1× 10−3 40 + 40 v ( 0.003) = −20 + 2 = −20 + = 12 V −6 2.5 ×10 2.5 ( )( ) ( )( ) (When calculating the value of v(0.003), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 3 ms = 0.003 s.) 1 1 40 ×10−3 2 × 10−3 + 40 ×10−3 2 × 10−3 + 40 × 10−3 3 × 10−3 2 v ( 0.007 ) = −20 + 2 = 52 V −6 2.5 ×10 ( )( ) ( )( ) ( )( ) (When calculating the value of v(0.007), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 7 ms = 0.007 s.) Section 7-3: Energy Storage in a Capacitor P 7.3-1 The current, i, through a capacitor is shown in Figure P 7.3-1. When v(0) = 0 and C = 0.5 F, determine and plot v(t), p(t), and w(t) for 0 s < t < 6 s. i(A) 1.0 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 t (s) Figure P 7.3-1 Solution: 0 t<2 ⎧ ⎪ i ( t ) = ⎨0.2 ( t − 2 ) 2 < t < 6 ⎪ 0.8 t >6 ⎩ Given The capacitor voltage is given by v (t ) = t 1 t i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ + v ( 0 ) ∫ 0 0.5 0 t v ( t ) = 2 ∫ 0 dτ + 0 = 0 For t < 2 0 In particular, v ( 2 ) = 0. For 2 < t < 6 v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V t t 2 2 In particular, v ( 6 ) = 3.2 V. For 6 < t t v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V t 6 Now the power and energy are calculated as 0 ⎧ ⎪ 2 p ( t ) = v ( t ) i ( t ) = ⎨0.04 ( t − 2 ) ⎪ 1.28 ( t − 4 ) ⎩ t<2 2<t <6 6<t and W (t ) = ∫ t 0 0 ⎧ t<2 ⎪⎪ 4 2<t<6 p (τ ) dτ = ⎨ 0.01( t − 2 ) ⎪ 2 6<t ⎪⎩0.8 ( t − 4 ) − 0.64 These plots were produced using three MATLAB scripts: capvol.m function v = CapVol(t) if t<2 v = 0; elseif t<6 v = 0.2*t*t - .8*t +.8; else v = 1.6*t - 6.4; end capcur.m function i = CapCur(t) if t<2 i=0; elseif t<6 i=.2*t - .4; else i =.8; end c7s4p1.m t=0:1:8; for k=1:1:length(t) i(k)=CapCur(k-1); v(k)=CapVol(k-1); p(k)=i(k)*v(k); w(k)=0.5*v(k)*v(k); end plot(t,i,t,v,t,p) text(5,3.6,'v(t), V') text(6,1.2,'i(t), A') text(6.9,3.4,'p(t), W') title('Capacitor Current, Voltage and Power') xlabel('time, s') % plot(t,w) % title('Energy Stored in the Capacitor, J') % xlabel('time, s') In a pulse power circuit the voltage of a 10-μF capacitor is zero for t < 0 and P 7.3-2 v = 5(1 − e −4000t ) V t ≥ 0 Determine the capacitor current and the energy stored in the capacitor at t = 0 ms and t = 10 ms. Solution: ⎧⎪ ic ( 0 ) = 0.2 A dv = (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒ ⎨ −19 dt ⎪⎩ ic (10ms ) = 8.5×10 A 1 W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0 2 −3 v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J ic = C P 7.3-3 If vc(t) is given by the waveform shown in Figure P 7.3-3, sketch the capacitor current for –1 s ≤ t ≤ 2 s. Sketch the power and the energy for the capacitor over the same time interval when C = 1 mF. vc (V) 20 –1 0 1 2 t (s) –20 Figure P 7.3-3 Solution: dvc so read off slope of vc (t ) to get i (t ) dt p (t ) = vc (t ) i (t ) so multiply vc (t ) & i (t ) curves to get p (t ) i (t ) = C P 7.3-4 The current through a 2-μF capacitor is 50 cos(10t + π/6) μA for all time. The average voltage across the capacitor is zero. What is the maximum value of the energy stored in the capacitor? What is the first nonnegative value of t at which the maximum energy is stored? Solution: π⎞ 1 t 1 t ⎛ i dτ = vc ( 0 ) + ∫0 50 cos⎜10t + ⎟ dτ = ∫ 0 2 6⎠ C ⎝ π 5 Now since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t ) 2 6 vc ( t ) = vc ( 0 ) + ( 2×10 ) ( 2.5) = 6.25 μ J 1 = C v2 = c max 2 2 −6 ∴ Wmax π⎞ 5 π⎤ 5 ⎡ ⎛ ⎢⎣ vc ( 0 ) − 2 sin 6 ⎦⎥ + 2 s in ⎝⎜10t + 6 ⎠⎟ π⎞ 5 ⎛ = sin ⎜10t + ⎟ V 2 6⎠ ⎝ 2 First non-negative t for max energy occurs when: 10t + π 6 = π 2 ⇒t = π 30 = 0.1047 s P 7.3-5 A capacitor is used in the electronic flash unit of a camera. A small battery with a constant voltage of 6 V is used to charge a capacitor with a constant current of 10 μA. How long does it take to charge the capacitor when C = 10 μF? What is the stored energy? Solution: Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 μ C Δq 60×10−6 = = 6 sec to charge i 10×10−6 1 1 2 stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 μ J 2 2 Δt = P 7.3-6 The initial capacitor voltage of the circuit shown in Figure P 7.3-6 is vc(0–) = 3 V. Determine (a) the voltage v(t) and (b) the energy stored in the capacitor at t = 0.2 s and t = 0.8 s when ⎧3e5t A 0 < t < 1 i (t ) = ⎨ t ≥ 1s ⎩0 Answer: (a) 18e5t V, 0 ≤ t < 1 (b) w(0.2) = 6.65 J and w(0.8) = 2.68 kJ t=0 + 5Ω v i(t) + 0.2 F – vc – Figure P 7.3-6 Solution: We have v (0+ ) = v (0− ) = 3 V vc ( t ) = t 1 t i (t ) dt + vc (0) = 5 ∫ 0 3 e5t dt + 3 = 3 ( e5t −1)+ 3 = 3 e5t V, 0<t <1 ∫ 0 C a) v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1 b) ⎧⎪ W (t ) t =0.2 s = 6.65 J 2 W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒ ⎨ 2 2 ⎪⎩W ( t ) t =0.8 s = 2.68 kJ P7.3-7. (a) Determine the energy stored in the capacitor in the circuit shown in Figure P7.3-7 when the switch is closed and the circuit is at steady state. (b) Determine the energy stored in the capacitor when the switch is open and the circuit is at steady state. Figure P7.3-7 Solution: The capacitor acts like an open circuit when this circuit is at steady state. (a) When the switch is closed and the circuit is at steady state, v ( t ) = 6 V . The energy stored by 1 ( 2.2 ×10 −9 )( 62 ) = 39.6 μJ . 2 (b) When the switch is closed and the circuit is at steady state, v ( t ) = 12 V . The energy stored by the capacitor is W = the capacitor is W = 1 2.2 × 10 −9 )(122 ) = 158.4 μJ . ( 2 Section 7-4: Series and Parallel Capacitors P 7.4-1 Find the current i(t) for the circuit of Figure P 7.4-1. Answer: i(t) = –1.2 sin 100t mA i(t) 3 μF + – 6 cos 100t V 2 μF 4 μF Figure P 7.4-1 Solution: 2μ F 4 μ F = 6μ F 6μ F in series with 3μ F = i (t ) = 2 μ F 6μ F⋅3μ F = 2μ F 6μ F+3μ F d (6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA dt P 7.4-2 Find the current i(t) for the circuit of Figure P 7.4-2. Answer: i(t) = –1.5e–250t mA i(t) 5 + 3e–250t V + – 4 μF 4 μF 2 μF 4 μF Figure P 7.4-2 Solution: 4 μ F in series with 4 μ F = 4 μ F×4 μ F = 2 μF 4 μ F+4 μ F 2 μF 2 μF = 4 μF 4 μ F in series with 4 μ F = 2 μ F i (t ) =(2×10−6 ) d (5+ 3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA dt P 7.4-3 The circuit of Figure P 7.4-3 contains five identical capacitors. Find the value of the capacitance C. Answer: C = 10 μF i(t) = 25 cos 250t mA C 14 sin 250t V + – C C C C Figure P 7.4-3 Solution: C in series with C = C C C ⋅C C = C +C 2 5 C = C 2 2 C⋅ 5 C 5 5 2 C in series with C = = C 2 7 C+5 C 2 ⎛5 ⎞ d ⎛5 ⎞ (25×10−3 ) cos 250t = ⎜ C ⎟ (14sin 250t ) = ⎜ C ⎟(14)(250) cos 250t ⎝ 7 ⎠ dt ⎝7 ⎠ −3 −6 so 25×10 = 2500 C ⇒ C = 10×10 = 10 μF P7.4-4 The circuit shown in Figure P 7.4-4 contains seven capacitors, each having capacitance C. The source voltage is given by v(t ) = 4 cos(3t ) V Find the current i(t) when C = 1 F. i(t) C + – v(t) C C C C C C Figure P 7.4-4 Solution: Replacing series and parallel capacitors by equivalent capacitors, the circuit can be reduced as follows: Then i (t ) = 8C d 8C d 8 ×1 32 v (t ) = 4 cos ( 3 t ) = ⎡⎣ −12sin ( 3 t ) ⎤⎦ = − sin ( 3 t ) V 21 d t 21 d t 21 7 (Checked using LNAP 6/25/04) P7.4-5 Determine the value of the capacitance C in the circuit shown in Figure P 7.4-5, given that Ceq = 8 F. Answer: C = 20 F 16 F A 12 F C 4F 12 F 10 F 30 F B Ceq Figure P 7.4-5 Solution: The 16 F capacitor is in series with a parallel combination of 4 F and 12 F capacitors. The capacitance of the equivalent capacitor is 16 ( 4 + 12 ) =8 F 16 + ( 4 + 12 ) The 30 F capacitor is in parallel with a short circuit, which is equivalent to a short circuit. After making these simplifications, we have Then 8 = C eq = 10 (12 + C + 8 ) ⇒ C = 20 F 10 + (12 + C + 8 ) (Checked using LNAP 6/26/04) P 7.4-6 Determine the value of the equivalent capacitance, Ceq, in the circuit shown in Figure P 7.4-6. Answer: Ceq = 10 F 15 F a 60 F 30 F 10 F 40 F b 60 F Ceq Figure P 7.4-6. Solution: C eq = 1 1 1 1 1 + + + 60 15 + 10 30 40 + 60 = 10 F (Checked using LNAP 6/26/04) P 7.4-7 The circuit shown in Figure P 7.4-7 consists of nine capacitors having equal capacitance, C. Determine the value of the capacitance C, given that Ceq = 50 mF. Answer: C = 90 mF C C C C Solution: First C Ceq C C C C Then 50 = C eq = 1 1 2 2 + + C 5C 5C (Checked using LNAP 6/26/04) ⇒ C = 90 mF Figure P 7.4-7 P 7.4-8 The circuit shown in Figure P 7.4-8 is at steady state before the switch opens at time t = 0. t=0 + – 20 Ω 5Ω 18 V + v(t) 60 mF 20 mF – Figure P 7.4-8 The voltage v(t) is given by ⎧ 3.6 V for t ≤ 0 v(t ) = ⎨ −2.5t for t ≥ 0 ⎩3.6e (a) (b) (c) (d) Determine the energy stored by each capacitor before the switch opens. Determine the energy stored by each capacitor 1 s after the switch opens. The parallel capacitors can be replaced by an equivalent capacitor. Determine the energy stored by the equivalent capacitor before the switch opens. Determine the energy stored by the equivalent capacitor 1 s after the switch opens. Solution: (a) The energy stored in the 60 mF capacitor is w1 = 1 ( 0.060 ) 3.62 = 0.3888 W and the 2 1 ( 0.020 ) 3.62 = 0.1296 J . 2 (b) One second after the switch opens, the voltage across the capacitors is 3.6e −2.5 = 0.2955 V . Then w1 = 2.620 mJ and w 2 = 0.873 mJ. energy stored in the 20 mF capacitor is w 2 = Next C eq = 0.06 + 0.02 = 80 mF. 1 ( 0.08) 3.62 = 0.5184 J = w1 + w 2 2 1 2 (d) w eq = ( 0.08 )( 0.2955 ) = 3.493 mJ = w1 + w 2 2 (c) w eq = P 7.4-9 The circuit shown in Figure P 7.4-9 is at steady state before the switch closes. The capacitor voltages are both zero before the switch closes (v1(0) = v2(0) = 0). t=0 5Ω + 10 mF + – 12 V v1 (t) – 25 Ω i(t) + 40 mF v2 (t) – Figure P 7.4-9 The current i(t) is given by for t < 0 ⎧ 0A i (t ) = ⎨ −30 t ⎩2.4e A for t > 0 (a) (b) Determine the capacitor voltages, v1(t) and v2(t), for t ≥ 0. Determine the energy stored by each capacitor 20 ms after the switch closes. The series capacitors can be replaced by an equivalent capacitor. (c) (d) Determine the voltage across the equivalent capacitor, + on top, for t ≥ 0. Determine the energy stored by the equivalent capacitor 20 ms after the switch closes. Solution: (a) v1 ( t ) = 1 t 240 −30t 2.4e −30τ dτ + 0 = ( e − 1) = +8 (1 − e−30τ ) V for t ≥ 0 ∫ 0.01 0 −30 1 t 2.4e −30τ dτ + 0 = 2 (1 − e−30t ) V for t ≥ 0 v 2 (t ) = ∫ 0 0.04 (b) When t = 20 ms, v1 ( 0.02 ) = 8 (1 − e −0.6 ) = 3.610 V and v 2 ( 0.02 ) = 2 (1 − e −0.6 ) = 0.902 V so the energy stored by the 10 mF capacitor is w1 = by the 40 mF capacitor is w 2 = Next C eq = 10 × 40 = 8 mF 10 + 40 1 ( 0.01) 3.6102 = 65.2 mJ and the energy stored 2 1 ( 0.04 ) 0.9022 = 16.3 mJ . 2 1 2.4e −30τ dτ = 10 (1 − e−30t ) V for t ≥ 0 ∫ 0.08 (d) When t = 20 ms, v ( 0.02 ) = 10 (1 − e −0.6 ) = 4.512 V so the energy stored by the equivalent (c) capacitor is w = v (t ) = 1 ( 0.008) 4.5122 = 81.4 mJ = w1 + w 2 . 2 P 7.4-10 Find the relationship for the division of current between two parallel capacitors as shown in Figure P 7.4-10. Answer: in = iCn/(C1 + C2), n = 1, 2 i i1 i2 C1 C2 Figure P 7.4-10 Solution: v1 = v2 ⇒ dv1 dv2 i i C = ⇒ 1 = 2 ⇒ i1 = 1 i2 dt dt C1 C2 C2 ⎛C ⎞ KCL: i = i1 + i2 = ⎜ 1 + 1⎟ i2 ⎝ C2 ⎠ ⇒ i2 = C2 i C1 +C2 Section 7-5: Inductors P 7.5-1 Nikola Tesla (1857–1943) was an American electrical engineer who experimented with electric induction. Tesla built a large coil with a very large inductance. The coil was connected to a source current is = 100sin 400t A so that the inductor current iL = is. Find the voltage across the inductor and explain the discharge in the air shown in the figure. Assume that L = 200 H and the average discharge distance is 2 m. Note that the dielectric strength of air is 3 × 106 V/m. Solution: di = 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V dt 8 ×1 0 6 V = 4 ×1 0 6 V ∴ v m ax = 8 ×1 0 6 V th u s h a v e a fie ld o f m m 2 6 w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m ∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d . F in d m a x . v o ltag e a c ro ss c o il: v (t ) = L P 7.5-2 The model of an electric motor consists of a series combination of a resistor and inductor. A current i(t) = 4te–t A flows through the series combination of a 10-Ω resistor and 0.1H inductor. Find the voltage across the combination. Answer: v(t) = 0.4e–t + 39.6te–t V Solution: v=L di + R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V dt P 7.5-3 The voltage, v(t), and current, i(t), of a 1-H inductor adhere to the passive convention. Also, v(0) = 0 V and i(0) = 0 A. (a) Determine v(t) when i(t) = x(t), where x(t) is shown in Figure P 7.5-3 and i(t) has units of A. (b) Determine i(t) when v(t) = x(t), where x(t) is shown in Figure P 7.5-3 and v(t) has units of V. Hint: x(t) = 4t – 4 when 1 < t < 2, and x(t) = –4t + 12 when 2 < t < 3. x 5 4 3 2 1 0 0 1 2 3 4 t(s) Figure P 7.5-3 Solution: (a) ⎧ 0 0 < t <1 ⎪ 4 1< t < 2 d ⎪ v(t ) = L i (t ) = ⎨ dt ⎪ −4 2 < t < 3 3<t ⎩⎪ 0 (b) i (t ) = t 1 t v (τ ) dτ + i ( 0 ) = ∫ v (τ ) dτ ∫ 0 L 0 For 0 < t < 1, v(t) = 0 V so t i ( t ) = ∫ 0 dτ + 0 = 0 A 0 For 1 < t < 2, v(t) = (4t − 4) V so t i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ =2 t 2 − 4 t + 2 A 0 1 ( t ( ) ) i (2) = 4 22 − 4 ( 2 ) + 2 = 2 A For 2 < t < 3, v(t) = −4t + 12 V so ( t i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ 2 ( ) i (3) = −2 32 + 12 ( 3) − 14 = 4 A t For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A 3 t ) 2 +2= ( −2 t 2 ) + 12 t − 14 A P 7.5-4 The voltage, v(t), across an inductor and current, i(t), in that inductor adhere to the passive convention. Determine the voltage, v(t), when the inductance is L = 250 mH and the current is i(t) = 120 sin (500t – 30°) mA. ⎛ π ⎞⎞ d d ⎛ Hint: A sin(ωt + θ ) = A cos(ωt + θ ) ⋅ (ωt + θ ) = Aω cos(ωt + θ ) = Aω sin ⎜ ωt + ⎜ θ + ⎟ ⎟ dt dt 2 ⎠⎠ ⎝ ⎝ Answer: v(t) = 15 sin(500t + 60°) V Solution: v (t ) = (250 × 10 −3 ) d (120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° ) dt = 15 cos(500t − 30° ) P 7.5-5 Determine iL(t) for t > 0 when iL(0) = –2 μA for the circuit of Figure P 7.5-5a when vs(t) is as shown in Figure P 7.5-5b. vs (mV) 4 iL vs + – 5 mH –1 1 (a) 2 3 t (μs) (b) Figure P 7.5-5 Solution: iL (t ) = for 1 5 ×10−3 0< t < 1 μ s iL (t ) = 1 5×10−3 t ∫ v (τ ) dτ 0 s − 2 ×10−6 vs (t ) = 4 mV ⎛ 4×10−3 ⎞ −3 −6 d τ t − 2×10−6 = 0.8− 2×10−6 A × − × = 4 10 2 10 ⎜ −3 ⎟ ∫0 ⎝ 5×10 ⎠ t ⎛ 4×10−3 ⎞ 6 iL (1μs) = ⎜ 1×10−6 ) ⎟ − 2×10−6 = − ×10−6 A =−1.2 A −3 ( 5 ⎝ 5×10 ⎠ for 1μ s <t < 3 μ s vs (t ) = −1 mV t 1 6 1×10−3 6 −3 −6 d τ − × − × =− 1 10 10 (t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A ( ) −3 ∫1μ s −3 5×10 5 5×10 5 −3 ⎛ 1×10 ⎞ iL ( 3μ s ) = ⎜ − + 3×10−6 ⎟ − 1×10−6 = −1.6 μA −3 ⎝ 5×10 ⎠ iL (t ) = for 3μ s < t vs (t ) = 0 so iL (t ) remains − 1.6 μA P 7.5-6 Determine v(t) for t > 0 for the circuit of Figure P 7.5-6a when iL(0) = 0 and is is as shown in Figure P 7.5-6b. is (mA) 1 + 2 kΩ is 4 mH v iL – 0 –1 t (ms) 0 (a) 5 3 1 7 8 (b) Figure P 7.5-6 Solution: v(t ) = In general ( 2 ×10 ) i 3 s (t ) + ( 4 ×10 ) dtd −3 is (t ) ⎛ 1×10−3 ⎞ d For 0<t <1 μ s is (t ) = (1) ⎜ t = 103 t ⇒ is (t ) = 1× 103 . Consequently −6 ⎟ 1 10 dt × ⎝ ⎠ v(t ) = (2 ×103 ) (1×103 ) t + 4 × 10−3 (1×103 ) = For 1μ s <t < 3μ s is (t ) = 1 mA ⇒ ( 2×10 6 t + 4) V d is (t ) = 0 . Consequently dt v(t ) = (2 ×103 ) (1×10−3 ) + ( 4×10−3 ) × 0 = 2 V ⎛ 1×10−3 ⎞ For 3μ s< t < 5μ s is (t ) = 4 ×10 − ⎜ t ⇒ −6 ⎟ ⎝ 1×10 ⎠ −3 d 1×10−3 = −103 . Consequently is (t ) = − −6 dt 1×10 v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t When 5μ s <t < 7μ s is (t ) = − 1×10−3 and d is (t ) = 0 . Consequently dt v(t ) = ( 2×103 )(10−3 ) = − 2 V ⎛ 1×10−3 ⎞ d t − 8 × 10−3 ⇒ is (t ) = 1× 103 When 7μ s< t < 8μ s is (t ) = ⎜ −6 ⎟ 1 10 × dt ⎝ ⎠ v(t ) = ( 2 × 103 )(103 t − 8 ×10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t When 8μ s < t , then is (t ) = 0 ⇒ d is (t ) = 0 . Consequently v(t ) = 0 . dt The voltage, v(t), and current, i(t), of a 0.5-H P 7.5-7 x inductor adhere to the passive convention. Also, 1.0 v(0) = 0 V and i(0) = 0 A. 0.8 (a) Determine v(t) when i(t) = x(t), where x(t) is shown 0.6 0.4 in Figure P 7.5-7 and i(t) has units of A. 0.2 (b) Determine i(t) when v(t) = x(t), where x(t) is shown 0.0 in Figure P 7.5-7 and v(t) has units of V. 0 Hint: x(t) = 0.2t – 0.4 when 2 < t < 6. 2 4 6 Figure P 7.5-7 Solution: (a) (b) ⎧ 0 0<t <2 d ⎪ v(t ) = L i (t ) = ⎨0.1 2 < t < 6 dt ⎪0 6<t ⎩ t 1 t i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ 0 L 0 t For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A 0 For 2 < t < 6, v(t) = 0.2 t − 0.4 V so t t i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A 2 2 ( ) i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A . For 6 < t, v(t) = 0.8 V so t i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A 6 8 t (s) P 7.5-8 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-8a when i(0) = 25 mA and vs(t) is the voltage shown in Figure P 7.5-8b. vs (V) 2 1 i(t) vs (t) + – 100 H 2 4 6 8 9 t(s) −2 −4 (b) (a) Figure P 7.5-8 Solution: i (t ) = so i (1) = 0.025 i (t ) = so i ( 3) = −0.055 i (t ) = so i ( 9 ) = 0.065 1 t 0 dt + 0.025 = 0.025 100 ∫ 0 −4 ( t − 1) 1 t − 4 d + 0.025 = τ 100 ∫1 100 0 < t <1 for for 2 ( t − 3) 1 t 2 dτ − 0.055 = − 0.055 ∫ 100 3 100 i (t ) = 1 t 0 dτ + 0.065 = 0.065 100 ∫9 for for 1< t < 3 3<t < 9 t >9 P 7.5-9 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-9a when i(0) = –2 A and vs(t) is the voltage shown in Figure P 7.5-9b. vs (V) i(t) + vs (t) – 5H 4 2 −1 1 (a) 2 3 t(s) (b) Figure P 7.5-9 Solution: i (t ) = i ( 0) + For 0 ≤ t ≤ 1 s 1 t 1 t 2 v τ d τ = − + v s (τ ) dτ ( ) s 5 ∫0 L ∫0 1 t 4 t 4 4 dτ = −2 + τ ∫ = −2 + t ∫ 5 0 5 0 5 6 8 ⎛1⎞ For example i ( 0 ) = −2, i (1) = − and i ⎜ ⎟ = − 5 5 ⎝2⎠ i ( t ) = −2 + For 1 ≤ t ≤ 3 s 1 1 1 t 4 1 4 dt + ∫ −1 dτ = −2 + − τ ∫ 5 0 5 1 5 5 6 7 8 For example i (1) = − , i ( 2 ) = − , i ( 3) = − 5 5 5 i ( t ) = −2 + Notice that i ( t ) = i (1) + For 3 ≤ t i (t ) = i ( 0) + t 0 6 1 t = − − ( t − 1) = −1 − 5 5 5 1 t 6 1 t −1 dτ = − − ( t − 1) = −1 − ∫ 5 1 5 5 5 1 1 1 3 1 t 8 4 dτ + ∫ −1 dτ + ∫ 0 dτ = − ∫ 5 0 5 1 5 3 5 In summary ⎧ −2 + 0.8t ⎪ i ( t ) = ⎨ −1 − 0.2t ⎪ ⎩ −1.6 0 ≤ t ≤1 1≤ t ≤ 3 3≤t P 7.5-10 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-10a when i(0) = 1 A and vs(t) is the voltage shown in Figure P 7.5-10b. vs (V) i(t) vs(t) + – 2 2H 2 4 6 t(s) −1 (a) (b) Figure P 7.5-10 Solution: 1 ⎧ ⎪ t ⎪ ∫ 2 dτ + 1 = ( t − 2 ) + 1 = t − 1 1 t ⎪ i ( t ) = ∫ v ( t ) dt + 1 = ⎨ 0 1 t 1 2 ⎪ dτ + 3 = − t + 5 ∫ 4 2 ⎪ 2 ⎪⎩ 2 t≤2 2≤t ≤4 4≤t ≤6 6≤t P 7.5-11 Determine i(t) for t ≥ 0 for the circuit of Figure P 7.5-11a when i(0) = 25 mA and vs(t) is the voltage shown in Figure P 7.5-11b. vs (V) 1 i(t) vs(t) + 200 H – 1 2 3 5 6 7 8 9 t(s) −1 −2 (a) (b) Figure P 7.5-11 Solution: i (t ) = i (t ) = 1 t −t −dτ + 0.025 = + 0.025 ∫ 0 200 200 for −2 ( t − 1) 1 t − 2 d + 0.02 = + 0.02 τ 200 ∫1 200 i (t ) = 1 t t−4 dτ − 0.01 = − 0.01 ∫ 4 200 200 i ( t ) = 0.015 = 15 mA for for t <9 0 < t <1 1< t < 4 4<t<9 The inductor current in the circuit shown in Figure P 7.5-12 is given by P 7.5-12 i (t ) = 6 + 4e −8t A for t ≥ 0 Determine v(t) for t > 0. + v(t) – i(t) 2Ω 12 V + – 8Ω 0.2 H Figure P 7.5-12 Solution: d i (t ) dt = −6.4e −8t V for t > 0 v L ( t ) = 0.2 Use KVL to get v ( t ) = 12 − ( −6.4e −8t ) = 12 + 6.4e −8t V for t > 0 (checked: LNAP 6/25/04) P 7.5-13 The inductor current in the circuit shown in Figure P 7.5-13 is given by i (t ) = 5 − 3e−4t A for t ≥ 0 Determine v(t) for t > 0. + v(t) i(t) – 24 Ω 10 A 24 Ω 24 Ω 4H Figure P 7.5-13 Solution: We’ll write and solve a mesh equation. Label the meshes as shown. Apply KVL to the center mesh to get i ( t ) + 10 24i a + 24 ( i a − i ( t ) ) + 24 ( i a − 10 ) = 0 ⇒ ia = = 5 − e − 4t A for t > 0 3 v ( t ) = 24i a = 120 − 24e −4t V for t > 0 Then (checked: LNAP 6/25/04) P 7.5-14 The inductor current in the circuit shown in Figure P 7.5-14 is given by i (t ) = 3 + 2e −3t A for t ≥ 0 Determine v(t) for t > 0. 6Ω 5A + v(t) i(t) 9Ω 5H – Figure P 7.5-14 Solution: Apply KVL to get d d v ( t ) = 6i ( t ) + 5 i ( t ) = 6 ( 3 + 2e−3t ) + 5 ( 3 + 2e−3t ) = 18 (1 − e−3t ) V for t > 0 dt dt (checked: LNAP 6/25/04) vs(t) (V) 4 P 7.5-15 Determine the current i(t) for t > 0 2 for the circuit of Figure P 7.5-15b when vs(t) is the voltage shown in Figure P 7.5-15a. The –4 inductor current at time t = 0 is i(0) = –12 A. –2 2 4 6 8 –2 (a) i(t) + – vs(t) 1 3 H (b) Figure P 7.5-15 Solution: i (t ) = 1 t 1 t v s (τ ) dτ + i (t 0 ) = ∫ v s (τ ) dτ − 12 ∫ 1 0 L t0 3 t i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4 0 t i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10 4 t i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t 10 In particular, i(4) = 36 A. In particular, i(10) = 0 A. 10 t (s) 2.5 H i(t) The initial current in the inductor is i ( 0 ) = 2 A . Determine + P7.5-16 The input to the circuit shown in Figure P7.5-16 is the voltage v ( t ) = 15 e −4 t V for t > 0 the inductor current, i ( t ) , for t > 0 . v (t) – Figure P7.5-16 Solution: The inductor current is related to the inductor voltage by i (t ) = 1 t v (τ ) dτ + i ( 0 ) L ∫0 That is i (t ) = 1 t 15 e −4τ 15 e −4τ dτ + 2 = ∫ 0 2.5 2.5 ( −4 ) t 0 + 2 = −1.5 ( e−4 t − 1) + 2 = 3.5 − 1.5 e −4 t A for t > 0 P7.5-17 The input to the circuit shown in Figure P7.5-17 is the voltage v ( t ) = 4 e −20 t V for t > 0 The output is the current i ( t ) = −1.2 e −20 t − 1.5 A for t > 0 The initial inductor current is i L ( 0 ) = −3.5 A . Determine the Figure P7.5-17 values of the inductance, L, and resistance, R. Solution: Apply KCL at either node to get i (t ) = v (t ) v (t ) ⎡ 1 t ⎤ + i L (t ) = + ⎢ ∫ v (τ ) dτ + i ( 0 ) ⎥ 0 R R ⎣L ⎦ That is −1.2 e −20 t 4 e −20 t 1 t −20τ 4 e −20 t 4 − 1.5 = + ∫ 4e + dτ − 3.5 = e −20 t − 1) − 3.5 ( 0 R L R L ( −20 ) ⎛ 4 1 ⎞ −20 t 1 =⎜ − − 3.5 ⎟e + 5L ⎝ R 5L ⎠ Equating coefficients gives −1.5 = 1 − 3.5 ⇒ L = 0.1 H 5L and −1.2 = 4 1 4 1 4 − = − = −2 ⇒ R =5 Ω R 5 L R 5 ( 0.1) R Figure P7.5-18 P7.5-18 . The source voltage the circuit shown in Figure P7.5-18 is v ( t ) = 8 e − 400 t V after time t = 0. The initial inductor current is i L ( 0 ) = 210 mA . Determine the source current i (t) for t > 0. Answer: i ( t ) = 360 e − 400 t − 190 mA for t > 0 Solution: Label the resistor current as shown. The resistor, inductor and voltage source are connected in parallel so the voltage across each is v ( t ) = 2.5 e − 400 t V . Notice that the labeled voltage and current of both the resistor and inductor do not adhere to the passive convention. 8 e − 400 t = −40 e −400 t mA 200 1 t −8 e − 400τ dτ A The inductor current is i L ( t ) = 0.21 + 0.05 ∫0 t −8 i L ( t ) = 0.21 + e − 400τ dτ A ∫ 0.05 ( −400 ) 0 The resistor current is i R ( t ) = − ( ) = 0.21 + 0.4 e − 400 t − 1 A = 400 e − 400 t − 190 mA Using KCL i ( t ) = 360 e − 400 t − 190 mA for t > 0 Figure P7.5-19 P7.5-19 . The input to the circuit shown in Figure P7.5-19 is the current i ( t ) = 5 + 2 e −7 t A for t > 0 v ( t ) = 75 − 82 e −7 t V for t > 0 The output is the voltage: Determine the values of the resistance and inductance. Solution: d 5 + 2 e −7 t ) ( dt −7 t = R ( 5 + 2 e ) + L ( ( −7 ) 2 e −7 t ) = 5 R + ( 2 R − 14 L ) e−7 t v ( t ) = 75 − 82 e −7 t = R ( 5 + 2 e−7 t ) + L 75 = 5 R ⇒ R = 15 Ω and 82 + 30 −82 = 2 R − 14 L = 30 − 14 L ⇒ L = =8 H 14 Equating coefficients gives and P7.5-20 Consider the inductor shown in Figure P7.5-20. The current and voltage are given by ⎧5 t − 4.6 0 ≤ t ≤ 0.2 ⎧12.5 0 < t < 0.2 ⎪ ⎪ i ( t ) = ⎨ a t + b 0.2 ≤ t ≤ 0.5 and v ( t ) = ⎨ 25 0.2 < t < 0.5 ⎪ c ⎪ 0 t ≥ 0.5 t > 0.5 ⎩ ⎩ where a, b and c are real constants. (The current is given in Amps, the voltage in Volts and the time in seconds.) Determine the values of a, b and c. Answer: a = 10 A/s, b = −5.6 A and c = −0.6 A Solution: At t = 0.2 s Figure P7.5-20 i ( 0.2 ) = 5 ( 0.2 ) − 4.6 = −3.6 A For 0.2 ≤ t ≤ 0.5 i (t ) = 1 t 25 dτ − 3.6 = 10τ 2.5 ∫ 0.2 At t = 0.5 s t 0.2 − 3.6 = 10 ( t − 0.2 ) − 3.6 = 10 t − 5.6 A i ( 0.5 ) = 10 ( 0.5 ) − 5.6 = −0.6 A For t ≥ 0.5 i (t ) = Checks: At t = 0.2 s 1 t 0 dτ − 0.6 = −0.6 2.5 ∫ 0.5 i ( 0.2 ) = 10 ( 0.2 ) − 5.6 = −3.6 A For 0.2 ≤ t ≤ 0.5 v ( t ) = 2.5 √ d d i ( t ) = 2.5 (10 t − 5.6 ) = 2.5 (10 ) = 25 V √ dt dt −0.6 − ( −3.6 ) = i ( 0.5 ) − i ( 0.2 ) = 1 0.5 25 dτ = 10 ( 0.5 − 0.2 ) = 3 A 2.5 ∫ 0.2 √ P7.5-21 At time t =0, the current in the inductor shown in Figure P7.5-21 is i(0) = 45 mA. Determine the values of the inductor current at times 1 ms, 4 ms and 6 ms. Figure P7.5-21 Solution: i (t ) = i ( 0) + 1 t "area under the curve" "area under the curve" v (τ ) dτ = i ( 0 ) + = 0.045 + ∫ 0.250 L 0 L i ( 0.001) = 0.045 + 20 ( 0.001) = 0.125 A = 125 mA , 0.250 1 20 ( 0.002 ) + 20 ( 0.002 ) 2 i ( 0.004 ) = 0.045 + = 0.285 A = 285 mA 0.250 1 20 ( 0.002 ) + 20 ( 0.002 ) + 0 2 i ( 0.006 ) = 0.045 + = 0.285 A = 285 mA 0.250 P7.5-22 One of the three elements shown in Figure P7.5-22 is a resistor, one is a capacitor and one is an inductor. Given i ( t ) = 0.25cos ( 2 t ) A , and v a ( t ) = −10sin ( 2 t ) V , v b ( t ) = 10sin ( 2 t ) V and v c ( t ) = 10 cos ( 2 t ) V Determine the resistance of the resistor, the capacitance of the capacitor and the inductance of the inductor. (We require positive values of resistor, capacitance and inductance.) Answers: resistance = 64Ω, capacitance = 0.0125 F and inductance = 20 H Figure P7.5-22 Solution: d d i ( t ) = ( 0.25cos ( 2 t ) ) = − ( 0.25 )( 2 ) sin ( 2 t ) = −0.5sin ( 2 t ) dt dt The voltage of an inductor is proportional to the derivative of the current. The constant of d proportionality is the inductance. We see that v a ( t ) is proportional to i ( t ) and the constant of dt proportionality is positive. Consequently, element a is the inductor. Then First, v a (t ) −10sin ( 2 t ) = = 20 H d i ( t ) −0.5sin ( 2 t ) dt t t 0.25sin ( 2τ ) Next = 0.125sin ( 2τ ) ∫ −∞ i (τ ) dτ = ∫ −∞ 0.25cos ( 2τ ) dτ = 2 The voltage of a capacitor is proportional to the integral of the current. The constant of proportionality is the reciprocal of the capacitance. We see that v b ( t ) is proportional to L= t ∫ −∞ i (τ ) dτ and the constant of proportionality is positive. Consequently, element b is the capacitor. Then 1 = C v b (t ) t ∫ −∞ i (τ ) dτ = 10sin ( 2 t ) 1 = 80 ⇒ C = = 0.0125 F 0.125sin ( 2 t ) 80 Finally, the voltage of element c is proportional to the current and the constant of proportionality v c (t ) is positive. Consequently, element c is the resistor and R = = 40 Ω . i (t ) P7.5-23 One of the three elements shown in Figure P7.5-23 is a resistor, one is a capacitor and one is an inductor. Given v ( t ) = 24 cos ( 5 t ) V , and i a ( t ) = 3cos ( 5 t ) A , i b ( t ) = 12sin ( 5 t ) A and i c ( t ) = −1.8sin ( 5 t ) A Determine the resistance of the resistor, the capacitance of the capacitor and the inductance of the inductor. (We require positive values of resistor, capacitance and inductance.) Figure P7.5-23 Solution: First, the current of element a is proportional to the voltage and the constant of proportionality is v ( t ) 24 cos ( 5 t ) = =8 Ω. positive. Consequently, element a is the resistor and R = i a ( t ) 3cos ( 5 t ) d d v ( t ) = ( 24 cos ( 5 t ) ) = − ( 24 )( 5 ) sin ( 5 t ) = −120sin ( 5 t ) dt dt Next The current of a capacitor is proportional to the derivative of the voltage. The constant of d proportionality is the capacitance. We see that i c ( t ) is proportional to v ( t ) and the constant dt of proportionality is positive. Consequently, element c is the capacitor. Then ic (t ) −1.8sin ( 5 t ) = = 0.015 F = 15 mF d i ( t ) −120sin ( 5 t ) dt t t 24sin ( 5τ ) v τ d τ = ( ) ∫ −∞ ∫ −∞ 24 cos ( 5τ ) dτ = 5 = 4.8sin ( 5τ ) C= Next The voltage of an inductor is proportional to the integral of the current. The constant of proportionality is the reciprocal of the inductance. We see that i b ( t ) is proportional to t ∫ −∞ v (τ ) dτ and the constant of proportionality is positive. Consequently, element b is the inductor. Then 1 = L ib (t ) t ∫ −∞ v (τ ) dτ = 12sin ( 5 t ) 1 = 2.5 ⇒ L = = 0.4 H 4.8sin ( 5 t ) 2.5 Section 7-6: Energy Storage in an Inductor P 7.6-1 The current, i(t), in a 100-mH inductor connected in a telephone circuit changes according to t≤0 ⎧0 ⎪ i (t ) = ⎨4t 0 ≤ t ≤ 1 ⎪4 t ≥1 ⎩ where the units of time are seconds and the units of current are amperes. Determine the power, p(t), absorbed by the inductor and the energy, w(t), stored in the inductor. t≤0 ⎧ 0 ⎧ 0 ⎪ ⎪ Answer: p (t ) = ⎨1.6t 0 < t < 1 and w(t ) = ⎨0.8t 2 ⎪ 0 ⎪ 0.8 t ≥1 ⎩ ⎩ t≤0 0 < t <1 t ≥1 The units of p(t) are W and the units of w(t) are J. Solution: v( t ) =100×10 −3 ⎧0 d ⎪ i ( t ) = ⎨0.4 dt ⎪0 ⎩ t<0 0≤ t ≤1 t>1 t <0 ⎧0 ⎪ p ( t ) =v( t ) i ( t ) = ⎨1.6t 0≤t ≤1 ⎪0 t >1 ⎩ W (t ) = ∫ t 0 ⎧ 0 ⎪ p(τ ) dτ = ⎨0.8t 2 ⎪ 0.8 ⎩ t <0 0<t <1 t >1 The current, i(t), in a 5-H inductor is P 7.6-2 t≤0 ⎧ 0 i (t ) = ⎨ ⎩4sin 2t t ≥ 0 where the units of time are s and the units of current are A. Determine the power, p(t), absorbed by the inductor and the energy, w(t), stored in the inductor. Hint: 2 (cos A)(sin B) = sin (A + B) + sin (A – B) Solution: ⎡ d ⎤ p (t ) = v (t ) i (t ) = ⎢5 (4sin 2t ) ⎥ (4sin 2t ) ⎣ dt ⎦ = 5 (8cos 2t ) (4sin 2t ) = 80 [2 cos 2t sin 2t ] = 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W t t 80 W (t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = − [cos 4τ |t0 ] = 20 (1 − cos 4t ) 0 0 4 The voltage, v(t), across a 25-mH inductor used in a fusion power experiment is P 7.6-3 0 t≤0 ⎧ v(t ) = ⎨ ⎩6 cos100t t ≥ 0 where the units of time are s and the units of voltage are V. The current in this inductor is zero before the voltage changes at t = 0. Determine the power, p(t), absorbed by the inductor and the energy, w(t), stored in the inductor. Hint: 2(cos A)(sin B) = sin(A + B) + sin(A – B) Answer: p(t) = 7.2 sin 200t W and w(t) = 3.6[1 – cos 200t] mJ Solution: i(t ) = 1 25×10−3 ∫ t 0 6 cos 100τ dτ + 0 = 6 [sin 100τ | 0t ] = 2.4sin100 t (25×10−3 )(100) p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t ) = 7.2 [ 2(cos100 t )(sin100 t ) ] = 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t t t 7.2 ⎡⎣cos 200τ |t0 ⎤⎦ W (t ) = ∫0 p (τ ) dτ = 7.2∫0 sin 200τ dτ = − 200 = 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJ P 7.6-4 The current in an inductor, L = 1/4 H, is i = 4te–t A for t ≥ 0 and i = 0 for t < 0. Find the voltage, power, and energy in this inductor. Partial Answer: w = 2t 2 e −2t J Solution: v = L di ⎛1⎞ d = ⎜ ⎟ 4 t e − t ) = (1−t ) e− t V ( dt ⎝ 4 ⎠ dt P = vi = ⎡⎣ (1−t ) e − t ⎤⎦ ( 4 t e− t ) = 4 t (1−t ) e −2t W 2 1 1⎛1⎞ W = Li 2 = ⎜ ⎟ ( 4 t e − t ) = 2 t 2 e − 2t J 2 2⎝4⎠ P 7.6-5 The current through the inductor of a television tube deflection circuit is shown in Figure P 7.6-5 when L = 1/2 H. Find the voltage, power, and energy in the inductor. i (A) 2 Partial Answer: p = 2t for 0 ≤ t <1 = 2(t − 2) for 1< t < 2 = 0 for other t 0 1 2 Figure P 7.6-5 Solution: v (t ) = L di 1 di = dt 2 dt t <0 ⎧ 0 ⎧ 0 ⎪ 2t ⎪ 0<t <1 ⎪ ⎪ 1 and i ( t ) = ⎨ ⇒ v( t ) = ⎨ ⎪−2( t − 2 ) 1<t < 2 ⎪ −1 ⎪ 0 ⎪⎩ 0 t >2 ⎩ p (t ) = v (t ) i (t ) t <0 ⎧ 0 ⎪2t 0<t <1 ⎪ = ⎨ ⎪2( t − 2 ) 1<t < 2 ⎪ 0 t >2 ⎩ W ( t ) = W ( t0 ) + ∫ t0 p ( t ) dt t i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 0 0 < t < 1: W ( t ) = ∫ 2 t dt = t 1<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t t 2 0 t 1 t >2 : W (t ) = W ( 2) = 0 2 − 4t + 4 t <0 0<t <1 1<t < 2 t >2 t (s) Section 7-7: Series and Parallel Inductors P 7.7-1 Find the current i(t) for the circuit of Figure P 7.7-1. Answer: i(t) = 15 sin 100t mA i(t) 2H 6 cos 100t V + – 6H 3H Figure P 7.7-1 Solution: 6×3 = 2 H and 2 H + 2 H = 4 H 6+3 1 t 6 ⎡sin100τ | t0 ⎤⎦ = 0.015sin100 t A = 15sin100 t mA i (t ) = ∫ 0 6 cos100τ dτ = 4 4×100 ⎣ 6H 3H = P 7.7-2 Find the voltage v(t) for the circuit of Figure P 7.7-2. Answer: v(t) = –6e–250t mV 5 + 3e–250t A + v(t) – 4 mH 4 mH 8 mH 4 mH Figure P 7.7-2 Soluton: (8×10 )×(8×10 ) −3 4 mH + 4 mH = 8 mH , 8mH 8mH = −3 8×10−3 +8×10−3 = 4 mH and 4 mH + 4 mH = 8 mH d (5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V v(t ) = (8×10−3 ) dt P 7.7-3 The circuit of Figure P 7.7-3 contains four identical inductors. Find the value of the inductance L. Answer: L = 2.86 H i(t) = 14 sin 250t mA L 25 cos 250t V + – L L L Figure P 7.7-3 Solution: L⋅ L L L 5 = and L + L + = L 2 2 2 L+ L ⎛5 ⎞ d ⎛5 ⎞ 25cos 250 t = ⎜ L ⎟ 14×10−3 ) sin 250 t = ⎜ L ⎟(14×10−3 )(250) cos 250 t ( ⎝ 2 ⎠ dt ⎝2 ⎠ 25 so L = = 2.86 H 5 −3 (14×10 ) (250) 2 L L = ( ) P 7.7-4 The circuit shown in Figure P 7.7-4 contains seven inductors, each having inductance L. The source voltage is given by v(t ) = 4 cos(3t ) V Find the current i(t) when L = 4 H. i(t) L + v(t) – L L L L L L Figure P 7.7-4 ⎛ L×2 L ⎞ + L⎟× L ⎜ L+2L 21 ⎠ +2L = L Solution: The equivalent inductance is: ⎝ 8 ⎛ L×2 L ⎞ + L⎟ + L ⎜ ⎝ L+2L ⎠ t 1 8 4 Then 4 cos ( 3τ ) dτ = i (t ) = × sin ( 3t ) = 127 sin ( 3t ) mA ∫ −∞ 21 21× 4 3 L 8 (Checked using LNAP 6/26/04) P 7.7-5 Determine the value of the inductance L in the circuit shown in Figure P 7.7-5, given that Leq = 18 H. Answer: L = 20 H 25 H A L 20 H 20 H 60 H 10 H 30 H B Leq Figure P 7.7-5 Solution: The 25 H inductor is in series with a parallel combination of 20 H and 60 H inductors. The inductance of the equivalent inductor is 25 + 60 × 20 = 40 H 60 + 20 The 30 H inductor is in parallel with a short circuit, which is equivalent to a short circuit. After making these simplifications, we have Then 18 = L eq = 10 + 1 1 1 1 + + 20 L 40 ⇒ 1 1 1 1 + + = ⇒ L = 20 H 20 L 40 8 (Checked using LNAP 6/26/04) P 7.7-6 Determine the value of the equivalent inductance, Leq, for the circuit shown in Figure P 7.7-6. Answer: Leq = 120 H a 60 H 15 H 30 H 10 H 40 H b 60 H Leq Figure P 7.7-6 Solution: L eq = 60 + 15 × 10 40 × 60 + 30 + = 60 + 6 + 30 + 24 = 120 H 15 + 10 40 + 60 (Checked using LNAP 6/26/04) P 7.7-7 The circuit shown in Figure P 7.7-7 consists of 10 inductors having equal inductance, L. Determine the value of the inductance L, given that Leq = 12 mH. Answer: L = 35 mH L L L L Leq L L L L L L Figure P 7.7-7 Solution: First Then 12 = L eq ⎛2 ⎞ ⎛2 ⎞ ⎜ L ⎟ × ⎜ L + 2 L ⎟ 12 5 ⎠ ⎝5 ⎠ = L ⇒ L = 35 mH =⎝ ⎛2 ⎞ ⎛2 ⎞ 35 ⎜ L⎟ + ⎜ L + 2 L⎟ ⎝5 ⎠ ⎝5 ⎠ (Checked using LNAP 6/26/04) P 7.7-8 The circuit shown in Figure P 7.7-8 is at steady state before the switch closes. The inductor currents are both zero before the switch closes (i1(0) = i2(0) = 0). t=0 24 Ω i1(t) + – i2(t) 12 V + v(t) − 8H 12 Ω 2H Figure P 7.7-8 The voltage v(t) is given by for t < 0 ⎧ 0V v(t ) = ⎨ −5t ⎩ 4e V for t > 0 (a) Determine the inductor currents, i1(t) and i2(t), for t ≥ 0. (b) Determine the energy stored by each inductor 200 ms after the switch closes. The parallel inductors can be replaced by an equivalent inductor. (c) Determine the current in the equivalent inductor, directed downward, for t ≥ 0. (d) Determine the energy stored by the equivalent inductor 200 ms after the switch closes. Solution: (a) i1 (t ) = 1 t −5τ 1 e −5t − 1) = 0.1(1 − e −5t ) A for t ≥ 0 4e dτ + 0 = ( ∫ 0 −10 8 i2 (t ) = 1 t −5τ 4e dτ = 0.4 (1 − e−5t ) A for t ≥ 0 ∫ 0 2 (b) When t = 0.25, i1 ( 0.2 ) = 0.1(1 − e −1 ) = 63.2 mA and i 2 ( 0.2 ) = 0.4 (1 − e −1 ) = 252.8 mA so the energy stored by the 8 H inductor is w1 = the 2 H indictor is w 2 = 63.9 mJ . L eq = (c) i (t ) = 1 ( 8) 0.06322 = 16.0 mJ and the energy stored by 2 8.2 = 1.6 H 8+2 1 t −5τ 4e dτ = 0.5 (1 − e−5t ) A for t ≥ 0 1.6 ∫ 0 (d) When t = 0.2 s, i ( 0.2 ) = 0.5 (1 − e −1 ) = 316 mA so the energy stored by the equivalent inductor is w = 1 (1.6 ) 0.3162 = 79.9 mJ = w1 + w 2 . 2 P 7.7-9 The circuit shown in Figure P 7.7-9 is at steady state before the switch opens at time t = 0. The current i(t) is given by for t ≤ 0 ⎧ 0.8 A i (t ) = ⎨ −2 t ⎩0.8e A for t ≥ 0 t=0 15 Ω 0.5 H + – i(t) 12 V 5Ω 2H Figure P 7.7-9 (a) Determine the energy stored by each inductor before the switch opens. (b) Determine the energy stored by each inductor 200 ms after the switch opens. The series inductors can be replaced by an equivalent inductor. (c) Determine the energy stored by the equivalent inductor before the switch opens. (d) Determine the energy stored by the equivalent inductor 200 ms after the switch opens. Solution: (a) The energy stored by the 0.5 H inductor is w1 = 1 ( 0.5) ( 0.82 ) = 0.16 J and the energy stored 2 1 ( 2 ) ( 0.82 ) = 0.64 J . 2 (b) 200 ms after the switch opens the current in the inductors is 0.8e−0.4 = 0.536 A . Then 1 1 w1 = ( 0.5 ) ( 0.5362 ) = 71.8 mJ and w 2 = ( 2 ) ( 0.5352 ) = 287.3 mJ. 2 2 by the 2 H inductor is w 2 = Next, L eq = 2 + 0.5 = 2.5 H . 1 ( 2.5) ( 0.82 ) = 0.8 J = w1 + w 2 2 1 (d) w eq = ( 2.5 ) ( 0.5362 ) = 359.12 mJ = w1 + w 2 2 (c) w eq = P 7.7-10 Determine the current ratio i1/i for the circuit shown in Figure P 7.7-10. Assume that the initial currents are zero at t0. i L1 Answer: 1 = i L1 + L2 L1 i1 i i2 L2 Figure P 7.7-10 Solution: i1 = 1 t ∫ v dt + i1 ( t0 ) , L1 t 0 i2 = 1 t ∫ v dt + i 2 ( t0 ) L 2 t0 but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0 ⎛1 1 t 1 ⎞ t 1 t t ∫ t v dt + ∫ t v dt = ⎜ + ∫ v dt ⎟ ∫ t v dt = L1 0 LP t 0 0 ⎝ L1 L2 ⎠ 0 1 t 1 ∫ t v dt i L2 L 0 L1 ∴1 = 1 = = 1 t 1 1 i L1 + L2 + ∫ t v dt LP 0 L1 L2 i = i1 + i2 = P7.7-11 Consider the combination of circuit elements shown in Figure P7.7-11. (a) Suppose element A is a 20 µF capacitor, element B is a 5 µF capacitor and element C is a 20 µF capacitor. Determine the equivalent capacitance. (b) Suppose element A is a 50 mH inductor, element B is a 30 mH inductor and element C is a 20 mH inductor. Determine the equivalent inductance. (c) Suppose element A is a 9 kΩ resistor, element B is a 6 kΩ resistor and element C is a 10 kΩ resistor. Determine the equivalent resistance. Answers: (a) Ceq = 20 µF, (b) Leq = 16 mH, (c) Req = 6 kΩ Figure P7.7-11 Solution: (a) C eq = 20 ( 5 ) ( 50 + 30 )( 20 ) = 16 mH (c) R = ( 9 + 6 )(10 ) = 6 kΩ + 20 = 24 μF (b) L eq = eq 20 + 5 ( 50 + 30 ) + 20 ( 9 + 6 ) + 10 P7.7-12 Consider the combination of circuit elements shown in Figure P7.7-12. (a) Suppose element A is a 8 µF capacitor, element B is a 16 µF capacitor and element C is a 12 µF capacitor. Determine the equivalent capacitance. (b) Suppose element A is a 20 mH inductor, element B is a 5 mH inductor and element C is a 8 mH inductor. Determine the equivalent inductance. (c) Suppose element A is a 20 kΩ resistor, element B is a 30 kΩ resistor and element C is a 16 kΩ resistor. Determine the equivalent resistance. Answers: (a) Ceq = 8 µF, (b) Leq = 12 mH, (c) Req = 28 kΩ Figure P7.7-11 Solution: (a) C eq = ( 8 + 16 )(12 ) = 8 μF ( 8 + 16 ) + 12 (b) L eq = ( 20 )( 5) + 8 = 12 mH 20 + 5 (c) R eq = ( 20 )( 30 ) + 16 = 28 kΩ 20 + 30 Section 7-8: Initial Conditions of Switched Circuits P 7.8-1 The switch in Figure P 7.8-1 has been open for a long time before closing at time t = 0. Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor current after the switch has been closed for a long time. Find vc(∞) and iL(∞). Answer: vc(0+) = 12 V, iL(0+) = 0, vc(∞) = 4 V, and iL(∞) = 1 mA 8 kΩ t=0 iL(t) 25 mH 12 V + – 2 μF + vc(t) – 4 kΩ Figure P 7.8-1 Solution: Then Next i L ( 0+ ) = i L ( 0 − ) = 0 and v C ( 0+ ) = v C ( 0 − ) = 12 V P 7.8-2 The switch in Figure P 7.8-2 has been open for a long time before closing at time t = 0. Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor current after the switch has been closed for a long time. Find vc(∞) and iL(∞). Answer: vc(0+) = 6 V, iL(0+) = 1 mA, vc(∞) = 3 V, and iL(∞) = 1.5 mA 6 kΩ 12 V + – 2 μF iL(t) 25 mH + vc(t) – t=0 6 kΩ 3 kΩ Figure P7.8-2 Solution: Then Next i L ( 0 + ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 V P 7.8-3 The switch in Figure P 7.8-3 has been open for a long time before closing at time t = 0. Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor current after the switch has been closed for a long time. Find vc(∞) and iL(∞). Answer: vc(0+) = 0 V, iL(0+) = 0, vc(∞) = 8 V, and iL(∞) = 0.5 mA t=0 8 kΩ 12 V + – 2 μF iL(t) 25 mH + vc(t) – 16 kΩ Figure P 7.8-3 Solution: Then Next i L ( 0 + ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 V P7.8-4 . The switch in the circuit shown in Figure P7.8-4 has been closed for a long time before it opens at time t = 0. Determine the values of vR(0-) and vL(0-), the voltage across the 4 Ω resistor and the inductor immediately before the switch opens and the values of vR(0+) and vL(0+), the voltage across the 4 Ω resistor and the inductor immediately after the switch opens. Figure P7.8-4 Solution: The circuit is at steady state immediately before the switch opens. We have The inductor acts like a short circuit so v L ( 0 − ) = 0 . Noticing that the 80 Ω and 20 Ω are connected in parallel and using voltage division: 4 4 ( 24 ) = ( 24 ) = 4.8 V 4 + ( 80 || 20 ) 4 + 16 Using current division: 24 1 ⎛ 24 ⎞ ⎛ 20 ⎞ i L ( 0) = ⎜ = ⎜ ⎟ ⎟ = 0.24 A ⎝ 80 + 20 ⎠ 4 + ( 80 || 20 ) 5 ⎝ 4 + 16 ⎠ vR (0 −) = The inductor current does not change instantaneously so i L ( 0 + ) = i L ( 0 − ) i L ( 0 ) . Immediately after the switch opens we have: v R ( 0 + ) = 4 i L ( 0 ) = 4 ( 0.24 ) = 0.96 V Using KVL: v R ( 0 + ) + v L ( 0 + ) + 80 i L ( 0 ) − 24 = 0 0.96 + v L ( 0 + ) + 80 ( 0.24 ) − 24 = 0 v L ( 0 + ) = 3.84 V v ( t ) = 75 − 82 e −7 t = R ( 5 + 2 e−7 t ) + L d 5 + 2 e −7 t ) ( dt −7 t = R ( 5 + 2 e ) + L ( ( −7 ) 2 e −7 t ) = 5 R + ( 2 R − 14 L ) e−7 t 75 = 5 R ⇒ R = 15 Ω and 82 + 30 −82 = 2 R − 14 L = 30 − 14 L ⇒ L = =8 H 14 Equation coefficients gives and P7.8-5 . The switch in the circuit shown in Figure P7.8-5 has been open for a long time before it closes at time t = 0. Determine the values of iR(0-) and iC(0-), the current in one of the 20 Ω resistors and in the capacitor immediately before the switch closes and the values of iR(0+) and iC(0+), the current in that 20 Ω resistor and in the capacitor immediately after the switch closes. Figure P7.8-5 Solution: The circuit is at steady state immediately before the switch closes. We have The capacitor acts like an open circuit so i C ( 0 − ) = 0 . Noticing that two 20 Ω are connected in series and using current division: iR (0 −) = ( 20 + 20 ) 120 = 2 120 = 80 mA ( ) ( ) 20 + ( 20 + 20 ) 3 Using current division and Ohm’s law: ⎡ ⎤ 20 v C ( 0) = ⎢ (120 )⎥ ( 20 ) = 0.8 V ⎣ 20 + ( 20 + 20 ) ⎦ The capacitor does not change instantaneously so v C ( 0 + ) = v C ( 0 − ) v C ( 0 ) . Immediately after the switch closes we have: Applying KVL to the loop consisting of the closed switch, the capacitor and a 20 Ω resistor gives 0 + v C ( 0 ) − 20 i R ( 0 + ) = 0 0 + 0.8 = 20 i R ( 0 + ) i R ( 0 + ) = 40 mA Applying KCL at the node at the right side of the circuit gives: vC (0 +) + i C ( 0 + ) + i R ( 0 + ) = 0.120 20 0.8 + i C ( 0 + ) + 0.04 = 0.120 20 i C ( 0 + ) = 0.04 = 40 mA Figure P7.8-6 P7.8-6. The switch in the circuit shown in Figure P7.8-6 has been open for a long time before it closes at time t = 0. Determine the values of vL(0-), the voltage across the inductor immediately before the switch closes and vL(0+), the voltage across the inductor immediately after the switch closes. Solution: The circuit is at steady state immediately before the switch closes. We have The inductor acts like a short circuit so vL (0 −) = 0 . The inductor current is the negative of the current source current: i L ( 0 ) = −120 mA The inductor current does not change instantaneously so i L ( 0 + ) = i L ( 0 − ) i L ( 0 ) . Immediately after the switch closes we have: Applying KVL to the left mesh gives: v L ( 0 + ) + 20 i L ( 0 ) = 0 v L ( 0 + ) + 20 ( −0.12 ) = 0 v L ( 0 + ) = 2.4 Figure P7.8-7 P7.8-7. The switch in the circuit shown in Figure P7.8-7 has been closed for a long time before it opens at time t = 0. Determine the values of iC(0-), the current in the capacitor immediately before the switch opens and iC(0+), the current in the capacitor immediately after the switch opens. Solution: The circuit is at steady state immediately before the switch opens. We have The capacitor acts like an open circuit so iC (0 −) = 0 . The capacitor voltage is equal to the voltage source voltage: v C ( 0 ) = 20 V The capacitor does not change instantaneously so v C ( 0 + ) = v C ( 0 − ) v C ( 0 ) . Immediately after the switch opens we have: Applying KCL at the top node of the capacitor, we see that: v C ( 0) iC (0 +) + =0 20 v C ( 0) iC (0 +) = − = −1 A 20 i3(t) P 7.8-8 3Ω The circuit shown in Figure P 7.8-8 is at steady state when the switch opens at time v1(t) − 6H + 12 V – 6Ω t=0 i2(t) Figure P 7.8-8 Solution: The capacitor voltage and inductor current don’t change instantaneously and so are the keys to solving this problem. Label the capacitor voltage and inductor current as shown. Before t = 0, with the switch closed and the circuit at steady state, the inductor acts like a short circuit and the capacitor acts like an open circuit. i3 (0 −) = i (0 −) = + 50 mF t = 0. Determine v1(0–), v1(0+), i2(0–), i2(0+), i3(0–), i3(0+), v4(0–), and v4(0+). + 12 = 1.33 A 9 v 4 (0 −) = v (0 −) = 6 i (0 −) = 8 V v1 ( 0 − ) = 0 V and i 2 ( 0 − ) = 0 A The capacitor voltage and inductor current don’t change instantaneously so v ( 0 + ) = v ( 0 − ) = 8 V and i ( 0 + ) = i ( 0 − ) = 1.33 A v4(t) − After the switch opens the circuit looks like this: From KCL: From KVL: i 3 ( t ) = 0 A and i 2 ( t ) = −i ( t ) v1 ( t ) + 6 i ( t ) = v ( t ) From Ohm’s Law: At t = 0 + v 4 (t ) = 6 i (t ) i 3 ( 0 + ) = 0 A and i 2 ( 0 + ) = −i ( 0 + ) = −1.33 A v1 ( 0 + ) = v ( 0 + ) − 6 i ( 0 + ) = 8 − 6 (1.333) = 0 V v 4 (0 +) = 6 i (0 +) = 8 V P 7.8-9 3Ω The circuit shown in Figure P 7.8-9 is at v1(t) steady state when the switch opens at time t = 0. Determine v1(0–), v1(0+), i2(0–), and i2(0+). + 6Ω 6H − + – 12 V 50 mF t=0 Hint: Modeling the open switch as an open circuit leads us to conclude that the inductor current changes instantaneously, which would require an infinite voltage. We can use a more accurate model of the open switch, a large resistance, to avoid the infinite voltage. Solution: The capacitor voltage and inductor current don’t change instantaneously and so are the keys to solving this problem. Label the capacitor voltage and inductor current as shown. i2(t) 6Ω Figure P 7.8-9 Before t = 0, with the switch closed and the circuit at steady state, the inductor acts like a short circuit and the capacitor acts like an open circuit. i 2 (0 −) = 0 i (0 −) = 12 = 1.333 A 9 v1 ( 0 − ) = 0 V v (0 −) = 6 i (0 −) = 8 V After the switch opens we model the open switch as a large resistance, R. From KVL: ( ) 12 = 3 i ( t ) + i 2 ( t ) + v1 ( t ) + ( R + 6 ) i ( t ) and v1 ( t ) + ( R + 6 ) i ( t ) = 6 i 2 ( t ) + v ( t ) The capacitor voltage and inductor current don’t change instantaneously so v (0 +) = v (0 −) = 8 V and i ( 0 + ) = i ( 0 − ) = 1.333 A At t = 0 + ( ) 12 = 3 i ( 0 + ) + i 2 ( 0 + ) + v 1 ( 0 + ) + ( R + 6 ) i ( 0 + ) ⎫⎪ 4 ⎬ ⇒ v1 ( 0 + ) = R and i 2 ( 0 + ) = 0 3 v 1 ( 0 + ) + ( R + 6 ) i ( 0 + ) = 6 i 2 ( 0 + ) + v ( 0 + ) ⎪⎭ As expected lim v 1 ( 0 + ) = ∞ . R →∞ P 7.8-10 The circuit shown in Figure P 7.8-10 is at steady state when the switch closes at time t = 0. Determine v1(0–), v1(0+), i2(0–), and i2(0+). 30 Ω i2(t) t=0 + + – 10 Ω 24 V 15 Ω 50 mF 15 Ω v1(t) − 3.5 H Figure P 7.8-10 Solution: The capacitor voltage and inductor current don’t change instantaneously and so are the keys to solving this problem. Label the capacitor voltage and inductor current as shown. Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit and the capacitor acts like an open circuit. i 2 (0 −) = i (0 −) = 24 = −0.4 A 60 v1 ( 0 − ) = 0 V v ( 0 − ) − 15 i ( 0 − ) = v 1 ( 0 − ) ⇒ v ( 0 − ) = −6 V v 3 ( 0 − ) = 15 i ( 0 − ) = −6 V The capacitor voltage and inductor current don’t change instantaneously so v ( 0 + ) = v ( 0 − ) = −6 V and i ( 0 + ) = i ( 0 − ) = −0.4 A After the switch closes the circuit looks like this: From Ohm’s Law: i 2 (t ) = − From KVL: From KCL: v (t ) 30 v 1 ( t ) = v 3 ( t ) + 24 v1 ( t ) 10 + v 3 (t ) 15 = i (t ) At t = 0 + i 2 (0 +) = − v (0 +) = 0.2 A 30 v 1 ( 0 + ) = v 3 ( 0 + ) + 24 ⎫ ⎪ ⎬ ⇒ v 1 ( 0 + ) = 7.2 V and v 3 ( 0 + ) = −16.8 V v1 ( 0 + ) v 3 ( 0 + ) + = i ( 0 + )⎪ 10 15 ⎭ P7.8-11 The circuit shown in Figure 7.8-11 has reached steady state before the switch opens at time t = 0. Determine the values of i L ( t ) , v C ( t ) and v R ( t ) immediately before the switch opens and the value of v R ( t ) immediately after the switch opens. Answers: i L ( 0 − ) = 1.25 A, v C ( 0 − ) = 20 V, v R ( 0 − ) = −5 V and v R ( 0 + ) = −4 V Figure 7.8-11 Solution: Because • This circuit has reached steady state before the switch opens at time t = 0. • The only source is a constant voltage source. At t=0−, the capacitor acts like an open circuit and the inductor acts like a short circuit. From the circuit 25 25 i1 ( 0 − ) = = = 1.25 A , 4 + ( 20 || 80 ) 4 + 16 ⎛ 80 ⎞ iL (0 −) = ⎜ ⎟ i1 ( 0 − ) = 1 A , ⎝ 20 + 80 ⎠ v C ( 0 − ) = 20 i L ( 0 − ) = 20 V and v R ( 0 − ) = −4 i 1 ( 0 − ) = −5 V The capacitor voltage and inductor current don’t change instantaneously so v C ( 0 + ) = v C ( 0 − ) = 20 V and iL (0 +) = iL (0 −) = 1 A Apply KCL at the top node to see that i1 ( 0 + ) = i L ( 0 + ) = 1 A From Ohm’s law v R ( 0 + ) = −4 i1 ( 0 + ) = −4 V (Notice that the resistor voltage did change instantaneously.) P7.8-12. The circuit shown in Figure 7.8-12 has reached steady state before the switch closes at time t = 0. Figure 7.8-12 (a) Determine the values of i L ( t ) , v C ( t ) and v R ( t ) immediately before the switch closes. (b) Determine the value of v R ( t ) immediately after the switch closes. Solution: Because • This circuit has reached steady state before the switch closes at time t = 0. • The only source is a constant voltage source. At t=0−, the capacitor acts like an open circuit and the inductor acts like a short circuit. From the circuit And iL (0 −) = 35 = 0.5 A , v R ( 0 − ) = 40 i L ( 0 − ) = 20 V , 30 + 40 v C ( 0 − ) = v R ( 0 − ) = 20 V The capacitor voltage and inductor current don’t change instantaneously so v C ( 0 + ) = v C ( 0 − ) = 20 V and i L ( 0 + ) = i L ( 0 − ) = 0.5 A ⎛ 40 ⎞ v R ( 0 + ) = 40 ⎜ ⎟ i L ( 0 + ) = 10 V ⎝ 40 + 40 ⎠ (Notice that the resistor voltage did change instantaneously.) P7.8-13 The circuit shown in Figure 7.8-12 has reached steady state before the switch opens at time t = 0. Determine the values of i L ( t ) , v C ( t ) and v R ( t ) immediately before the switch opens and the value of v R ( t ) immediately after the switch opens. Answer: i L ( 0 − ) = 0.4__A, v C ( 0 − ) = 16 V, v R ( 0 − ) = 0 V and v R ( 0 + ) = −12 V Solution: Because • This circuit has reached steady state before the switch opens at time t = 0. • The only source is a constant voltage source. At t=0−, the capacitor acts like an open circuit and the inductor acts like a short circuit. The current in the 30 Ω resistor is zero so v R ( 0 − ) = 0 V . Next 24 = 0.4 A and 20 + 40 v C ( 0 − ) = 40 i L ( 0 − ) = 16 V i1 ( 0 − ) = The capacitor voltage and inductor current don’t change instantaneously so v C ( 0 + ) = v C ( 0 − ) = 16 V and i L ( 0 + ) = i L ( 0 − ) = 0.4 A Apply KCL at the bottom node and then Ohm’s law to get v R ( 0 + ) = −30 i L ( 0 + ) = −12 V (Notice that the resistor voltage did change instantaneously.) Figure 7.8-13 Section 7-9: Operational amplifier Circuits and Linear Differential Equations P 7.9-1 Design a circuit with one input, x(t), and one output, y(t), that are related by this differential equation: 1 d2 d 5 y (t ) + 4 y (t ) + y (t ) = x(t ) 2 2 dt 2 dt Solution: P 7.9-2 Design a circuit with one input, x(t), and one output, y(t), that are related by this differential equation: 1 d2 5 y (t ) + y (t ) = − x(t ) 2 2 dt 2 Solution: P 7.9-3 Design a circuit with one input, x(t), and one output, y(t), that are related by this differential equation: d3 d2 d 2 3 y (t ) + 16 2 y (t ) + 8 y (t ) + 10 y (t ) = −4 x(t ) dt dt dt Solution: P 7.9-4 Design a circuit with one input, x(t), and one output, y(t), that are related by this differential equation: d3 d2 d y (t ) + 16 2 y (t ) + 8 y (t ) + 10 y (t ) = 4 x(t ) 3 dt dt dt Solution: Section 7.11 How Can We Check…? P 7.11-1 A homework solution indicates that the current and voltage of a 100-H inductor are t <1 ⎧ 0.025 ⎪ t t <1 ⎧0 ⎪− + 0.065 1 < t < 3 ⎪ −4 1 < t < 3 ⎪ 25 ⎪ i (t ) = ⎨ and v(t ) = ⎨ ⎪ t − 0.115 3 < t < 9 ⎪ 2 3<t <9 ⎪⎩ 0 ⎪ 50 t >9 ⎪ 0.065 t <9 ⎩ where the units of current are A, the units of voltage are V, and the units of time are s. Verify that the inductor current does not change instantaneously. Solution: We need to check the values of the inductor current at the ends of the intervals. at t = 1 at t = 3 ? − 1 + 0.065 = 0.025 0.025 = 25 − 3 3 + 0.065 =? − 0.115 25 50 −0.055 = −0.055 at t = 9 ( Yes!) ( Yes!) 9 − 0.115 =? 0.065 50 0.065 = 0.065 ( Yes!) The given equations for the inductor current describe a current that is continuous; as must be the case since the given inductor voltage is bounded. P 7.11-2 A homework solution indicates that the current and voltage of a 100-H inductor are ⎧ t t <1 ⎪− 200 + 0.025 ⎪ ⎪⎪ − t + 0.03 1 < t < 4 i (t ) = ⎨ 100 ⎪ t − 0.03 4<t <9 ⎪ ⎪ 100 ⎪⎩ 0.015 t <9 and t <1 ⎧ −1 ⎪ −2 1 < t < 4 ⎪ v(t ) = ⎨ ⎪ 1 4<t <9 ⎪⎩ 0 t >9 where the units of current are A, the units of voltage are V, and the units of time are s. Is this homework solution correct? Justify your answer. P7.11-2 We need to check the values of the inductor current at the ends of the intervals. 1 ? − 1 + 0.03 ( Yes!) + 0.025 = at t = 1 − 200 100 4 ? 4 − 0.03 ( No!) + 0.03 = 100 100 The equation for the inductor current indicates that this current changes instantaneously at t = 4s. This equation cannot be correct. at t = 4 − Design Problems DP 7-1 Consider a single-circuit element, that is, a single resistor, capacitor, or inductor. The voltage, v(t), and current, i(t), of the circuit element adhere to the passive convention. Consider the following cases: (a) v(t) = 4 + 2e–3t V and i(t) = –3e–3t A for t > 0 (b) v(t) = –3e–3t V and i(t) = 4 + 2e–3t A for t > 0 (c) v(t) = 4 + 2e–3t V and i(t) = 2 + e–3t A for t > 0 For each case, specify the circuit element to be a capacitor, resistor, or inductor and give the value of its capacitance, resistance, or inductance. Solution: i (t ) d = 0.5 F . v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C = d dt v (t ) dt v (t ) d b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L = = 0.5 H . d dt i (t ) dt v (t ) c) v(t) is proportional to i(t) so the element is a resistor. R = = 2 Ω. i (t ) a) DP 7-2 Figure DP 7.2 shows a voltage source and unspecified circuit elements. Each circuit element is a single resistor, capacitor, or inductor. Consider the following cases: (a) i(t) = 1.131 cos (2t + 45°) A (b) i(t) = 1.131 cos (2t – 45°) A For each case, specify each circuit element to be a capacitor, resistor, or inductor and give the value of its capacitance, resistance, or inductance. Hint: cos (θ + φ) = cos θcos φ – sin θsin φ Solution: (a) 1.131cos ( 2t + 45° ) = 1.131 ⎡⎣ cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) ⎤⎦ = 0.8 cos 2 t − 0.8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that ∫ t t −∞ v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with a capacitor to get the minus sign. Then 4 cos 2 t 4 cos 2 t R= = = 5 Ω and i1 (t ) 0.8 cos 2 t −0.8 sin 2 t i2 (t ) = = 0.1 F C= d 8 sin 2 − t 4 cos 2 t dt (b) 1.131cos ( 2t − 45° ) = 1.131 ⎡⎣ cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) ⎤⎦ = 0.8 cos 2 t + 0.8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. Then noticing that t t −∞ −∞ ∫ v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t d d v ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with an inductor to get the plus sign. Then 4 cos 2 t 4 cos 2 t = = 5 Ω and i1 (t ) 0.8 cos 2 t R= ∫ L= t −∞ 4 cos 2 t dτ i2 (t ) = 2 sin 2 t = 2.5 H 0.8 sin 2 t 4 cos 2t V +– i(t) Figure DP 7.2 DP 7-3 Figure DP 7.3 shows a voltage source and unspecified circuit elements. Each circuit element is a single resistor, capacitor, or inductor. Consider the following cases: (a) v(t) = 11.31 cos (2t + 45°) V (b) v(t) = 11.31 cos (2t – 45°) V + v(t) – 4 cos 2t A Figure DP 7.3 For each case, specify each circuit element to be a capacitor, resistor, or inductor and give the value of its capacitance, resistance, or inductance. Hint: cos (θ + φ) = cos θcos φ – sin θsin φ Solution: a) 11.31cos ( 2t + 45° ) = 11.31 ⎡⎣ cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) ⎤⎦ = 8 cos 2 t − 8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that t t −∞ −∞ ∫ i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with an inductor to get the minus sign. Then v (t ) 8 cos 2 t −8 sin 2 t v2 (t ) R= 1 = = 2 Ω and L = = =1H d 4 cos 2 t 4 cos 2 t − 8 sin 2 t 4 cos 2 t dt b) 11.31cos ( 2t + 45° ) = 11.31 ⎡⎣ cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) ⎤⎦ = 8 cos 2 t + 8 sin 2 t The first term is proportional to the voltage. Associate it with the resistor. The noticing that ∫ t −∞ t i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t −∞ d d i ( t ) = 4 cos 2 t = −8 sin 2 t dt dt associate the second term with a capacitor to get the minus sign. Then 8 cos 2 t v (t ) R= 1 = = 2 Ω and C = 4 cos 2 t 4 cos 2 t ∫ t −∞ 4 cos 2 t dτ v2 (t ) = 2 sin 2 t = 0.25 F 8 sin 2 t VB DP 7-4 A high-speed flash unit for sports photography requires a flash voltage v(0+) = 3 V and dv(t ) = 24 V/s dt t =0 3Ω –+ t=0 Switch 1 Switch 2 1Ω The flash unit uses the circuit shown in Figure DP 7.4. Switch 1 has been closed a long time, and switch 2 has been open a long time at t = 0. Actually, the long time in this case is 3 s. Determine the required battery voltage, VB, when C = 1/8 F. 1 2 C Flash voltage H + v – 3Ω t=0 + – Figure DP 7.4 Solution: at t=0− iL ( 0− ) = 0 VB 4 − We require vC ( 0 ) = 3 V so VB = 12 V By voltage division: vC ( 0− ) = at t=0+ Now we will check vC ( 0+ ) = vC ( 0 − ) = 3 V and Apply KCL at node a: iL ( 0 ) + i (0 ) = + C Finally as required. dvC dt t = 0+ VB − vC ( 0+ ) 3 12 − 3 ⇒ iC ( 0+ ) = 3 A 3 iC ( 0+ ) 3 V = = = 24 C 0.125 s 0 + iC ( 0 + ) = t = 0+ iL ( 0+ ) = iL ( 0− ) = 0 First: + dvC dt VB DP 7-5 For the circuit shown in Figure DP 7.5, select a value of R so that the energy stored in the inductor is equal to the energy stored in the capacitor at steady state. 20 Ω 10 mH 10 V + – 1 μF R Figure DP 7.5 1 1 L i L2 = C v C2 where iL and vC are the steady-state inductor current and 2 2 v capacitor voltage. At steady state, i L = C . Then R Solution: We require 2 ⎛v ⎞ L ⎜ C ⎟ = C vC2 ⎝R⎠ ⇒ C= L R2 ⇒ R = L = C 10 −2 = 10 −6 10 4 = 10 2 Ω Chapter 8 Exercises Exercise 8.3-1 The circuit shown in Figure E 8.3-1 is at steady state before the switch closes at time t = 0. Determine the capacitor voltage, v(t), for t ≥ 0. Answer: v(t) = 2 + e–2.5t V for t > 0 Figure E 8.3-1 Solution: Before the switch closes: After the switch closes: Therefore Finally, 2 = 8 Ω so τ = 8 ( 0.05 ) = 0.4 s . 0.25 v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0 Rt = Exercise 8.3-2 The circuit shown in Figure E 8.3-2 is at steady state before the switch closes at time t = 0. Determine the inductor current, i(t), for t > 0. 1 1 Answer: i (t ) = + e −1.33t A for t > 0 4 12 Figure E 8.3-2 Solution: Before the switch closes: After the switch closes: Therefore Finally, 2 6 = 8 Ω so τ = = 0.75 s . 0.25 8 1 1 i (t ) = isc + (i (0) − isc ) e−t τ = + e−1.33 t A for t > 0 4 12 Rt = Exercise 8.7-1 The electrical power plant for the orbiting space station shown in Figure E 8.71a uses photovoltaic cells to store energy in batteries. The charging circuit is modeled by the circuit shown in Figure E 8.7-1b, where vs = 10 sin 20t V. If v(0–) = 0, find v(t) for t > 0. Answer: v = 4 e–10t – 4 cos 20t + 2 sin 20t V Figure E 8.7-1 Solution: Apply KVL: vs ( t ) = 10sin 20t V Natural Response: Forced Response: ⎛ d v( t ) ⎞ −10sin 20t + 10 ⎜ .01 ⎟ + v (t ) = 0 dt ⎠ ⎝ d v( t ) ⇒ + 10 v ( t ) = 100sin 20t dt vn (t ) = Ae −t τ where τ = Rt C ∴ vn (t ) = Ae−10t try v f (t ) = B1 cos 20t + B2 sin 20 t Plugging v f (t ) into the differential equation and equating like terms yields: −20 B1 sin 20t + 20 B2 cos 20t + 10 B1 cos 20t + 10 B2 sin 20t = 100 sin 20t 20 B2 + 10 B1 = 0 and − 20 B1 + 10 B2 = 100 B1 = −4 and B2 = 2 Complete Response: v(t ) = vn (t ) + v f (t ) = Ae −10t − 4 cos 20t + 2 sin 20t Now v(0− ) = v(0+ ) = 0 = A − 4 ⇒ A = 4 v(t ) = 4 e−10t − 4 cos 20t + 2sin 20 t V Section 8.3: The Response of a First Order Circuit to a Constant Input P 8.3-1 The circuit shown in Figure P 8.3-1 is at steady state before the switch closes at time t = 0. The input to the circuit is the voltage of the voltage source, 12 V. The output of this circuit is the voltage across the capacitor, v(t). Determine v(t) for t > 0. Answer: v(t) = 6 – 2e–1.33t V for t > 0 Figure P 8.3-1 Solution: Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the initial capacitor voltage, v (0). By voltage division v (0) = 6 (12 ) = 4 V 6+6+6 Next, consider the circuit after the switch closes. The closed switch is modeled as a short circuit. We need to find the Thevenin equivalent of the part of the circuit connected to the capacitor. Here’s the circuit used to calculate the open circuit voltage, Voc. Voc = 6 (12 ) = 6 V 6+6 Here is the circuit that is used to determine Rt. A short circuit has replaced the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. ( 6 )( 6 ) = 3 Ω Rt = 6+6 Then τ = R t C = 3 ( 0.25 ) = 0.75 s Finally, v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0 P 8.3-2 The circuit shown in Figure P 8.3-2 is at steady state before the switch opens at time t = 0. The input to the circuit is the voltage of the voltage source, 12 V. The output of this circuit is the current in the inductor, i(t). Determine i(t) for t > 0. Answer: i(t) = 1 + e–0.5t A for t > 0 Figure P 8.3-2 Solution: Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. An inductor in a steady-state dc circuit acts like an short circuit, so a short circuit replaces the inductor. The current in that short circuit is the initial inductor current, i(0). 12 i ( 0) = = 2 A 6 Next, consider the circuit after the switch opens. The open switch is modeled as an open circuit. We need to find the Norton equivalent of the part of the circuit connected to the inductor. Here’s the circuit used to calculate the short circuit current, Isc. 12 =1 A 6+6 Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. ( 6 + 6 )( 6 ) = 4 Ω R t = 6 || ( 6 + 6 ) = ( 6 + 6) + 6 L 8 Then τ= = =2 s Rt 4 I sc = Finally, i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e−0.5 t A for t > 0 P 8.3-3 The circuit shown in Figure P 8.3-3 is at steady state before the switch closes at time t = 0. Determine the capacitor voltage, v(t), for t > 0. Answer: v(t) = – 6 + 18e–6.67t V for t > 0 Figure P 8.3-3 Solution: Before the switch closes: After the switch closes: −6 = 3 Ω so τ = 3 ( 0.05 ) = 0.15 s . −2 Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0 Therefore R t = P 8.3-4 The circuit shown in Figure P 8.3-4 is at steady state before the switch closes at time t = 0. Determine the inductor current, i(t), for t > 0. 10 Answer: i (t ) = −2 + e −0.5t A for t > 0 3 Figure P 8.3-4 Solution: Before the switch closes: After the switch closes: −6 6 = 3 Ω so τ = = 2 s . −2 3 t − 10 Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0 3 Therefore R t = P 8.3-5 The circuit shown in Figure P 8.3-5 is at steady state before the switch opens at time t = 0. Determine the voltage, vo(t), for t > 0. Answer: vo(t) = 10 – 5e–12.5t V for t > 0 Figure P 8.3-5 Solution: Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get: Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s . Next, v C (t ) = voc + (v (0) − voc ) e − t τ = 10 − 5 e−12.5 t V for t > 0 Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0 P 8.3-6 The circuit shown in Figure P 8.3-6 is at steady state before the switch opens at time t = 0. Determine the voltage, vo(t), for t > 0. Answer: vo(t) = 5e–4000t V for t > 0 Figure P 8.3-6 5 = 0.25 mA ⇒ i o ( 0 ) = 0.25 mA . After the switch Solution: Before the switch opens, i o ( t ) = 20 × 103 opens the part of the circuit connected to the inductor can be replaced by it's Norton equivalent circuit to get: Therefore τ = 5 = 0.25 ms . 20 × 103 Next, i L (t ) = isc + (i L (0) − isc ) e Finally, vo ( t ) = 5 − t τ = 0.5 − 0.25 e−4000 t mA for t > 0 d i L ( t ) = 5 e −4000 t V dt for t > 0 P 8.3-7 Figure P 8.3-7a shows astronaut Dale Gardner using the manned maneuvering unit to dock with the spinning Westar VI satellite on November 14, 1984. Gardner used a large tool called the apogee capture device (ACD) to stabilize the satellite and capture it for recovery, as shown in Figure P 8.3-7a. The ACD can be modeled by the circuit of Figure P 8.3-7b. Find the inductor current iL for t > 0. Answer: iL(t) = 6e–20t A Figure P 8.3-7 Solution: At t = 0− (steady-state) Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit is at steady-state: for t > 0 iL ( t ) = iL ( 0 ) e− ( R L ) t = 6 e−20t A P 8.3-8 The circuit shown in Figure P 8.3-8 is at steady state before the switch opens at time t = 0. The input to the circuit is the voltage of the voltage source, Vs. This voltage source is a dc voltage source; that is, Vs is a constant. The output of this circuit is the voltage across the capacitor, vo(t). The output voltage is given by vo(t) = 2 + 8e–0.5t V for t > 0 Determine the values of the input voltage, Vs, the capacitance, C, and the resistance, R. Figure P 8.3-8 Solution: Before the switch opens, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the capacitor voltage, will have constant values. Opening the switch disturbs the circuit. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch opened. Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the capacitor voltage, vo(t). Because the circuit is at steady state, the value of the capacitor voltage will be constant. This constant is the value of the capacitor voltage just before the switch opens. In the absence of unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor voltage immediately after the switch opens is equal to the value immediately before the switch opens. This value is called the initial condition of the capacitor and has been labeled as vo(0). There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore vo ( 0 ) = Vs The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives vo ( 0 ) = 2 + 8 e0 = 10 V Consequently, Vs = 10 V Next, consider the circuit after the switch opens. Eventually (certainly as t →∞) the circuit will again be at steady state. Here is the circuit at t = ∞, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the steady-state capacitor voltage, vo(∞). There is no current in the horizontal resistor and vo(∞) is equal to the voltage across the vertical resistor. Using voltage division, 10 vo ( ∞ ) = (10 ) R + 10 The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives vo ( ∞ ) = 2 + 8 e −∞ = 2 V Consequently, 10 (10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω R + 10 −t τ Finally, the exponential part of vo(t) is known to be of the form e where τ = R t C and Rt is 2= the Thevenin resistance of the part of the circuit connected to the capacitor. Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. R t = 10 + so ( 40 )(10 ) = 18 40 + 10 Ω τ = R t C = 18 C From the equation for vo(t) t −0.5 t = − ⇒ τ =2s τ Consequently, 2 = 18 C ⇒ C = 0.111 = 111 mF P 8.3-9 The circuit shown in Figure P 8.3-9 is at steady state before the switch closes at time t = 0. The input to the circuit is the voltage of the voltage source, 24 V. The output of this circuit, the voltage across the 3-Ω resistor, is given by vo(t) = 6 – 3e–0.35t V when t > 0 Determine the value of the inductance, L, and of Figure P 8.3-9 the resistances, R1 and R2. Solution: Before the switch closes, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch closed. The inductor current is equal to the current in the 3 Ω resistor. Consequently, − 0.35 t vo (t ) 6 − 3 e − 0.35 t = = 2− e i (t ) = A when t > 0 3 3 In the absence of unbounded voltages, the current in any inductor is continuous. Consequently, the value of the inductor current immediately before t = 0 is equal to the value immediately after t = 0. Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(0). Apply KVL to the loop to get R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0 24 R1 + R 2 + 3 The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives ⇒ i ( 0) = i ( 0 ) = 2 − e0 = 1 A Consequently, 1= 24 ⇒ R1 + R 2 = 21 R1 + R 2 + 3 Next, consider the circuit after the switch closes. Here is the circuit at t = ∞, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R1. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to the loop to get 24 R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) = R2 + 3 The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives i ( ∞ ) = 2 − e −∞ = 2 A Consequently 2= 24 ⇒ R2 = 9 Ω R2 + 3 Then R1 = 12 Ω Finally, the exponential part of i(t) is known to be of the form e −t τ where τ = L and Rt is the Rt Thevenin resistance of the part of the circuit that is connected to the inductor. Here is shows the circuit that is used to determine Rt. A short circuit has replaced combination of resistor R1 and the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. R t = R 2 + 3 = 9 + 3 = 12 Ω so τ= L L = R t 12 From the equation for i(t) −0.35 t = − t τ ⇒ τ = 2.857 s Consequently, 2.857 = L ⇒ L = 34.28 H 12 P 8.3-10 A security alarm for an office building door is modeled by the circuit of Figure P 8.3-10. The switch represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t > 0 for the circuit of Figure P 8.3-10. The switch has been closed for a long time at t = 0–. Figure P 8.3-10 Solution: First, use source transformations to obtain the equivalent circuit for t < 0: for t > 0: So iL ( 0 ) = 2 A, I sc 1 L 1 s = 0, Rt = 3 + 9 = 12 Ω, τ = = 2 = 24 Rt 12 and iL ( t ) = 2e−24t t >0 Finally v ( t ) = 9 iL ( t ) = 18 e−24t t >0 P 8.3-11 The voltage v(t) in the circuit shown in Figure P 8.3-11 is given by v(t) = 8 + 4e–2t V for t > 0 Determine the values of R1, R2, and C. Figure P 8.3-11 Solution: As t → ∞ the circuit reaches steady state and the capacitor acts like an open circuit. Also, from the given equation, v ( t ) → 8 V , as labeled on the drawing to the right, then 8= After t = 0 4 24 R2 + 4 ⇒ R2 = 8 Ω v C ( t ) = 24 − v ( t ) = 16 − 4e −2t Immediately after t = 0 v C ( 0 + ) = 16 − 4 = 12 V The capacitor voltage cannot change instantaneously so v ( 0 − ) = 12 V The circuit is at steady state just before the switch closes so the capacitor acts like an open circuit. Then 12 = 8 24 R1 + 4 + 8 ⇒ R1 = 4 Ω After t = 0 the Thevenin resistance seen by the capacitor is 8 Rt = 8 & 4 = Ω 3 1 3 2= ⇒ C= F so 8 16 C 3 Figure P 8.3-12 P 8.3-12 The circuit shown in Figure P 8.3-12 is at steady state when the switch opens at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch closed and the circuit at steady state, the inductor acts like a short circuit so we have Using superposition 9 − 5 ×10−3 = −2 mA 3000 The inductor current is continuous so i ( 0 + ) = i ( 0 − ) = −2 mA . i (0 −) = After t = 0, the switch is open. Determine the Norton equivalent circuit for the part of the circuit connected to the inductor: 9 = 3 mA 3000 R t = 3000 & 6000 = 2000 Ω i sc = The time constant is given by τ = L 5 1 = = 0.0025 so = 400 . R t 2000 τ The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −0.002 − 0.003) e−400 t + 0.003 = 3 − 5e−400 t mA for t ≥ 0 (checked: LNAP 6/29/04) Figure P 8.3-13 P 8.3-13 The circuit shown in Figure P 8.3-13 is at steady state when the switch opens at time t = 0. Determine v(t) for t ≥ 0. Solution: Before t = 0, with the switch closed and the circuit at the steady state, the capacitor acts like an open circuit so we have Using superposition v (0 −) = 60 & 60 60 & 30 ⎛1⎞ ⎛1⎞ 6+ 36 = ⎜ ⎟ 6 + ⎜ ⎟ 36 = 12 V 30 + ( 60 & 60 ) 60 + ( 60 & 30 ) ⎝2⎠ ⎝4⎠ The capacitor voltage is continuous so v ( 0 + ) = v ( 0 − ) = 12 V . After t = 0 the switch is open. Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor: v oc = 60 6=4V 60 + 30 R t = 30 & 60 = 20 kΩ The time constant is τ = R t C = ( 20 × 103 )( 5 × 10−6 ) = 0.1 s so 1 τ = 10 1 . s The capacitor voltage is given by v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−10t + 4 = 4 + 8 e −10t V for t ≥ 0 (checked: LNAP 6/29/04) Figure P 8.3-14 P 8.3-14 The circuit shown in Figure P 8.3-14 is at steady state when the switch closes at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit so we have i (t ) = − ⎤ 18 ⎡ 5 2 ⎥ = 0.29 A ⎢ 4 + 18 ⎣ 5 + 20 + (18 & 4 ) ⎦ After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent circuit. First, performing a couple of source transformations reduces the circuit to Next, replace the series voltage sources by an equivalent voltage source, replace the series resistors by an equivalent resistor and do a couple of source transformations to get so The current is given by 2 1 1 = 0.25 ⇒ =5 10 s τ −5t i ( t ) = [ 0.29 − 0.4] e + 0.4 = 0.4 − 0.11e −5t A for t ≥ 0 τ= P 8.3-15 The circuit in Figure P 8.3-15 is at steady state before the switch closes. Find the inductor current after the switch closes. Hint: i(0) = 0.1 A, Isc = 0.3 A, Rt = 40 Ω Answer: i(t) = 0.3 – 0.2e–2t A t ≥ 0 Solution: At steady-state, immediately before t = 0: ⎞ 12 ⎛ 10 ⎞ ⎛ i ( 0) = ⎜ ⎟ ⎜ ⎟ = 0.1 A ⎝ 10 + 40 ⎠ ⎝ 16 + 40||10 ⎠ After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be so I sc = 0.3 A, Rt = 40 Ω, τ = Finally: L 20 1 = = s Rt 40 2 i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A P8.3-16. Consider the circuit shown in Figure P8.316. a) Determine the time constant, τ, and the steady state capacitor voltage when the switch is open. b) Determine the time constant, τ, and the steady state capacitor voltage when the switch is closed. Figure P8.3-16. Solution: Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. ⎛ 120 ⎞ a.) When the switch is open, v oc = ⎜ ⎟ 24 = 16 V and R t = 120 || 60 = 40 Ω . The steady state ⎝ 60 + 120 ⎠ capacitor voltages is voc =16 V. The time constant is τ = (40)(0.02) = 0.8 s. ⎛ 120 ⎞ b.) When the switch is closed, v oc = ⎜ ⎟ 24 = 19.2 V and R t = 120 || 60 || 60 = 24 Ω . The steady ⎝ 30 + 120 ⎠ state capacitor voltages is voc =19.2 V. The time constant is τ = (24)(0.02) = 0.48 s. LNAPTR, 11/1/07 Figure P 8.3-17 P 8.3-17 The circuit shown in Figure P 8.3-17 is at steady state before the switch closes. The response of the circuit is the voltage v(t). Find v(t) for t > 0. Hint: After the switch closes, the inductor current is i(t) = 0.2(1 – e–1.8t) A Answer: v(t) = 8 + e–1.8t V Solution: Immediately before t = 0, i (0) = 0. After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the inductor current: I sc = 0.2 A, Rt = 45 Ω, τ = So i (t ) = 0.2 (1 − e−1.8t ) A Now that we have the inductor current, we can calculate v(t): d i (t ) dt = 8(1 − e −1.8t ) + 5(1.8)e −1.8t v(t ) = 40 i (t ) + 25 = 8 + e −1.8t V for t > 0 L 25 5 = = Rth 45 9 Figure P 8.3-18 P 8.3-18 The circuit shown in Figure P 8.3-18 is at steady state before the switch closes. The response of the circuit is the voltage v(t). Find v(t) for t > 0. Answer: v(t) = 37.5 – 97.5e–6400t V Solution: At steady-state, immediately before t = 0 so i(0) = 0.5 A. After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get I sc = 93.75 mA, Rt = 640 Ω, τ = L .1 1 = = s Rt 640 6400 i (t ) = 406.25 e−6400t + 93.75 mA Finally: v (t ) = 400 i (t ) + 0.1 d i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t ) dt = 37.5 − 97.5e −6400t V Figure P 8.3-19 P 8.3-19 ≥ 0. The circuit shown in Figure P 8.3-19 is at steady state before the switch closes. Find v(t) for t Solution: Before the switch closes v(t) = 0 so v ( 0 + ) = v ( 0 − ) = 0 V . For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor, i.e. the part of the circuit to the left of the terminals a – b. Write mesh equations to find voc: Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i1 = 18 36 i1 = 18 ⇒ i1 = i2 = Using KVL, Find Rt: ⎛7⎞ ⎛1⎞ voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V ⎝ 3⎠ ⎝ 2⎠ 14 ⎛ 1 ⎞ 7 ⎜ ⎟= A 3 ⎝2⎠ 3 1 A 2 Rt = Then and 12 (10 + 2 ) =6Ω 12 + (10 + 2 ) 1 1 ⎛ 1 ⎞ 1 =4 ⎟= s ⇒ s τ ⎝ 24 ⎠ 4 −4t + v oc = ( 0 − 12 ) e + 12 = 12 1 − e−4t τ = R tC = 6 ⎜ v ( t ) = ( v ( 0 + ) − v oc ) e −t τ ( ) V for t ≥ 0 (checked: LNAP 7/15/04) P 8.3-20 The circuit shown in Figure P 8.3-20 is at steady state before the switch closes. Determine i(t) for t ≥ 0. Figure P 8.3-20 Solution: Before the switch closes the circuit is at steady state so the inductor acts like a short circuit. We have ⎞ 1⎛ 24 i ( t ) = ⎜⎜ ⎟ = 0.8 A 2 ⎝ 5 + ( 20 & 20 ) ⎟⎠ so i ( 0 + ) = i ( 0 − ) = 0.8 A After the switch closes, find the Thevenin equivalent circuit for the part of the circuit connected to the inductor. Using voltage division twice ⎛ 20 1 ⎞ v oc = ⎜ − ⎟ 24 = 7.2 V ⎝ 25 2 ⎠ R t = ( 5 & 20 ) + ( 20 & 20 ) = 14 Ω i sc = v oc Rt = 7.2 = 0.514 A 14 Then τ= and L 3.5 1 = = s R t 14 4 ⇒ 1 τ =4 1 s i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0.8 − 0.514 ) e −4t + 0.514 = 0.286e−4t + 0.514 A for t ≥ 0 (checked: LNAP 7/15/04) P8.3-21 The circuit in Figure P8.3-21 at steady state before the switch closes. Determine an equation that represents the capacitor voltage after the switch closes. Figure P8.3-21 Solution: For t < 0, the switch is open and the capacitor acts like an open circuit because the circuit is at steady state. Consequently, the current in the 10 Ω resistor is 0A and so the voltage across this resistor is 0 V. KVL gives v ( t ) = 18 V . Immediately before the switch opens we have v ( 0 − ) = 18 V . The capacitor voltage does not change instantaneously so v ( 0 + ) = v ( 0 − ) = 18 V . For t > 0, the Thevenin equivalent of the part of the circuit connected to the capacitor is characterized by 40 R t = 10 || 40 = 8 Ω and, using voltage division, voc = (18) = 14.4 V . 10 + 40 A = voc = 14.4 V , B = v ( 0 + ) − voc = 18 − 14.4 = 3.6 V and a = 1 τ = 1 1 1 = =5 R t C 8 ( 0.025 ) s LNAPTR, 11/3/07 Figure P 8.3-22 P 8.3-22 The circuit shown in Figure P 8.3-22 is at steady state when the switch closes at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit so we have i (0 +) = i (0 −) = 4 A After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent circuit. Using superposition, the short circuit current is given by ⎛ ⎞ ⎛ 3+8 ⎞ 8 i sc = ⎜⎜ ⎟⎟ 2 + ⎜⎜ ⎟⎟ 4 = 3.75 A ⎝ 8 + ( 5 + 3) ⎠ ⎝ ( 3 + 8 ) + 5 ⎠ R t = 8 + 3 + 5 = 16 Ω so τ= 2 1 1 = 0.125 s ⇒ =8 16 s τ The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 4 − 3.75 ) e−8t + 3.75 = 3.75 − 0.25 e −8t A for t ≥ 0 (checked: LNAP 7/15/04) P8.3-23 The circuit in Figure P8.3-23 at steady state before the switch closes. Determine an equation that represents the inductor current after the switch closes. Figure P8.3-23 Solution: For t < 0, the switch is open and the inductor acts like short circuit because the circuit is at steady state. Consequently, the current in inductor just before the switch closes is i ( 0 − ) = 7 Amps . The inductor current does not change instantaneously so i ( 0 + ) = i ( 0 − ) = 7 Amps . For t > 0, the Thevenin equivalent of the part of the circuit connected to the inductor is characterized by 20 R t = 20 + 60 = 80 Ω and, using current division, i sc = ( 7 ) = 1.75 Amps . 20 + 60 1 R t 80 1 = = 32 A = i sc = 1.75 Amps , B = i ( 0 + ) − i sc = 7 − 1.75 = 5.25 Amps and a = = τ L 2.5 s LNAPTR, 11/3/07 P8.3-24 Consider the circuit shown in Figure 8.3-24a and corresponding plot of the inductor current shown in Figure 8.3-24b. Determine the values of L, R1 and R 2 . Answer: L = 4.8 H, R1 = 200 Ω and R 2 = 300 Ω. Hint: Use the plot to determine values of D, E, F and a such that the inductor current can be represented as ⎧ D for t ≤ 0 i (t ) = ⎨ − at ⎩ E + F e for t ≥ 0 Figure 8.3-24 Solution: From the plot D = i(t) for t < 0 =120 mA = 0.12 A, E + F = i(0+) = 120 mA = 0.12 A and E = lim i ( t ) = 200 mA = 0.2 A . t →∞ The point labeled on the plot indicates that i(t) = 160 mA when t = 27.725 ms = 0.027725 s. Consequently 160 = 200 − 80 e − a ( 0.027725) Then ⎧120 mA for t ≤ 0 i (t ) = ⎨ −25 t mA for t ≥ 0 ⎩ 200 − 80 e ⎛ 160 − 200 ⎞ ln ⎜ ⎟ 80 ⎠ = 25 1 ⇒ a= ⎝ −0.027725 s When t < 0, the circuit is at steady state so the inductor acts like a short circuit. 24 R1 = = 200 Ω 0.12 As t → ∞, the circuit is again at steady state so the inductor acts like a short circuit. R1 || R 2 = 120 = 200 || R 2 24 = 120 Ω 0.2 ⇒ R 2 = 300 Ω Next, the inductance can be determined using the time constant: 25 = a = 1 τ = R1 || R 2 L = 120 120 ⇒ L= = 4.8 H 25 L P8.3-25 Consider the circuit shown in Figure P8.3-25a and corresponding plot of the voltage across the 40 Ω resistor shown in Figure P8.3-25b. Determine the values of L and R 2 . Answer: L = 8 H and R 2 = 10 Ω. Hint: Use the plot to determine values of D, E, F and a such that the voltage can be represented as ⎧ D for t < 0 v (t ) = ⎨ − at ⎩ E + F e for t > 0 Figure P8.3-25 Solution: From the plot D = v(t) for t < 0 =20 V, E + F = v(0+) = 100 V and E = lim v ( t ) = 20 V . The t →∞ point labeled on the plot indicates that v(t) = 60 V when t = 0.14 s. Consequently 60 = 20 + 80 e − a ( 0.14) ⎛ 60 − 20 ⎞ ln ⎜ ⎟ 1 80 ⎠ ⇒ a= ⎝ =5 −0.14 s Then ⎧20 V for t ≤ 0 v (t ) = ⎨ −5t ⎩ 20 + 80 e V for t ≥ 0 At t = 0+, i (0 +) = 100 = 2.5 A 40 When t < 0, the circuit is at steady state so the inductor acts like a short circuit. 40 || R 2 = 20 = 8 Ω ⇒ R 2 = 10 Ω 2.5 Next, the inductance can be determined using the time constant: 1 40 40 5=a= = ⇒ L= =8 H 5 τ L Figure P8.3-26 P8.3-26 Determine vo(t) for t > 0 for the circuit shown in Figure P8.3-26. Solution: Replace the series inductors with an equivalent inductor and label the current in the inductor: We will determine the inductor current, i(t) ,first and then use it to determine vo(t) . Determine the initial condition, i(o), by considering the circuit when t < 0 and the circuit is at steady state. Since an inductor in a dc circuit acts like a short circuit , we have Using current division, we have ⎛ 18 ⎞ i ( 0) = ⎜ ⎟ 2.4 = 1.2 mA ⎝ 18 + 18 ⎠ Next, consider the circuit when t > 0 and the circuit is not at steady state: To find the Norton equivalent of the part of the circuit connected to the inductor we determine both the Thevenin resistance and the short circuit current: and τ= The time constant is: L 1.5 1 = = = 0.0556 second R t 27 18 The inductor current is given by i ( t ) = ( i ( 0 ) − i sc ) e −t /τ + i sc = (1.2 − 0.8 ) e −18t − 0.8 mA for t ≥ 0 Using KCL Finally vo (t ) 9 + ⎡⎣(1.2 − 0.8 ) e −18t − 0.8⎤⎦ ×10−3 = 2.4 × 10−3 v o ( t ) = 14.4 − 3.6 e −18t mV for t > 0 P8.3-27 The circuit shown in Figure P8.3-27 is at steady state before the switch closes at time t = 0. After the switch closes, the inductor current is given by i ( t ) = 0.6 − 0.2 e −5t A for t ≥ 0 Determine the values of R1 , R 2 and L. Answers: R1 = 20 Ω , R 2 = 10 Ω and L = 5 H Figure P8.3-27 Solution: The steady state current before the switch closes is equal to i ( 0 ) = 0.6 − 0.2 e−5( 0 ) = 0.4 A . The inductor will act like a short circuit when this circuit is at steady state so 0.4 = i ( 0 ) = 12 R1 + R 2 ⇒ R1 + R 2 = 30 Ω After the switch has been open for a long time, the circuit will again be at steady state. The steady state inductor current will be i ( ∞ ) = 0.6 − 0.2 e−5( ∞ ) = 0.6 A The inductor will act like a short circuit when this circuit is at steady state so 0.6 = i ( ∞ ) = 12 R1 ⇒ R1 = 20 Ω Then R 2 = 10 Ω. After the switch is closed, the Thevenin resistance of the part of the circuit connected to the inductor is R t = R1 . Then 5= 1 τ = Rt L = R1 L = 20 ⇒ L=4H L P8.3-28 After time t = 0, a given circuit is represented by the circuit diagram shown in FigureP8.3-28. a.) Suppose that the inductor current is i ( t ) = 21.6 + 28.4 e −4 t mA for t ≥ 0 Determine the values of R1 and R 3 . Figure P8.3-28 b.) Suppose instead that R1 = 16 Ω , R 3 = 20 Ω and the initial condition is i(0) = 10 mA. Determine the inductor current for t ≥ 0. Solution: The inductor current is given by i ( t ) = i sc + ( i ( 0 ) − i sc ) e− at a. Comparing this to the given equation gives 21.6 = i sc = 4= Rt 2 ( R1 R1 + 4 ( 36 ) for t ≥ 0 where a = ⇒ R1 = 6 Ω and ) ⇒ R t = 8 Ω . Next 8 = R t = R1 + 4 || R 3 = 10 || R 3 ⇒ R 3 = 40 Ω . 10 16 = 5 s . also i sc = ( 36 ) = 28.8 mA . Then τ 2 16 + 4 = 28.8 + (10 − 28.8 ) e−5t = 28.2 − 18.8 e−5t . b. R t = (16 + 4 ) || 20 = 10 Ω so a = i ( t ) = i sc + ( i ( 0 ) − i sc ) e − at 1 = 1 τ = Rt L . P8.3-29 Consider the circuit shown in Figure P8.3-29. a.) Determine the time constant, τ, and the steady state capacitor voltage, v(∞), when the switch is open. b.) Determine the time constant, τ, and the steady state capacitor voltage, v(∞), when the switch is closed. Answers: a.) τ = 3 s and v(∞) = 24 V; b.) τ = 2.25 s and v(∞) = 2 V; Figure P8.3-29 Solution: a.) When the switch is open we have After replacing series and parallel resistors by equivalent resistors, the part of the circuit connected to the capacitor is a Thevenin equivalent circuit with R t = 33.33 Ω . The time constant is τ = R t C = 33.33 ( 0.090 ) = 3 s . Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state. Consequently, there is zero current in the 33.33 Ω resistor and KVL gives v(∞) = 24 V. b.) When the switch is closed we have This circuit can be redrawn as Now we find the Thevenin equivalent of the part of the circuit connected to the capacitor: So R t = 25 Ω and τ = R t C = 25 ( 0.090 ) = 2.25 s Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state. Consequently, there is zero current in the 25 Ω resistor and KVL gives v(∞) = 12 V. Section 8-4: Sequential Switching P 8.4-1 The circuit shown in Figure P 8.4-1 is at steady state before the switch closes at time t = 0. The switch remains closed for 1.5 s and then opens. Determine the capacitor voltage, v(t), for t > 0. Hint: Determine v(t) when the switch is closed. Evaluate v(t) at time t = 1.5 s to get v(1.5). Use v(1.5) as the initial condition to determine v(t) after the switch opens again. for 0 < t < 1.5 s ⎧ 5 + 5e −0.5t V Answer: v(t ) = ⎨ −2.5( t −1.5) V for 1.5 s < t ⎩10 − 2.64e Figure P 8.4-1 Solution: Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc = 5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore v (t ) = voc + (v (0) − voc ) e −t τ = 5 + 5e−5 t V for 0 < t < 1.5 s At t =1.5 s, v (1.5) = 5 + 5e ( τ = 8 × 0.05 = 0.4 s . Therefore −0.05 1.5) = 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so v (t ) = voc + (v (1.5) − voc ) e −( t −1.5 ) τ = 10 − 5 e −2.5 ( t −1.5 ) V for 1.5 s < t Finally ⎧⎪ 5 + 5 e−5 t V for 0 < t < 1.5 s v (t ) = ⎨ −2.5 ( t −1.5) for 1.5 s < t V ⎪⎩10 − 5 e P 8.4-2 The circuit shown in Figure P 8.4-2 is at steady state before the switch closes at time t = 0. The switch remains closed for 1.5 s and then opens. Determine the inductor current, i(t), for t > 0. for 0 < 1 < 1.5 s ⎧ 2 + e −0.5t A Answer: v(t ) = ⎨ −0..667( t −1.5) A for 1.5 s < t ⎩3 − 0.53e Figure P 8.4-2 Solution: Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get: Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 12 A and Rt = 6 Ω so τ = = 2 s . Therefore 6 i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s 12 −0.5 1.5 At t =1.5 s, i (1.5) = 2 + e ( ) = 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ = = 1.5 s . 8 Therefore −( t −1.5 ) τ −0.667 ( t −1.5 ) i (t ) = isc + (i (1.5) − isc ) e = 3 − 0.53 e V for 1.5 s < t Finally ⎧⎪ 2 + e−0.5 t A for 0 < t < 1.5 s i (t ) = ⎨ −0.667 ( t −1.5 ) for 1.5 s < t A ⎪⎩3 − 0.53 e P 8.4-3 Cardiac pacemakers are used by people to maintain regular heart rhythm when they have a damaged heart. The circuit of a pacemaker can be represented as shown in Figure P 8.4-3. The resistance of the wires, R, can be neglected since R < 1 mΩ. The heart’s load resistance, RL, is 1 kΩ. The first switch is activated at t = t0, and the second switch is activated at t1 = t0 + 10 ms. This cycle is repeated every second. Find v(t) for t0 ≤ t ≤ 1. Note that it is easiest to consider t0 = 0 for this calculation. The cycle repeats by switch 1 returning to position a and switch 2 returning to its open position. Hint: Use q = Cv to determine v(0–) for the 100-μF capacitor. Figure P 8.4-3 Solution: At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 μF capacitor is 3 V. The charge stored by the capacitor is q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C 0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at t = 0+). The voltage across the parallel capacitors is determined by considering charge conservation: q ( 0+ ) = (100 μ F) v ( 0+ ) + (400 μ F) v ( 0+ ) v (0 + ) = 100 ×10 ( ) q ( 0+ ) −6 + 400 ×10−6 v 0+ = 0.6 V 10 ms < t < l s: Combine 100 μF & 400 μF in parallel to obtain v(t ) = v ( 0+ ) e − (t −.01) RC 3 = 0.6e − (t −.01) (10 ) (5 x10 v(t ) = 0.6 e−2(t −.01) V −4 ) = q ( 0− ) 500 × 10−6 = 300 × 10−6 500 × 10−6 P 8.4-4 An electronic flash on a camera uses the circuit shown in Figure P 8.4-4. Harold E. Edgerton invented the electronic flash in 1930. A capacitor builds a steady-state voltage and then discharges it as the shutter switch is pressed. The discharge produces a very brief light discharge. Determine the elapsed time t1 to reduce the capacitor voltage to one-half of its initial voltage. Find the current, i(t), at t = t1. Solution: Figure P 8.4-4 v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s ∴ v ( t ) = 5 e −10 t V for t > 0 2.5 = 5 e −10 t1 i (t1 ) = v (t1 ) 100 ×10 3 t 1 = 0.0693 s = 2.5 = 25 μ A 100 × 103 P 8.4-5 The circuit shown in Figure P 8.4-5 is at steady state before the switch opens at t = 0. The switch remains open for 0.5 second and then closes. Determine v(t) for t ≥ 0. Figure P 8.4-5 Solution: The circuit is at steady state before the switch closes. The capacitor acts like an open circuit. The initial condition is ⎛ 40 ⎞ v (0 +) = v (0 −) = ⎜ ⎟ 24 = 12 V ⎝ 40 + 40 ⎠ After the switch closes, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. R t = 6.67 Ω and v oc = 4 V Recognize that The time constant is τ = R t C = ( 6.67 )( 0.05 ) = 0.335 s ⇒ 1 τ = 2.988 3 1 s The capacitor voltage is v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−3t + 4 = 4 + 8e−3t V for 0 ≥ t ≥ 0.5 s When the switch opens again at time t = 0.5 the capacitor voltage is −3 0.5 v ( 0.5 + ) = v ( 0.5 − ) = 4 + 8e ( ) = 5.785 V After time t = 0.5 s, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. Recognize that R t = 20 Ω and v oc = 12 V The time constant is τ = R t C = 20 ( 0.05 ) = 1 1 ⇒ τ =1 1 s The capacitor voltage is − t −0.5 τ −10 t − 0.5 v ( t ) = ( v ( 0.5 + ) − v oc ) e ( ) + v oc = ( 5.785 − 12 ) e ( ) + 12 = 12 − 6.215e −10( t − 0.5) V for t ≥ 0.5 s so ⎧ 12 V ⎪ v (t ) = ⎨ 4 + 8 e −3t V ⎪ −( t − 0.5) V ⎩12 − 6.215e for t ≥ 0 for 0 ≤ t ≤ 0.5 s for t ≥ 0.5 s Section 8.5 Stability of First-Order Circuits P 8.5-1 The circuit in Figure P 8.5-1 contains a current-controlled voltage source. What restriction must be placed on the gain, R, of this dependent source in order to guarantee stability? Answer: R < 400 Ω Figure P 8.5-1 Solution: This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as 100 ⎫ v T (400− R) 100 iT ⎪ = 100+ 400 ⎬ ⇒ Rt = iT 100+ 400 ⎪ vT = 400 i (t ) − R i (t ) ⎭ i (t ) = The circuit is stable when R < 400 Ω. P 8.5-2 The circuit in Figure P 8.5-2 contains a current-controlled current source. What restriction must be placed on the gain, B, of this dependent source in order to guarantee stability? Figure P 8.5-2 Solution: The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as Ohm’s law: i ( t ) = − vT ( t ) 6000 KCL: i ( t ) + B i ( t ) + i T ( t ) = vT ( t ) 3000 ⎛ v ( t ) ⎞ vT ( t ) ∴ i T ( t ) = − ( B + 1) ⎜ − T ⎟+ ⎝ 6000 ⎠ 3000 ( B + 3) vT ( t ) = 6000 Rt = vT ( t ) 6000 = iT ( t ) B + 3 The circuit is stable when B > −3 A/A. Section 8.6 The Unit Step Source P 8.6-1 The input to the circuit shown in Figure P 8.6-1 is the voltage of the voltage source, vs(t). The output is the voltage across the capacitor, vo(t). Determine the output of this circuit when the input is vs(t) = 8 – 15 u(t) V. Figure P 8.6-1 Solution: The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is The steady-state circuit for t < 0. vo ( 0 ) = A + B e − a ( 0) = A+ B Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives A+ B = 8 V Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The steady-state circuit for t > 0. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e − a (∞) =A Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives Therefore A = -7 V and B = 15 V The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = 6 Ω Therefore a= 1 1 = 2.5 −3 s ( 6 ) ( 66.7 ×10 ) (The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .) Putting it all together: 8 V for t ≤ 0 ⎧ vo ( t ) = ⎨ − 2.5 t V for t ≥ 0 ⎩−7 + 15 e P 8.6-2 The input to the circuit shown in Figure P 8.6-2 is the voltage of the voltage source, vs(t). The output is the voltage across the capacitor, vo(t). Determine the output of this circuit when the input is vs(t) = 3 + 3 u(t) V. Figure P 8.6-2 Solution: The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is The steady-state circuit for t < 0. vo ( 0 ) = A + B e − a ( 0) = A+ B Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives A+ B = 6 ( 3) = 2 V 3+ 6 Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is The steady-state circuit for t > 0. vo ( ∞ ) = A + B e − a (∞) =A Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives A= 6 ( 6) = 4 V 3+ 6 Therefore B = −2 V The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt = ( 3)( 6 ) = 2 3+ 6 Ω Therefore a= 1 ( 2 )(.5) =1 1 s (The time constant is τ = ( 2 )( 0.5 ) = 1 s .) Putting it all together: for t ≤ 0 ⎧ 2 V vo ( t ) = ⎨ − t ⎩4 − 2 e V for t ≥ 0 P 8.6-3 The input to the circuit shown in Figure P 8.6-3 is the voltage of the voltage source, vs(t). The output is the current across the inductor, io(t). Determine the output of this circuit when the input is vs(t) = – 7 + 13 u(t) V. Figure P 8.6-3 Solution: The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t for t > 0 where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = 0, will be equal to the steady state inductor current before the input changes. At time t = 0 the output current is The steady-state circuit for t < 0. io ( 0 ) = A + B e − a ( 0) = A+ B Consequently, the inductor current is labeled as A + B. Analysis of the circuit gives A+ B = −7 = −1.4 A 5 Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The steady-state circuit for t > 0. The value of the inductor current at time t = ∞, will be equal to the steady state inductor current after the input changes. At time t = ∞ the output current is io ( ∞ ) = A + B e − a (∞) =A Consequently, the inductor current is labeled as A. Analysis of the circuit gives A= 6 = 1.2 A 5 Therefore B = −2.6 V The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit connected to the inductor. 1 L =τ = a Rt Here is the circuit used to calculate Rt. Rt = ( 5) ( 4 ) = 2.22 5+ 4 2.22 1 a= = 1.85 Therefore 1.2 s 1.2 (The time constant is τ = = 0.54 s .) 2.22 Putting it all together: −1.4 A for t ≤ 0 ⎧ io ( t ) = ⎨ − 1.85 t A for t ≥ 0 ⎩1.2 − 2.6 e Ω P8.6-4 Determine vo(t) for t > 0 for the circuit shown in Figure P8.6-4. Figure P8.6-4 Solution: Determine the initial condition, vo(o), by considering the circuit when t < 0 and the circuit is at steady state. Since a capacitor in a dc circuit acts like a open circuit , we have Recognizing the inverting amplifier, we have ⎛ 45 ⎞ v o ( 0 ) = ⎜ − ⎟ 2.4 = −3.6 V ⎝ 30 ⎠ Next, consider the circuit when t > 0 and the circuit is not at steady state: To find the Thevenin equivalent of the part of the circuit connected to the capacitor we determine both the open circuit voltage and the short circuit current: and Now we calculate the Thevenin resistance: v oc −5.4 Rt = = = 20 kΩ i sc −0.27 × 10−3 and the time constant: τ = R t C = ( 20 ×103 )( 5 × 10−6 ) = 0.1 second The capacitor voltage is given by v o ( t ) = ( v o ( 0 ) − v oc ) e −t /τ + v oc = ( −3.6 − ( −5.4 ) ) e −10 t − 5.4 = 1.8 e −10 t − 5.4 V for t ≥ 0 P 8.6-5 The initial voltage of the capacitor of the circuit shown in Figure P 8.6-5 is zero. Determine the voltage v(t) when the source is a pulse, described by t <1 s ⎧0 ⎪ vs = ⎨4 V 1 < t < 2 s ⎪0 t>2s ⎩ Solution Figure P 8.6-5 τ = R C = ( 5 × 105 )( 2 × 10−6 ) = 1 s Assume that the circuit is at steady state at t = 1−. Then v ( t ) = 4 − 4 e − ( t −1) V for 1 ≤ t ≤ 2 so v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V and v ( t ) = 2.53 e − ( t − 2) V for t ≥ 2 Finally ⎧0 ⎪ v(t ) = ⎨4− 4e − (t −1) ⎪2.53e − (t − 2) ⎩ t ≤1 1≤t ≤ 2 t ≥2 P 8.6-6 Studies of an artificial insect are being used to understand the nervous system of animals. A model neuron in the nervous system of the artificial insect is shown in Figure P 8.6-6. A series of pulses, called synapses, is required. The switch generates a pulse by opening at t = 0 and closing at t = 0.5 s. Assume that the circuit is in steady state and that v(0–) = 10 V. Determine the voltage v(t) for 0 < t < 2 s. Figure P 8.6-6 Solution: The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0. For 0 < t < 0.5 s the switch is open: v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 × 1 1 = s 6 2 so v ( t ) = 10 e − 2 t V In particular, v ( 0.5 ) = 10 e For t > 0.5 s the switch is closed: − 2 ( 0.5) = 3.679 V v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω, 1 6 τ = 2× = so 1 s 3 v ( t ) = 10 + ( 3.679 − 10 ) e = 10 − 6.321 e − 3 ( t − 0.5 ) − 3 ( t − 0.5) V V P8.6-7 Determine the voltage v o ( t ) in the circuit shown in Figure P8.6-7. Figure P8.6-7 Solution: This is a first order circuit containing an inductor. First, determine i L ( t ) . Consider the circuit for time t < 0. Step 1: Determine the initial inductor current. The circuit will be at steady state before the source voltage changes abruptly at time t = 0 . The source voltage will be 2 V, a constant. The inductor will act like a short circuit. i L ( 0) = 2 2 = = 0.25 A 10 || ( 25 + 15 ) 8 Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time t = 0 . Determine the Norton equivalent circuit for the part of the circuit connected to the inductor. Replacing the resistors by an equivalent resistor, we recognize v oc = −6 V and R t = 8 Ω Consequently i sc = −6 = −0.75 A 8 t < 0 , at steady state: Step 3. The time constant of a first order circuit containing an inductor is given by τ= Consequently τ= L Rt L 4 1 1 = = 0.5 s and a = = 2 Rt 8 s τ Step 4. The inductor current is given by: i L ( t ) = i sc + ( i ( 0 ) − i sc ) e − at = −0.75 + ( 0.25 − ( −0.75 ) ) e −2 t = −0.75 + e −2 t for t ≥ 0 Step 5. Express the output voltage as a function of the source voltage and the inductor current. Using current division: iR = 10 i L = 0.2 i L 10 + ( 25 + 15 ) Then Ohm’s law gives v o = 15 i R = 3 i L Step 6. The output voltage is given by v o ( t ) = −2.25 + 3 e −2 t for t ≥ 0 P 8.6-8 Determine vc(t) for t > 0 for the circuit of Figure P 8.6-8. Figure P 8.6-8 Solution: For t < 0, the circuit is: After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: vc ( t ) = −15 + ( −6 − ( −15 ) ) e = −15 + 9 e − 5 t V − t / ( 4000×0.00005 ) P 8.6-9 The voltage source voltage in the circuit shown in Figure P 8.6-9 is vs(t) = 7 – 14u (t) V Determine v(t) for t > 0. Figure P 8.6-9 Solution: The input changes abruptly at time t = 0. The voltage v(t) may not be continuous at t = 0, but the capacitor voltage, vC(t) will be continuous. We will find vC(t) first and then use KVL to find v(t). The circuit will be at steady state before t = 0 so the capacitor will act like an open circuit. v (0 +) = v (0 −) = 5 7 = 4.375 V 5+3 After t = 0, we replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. 5 5 v oc = ( 7 − 14 ) = ( −7 ) = −4.375 V 8 8 R t = 5 & 3 = 1.875 Ω The time constant is τ = R t C = 0.8625 s 1 So v C ( t ) = ⎡⎣ 4.375 − ( −4.375 ) ⎤⎦ e −1.16t 1 s τ + ( −4.375 ) = −4.375 + 8.75e−1.16t V for t ≥ 0 = 1.16 Using KVL v ( t ) = v s ( t ) − vC ( t ) = −7 − ⎡⎣ −4.375 + 8.75e −1.16t ⎤⎦ = −2.625 − 8.75e−1.16t V for t > 0 P 8.6-10 Determine the voltage v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-10. Figure P 8.6-10 Solution: For t < 0 Using voltage division twice v (t ) = so 32 30 5− 5 = 0.25 V 32 + 96 120 + 30 v ( 0 − ) = 0.25 V and v ( 0 + ) = v ( 0 − ) = 0.25 V For t > 0, find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Using voltage division twice v oc = 32 30 20 − 20 = 5 − 4 = 1 V 32 + 96 120 + 30 R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω then τ = 48 × 0.0125 = 0.6 s so 1 τ Now = 1.67 1 s v ( t ) = [ 0.25 − 1] e −1.67t + 1 = 1 − 0.75e −1.67t V for t ≥ 0 (checked: LNAP 7/1/04) P 8.6-11 The voltage source voltage in the circuit shown in Figure P 8.6-11 is vs(t) = 5 + 20u (t) Determine i(t) for t ≥ 0. Figure P 8.6-11 Solution: For t > 0 the circuit is at steady state so the inductor acts like a short circuit: Apply KVL to the supermesh corresponding to the dependent source to get −5000i b + 1000 ( 3i b ) − 5 = 0 ⇒ i b = 0.2 mA Apply KVL to get i ( t ) = 3i b = 0.6 mA i ( 0 − ) = 0.6 mA so i ( 0 + ) = i ( 0 − ) = 0.6 mA and For t > 0, find the Norton equivalent circuit for the part of the circuit that is connected to the inductor. Apply KCL at the top node of the dependent source to see that i b = 0 A . Then ( ) v oc = 25 − 5000 i b = 25 V Apply KVL to the supermesh corresponding to the dependent source to get ( ) −5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA Apply KCL to get i sc = 3 i b = 3 mA Then Rt = v oc i sc = 8.33 kΩ τ= Then So Now 25 = 3 ms 8333 1 1 = 333 τ s i ( t ) = [ 0.6 − 3] e −333t + 3 = 3 − 2.4e −333t mA for t ≥ 0 (checked: LNAP 7/2/04) P 8.6-12 The voltage source voltage in the circuit shown in Figure P 8.6-12 is vs(t) = 12 – 6u (t) V Determine v(t) for t ≥ 0. Figure P 8.6-12 Solution: For t > 0, the circuit is at steady state so the capacitor acts like an open circuit. We have the following situation. Notice that v(t) is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −v(t) Apply KCL at node a: ⎛ 12 − v ( t ) ⎞ v ( t ) ⎛ 3 ⎞ −⎜ + ⎜ − v (t ) ⎟ = 0 ⎟+ 8 4 ⎝ 4 ⎠ ⎝ ⎠ −12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V So v ( 0 + ) = v ( 0 − ) = −4 V For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor, i.e. the part of the circuit to the left of terminals a – b. Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a: ⎛ 6 − voc ⎞ voc ⎛ 3 ⎞ −⎜ + ⎜ − voc ⎟ = 0 ⎟+ ⎝ 8 ⎠ 4 ⎝ 4 ⎠ −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces va = 0 Apply KCL at node a: ⎛ 6−0⎞ 0 ⎛ 3 ⎞ −⎜ ⎟ + + ⎜ − 0 ⎟ + i sc = 0 ⎝ 8 ⎠ 4 ⎝ 4 ⎠ i sc = Rt = 6 3 = A 8 4 voc −2 8 = =− Ω i sc 3 4 3 Then ⎛ 8 ⎞⎛ 3 ⎞ 1 τ = R tC = ⎜ − ⎟ ⎜ ⎟ = − s 5 ⎝ 3 ⎠ ⎝ 40 ⎠ and ⇒ 1 τ = −5 1 s v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( −4 − ( −2 ) ) e5t + ( −2 ) = −2 (1 + e5t ) V for t ≥ 0 Notice that v(t) grows exponentially as t increases. (checked: LNAP 7/8/04) P 8.6-13 Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-13. Figure P 8.6-13 Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit. The mesh equations are ( ) ( ) 2 i x + 4 i x + i (0 −) + 3i x + 6 i x + i (0 −) = 0 1i ( 0 − ) − 3 i x = 0 so i (0 +) = i (0 −) = 0 For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the inductor. First, simplify the circuit using a source transformation: Identify the open circuit voltage and short circuit current. Apply KVL to the mesh to get: (10 + 2 + 3) i x − 15 = 0 ⇒ ix = 1 A Then v oc = 3 i x = 3 V Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 and i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 = 4 i sc 3 so ⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠ The Thevenin resistance is Rt = The time constant is given by τ = 3 =3Ω 1 L 5 1 1 = = 1.67 s so = 0.6 . Rt 3 τ s The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0 − 1) e−0.6 t + 1 = 1 − e −0.6t A for t ≥ 0 P 8.6-14 Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-14. Figure P 8.6-14 Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit. The initial condition is 0 2 i (0 +) = i (0 −) = + = 0.02 A 150 100 For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the inductor. First, simplify the circuit using source transformations: i sc = 20 + 40 = 60 mA R t = 100 ||150 = The time constant is given by τ = 100 × 150 = 60 Ω 100 + 150 L 2 1 1 = = 0.0333 s so = 30 . R t 60 τ s The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 20 − 60 ) e −30 t + 60 = 60 − 40 e −30 t mA for t ≥ 0 (checked: LNAP 7/15/04) P 8.6-15 Determine v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-15. Figure P 8.6-15 Solution: When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. The 0 A current source also acts like an open circuit. The initial condition is v (0 + ) = v (0 −) = 0 V For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. 30 ⎡ 170 voc = ⎢ ( 20 ) ⎤⎥ 10 − ⎡⎢ ( 20 ) ⎤⎥ 50 ⎣170 + 30 ⎦ ⎣170 + 30 ⎦ = 170(20)(10) − 30(20)(50) 4000 = = 20 V 200 200 Rt = 8 + ( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 ) The time constant is τ = R t C = ( 50 ) (10−3 ) = 0.05 s s so 1 τ = 20 1 . s The capacitor voltage is given by ( v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 0 − 20 ) e −20t + 20 = 20 1 − e−20t ) V for t ≥ 0 P 8.6-16 Determine v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-16. Figure P 8.6-16 Solution: When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. The 0 A current source also acts like an open circuit. After a couple of source transformations, the initial condition is calculated as v (0 +) = v (0 −) = 18 16 = 10.667 V 9 + 18 For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Using source transformations, reduce the circuit as follows. Now recognize R t = 10 Ω and v oc = 4 V . The time constant is τ = R t C = (10 ) ( 20 ×10−3 ) = 0.2 s s so 1 τ =5 1 . s The capacitor voltage is given by v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = (10.667 − 4 ) e−5t + 4 = 4 + 6.667 e −5t V for t ≥ 0 (checked: LNAP 7/15/04) P 8.6-17 Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-17. Figure P 8.6-17 Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit. The 0 V voltage source also acts like a short circuit. After a replacing series and parallel resistors by equivalent resistors, the equivalent resistors, current source and short circuit are all connected in parallel. Consequently i (0 +) = i (0 −) = 2 A For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Replace series and parallel resistors by an equivalent resistor. 18 & (12 + 24 ) = 12 Ω Do a source transformation, then replace series voltage sources by an equivalent voltage source. Do two more source transformations. Now recognize R t = 8 Ω and i sc = 3 A . The time constant is given by τ= L 2 1 1 = = 0.25 s so = 4 . τ Rt 8 s The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 2 − 3) e −4t + 3 = 3 − e −4t A for t ≥ 0 (checked: LNAP 7/15/04) P 8.6-18 The voltage source voltage in the circuit shown in Figure P 8.6-18 is vs(t) = 8 + 12u(t) V Determine v(t) for t ≥ 0. Figure P 8.6-18 Solution: Simplify the circuit by replacing the parallel capacitors by an equivalent capacitor and by doing a couple of source transformations: For t < 0 the circuit is at steady state so the capacitor acts like an open circuit. The voltage source voltage is 6.4 V, so v ( 0 + ) = v ( 0 − ) = 6.4 V For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the capacitor. In this case we recognize the voc = 16 V and Rt = 50 Ω. The time constant is τ = R t C = ( 50 ) ( 8 × 10−3 ) = 0.4 s ⇒ 1 τ = 2.5 1 s Then −t v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6.4 − 16 ) e −2.5t + 16 = 16 − 9.6e −2.5t V for t ≥ 0 (checked: LNAP 7/12/04) P8.6-19 Determine the current i o ( t ) in the circuit shown in Figure P8.6-19. Figure P8.6-19 Solution: This is a first order circuit containing a capacitor. First, determine v C ( t ) . Consider the circuit for time t < 0. Step 1: Determine the initial capacitor voltage. t < 0 , at steady state: The circuit will be at steady state before the source voltage changes abruptly at time t = 0 . The source voltage will be 5 V, a constant. The capacitor will act like an open circuit. Apply KVL to the mesh to get: (10 + 2 + 3) i x − 5 = 0 Then ⇒ ix = 1 A 3 v C (0) = 3 i x = 1 V Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time t = 0 . Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. First, determine the open circuit voltage, v oc : Apply KVL to the mesh to get: (10 + 2 + 3) i x − 15 = 0 ⇒ ix = 1 A Then v oc = 3 i x = 3 V Next, determine the short circuit current, i sc : Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 = And 4 i sc 3 ⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠ so The Thevenin resistance is Rt = 3 =3Ω 1 Step 3. The time constant of a first order circuit containing an capacitor is given by τ = Rt C Consequently 1 1 ⎛1⎞ ⎟ = 0.25 s and a = = 4 τ s ⎝ 12 ⎠ τ = Rt C = 3⎜ Step 4. The capacitor voltage is given by: ( ) v C ( t ) = v oc + v C ( 0 ) − v oc e − at = 3 + (1 − 3) e−4 t = 3 − 2 e−4 t for t ≥ 0 Step 5. Express the output current as a function of the source voltage and the capacitor voltage. io (t ) = C d 1 d v C (t ) = vC (t ) dt 12 dt Step 6. The output current is given by io (t ) = 1 d 1 2 3 − 2 e −4 t = ( −2 )( −4 ) e −4 t = e −4 t for t ≥ 0 12 dt 12 3 ( ) P 8.6-20 The voltage source voltage in the circuit shown in Figure P 8.6-20 is vs(t) = 25u(t) – 10 V Determine i(t) for t ≥ 0. Figure P 8.6-20 Solution: Simplify the circuit by replacing the series inductors by an equivalent inductor. Then, after a couple of source transformations, we have For t < 0 the circuit is at steady state and so the inductor acts like a short circuit. The voltage source voltage is −6 V so i ( 0 + ) = i ( 0 − ) = −60 mA For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the inductor. In this case we recognize voc = 9V and Rt = 100 Ω so isc = 90 mA. The time constant is Then τ= L 20 = = 0.2 s R t 100 ⇒ 1 τ =5 1 s i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −60 − 90 ) e −5t + 90 = 90 − 150e−5t mA for t ≥ 0 (checked: LNAP 7/12/04) P 8.6-21 The voltage source voltage in the circuit shown in Figure P 8.6-21 is vs(t) = 30 – 24u(t) V Determine i(t) for t ≥ 0. Figure P 8.6-21 Solution: Simplify the circuit by replacing the parallel inductors by an equivalent inductor. Then, after doing a couple of source transformations, we have For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The voltage source voltage is 12 V so i ( 0 + ) = i ( 0 − ) = 0.2 A For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the inductor. In this case, we recognize voc = 2.4 V and Rt = 60 Ω so isc = 0.04 A. The time constant is τ= Then L 4 1 = = s R t 60 15 ⇒ 1 τ = 15 1 s i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0.2 − 0.04 ) e−15t + 0.04 = 40 + 160e −15t mA (checked: LNAP 7/13/04) P 8.6-22 The voltage source voltage in the circuit shown in Figure P 8.6-22 is vs(t) = 10 + 40u(t) V Determine v(t) for t ≥ 0. Figure P 8.6-22 Solution: Simplify the circuit by replacing the series capacitors by an equivalent capacitor. Then, after doing some source transformations, we have For t < 0 the circuit is at steady state so the capacitor acts like an open circuit. The voltage source voltage is 8 V so v (0 +) = v (0 −) = 8 V For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the capacitor. In this case we recognize voc = 40 V and Rt = 8 Ω. The time constant is τ = R t C = ( 8 ) ( 60 ×10−3 ) = 0.48 Then ⇒ 1 τ = 2.08 1 s v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 8 − 40 ) e −2.08t + 40 = 40 − 32e−2.08t V for t ≥ 0 (checked: LNAP 7/13/04) P 8.6-23 Determine v(t) for t > 0 for the circuit shown in Figure P 8.6-23. Figure P 8.6-23 Solution: The resistor voltage, v(t), may not be continuous at time t = 0. The inductor will be continuous. We will find the inductor current first and then find v(t). Label the inductor current as i(t). For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is i (0 +) = i (0 −) = 0 A For t > 0 use source transformations to simplify the part of the circuit connected to the inductor until is a Norton equivalent circuit. Recognize that R t = 2 Ω and i sc = 1.333 A The time constant is τ= L 3 = Rt 2 ⇒ 1 τ = 0.667 1 s Then ( i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = 1.333 1 − e−0.667t ) A for t ≥ 0 Returning to the original circuit we see that d 3 i (t ) v (t ) d = i ( t ) + dt = i (t ) + i (t ) dt 2 3 = 1.333 (1 − e −0.667 t ) + ( −0.667 )(1.333) ( −e −0.667 t ) = 1.333 − 0.4439e−0.667 Finally v ( t ) = 2.667 − 0.889e0.667 t V for t > 0 (checked: LNAP 7/14/04) P 8.6-24 The input to the circuit shown in Figure P 8.6-24 is the current source current is(t) = 2 + 4u(t) A The output is the voltage v(t). Determine v(t) for t > 0. Figure P 8.6-24 Solution: Label the inductor current, i(t). We will find i(t) first, then find v(t). For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is ⎛ 3 ⎞ i (0 +) = i (0 −) = ⎜ ⎟2 =1 A ⎝ 3+3⎠ For t > 0 use source transformations to simplify the part of the circuit connected to the inductor until it is a Norton equivalent circuit. Recognize that R t = 2 Ω and i sc = 3 A The time constant is Then τ= L 0.25 = = 0.125 s Rt 2 ⇒ 1 τ =8 1 s i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = (1 − 3) e−8t + 3 = 3 − 2e −8t A for t ≥ 0 Returning to the original circuit d ⎛ ⎞ 0.25 i ( t ) ⎟ ⎜ d d −8t −8t dt v (t ) = 3⎜ i (t ) + ⎟ + 0.25 i ( t ) = 3i ( t ) + 0.5 i ( t ) = 3 ( 3 − 2e ) + 0.5 (16e ) 3 dt dt ⎜⎜ ⎟⎟ ⎝ ⎠ −8t = 9 + 2 e V for t > 0 (checked: LNAP 7/26/04) P 8.6-25 The input to the circuit shown in Figure P 8.6-25 is the voltage source voltage vs = 6 + 6u (t) The output is the voltage vo(t). Determine vo(t) for t > 0. Figure P 8.6-25 Solution: Label the capacitor voltage, v(t). W will find v(t) first then find vo(t). For t < 0 the circuit is at steady state and the capacitor acts like an open circuit. The initial condition is v (0 +) = v (0 −) = 6 V For t > 0 we replace series and then parallel resistors by equivalent resistors in order to replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. We recognize R t = 4 Ω and v oc = 12 V The time constant is τ = R t C = 4 ( 0.125 ) = 0.5 s ⇒ 1 τ =2 1 s The capacitor voltage is given by −t v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6 − 12 ) e −2t + 12 = 12 − 6e−2t V for t ≥ 0 Returning to the original circuit and applying KCL we see 0.125 so v o ( t ) = 0.25 12 − v ( t ) v o ( t ) d v (t ) = + 6 2 dt v (t ) d = 0.25 (12e −2t ) − 4 + 4 − 2e−2t = e −2t V for t > 0 v (t ) − 4 + 3 dt (checked: LNAP 7/26/04) P 8.6-26 Determine v(t) for t > 0 for the circuit shown in Figure P 8.6-26. Figure P 8.6-26 Solution: Label the inductor current as i(t). We will find i(t) first then use it to find v(t). For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is i (0 +) = i (0 −) = 0 A For t > 0 we replace series and parallel resistors by equivalent resistors. Then the part of the circuit connected to the inductor will be a Thevenin equivalent circuit. We recognize R t = 2 Ω and v oc = 12 V so i sc = v oc Rt =6A The time constant is τ= L 0.5 = = 0.25 Rt 2 ⇒ 1 τ =4 1 s The inductor current is given by ( i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0 − 6 ) e−4t + 6 = 6 1 − e−4t ) A for t ≥ 0 Returning to the original circuit and applying KCL we see i (t ) + 0.5 d i ( t ) − 12 v t () dt = 3 4 so v ( t ) = 4i ( t ) + 2d ⎛ 2⎞ i ( t ) − 16 = 24 (1 − e −4t ) + ⎜ ⎟ ( 24e −4t ) − 16 = 8 − 8e −4t V for t > 0 3 dt ⎝ 3⎠ (checked: LNAP 7/26/04) P 8.6-27 When the input to the circuit shown in Figure P 8.6-27 is the voltage source voltage vs(t) = 3 – u(t) V the output is the voltage vo(t) = 10 + 5 e–50t V for t ≥ 0 Determine the values of R1 and R2. Figure P 8.6-27 Solution: Apply KCL at the inverting input of the op amp to get v o (t ) − v (t ) R2 = R2 ⎞ ⎛ v (t ) ⇒ v o ( t ) = ⎜1 + ⎟ v (t ) 1000 ⎝ 1000 ⎠ We will determine the capacitor voltage first and then use it to determine the output voltage. When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to get 3 − v (0 −) = 0 ⇒ v (0 −) = 3 V R1 The initial condition is v (0 +) = v (0 −) = 3 V For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. 2 − voc = 0 ⇒ voc = 2 V R1 2 = i sc R1 ⇒ Rt = voc = R1 i sc The time constant is τ = R t C = R t (10−6 ) . From the given equation for v o ( t ) , R t (10−6 ) = 1 τ = 50 1 , so s 6 1 10 ⇒ R1 = R t = = 20 kΩ 50 50 The capacitor voltage is given by v ( t ) = ( v ( 0 ) − v oc ) e −t τ + v oc = ( 3 − 2 ) e −50t + 2 = 2 + e−50t V for t ≥ 0 So v o (t ) = 5 v (t ) ⇒ 5 = 1 + R2 1000 ⇒ R 2 = 4 kΩ (checked LNAPTR 7/31/04) P8.6-28 The time constant of a particular circuit is τ = 0.25 s. In response to a step input, a capacitor voltage changes from −2.5 V to 4.2 V. How long did it take for the capacitor voltage to increase from −2.0 V to +2.0 V? Solution: The capacitor voltage can be represented by the equation v ( t ) = A + B e − 4t for t ≥ 0. Given that A + B = v(0) = −2.5 V and A = v(∞) = 4.2 V we determine v ( t ) = 4.2 − 6.7 e − 4 t . ⎛ −2 − 4.2 ⎞ ln ⎜ ⎟ −6.7 ⎠ Let t1 be the time at which v(t1) = −2.0 V. Then t 1 = ⎝ = 0.01939 s . −4 ⎛ 2 − 4.2 ⎞ ln ⎜ ⎟ −6.7 ⎠ Let t2 be the time at which v(t2) = 2.0 V. Then t 2 = ⎝ = 0.27841 s . −4 The transition requires 0.27841 − 0.01939 = 0.25902 s. Section 8.7 The Response of a First-Order Circuit to a Nonconstant Source P 8.7-1 Find vc(t) for t > 0 for the circuit shown in Figure P 8.7-1 when v1 = 8e–5tu(t) V. Assume the circuit is in steady state at t = 0–. Answer: vc(t) = 4e–9t + 18e–5t V Figure P 8.7-1 Solution: Assume that the circuit is at steady state before t = 0: KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A Then vc (0− ) = −12 ix = 22 V = vc (0+ ) After t = 0: KVL : 12i (t ) − 8e −5t + v ( t ) = 0 x c dv ( t ) 1 dvc ( t ) KCL : −ix ( t )−2ix ( t ) + (1 36) c = 0 ⇒ ix ( t ) = dt 108 dt ⎡ 1 dvc ( t ) ⎤ −5t ∴ 12 ⎢ + v (t ) = 0 ⎥ − 8e c dt 108 ⎣ ⎦ dv (t ) c + 9v (t ) = 72e −5t ⇒ v (t ) = Ae−9t c cn dt Try v (t ) = Be−5t & substitute into the differential equation ⇒ B = 18 cf ∴ v (t ) = Ae −9t + 18 e−5t c v (0) = 22 = A + 18 ⇒ A = 4 c ∴ v (t ) = 4e−9t + 18e−5t V c P 8.7-2 Find v(t) for t > 0 for the circuit shown in Figure P 8.7-2. Assume steady state at t = 0–. Answer: v(t) = 20e–10t/3 – 12e–2t V Figure P 8.7-2 Solution: Assume that the circuit is at steady state before t = 0: iL (0+ ) = iL (0− ) = 12 = 3A 4 After t = 0: v( t ) −12 v( t ) + iL ( t ) + = 6 e−2t 4 2 di ( t ) also : v ( t ) = (2 / 5) L dt di ( t ) ⎤ 3⎡ iL ( t ) + ⎢(2 / 5) L ⎥ = 3 + 6 e−2t 4⎣ dt ⎦ KCL : diL ( t ) 10 + iL ( t ) = 10 + 20 e−2t 3 dt ∴ in (t ) = Ae − (10 / 3)t , try i f (t ) = B + Ce−2t , substitute into the differential equation, and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t ∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15 ∴ iL (t ) = −15e− (10 / 3) t + 3 + 15e−2t Finally, v(t ) =( 2 / 5 ) diL = 20 e− (10 / 3) t −12 e −2t V dt P 8.7-3 Find vc(t) for t > 0 for the circuit shown in Figure P 8.7-3 when is = [2 cos 2t] u(t) mA. Figure P 8.7-3 Solution: Assume that the circuit is at steady state before t = 0: Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get: dvc ( t ) ⎞ ⎛ +v t =0 ⇒ KVL: − 10 cos 2t + 15 ⎜ 1 30 dt ⎟ c ( ) ⎝ ⎠ dvc ( t ) + 2vc ( t ) = 20 cos 2t dt vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e −2t + 5cos 2t + 5sin 2t V P 8.7-4 Many have witnessed the use of an electrical megaphone for amplification of speech to a crowd. A model of a microphone and speaker is shown in Figure P 8.7-4a, and the circuit model is shown in Figure P 8.74b. Find v(t) for vs = 10(sin 100t)u(t) which could represent a person whistling or singing a pure tone. Figure P 8.7-4 Solution: Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) = 0 A. After t = 0, we have: di ( t ) KVL : − 10sin100t + i ( t ) + 5 + v (t ) = 0 dt v( t ) Ohm's law : i ( t ) = 8 dv( t ) ∴ +18 v( t ) = 160sin100t dt ∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t ∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55 so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V P 8.7-5 A lossy integrator is shown in Figure P 8.7-5. The lossless capacitor of the ideal integrator circuit has been replaced with a model for the lossy capacitor, namely, a lossless capacitor in parallel with a 1-kΩ resistor. If and vs = 15e–2tu(t) V vo(0) = 10 V, find vo(t) for t > 0. Figure P 8.7-5 Solution: Assume that the circuit is at steady state before t = 0. vo ( t ) = −vc ( t ) vC (0+ ) = vC (0− ) = −10 V After t = 0, we have i (t ) = vs ( t ) 8 e−5 t = = 0.533 e−5 t mA 15000 15000 The circuit is represented by the differential dv ( t ) vC ( t ) + . Then equation: i ( t ) = C C dt R ( 0.533 ×10 ) e −3 −5 t = ( 0.25 ×10−6 ) dvc ( t ) + (10−3 ) vc ( t ) ⇒ dt dvc ( t ) + 4000 vc ( t ) = 4000 e−5t dt Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get ( d B e−5t dt ) + 4000 ( B e ) = 4000 e −5t −5t ⇒ B= 4000 = −1.00125 ≅ −1 −3995 vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae−4000t vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e −2t − 11 e−4000t V Finally vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0 P 8.7-6 Determine v(t) for the circuit shown in Figure P 8.7-6. Figure P 8.7-6 Solution: Assume that the circuit is at steady state before t = 0. 2 v (0+ ) = v (0− ) = 30 = 10 V 4+ 2 After t = 0 we have KVL : ⎛ 1 d v( t ) ⎞ 5 d v( t ) + v (t ) + 4 ⎜ −i ⎟ = 30 2 dt 2 dt ⎝ ⎠ ⎛ 1 d v( t ) ⎞ −3t 2 i ( t ) + 4⎜ i ( t ) − ⎟ + 30 = e 2 dt ⎝ ⎠ The circuit is represented by the differential equation d v( t ) 6 6 2 (10 + e −3t ) v (t ) = + 19 19 3 dt Take vn ( t ) = Ae −( 6 /19 ) t . Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get −3Ce −3t + 6 60 4 −3t ( B + Ce−3t ) = + e 19 19 19 Equate coefficients to get 4 4 ⇒ v f ( t ) = e−3t + Ae− (6 /19) t 51 51 4 v ( t ) = vn ( t ) + v f ( t ) = 10 − e−3t + Ae− (6 /19) t 51 4 4 vc (0+ ) = 10 V, ⇒ 10 = 10 − + A ⇒ A = 51 51 4 − (6 /19) t −3t ∴ vc (t ) = 10 + (e −e ) V 51 B = 10 , C = − Then Finally P 8.7-7 Determine v(t) for the circuit shown in Figure P 8.7-7a when vs varies as shown in Figure P 8.7-7b. The initial capacitor voltage is vc(0) = 0. Figure P 8.7-7 Solution: We are given v(0) = 0. From part b of the figure: ⎧5t 0 ≤ t ≤ 2 s vs ( t ) = ⎨ t > 2s ⎩10 Find the Thevenin equivalent of the part of the circuit that is connected to the capacitor: The open circuit voltage: The short circuit current: (ix=0 because of the short across the right 2 Ω resistor) Replace the part of the circuit connected to the capacitor by its Thevenin equivalent: dv( t ) + v ( t ) − vs ( t ) = 0 dt dv( t ) v ( t ) vs ( t ) + = dt 2 2 vn ( t ) = Ae−0.5 t 2 For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0, we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) . At t = 2 s, v( 2 ) = 10e−1 = 3.68 . Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e − (t −2) / 2 − (t − 2) / 2 . . Using v ( 2 ) = 3.68 , we P 8.7-8 The electron beam, which is used to “draw” signals on an oscilloscope, is moved across the face of a cathode-ray tube (CRT) by a force exerted on electrons in the beam. The basic system is shown in Figure P 8.7-8a. The force is created from a time-varying, ramp-type voltage applied across the vertical or the horizontal plates. As an example, consider the simple circuit of Figure P 8.7-8b for horizontal deflection where the capacitance between the plates is C. Derive an expression for the voltage across the capacitance. If v(t) = kt and Rs = 625 kΩ, k = 1000, and C = 2000 pF, compute vc as a function of time. Sketch v(t) and vc(t) on the same graph for time less than 10 ms. Does the voltage across the plates track the input voltage? Figure P 8.7-8 Solution: ⎡ d v (t ) ⎤ KVL: − kt + Rs ⎢C C ⎥ + vC ( t ) = 0 dt ⎦ ⎣ d vC ( t ) k 1 vC ( t ) = t ⇒ + Rs C Rs C dt vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t & plug into D.E. ⇒ B1 + 1 k t thus B0 = − kRs C , B1 = k . [ B0 + B1t ] = Rs C Rs C Now we have vc (t ) = Ae− t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C. ∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )] v(t) and vC(t) track well on a millisecond time scale. Section 8.10 How Can We Check…? P 8.10-1 Figure P 8.10-1 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the capacitor voltage at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot does indeed represent the voltage of the capacitor in this circuit. Figure P 8.10-1 Solution: First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor voltage is vc = 4 V. We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot. Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the capacitor is R t = 2000 + 4000 3 2 ⎛4 ⎞ τ = R t C = ⎜ × 103 ⎟ ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is 3 ⎝3 ⎠ vc (t ) = 4 e− t 0.67 +4 V where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then vc (t1 ) = So the plot is correct. ⎛ 1.33 ⎞ −⎜ ⎟ 4 e ⎝ .67 ⎠ + 4 = 4.541 ~ 4.5398 V P 8.10-2 Figure P 8.10-2 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the inductor current at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot does indeed represent the current of the inductor in this circuit. Figure P 8.10-2 Solution: The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit. Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the inductor is R t = 2000 + 4000 3 5 15 L ms . Thus inductor current is τ= = = R t 4 ×103 4 3 iL (t ) − 2 e − t 3.75 + 5 mA where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then iL (t1 ) = ⎛ 3.75 ⎞ −⎜ ⎟ −2 e ⎝ 3.75 ⎠ + 5 = 4.264 mA ≠ 4.7294 mA so the plot does not correspond to this circuit. P 8.10-3 Figure P 8.10-3 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the inductor current at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Specify that value of the inductance, L, required to cause the current of the inductor in this circuit to be accurately represented by this plot. Figure P 8.10-3 Solution: Notice that the steady-state inductor current does not depend on the inductance, L. The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit. After t = 0 So I sc = 5 mA and τ = The inductor current is given by iL (t ) = −2e −1333t has units of Henries. Let t 1 = 3.75 ms, then L + 5 mA , where t has units of seconds and L 4.836 = iL (t1 ) = −2 e−(1333)⋅(0.00375) L + 5 = −2e−5 L + 5 4.836−5 = e −5 L −2 so and is the required inductance. L= L 1333 −5 =2 H ⎛ 4.836−5 ⎞ ln ⎜ ⎟ ⎝ −2 ⎠ P 8.10-4 Figure P 8.10-4 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the capacitor voltage at that time. Assume that this circuit has reached steady state before time t = 0. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Specify values of A, B, R1, R2, and C that cause the voltage across the capacitor in this circuit to be accurately represented by this plot. Figure P 8.10-4 Solution: First consider the circuit. When t < 0 and the circuit is at steady-state: For t > 0 So Voc = R2 R1 R2 RRC ( A + B) , Rt = and τ = 1 2 R1 + R2 R1 + R2 R1 + R2 Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor voltage is (Voc =) 4 V, so vC (t ) = − 6e−t τ + 4 At t 1 = 1.333 ms 3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4 −0.001333 = 0.67 ms ⎛ −4+ 3.1874 ⎞ ln ⎜ ⎟ −6 ⎝ ⎠ τ = so Combining the information obtained from the circuit with the information obtained from the plot gives R2 R2 R1 R2C A = −2, ( A + B ) = 4, = 0.67 ms R1 + R2 R1 + R2 R1 + R2 There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is one convenient way. Pick R = 3000 and R = 6000. Then 1 1 2 2 2A = −2 ⇒ A = − 3 3 2( A+ B) = 4 ⇒ B −3 = 6 ⇒ B = 9 3 2 1 μF = C 2000 ⋅ C = ms ⇒ 3 3 Spice Problems SP 8-1 The input to the circuit shown in Figure SP 8.1 is the voltage of the voltage source, vi(t). The output is the voltage across the capacitor, vo(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t. Hint: Represent the voltage source using the PSpice part named VPULSE. Figure SP 8.1 Solution: SP 8-2 The input to the circuit shown in Figure SP 8.2 is the voltage of the voltage source, vi(t). The output is the current in the inductor, io(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, io(t), as a function of t. Hint: Represent the voltage source using the PSpice part named VPULSE. Figure SP 8.2 Solution: SP 8-3 The circuit shown in Figure SP 8.3 is at steady state before the switch closes at time t = 0. The input to the circuit is the voltage of the voltage source, 12 V. The output of this circuit is the voltage across the capacitor, v(t). Use PSpice to plot the output, v(t), as a function of t. Use the plot to obtain an analytic representation of v(t) for t > 0. Hint: We expect v(t) = A + Be–t/τ for t > 0, where A, B, and τ are constants to be determined. Figure SP 8.3 Solution: v(t ) = A + B e −t / τ 7.2 = v(0) = A + B e 0 8.0 = v(∞) = A + B e −∞ for t > 0 ⇒ 7.2 = A + B ⎫⎪ ⎬ ⇒ B = −0.8 V ⇒ A = 8.0 V ⎪⎭ 7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ ⎛ 8 − 7.7728 ⎞ = ln ⎜ ⎟ = −1.25878 τ 0.8 ⎝ ⎠ 0.05 ⇒ τ= = 39.72 ms 1.25878 ⇒ − 0.05 Therefore v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0 SP 8-4 The circuit shown in Figure SP 8.4 is at steady state before the switch closes at time t = 0. The input to the circuit is the current of the current source, 4 mA. The output of this circuit is the current in the inductor, i(t). Use PSpice to plot the output, i(t), as a function of t. Use the plot to obtain an analytic representation of i(t) for t > 0. Hint: We expect i(t) = A + Be–t/τ for t > 0, where A, B, and τ are constants to be determined. Solution: Figure SP 8.4 i (t ) = A + B e−t / τ 0 = i (0) = A + B e 0 ⇒ 0 = A+ B 4 × 10−3 = i (∞) = A + B e −∞ ⇒ A = 4 × 10−3 2.4514 ×10 = v(5 ×10 ) = ( 4 ×10 −3 ⇒ − −6 5 × 10−6 ⇒ τ= Therefore for t > 0 τ −3 ⎫⎪ −3 ⎬ ⇒ B = −4 × 10 A A ⎪⎭ ) − ( 4 ×10 ) e ( −3 ) − 5×10−6 / τ ⎛ ( 4 − 2.4514 ) × 10−3 ⎞ = ln ⎜ ⎟ = −0.94894 −3 4 × 10 ⎝ ⎠ 5 × 10−6 = 5.269 μ s 0.94894 i (t ) = 4 − 4 e−t / 5.269×10 −6 mA for t > 0 DESIGN PROBLEMS DP 8-1 Design the circuit in Figure DP 8.1 so that v(t) makes the transition from v(t) = 6 V to v(t) = 10 V in 10 ms after the switch is closed. Assume that the circuit is at steady state before the switch is closed. Also assume that the transition will be complete after 5 time constants. Figure DP 8.1 Solution: Steady-state response when the switch is open: 6 = R3 R1 + R 2 + R 3 Steady-state response when the switch is closed: 10 = 10 ms = 5 τ = ( R 1 || R 3 ) C = R3 6 R3 R1 + R 3 12 ⇒ R1 + R 2 = R 3 . 12 ⇒ R1 = R3 5 . C Let C = 1 μF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ. DP 8-2 Design the circuit in Figure DP 8.2 so that i(t) makes the transition from i(t) = 1 mA to i(t) = 4 mA in 10 ms after the switch is closed. Assume that the circuit is at steady state before the switch is closed. Also assume that the transition will be complete after 5 time constants. Figure DP 8.2 Solution: steady state response when the switch is open: 0.001 = 12 R1+R 2 steady state response when the switch is closed: 0.004 = 12 R1 ⇒ R + R = 12 kΩ . 1 2 ⇒ R1 = 3 kΩ . Therefore, R 2 = 9 kΩ. ⎛ L ⎞ L 10 ms = 5 τ = 5 ⎜ ⇒ L = 240 H ⎟= ⎜ R1 + R 2 ⎟ 2400 ⎝ ⎠ DP 8-3 The switch in Figure DP 8.3 closes at time 0, 2Δt, 4Δt, …2kΔt and opens at times Δt, 3Δt, 5Δt….(2k + 1)Δt. When the switch closes, v(t) makes the transition from v(t) = 0 V to v(t) = 5 V. Conversely, when the switch opens, v(t) makes the transition from v(t) = 5 V to v(t) = 0 V. Suppose we require that Δt = 5τ so that one transition is complete before the next one begins. (a) Determine the value of C required so that Δt = 1 μs. (b) How large must Δt be when C = 2 μF? Answer: (a) C = 4 pF; (b) Δt = 0.5s Figure DP 8.3 Solution: Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. 10−6 (a) Δt = 5 Rt C ⇒ C = = 4 pF 5 50×103 ( (b) Δ t = 5 50×10 3 )( 2×10 −6 ( ) ) = 0.5 s DP 8-4 The switch in Figure DP 8.3 closes at time 0, 2Δt, 4Δt, …2kΔt and opens at times Δt, 3Δt, 5Δt….(2k + 1)Δt. When the switch closes, v(t) makes the transition from v(t) = 0 V to v(t) = 5 V. Conversely, when the switch opens, v(t) makes the transition from v(t) = 5 V to v(t) = 0 V. Suppose we require that one transition be 95 percent complete before the next one begins. (a) Determine the value of C required so that Δt = 1 μs. (b) How large must Δt be when C = 2 μF? Hint: Show that Δt = – τ ln(1 – k) is required for the transition to be 100 k percent complete. Answer: (a) C = 6.67 pF; (b) Δt = 0.3 s Solution: Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. Δt −Δt τ When the switch is open: 5 e = (1 − k ) 5 ⇒ ln (1− k ) = − ⇒ Δ t = −τ ln (1− k ) When the switch is open: 5 − 5 e (a) C = −Δt τ 10−6 = 6.67 pF − ln (1−.95 ) ( 50×103 ) τ = k 5 ⇒ Δ t = −τ ln (1− k ) (b) Δ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s DP 8-5 A laser trigger circuit is shown in Figure DP 8.5. In order to trigger the laser, we require 60 mA < |i| < 180 mA for 0 < t < 200 μs. Determine a suitable value for R1 and R2. Figure DP 8.5 Solution: i (0) = R1 20 × 40 R1 R1 + 40 40 + 40 + R1 For t > 0: i (t ) = i (0) e −t τ where τ = L 10−2 = R t 40+ R 2 At t < 200μ s we need i ( t ) > 60 mA and i ( t ) <180 mA First let's find a value of R 2 to cause i (0) < 180 mA. 1 t A = 166.7 mA so i (t ) = 0.1667 e− τ . 6 Next, we find a value of R 2 to cause i (0.0002) > 60 mA. Try R 2 = 40 Ω . Then i (0) = 10−2 1 s. = 0.2 ms = 50 5000 i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mA Try R 2 = 10Ω, then τ = DP 8-6 Fuses are used to open a circuit when excessive current flows (Wright, 1990). One fuse is designed to open when the power absorbed by R exceeds 10 W for 0.5 s. Consider the circuit shown in Figure DP 8.6. The input is given by vs = A[u(t) – u(t – 0.75)] V. Assume that iL(0–) = 0. Determine the largest value of A that will not cause the fuse to open. Solution: The current waveform will look like this: Figure DP 8.6 We only need to consider the rise time: iL (t ) = −t −t Vs A τ (1 − e τ ) = (1 − e ) R+2 R+2 where L 0.2 1 s = = Rt 3 15 A ∴ iL (t ) = (1 − e −15t ) 3 τ = Now find A so that iL2 R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s ∴ we want [iL2 (0.25)]R fuse = 10 W ⇒ A2 (1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V 9 Figure DP 8-7 DP 8-7 Design the circuit in Figure DP 8-7(a) to have the response in Figure DP 8-7 (b) by specifying the values of C, R1 and R2. Solution: D ⎧ The voltage v(t) is represented by an equation of the form v ( t ) = ⎨ − at ⎩E + F e for t < 0 for t > 0 where D, E, F and a are unknown constants. The constants D, E and F are described by D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t ) t →∞ t →0+ From the plot, we see that D = 10, E = 20, and E + F = 10 V Consequently, 10 ⎧ v (t ) = ⎨ − at ⎩ 20 − 10 e for t < 0 for t > 0 To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot. We see v ( 3.22 ) = 18 V , that is, the value of the voltage is 18 volts at time 3.22 seconds. Substituting these into the equation for v ( t ) gives − a 3.22 18 = 20 − 10 e ( ) ⇒ a= ln ( 0.2 ) = 0.5 −3.22 Consequently 10 ⎧ v (t ) = ⎨ −0.5 t ⎩20 − 10 e for t < 0 for t > 0 Now let’s turn our attention to the circuit. When the circuit is at steady state, the capacitor acts like an open circuit. After t = 0, the switch is closed and the steady state voltage is determined from the plot to be v(t) = E = 20 V. On the right we see the circuit that results from replacing the capacitor by an open circuit and the switch by a short circuit. Using voltage division gives 80 20 = ( 25) ⇒ R 2 = 20 Ω 80 + R 2 Before t = 0, the switch is open and the steady state voltage is determined from the plot to be v(t) = E + F = 10 V. On the right we see the circuit that results from replacing the capacitor by an open circuit and the switch by an open circuit. Using voltage division gives 10 = 80 80 ( 25 ) = ( 25) ⇒ R1 = 100 Ω 80 + R1 + R 2 100 + R1 Recalling that a = 0.5 from the plot, consider the time 1 constant 2 = = τ = C R t . After t = 0, the Thevenin a resistance of the part of the circuit connected to the capacitor is R t = 80 || R 2 = 80 || 20 = 16 Ω . Then C = 2 2 = = 0.125 F . R t 16 Figure DP 8-8 DP 8-8 Design the circuit in Figure DP 8-8(a) to have the response in Figure DP 8-8 (b) by specifying the values of L, R1 and R2. Solution: D ⎧ The voltage v(t) is represented by an equation of the form v ( t ) = ⎨ − at ⎩E + F e for t < 0 for t > 0 where D, E, F and a are unknown constants. The constants D, E and F are described by D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t ) t →∞ t →0+ From the plot, we see that D = 2, E = 8, and E + F = 2 V Consequently, 2 ⎧ v (t ) = ⎨ − at ⎩8 − 6 e for t < 0 for t > 0 To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot. We see v ( 0.358 ) = 7 V , that is, the value of the voltage is 7 volts at time 0.358 seconds. Substituting these into the equation for v ( t ) gives 7 = 8− 6e − a ( 0.358) ⇒ a= ( 6) = 5 ln 1 −0.358 Consequently 2 ⎧ v (t ) = ⎨ −5 t ⎩8 − 6 e for t < 0 for t > 0 Now let’s turn our attention to the circuit. When the circuit is at steady state, the inductor acts like a short circuit. After t = 0, the switch is closed and the steady state voltage is determined from the plot to be v(t) = E = 8 V. On the right we see the circuit that results from replacing the inductor by a short circuit and the switch by a short circuit. Using voltage division gives 8 8= (12 ) ⇒ R1 = 4 Ω 8 + R1 Before t = 0, the switch is open and the steady state voltage is determined from the plot to be v(t) = E + F = 2 V. On the right we see the circuit that results from replacing the inductor by a short circuit and the switch by an open circuit. Using voltage division gives 8 8 2= (12 ) = (12 ) ⇒ R 2 = 36 Ω 8 + R1 + R 2 12 + R 2 Recalling that a = 5 from the plot, consider the time constant 1 L 0.2 = = τ = . After t = 0, the Thevenin resistance of the a Rt part of the circuit connected to the capacitor is R t = 8 + R 2 = 8 + 4 = 12 Ω . Then L = 0.2 R t = 2.4 H . Figure DP 8-9 DP 8-9 Design the circuit in Figure DP 8-9(a) to have the response in Figure DP 8-9 (b) by specifying the values of C, R1 and R2. Solution: D ⎧ The voltage v(t) is represented by an equation of the form v ( t ) = ⎨ − at ⎩E + F e for t < 0 for t > 0 where D, E, F and a are unknown constants. The constants D, E and F are described by D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t ) t →∞ From the plot, we see that Consequently, t →0 + D = 9, E = 5, and E + F = 9 V 9 ⎧ v (t ) = ⎨ − at ⎩5 + 4 e for t < 0 for t > 0 To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot. We see v ( 0.173) = 6 V , that is, the value of the voltage is 6 volts at time 0.173 seconds. Substituting these into the equation for v ( t ) gives − a 0.173) 6 = 5+ 4e ( ⇒ a= ln ( 0.25 ) =8 −0.173 Consequently 9 ⎧ v (t ) = ⎨ −8 t ⎩5 + 4 e for t < 0 for t > 0 Now let’s turn our attention to the circuit. When the circuit is at steady state, the capacitor acts like an open circuit. Before t = 0, the switch is open and the steady state voltage is determined from the plot to be v(t) = E + F = 9 V. On the right we see the circuit that results from replacing the capacitor by an open circuit and the switch by an open circuit. Using voltage division gives 9= R2 240 + R 2 (15 ) ⇒ R 2 = 360 Ω After t = 0, the switch is closed and the steady state voltage is determined from the plot to be v(t) = E = 5 V. On the right we see the circuit that results from replacing the capacitor by an open circuit and the switch by a short circuit. Using voltage division gives 5= R1 || R 2 240 + R1 || R 2 (15) ⇒ R1 || R 2 = 120 Ω R1 || R 2 = R1 || 360 = 120 Ω ⇒ R1 = 180 Ω Recalling that a = 8 from the plot, consider the 1 time constant 0.125 = = τ = C R t . After t = 0, a the Thevenin resistance of the part of the circuit connected to the capacitor is R t = 240 || R1 || R 2 = 240 ||180 || 360 = 80 Ω . Then 0.125 0.125 C= = = 0.0015625 F = 1.5625 mF . Rt 80 Figure DP 8-10 DP 8-10 Design the circuit in Figure DP 8-10(a) to have the response in Figure DP 8-10 (b) by specifying the values of L, R1 and R2. Solution: D ⎧ The current i(t) is represented by an equation of the form i ( t ) = ⎨ − at ⎩E + F e for t < 0 for t > 0 where D, E, F and a are unknown constants. The constants D, E and F are described by D = i ( t ) when t < 0 , E = lim i ( t ) , E + F = lim i ( t ) t →∞ t →0 + From the plot, we see that D = 120, E = 60, and E + F = 120 mA Consequently, 120 mA for t < 0 ⎧ i (t ) = ⎨ − at ⎩60 + 60 e mA for t > 0 To determine the value of a , we pick a time when the circuit is not at steady state. One such point is labeled on the plot. We see i ( 0.1195 ) = 70 mA , that is, the value of the current is 70 mA at time 0.1195 seconds. Substituting these into the equation for i ( t ) gives 70 = 60 + 60 e − a ( 0.1195) ⇒ a= ( 6 ) = 15 ln 1 −0.1195 Consequently 120 mA for t < 0 ⎧ i (t ) = ⎨ −15 t mA for t > 0 ⎩60 + 60 e Now let’s turn our attention to the circuit. When the circuit is at steady state, the inductor acts like a short circuit. Before t = 0, the switch is closed and the steady state current is determined from the plot to be i(t) = E + F = 120 mA. On the right we see the circuit that results from replacing the inductor by a short circuit and the switch by a short circuit. Using current division gives 28 120 = (180 ) ⇒ R1 = 14 Ω 28 + R1 After t = 0, the switch is open and the steady state voltage is determined from the plot to be v(t) = E + F = 2 V. On the right we see the circuit that results from replacing the inductor by a short circuit and the switch by an open circuit. Using current division gives 28 28 60 = (180 ) = (180 ) ⇒ R 2 = 42 Ω 28 + R1 + R 2 28 + 14 + R 2 Recalling that a = 15 from the plot, consider the time constant 1 1 L = =τ = . After t = 0, the Thevenin resistance of the part of the 15 a Rt circuit connected to the capacitor is R t = 28 + R1 + R 2 = 28 + 14 + 42 = 84 Ω . Then L = Rt 15 = 84 = 5.6 H . 15 Chapter 9 - Complete Response of Circuits with Two Energy Storage Elements Exercises 2H Exercise 9.2-1 Find the second-order differential equation for the circuit shown in Figure E 9.2-1 in terms of i using the direct method. Answer: is 1Ω 1 2F i d 2i 1 di 1 dis + +i = 2 dt 2 dt 2 dt Figure E 9.2-1 Solution: KVL a : 2 di dt + v + 1(i − is ) = 0 ⇒ v = − 2 di i = 1 = 1 ∴ 2 2 dv dis dt dt = 1 −1 2 2 di d dt dt dt − d 2i 1 di 1 dis + +i= 2 2 dt 2 dt dt − i + is ( −2di dt −i +i ) s d 2i dt 2 Exercise 9.2-2 Find the second-order differential equation for the circuit shown in Figure E 9.2-2 in terms of v using the operator method. v is 1Ω di d 2v di Answer: 2 + 2 + 2v = 2 s dt dt dt 1H Ground Figure E 9.2-2 Solution: KCL at v : using s = d v + i + 1 sv = is 2 1 also v = si (2) Solving for i in (1) & plugging into (2) di d 2v dv +2 + 2v = 2 s yields s 2v + 2 sv + 2v = 2 sis or dt dt dt dt (1) 1 2F v Exercise 9.3-1 Find the characteristic equation and the natural frequencies for the circuit shown in Figure E 9.3-1. Answer: 6Ω is s2 + 7s + 10 = 0; s1 = –2, s2 = –5 4Ω 1 4 F 1H Ground Figure E 9.3-1 Solution: v + iL + 1 dv = i 4 dt s 4 di KVL for the right mesh : v = 6iL + L dt KCL at the top node : (1) (2) Plugging (2) into (1) yields d 2i di L + 7 L + 10i = 4i L s dt dt 2 ∴ characteristic equation ⇒ s 2 + 7 s + 10 = 0 & natural frequencies ⇒ s = − 2, − 5 v Exercise 9.4-1 Find the natural response of the RLC circuit of Figure 9.4-1 when R = 6 Ω, L = 7 H, and C = 1/42 F. The initial conditions are v(0) = 0 and i(0) = 10 A. –t L R i –6t Answer: vn(t) = – 84(e – e ) V Figure 9.4-1 Solution: This is a parallel RLC circuit with α = 1 1 = = 7 2 2 RC 2(6)(1/ 42) 1 1 = = 6 LC (7)(1/ 42) and ω o2 = The roots of the characteristic equation are s = −α ± α 2 −ω o2 = − 7 2 ± ( 7 2 ) −6 2 = − 1, − 6 so the natural response is vn (t ) = A 1e− t + A2 e−6t Need vn (0) and dvn dt t =0 to evaluate A1 & A2 . We are given vn (0) = 0. Apply KCL at the top node to get i (t ) + v (t ) 1 d + v (t ) = 0 ⇒ 6 42 dt d v ( t ) = − ( 7 v ( t ) + 42 i ( t ) ) dt At time t=0, d V v ( t ) = − ( 7 v ( 0 ) + 42 i ( 0 ) ) = − ( 7 × 0 + 42 × 10 ) = −420 dt s t =0 So vn (0) = 0 = A1 + A2 ⎫ ⎪ dvn ⎬ A = − 84 , A2 = 84 =−420= − A1 − 6 A2 ⎪ 1 dt t =0 ⎭ and vn (t ) = −84e− t + 84e −6t V C Exercise 9.5-1 A parallel RLC circuit has R = 10 Ω, C = 1 mF, L = 0.4 H, v(0) = 8 V, and i(0) = 0. Find the natural response vn(t) for t < 0. Answer: vn(t) = e –50t(8 – 400t) V Solution: For parallel RLC 1 1 1 1 α = = = 50, ω o2 = = = 2500 −3 LC 2 RC 2(10)(10 ) (0.4)(10−3 ) ∴ s = −50 ± (50)2 − 2500 = −50, −50 ∴ vn (t ) = A1e −50t + A2te−50t + ) = 0 & v(0+ ) = 8 V −v(0+ ) i (0+ ) = = −0.8 V C 10Ω i (0+ ) dv V = C = − 800 dt t = 0+ C s with i (0 At t = 0+ vn (0) = 8 = A1 ⇒ vn (t ) = 8e −50t + A2te−50t dv(0) = −800 = − 400 + A2 ⇒ A2 = − 400 dt ∴ vn (t ) = 8e −50t − 400t e−50t V Exercise 9.6-1 A parallel RLC circuit has R = 62.5 Ω, L = 10 mH, C = 1 μF, v(0) = 10 V, and i(0) = 80 mA. Find the natural response vn(t) for t > 0. Answer: vn(t) = e–8000t [10 cos 6000t – 26.7 sin 6000t] V Solution: 1 1 = = 8000 2 RC 2(62.5)(10−6 ) 1 1 = = 108 ω2 = 0 LC − 6 (.01)(10 ) α= ∴ s = − α ± α 2 − ω 02 = − 8000 ± (8000) 2 − 108 = − 8000 ± j 6000 ∴ vn (t ) = e−8000t [ A1 cos 6000 t + A2 sin 6000t ] KCL at top : 0.08 + 10 + iC = 0 62.5 ⇒ iC (0+ ) = − 0.24 A ∴ dv(0+ ) iC (0+ ) = = −2.4 ×10+5 V/s dt C vn (0) = 10 = A1 dvn (0) = − 2.4×105 = 6000 A2 − 8000(10) ⇒ A2 = − 26.7 dt ∴ vn (t ) = e −8000t [10 cos 6000 t − 26.7 sin 6000t ] V Exercise 9.7-1 A circuit is described for t > 0 by the equation d 2i di + 9 + 20i = 6is 2 dt di where is = 6 + 2t A. Find the forced response if for t > 0. Answer: if = 1.53 + 0.6t A Solution: i "+ 9i '+ 20i = 36 + 12t Try i f = A + Bt & plug into above 0 + 9 B + 20( A + Bt ) = 36 + 12t Equating the constant coefficients and the coefficients of t gives 20 Bt = 12t ⇒ B = 0.6 and 9 B + 20 A = 36 ⇒ A = 1.53 ∴ if = 1.53 + 0.6t A i Exercise 9.9-1 Find v2(t) for t > 0 for the circuit of Figure E 9.9-1. Assume there is no initial stored energy. 3 10 + 1Ω v1 + v2 – 1 12 F – Answer: v2(t) = – 15e–2t + 6e– 4t – e–6t + 10 V H 5 6 F 10u(t) A Figure E 9.9-1 Solution: no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 t = 0+ KVL : − 0 + 3 di(0+ ) di(0+ ) +0=0 ⇒ =0 10 dt dt 0V KCL at A : + i1 (0+ )+0 = 0 1Ω dv1 (0+ ) =0 ⇒ dt KCL at B : − 0 + i 2 (0+ ) − 10 = 0 ⇒ i 2 (0+ ) = 5 6 dv 2 (0+ ) dv 2 (0+ ) = 12 V s = 10 ⇒ dt dt t>0 v1 1 + v1 ' + i = 0 1 12 KCL at B : − i + (5 / 6)v '2 =10 KCL at A : (1) KVL : − v1 + ( 3 /10 ) i '+ v2 = 0 ( 2) ( 3) Eliminating i from (1) & (3) yields 1 v '1 + (5 / 6)v2 '− 10 = 0 12 3 ⎛5 ⎞ −v1 + ⎜ v2" ⎟ + v2 = 0 10 ⎝ 6 ⎠ v1 + (4) ( 5) From (5) 1 v1 = v2 + v2" ⇒ v1' = v2" + (1/ 4)v2" 4 Now substituting into (4) yields 1 1⎛ 1 ⎞ 5 v2 ' + v2" + ⎜ v2 ' + v2"' ⎟ + v2 ' = 10 4 12 ⎝ 4 ⎠ 6 v2"' + 12v2" + 44v2 ' + 48v2 = 480 Natural Response: Forced Response: Complete Response: v2 n : s 3 + 12 s 2 + 44s + 48 = 0 ⇒ s = − 2, −4, −6 ∴ v2 n = A1e −2t + A2 e −4t + A3e−6t v2 f : try v2 f = B and plug into Diff. Eq. ⇒ B = 10 v2 (t ) = A1e −2t + A2e −4t + A3e −6t + 10 dv2 (0+ ) d 2 v2 (0+ ) Recall v2 (0 ) = 0, = 12 V/s, then from (5) = 4[v1 (0+ ) − v2 (0+ )] = 0 . 2 dt dt + v2 (0 ) = 0 = A1 + A2 + A3 + 10 (6) + dv2 (0+ ) = 12 = − 2 A1 − 4 A2 − 6 A3 (7) dt d 2 v2 (0+ ) = 0 = 4 A1 + 16 A2 + 36 A3 (8) dt 2 Solving (6)-(80 simultaneously gives A1 = −15, A2 = 6, A3 = −1 Then v2 (t ) = − 15e −2t + 6e −4t − e−6t + 10 V jω Exercise 9.10-1 A parallel RLC circuit has L = 0.1 H and C = 100 mF. Determine the roots of the characteristic equation and plot them on the s-plane when (a) R = 0.4 Ω and (b) R = 1.0 Ω. × –20 × –5 0 Answer: (a) s = – 5, – 20 (Figure E 9.10-1) Figure E 9.10-1 Solution: s2 + 1 1 s+ = 0 and L = 0.1, C = 0.1 ⇒ RC LC a) R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0 s = − 5, − 20 b) R = 1Ω ⇒ s 2 + 10s + 100 = 0 s = − 5 ± j5 3 s2 + 10 s + 100 = 0 R σ Section 9-2: Differential Equations for Circuits with Two Energy Storage Elements 2Ω P 9.2-1 Find the differential equation for the circuit shown in Figure P 9.2-1 using the direct method. 1 mH + – vs 100 Ω Figure P 9.2-1 Solution: KCL: i L = v dv +C R2 dt KVL: Vs = R 1i L + L L v + C dv OP + L dv + LC d v + v = R M N R dt Q R dt dt L R + 1OPv + LMR C + L OP dv + [LC] d v = M dt N R Q N R Q dt 2 vs 1 2 2 2 2 vs 1 1 2 2 2 R 1 = 2Ω, R 2 = 100Ω, L = 1mH, C = 10μF 2 dv −8 d v . v +.00003 + 1 × 10 v s = 102 dt dt 2 dv d 2 v 8 8 1 × 10 v s = 102 . × 10 v + 3000 + 2 dt dt di L +v dt 10 μ F P 9.2-2 Find the differential equation for the circuit shown in Figure P 9.2-2 using the operator method. Answer: d2 d i (t ) + 11, 000 iL (t ) + 1.1 × 108 iL (t ) = 108 is (t ) 2 L dt dt 10 Ω is 10 μ F 100 Ω 1 mH iL Figure P 9.2-2 Solution: v + i L + Csv R1 KVL: v = R 2 i L + Lsi L KCL: i s = Solving Cramer's rule for i L : iL = is R 2 Ls + + R 2 Cs + LCs2 +1 R1 R1 LM1 + R OPi + LM L + R COPsi N R Q NR Q 2 L 1 2 L + LC s2 i L = i s 1 R 1 = 100Ω, R 2 = 10Ω, L = 1mH, C = 10μF 1.1i L +.00011si L + 1 × 10 −8 s2 i L = i s 11 . × 108 i L + 11000si L + s2 i L = 1 × 108 i s P 9.2-3 Find the differential equation for iL(t) for t > 0 for the circuit of Figure P 9.2-3. iL is R1 L t=0 + – vs C + vc – Figure P 9.2-3 Solution: t>0 KCL: i L + C dv c v s + v c + = 0 dt R2 KVL: R 1i s + R 1i L + L di L − vc − vs = 0 dt Solving for iL : − R1 d 2iL ⎡ R1 1 ⎤ diL ⎡ R1 1 ⎤ R1 dis 1 dvs + + + + = − + i i s ⎢ ⎥ ⎢ ⎥L dt 2 ⎣ L R2C ⎦ dt ⎣ LR2C LC ⎦ LCR2 L dt L dt R2 P 9.2-4 The input to the circuit shown in Figure P 9.2-4 is the voltage of the voltage source, Vs. The output is the inductor current i(t). Represent the circuit by a secondorder differential equation that shows how the output of this circuit is related to the input, for t > 0. t=0 + – Hint: Use the direct method. R1 i(t) + L Vs v(t) C – Figure P 9.2-4 Solution: After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠ Apply KVL to get v (t ) = L Substituting v ( t ) into the first equation gives d i (t ) + R2 i (t ) dt ⎛ d⎛ d d ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ then R1 C L d2 dt 2 ( i ( t ) + R1 C R 2 + L ) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs Dividing by R1 C L : ⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs + + = i t i t i t ⎜ ⎟ ⎜ ⎟ ( ) ( ) ( ) ⎜ R1 C L ⎟ dt ⎜ R1 C L ⎟ R1 C L dt 2 ⎝ ⎠ ⎝ ⎠ d2 R3 R2 P 9.2-5 The input to the circuit shown in Figure P 9.2-5 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Use the direct method. i(t) t=0 R1 L + – vs + R2 C v(t) – Figure P 9.2-5 Solution: After the switch closes, use KCL to get i (t ) = v (t ) d + C v (t ) R2 dt Use KVL to get v s = R1 i ( t ) + L d i (t ) + v (t ) dt Substitute to get vs = R1 R2 = CL v ( t ) + R1C d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt ⎛ R1 + R 2 d2 L ⎞d v t + R1C + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ ⎟ dt dt R R CL 2 2 ⎝ ⎠ Finally, vs CL = ⎛ R1 R1 + R 2 d2 1 ⎞d + + v t v (t ) + v (t ) ( ) ⎜ ⎟ 2 ⎜ L R 2C ⎟ dt dt R 2CL ⎝ ⎠ t=0 P 9.2-6 The input to the circuit shown in Figure P 9.2-6 is the voltage of the voltage source, vs. The output is the inductor current i(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. R1 R2 + – + C vs L i(t) Hint: Use the direct method. Figure P 9.2-6 Solution: After the switch closes use KVL to get R2 i (t ) + L d i (t ) = v (t ) dt Use KCL and KVL to get d ⎛ ⎞ v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) dt ⎝ ⎠ Substitute to get d d2 d i ( t ) + R1CL 2 i ( t ) + R 2i ( t ) + L i ( t ) dt dt dt 2 d d = R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t ) dt dt v s = R1i ( t ) + R1CR 2 Finally vs R1CL = ⎛ R2 R1 + R 2 d2 1 ⎞d + i t + i (t ) + i (t ) ⎟ 2 ( ) ⎜ ⎜ L R1C ⎟ dt dt R CL 1 ⎝ ⎠ v(t) – P 9.2-7 The input to the circuit shown in Figure P 9.2-7 is the voltage of the voltage source, vs. The output is the inductor current i2(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Use the operator method. R2 t=0 L1 R1 + – i1(t) L2 i2(t) vs R3 Figure P 9.2-7 Solution: After the switch opens, KVL gives L1 d d i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t ) dt dt L1 d i1 ( t ) + R1 ( i1 ( t ) + i 2 ( t ) ) = 0 dt KVL and KCL give Use the operator method to get L1s i1 = R 2 i 2 + L 2 s i 2 L1s i1 + R1 ( i1 + i 2 ) = 0 L1s 2i1 + R1s i1 + R1s i 2 = 0 s ( R 2i 2 + L 2 s i 2 ) + R1 L1 (R i 2 2 + L 2 s i 2 ) + R1s i 2 = 0 ⎛ ⎞ L2 R1 R 2 L 2 s 2 i 2 + ⎜ R 2 + R1 + R1 ⎟ s i 2 + i2 = 0 ⎜ ⎟ L L 1 1 ⎝ ⎠ ⎛ R 2 R1 R1 ⎞ R1 R 2 s 2i 2 + ⎜ + + ⎟ s i2 + i2 = 0 ⎜ L 2 L 2 L1 ⎟ L1 L 2 ⎝ ⎠ so ⎛ R 2 R1 R1 ⎞ d R1 R 2 d2 i t + + + ⎟ i 2 (t ) + i 2 (t ) = 0 2 2( ) ⎜ ⎜ L 2 L 2 L1 ⎟ dt dt L L 1 2 ⎝ ⎠ t=0 P 9.2-8 The input to the circuit shown in Figure P 9.2-8 is the voltage of the voltage source, vs. The output is the capacitor voltage v2(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. R2 R1 + + + – Hint: Use the operator method. C1 vs v1(t) C2 – R3 Figure P 9.2-8 Solution: After the switch closes, KVL and KCL give d ⎛ d ⎞ v1 ( t ) + R 3 ⎜ C 1 v1 ( t ) + C 2 v 2 ( t ) ⎟ = v s dt ⎝ dt ⎠ KVL gives v 1 ( t ) = R 2 C2 Using the operator method d v 2 (t ) + v 2 (t ) dt v1 + R 3 ( C 1sv1 + C 2 sv 2 ) = v s v1 = R 2C 2 sv 2 + v 2 so (1 + R C s ) v 2 Then 2 v1 = (1 + R 2C2 s ) v 2 2 + R 3C 1s (1 + R 2C 2 s ) v 2 + R 3C 2 sv 2 = v s R 2 R 3C 1C 2 s 2 v 2 + ( R 2C 2 + R 3C 1 + R 3C 2 ) sv 2 + v 2 = v s s 2v 2 + R 2C 2 + R 3C 1 + R 3C 2 R 2 R 3C 1C 2 sv 2 + vs 1 v2 = R 3 R 2C 1C 2 R 2 R 3C 1C 2 ⎛ 1 vs 1 1 ⎞ 1 + + s 2v 2 + ⎜ sv 2 + v2 = ⎟ ⎜ R 3C 1 R 2 C 2 R 2 C 1 ⎟ R 2 R 3C 1C 2 R 2 R 3C1C 2 ⎝ ⎠ so ⎛ 1 1 1 ⎞d 1 d2 = 2 v 2 (t ) + ⎜ + + v (t ) ⎟⎟ v ( t ) + ⎜ R 2 R 3C 1C 2 dt R 2 R 3C 1C 2 ⎝ R 3C1 R 2C 2 R 2C 1 ⎠ dt vs v2(t) – t=0 P 9.2-9 The input to the circuit shown in Figure P 9.2-9 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. L + – vs R1 R2 Hint: Use the direct method. C Figure P 9.2-9 Solution: After the switch closes i (t ) = C d v (t ) dt KCL and KVL give ⎛ 1 ⎛ d d ⎞⎞ vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t ) ⎜ R1 ⎝ dt dt ⎠ ⎟⎠ ⎝ Substituting gives ⎛ R2 ⎞ ⎛ R2 ⎞ d2 d v s = ⎜1 + LC v ( t ) + R 2C v ( t ) + ⎜ 1 + ⎟ ⎟⎟ v ( t ) 2 ⎜ ⎟ ⎜ R dt dt R 1 ⎠ 1 ⎠ ⎝ ⎝ ⎛ R2 ⎞ ⎛ R2 ⎞ d2 d = ⎜1 + LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 + ⎟ ⎟ v (t ) ⎜ ⎜ R1 ⎟⎠ dt dt R1 ⎟⎠ ⎝ ⎝ Finally R1v s LC ( R1 + R 2 ) = R1 R 2 1 d2 d v (t ) + v (t ) + v (t ) dt LC L ( R1 + R 2 ) dt i(t) + v(t) – P 9.2-10 The input to the circuit shown in Figure P 9.2-10 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Find a Thévenin equivalent circuit. ia i(t) t=0 L R1 + – vs bia + R2 C v(t) – Figure P 9.2-10 Solution: Find the Thevenin equivalent circuit for the part of the circuit to the left of the inductor. v s − v oc ⎫ ⎪ R1 ⎪ v s R 2 (1 + b ) ⎬ ⇒ v oc = v oc ⎪ R1 + R 2 (1 + b ) ia + bia = R 2 ⎪⎭ ia = i sc = i a (1 + b ) = vs R1 (1 + b ) v s R 2 (1 + b ) Rt = v oc i sc = R1 + R 2 (1 + b ) R1 R 2 = vs R1 + R 2 (1 + b ) (1 + b ) R1 Rt i (t ) + L d i (t ) + v ( t ) − v oc = 0 dt i (t ) = C d v (t ) d 2 v (t ) Rt C + LC + v ( t ) = v oc dt d t2 Finally, ⇒ d v (t ) dt d 2 v (t ) Rt d v (t ) v (t ) 1 + + v t = ( ) d t2 L dt LC LC R1 R 2 d 2 v (t ) d v (t ) v (t ) 1 + + v (t ) = 2 dt LC LC L ( R1 + R 2 (1 + b ) ) d t P 9.2-11 The input to the circuit shown in Figure P 9.2-11 is the voltage of the voltage source, vs(t). The output is the voltage v2(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input. – R1 R2 + + + – vs(t) v1(t) C1 – Hint: Use the direct method. Figure P 9.2-11 Solution: KCL gives v s ( t ) − v1 ( t ) = C1 d v1 ( t ) dt ⇒ v s ( t ) = R1C 1 d v1 ( t ) + v1 ( t ) dt = C2 d v 2 (t ) dt ⇒ v1 ( t ) = R 2C 2 d v 2 (t ) + v 2 (t ) dt R1 and v1 ( t ) − v 2 ( t ) R2 Substituting gives v s ( t ) = R1C 1 d ⎡ d d ⎤ R 2C 2 v 2 ( t ) + v 2 ( t ) ⎥ + R 2 C 2 v 2 ( t ) + v 2 ( t ) ⎢ dt ⎣ dt dt ⎦ so ⎛ 1 1 d2 1 ⎞ 1 vs (t ) = 2 v 2 (t ) + ⎜ + v 2 (t ) ⎟⎟ v 2 ( t ) + ⎜ R1 R 2C 1C 2 dt R1 R 2C 1C 2 ⎝ R1C 1 R 2C 2 ⎠ + C2 v2(t) – C2 P 9.2-12 The input to the circuit shown in Figure P 9.2-12 is the voltage of the voltage source, vs(t). The output is the voltage vo(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input. – v2(t) + R1 R2 C1 – Hint: Use the operator method. + – + v1(t) – + vs(t) + vo(t) – Figure P 9.2-12 Solution: d v1 ( t ) + v1 ( t ) dt v 2 (t ) d d C 1 v1 ( t ) + C 2 v 2 ( t ) + =0 dt dt R2 v s ( t ) = R1C 1 KVL gives KCL gives v o (t ) = v 2 (t ) KVL gives Using the operator method v s = R1C 1sv1 + v1 C 1sv1 + C 2 sv 2 + v2 R2 =0 Solving ⎛ C2 ⎞ 1 v1 = − ⎜ v2 + v2 ⎟ ⎜ C1 R 2C 1s ⎟⎠ ⎝ ⎛ C2 1 ⎞ sv s = ( sR1C 1 + 1) ⎜ s+ ⎟ vo ⎜ C1 R 2C 1 ⎟⎠ ⎝ ⎛ 1 1 1 ⎞ 1 sv s = s 2 v o + ⎜ vo + ⎟⎟ sv o + ⎜ R1C 2 R C R C R R C C 2 2 ⎠ 1 2 1 2 ⎝ 1 1 The corresponding differential equation is ⎛ 1 1 d d2 1 ⎞d 1 vs (t ) = 2 v o (t ) + ⎜ + v o (t ) ⎟⎟ v o ( t ) + ⎜ R1C 2 dt dt R1 R 2C 1C 2 ⎝ R1C 1 R 2C 2 ⎠ dt t=0 P 9.2-13 The input to the circuit shown in Figure P 9.2-13 is the voltage of the voltage source, vs(t). The output is the voltage vo(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input. Hint: Use the direct method. – v(t) + C R1 i(t) – + – + vs(t) + L R2 vo(t) – Figure P 9.2-13 Solution: After the switch opens, KCL gives vs (t ) R1 +C d v (t ) = 0 dt KVL gives v (t ) − v o (t ) = L and Ohm’s law gives d i (t ) dt v o (t ) = R2 i (t ) so d 1 v (t ) = − vs (t ) dt R1 C and d d d2 v (t ) − v o (t ) = L 2 i (t ) dt dt dt Then − 1 d d2 d v s (t ) = v (t ) = L 2 i (t ) + R2 i (t ) R1C dt dt dt or R2 d 1 d2 − vs (t ) = 2 i (t ) + i (t ) R1CL dt L dt R2 P 9.2-14 The input to the circuit shown in Figure P 9.2-14 is the voltage of the voltage source, vs(t). The output is the voltage v2(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input. + v1(t) – C1 R1 – + – Hint: Use the direct method. R3 + + vs(t) C2 v2(t) – Figure P 9.2-14 Solution: KCL gives vs (t ) R1 and = v1 ( t ) R2 v 2 ( t ) + v1 ( t ) R3 + C1 + C2 d v1 ( t ) dt d v 2 (t ) = 0 dt so v1 ( t ) + R 2C 1 R2 d v1 ( t ) = vs (t ) dt R1 and d ⎛ ⎞ v 1 ( t ) = − ⎜ v 2 ( t ) + R 3C 2 v 2 ( t ) ⎟ dt ⎝ ⎠ Substituting gives R2 ⎡ d d ⎡ d ⎤⎤ ⎢ v 2 ( t ) + R 3C 2 dt v 2 ( t ) + R 2C 1 dt ⎢v 2 ( t ) + R 3C 2 dt v 2 ( t ) ⎥ ⎥ = − R v s ( t ) ⎣ ⎦⎦ ⎣ 1 or ⎛ 1 d2 1 ⎞d 1 1 v t + + v 2 (t ) + v 2 (t ) = − v s (t ) ⎟ 2 2( ) ⎜ ⎜ R 2C 1 R 3C 2 ⎟ dt dt R R C C R R C C 2 3 1 2 1 3 1 2 ⎝ ⎠ P 9.2-15 Find the second-order differential equation for i2 for the circuit of Figure P 9.2-15 using the operator method. Recall that the operator for the integral is 1/s. Answer: 3 1Ω vs + – i1 d 2 vs d 2i2 di2 + + i = 4 2 2 dt 2 dt dt 2 Apply KVL to the left mesh : i1 + s(i1 −i2 ) = vs (1) dt ⎛1⎞ Apply KVL to the right mesh : 2i2 + 2 ⎜ ⎟ i2 ⎝s⎠ ⎛1⎞ ⇒ i1 = 2⎜ ⎟i2 + ⎝s⎠ + s(i2 − i1 ) = 0 ⎛1⎞ 2⎜ 2 ⎟i2 + i2 ⎝s ⎠ (2) Plugging (2) into (1) yields 3s 2i + 4si + 2i = s 2v 2 2 2 s or 1H i2 Figure P 9.2-15 Solution: where s = d 2Ω d 2v d 2i di s 2 2 + 4 + 2i = 3 2 2 2 dt dt dt 1 2 F Section 9-3: Solution of the Second Order Differential Equation - The Natural Response P 9.3-1 Find the characteristic equation and its roots for the circuit of Figure P 9.2-2. 10 Ω is 10 μ F 100 Ω 1 mH iL Figure P 9.2-2 Solution: From Problem P 9.2-2 the characteristic equation is −11000 ± (11000)2 − 4(1.1×108 ) 1.1× 108 + 11000s + s 2 = 0 ⇒ s , s = = −5500 ± j8930 1 2 2 100 mH P 9.3-2 Find the characteristic equation and its roots for the circuit of Figure P 9.3-2. 2 4 Answer: s + 400s + 3 × 10 = 0 roots: s = – 300, – 100 iL is 40 Ω + vc – 1 3 mF Figure P 9.3-2 Solution: KVL: 40 (is − iL ) = 100 ×10−3 diL + vc dt ⎛1 ⎞ dv i L = i c = ⎜ × 10−3 ⎟ c ⎝3 ⎠ dt di 40 di 100 d 2iL 40 × 10−3 s − ×10−3 L − × 10−6 dt 3 dt 3 3 dt 2 di d 2iL di + 400 L +30000i L = 400 s 2 dt dt dt 2 s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100, iL = s2 = −300 P 9.3-3 Find the characteristic equation and its roots for the circuit shown in Figure P 9.3-3. 2Ω 1Ω vs + vc – + – 1 mH iL Figure P 9.3-3 Solution: v − vs dv + i L + 10 × 10−6 = 0 1 dt di KVL: v = 2i L +10−3 L dt KCL: 2 diL diL −6 −6 −3 d iL 0 = 2iL + 10 − vs + iL + 10 × 10 ⋅ 2 + 10 × 10 × 10 dt dt dt 2 di d i vs = 3iL + .00102 L + 1× 10−8 2L dt dt 2 d iL di + 102000 L + 3 × 10−8 iL = 1× 108 vs dt dt 2 s + 102000s + 3 ×108 = 0, ∴ s1 = 3031, s2 = − 98969 −3 10 μ F P 9.3-4 German automaker Volkswagen, in its bid to make more efficient cars, has come + 10u(t) V – 1 2H up with an auto whose engine saves energy by 5 mF shutting itself off at stoplights. The “stop– start” system springs from a campaign to 10 Ω + – 7u(t) V develop cars in all its world markets that use less fuel and pollute less than vehicles now on the road. The stop–start transmission control Figure P 9.3-4 has a mechanism that senses when the car does not need fuel: coasting downhill and idling at an intersection. The engine shuts off, but a small starter flywheel keeps turning so that power can be quickly restored when the driver touches the accelerator. A model of the stop–start circuit is shown in Figure P 9.3-4. Determine the characteristic equation and the natural frequencies for the circuit. Answer: s2 + 20s + 400 = 0 s = – 10 ± j17.3 Solution: Assume zero initial conditions 1 di1 1 di2 − = 10 − 7 2 dt 2 dt 1 di1 1 di2 + + 200 ∫ i2 dt = 7 loop 2 : − 2 dt 2 dt ⎡⎛ ⎤ 1 ⎞ 1 − s ⎢⎜ 10 + 2 s ⎟ ⎥ 2 ⎝ ⎠ ⎥ determinant : ⎢ ⎢ 1 1 200 ⎛ ⎞⎥ ⎜ s+ ⎟ ⎢ −2s s ⎠ ⎥⎦ ⎝2 ⎣ s 2 + 20s + 400 = 0, ∴ s = − 10 ± j 17.3 loop 1 : 10i1 + Section 9.4: Natural Response of the Unforced Parallel RLC Circuit P 9.4-1 Determine v(t) for the circuit of Figure P 9.4-1 when L = 1 H and vs = 0 for t ≥ 0. The initial conditions are v(0) = 6 V and dv/dt(0) = – 3000 V/s. L vs(t) + – + 80 Ω v(t) – Answer: v(t) = – 2e–100t + 8e–400t V Figure P 9.4-1 Solution: v ( 0 ) = 6, dv ( 0 ) = −3000 dt Using operators, the node equation is: Csv + v + R So the characteristic equation is: s 2 + So v ( t ) = Ae −100t + Be−400t ( v −vs ) sL L ⎛ ⎞ = 0 or ⎜ LCs 2 + s + 1⎟ v = v s ⎝ R ⎠ 1 1 s+ = 0 RC LC ⇒ s1,2 = − 250 ± 2502 − 40, 000 = − 100, − 400 v ( 0) = 6 = A + B dv ( 0 ) ⎫ A = −2 = − 3000 = − 100 A − 400 B ⎬ dt ⎭ B = 8 ∴ v ( t ) = − 2e −100t + 8e−400t t>0 25μ F P 9.4-2 An RLC circuit is shown in Figure P 9.4-2, where v(0) = 2 V. The switch has been open for a long time before closing at t = 0. Determine and plot v(t). t=0 + 1 3F v(t) – 3 4 Ω Figure P 9.4-2 Solution: v ( 0 ) = 2, i ( 0 ) = 0 Characteristic equation s 2 + v ( t ) = Ae − t + Be−3t 1 1 s+ = 0 ⇒ s 2 + 4s + 3 = 0 ⇒ s = − 1, − 3 RC LC Use eq. 9.5 − 12 ⇒ s1 A + s2 B = − −1A − 3B = − also have v ( 0 ) = 2 = A + B 2 −0 = − 8 1 4 From (1) & ( 2 ) get A = −1, B = 3 ∴ v ( t ) = −e − t + 3 e −3t V v ( 0) i ( 0) − RC C (1) ( 2) 1H 2H P 9.4-3 Determine i1(t) and i2(t) for the circuit of Figure P 9.4-3 when i1(0) = i2(0) = 11 A. 1 Ω i2 i1 3H 2 Ω Figure P 9.4-3 Solution: di1 di −3 2 = 0 dt dt di di KVL : − 3 1 + 3 2 + 2i2 = 0 dt dt KVL : i1 + 5 (1) ( 2) in operator form (1 + 5s ) i1 + ( −3s ) i2 = 0⎫⎪ ⎬ ( −3s ) i1 + ( 3s + 2 ) i2 = 0 ⎪⎭ Thus i1 ( t ) = Ae i 2 ( t ) = Ce -t −t 6 6 thus Δ = (1 + 5s )( 3s + 2 ) − 9s 2 = 6s 2 + 13s + 2 = 0 + Be −2t + De-2t Now i1 ( 0 ) = 11 = A + B; i2 ( 0 ) = 11 = C + D from (1) & ( 2 ) get di1 ( 0 ) di2 ( 0 ) A C 33 143 = − = − − 2B ; = − = − − 20 dt 2 6 dt 6 6 which yields A = 3, B = 8, C = − 1, 0 = 12 i1 (t ) = 3e − t /6 + 8e−2t A & i2 (t ) = − e− t /6 + 12e−2t A ⇒ s = − 1 −2 6, P 9.4-4 The circuit shown in Figure P 9.4-4 contains a switch that is sometimes open and sometimes closed. Determine the damping factor, α, the resonant frequency, ω0, and the damped resonant frequency, ωd, of the circuit when (a) the switch is open and (b) the switch is closed. 10 Ω 40 Ω i(t) 2H + – + 50 Ω 20 V v(t) 5 mF – Figure P 9.4-4 Solution: Represent this circuit by a differential equation. (R1 = 50 Ω when the switch is open and R1 = 10 Ω when the switch is closed.) Use KCL to get i (t ) = v (t ) d + C v (t ) R2 dt Use KVL to get v s = R1 i ( t ) + L d i (t ) + v (t ) dt Substitute to get vs = R1 R2 v ( t ) + R1C d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt ⎛ R1 + R 2 d2 L ⎞d = CL 2 v ( t ) + ⎜ R1C + v (t ) + v (t ) ⎟ ⎜ dt R 2 ⎟⎠ dt R2 ⎝ Finally, Compare to to get vs CL = ⎛ R1 R1 + R 2 d2 1 ⎞d v t + + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ L R 2C ⎟ dt dt R CL 2 ⎝ ⎠ d2 d v ( t ) + ω 02 v ( t ) = f (t ) dt dt R1 R1 + R 2 1 + 2α = and ω 0 2 = L R 2C R 2CL v t + 2α 2 ( ) (a) When the switch is open α = 14.5 , ω 0 = 14.14 rad/s and ω d = j3.2 (the circuit is overdamped). (b) When the switch is closed α = 4.5 , ω 0 = 10.954 rad/s and ω d = 9.987 (the circuit is underdamped). P 9.4-5 The circuit shown in Figure P 9.4-5 is used to detect smokers in airplanes who surreptitiously light up before they can take a single puff. The sensor activates the switch, and the change in the voltage v(t) activates a light at the flight attendant’s station. Determine the natural response v(t). Answer: v(t) = – 1.16e–2.7t + 1.16e–37.3t V Sensor t=0 Light bulb 0.4 H 1Ω 1A + 1 40 F v(t) – Figure P 9.4-5 Solution: 1 1 s + = 0 RC LC s 2 + 40 s + 100 = 0 s2 + s = − 2.7 , − 37.3 The initial conditions are v(0) = 0 , i (0) 1 A . vn = A1e −2.7 t + A2 e−37.3t , v(0) = 0 = A1 + A2 KCL at t = 0+ yields : (1) v(0+ ) 1 dv(0+ ) + i (0+ ) + =0 1 40 dt dv(0+ ) = − 40v(0+ ) − 40i (0+ ) = − 40(1) = − 2.7 A1 − 37.3 A2 dt from (1) and (2) ⇒ A1 = −1.16 , A2 = 1.16 ∴ So v(t ) = vn (t ) = − 1.16e −2.7 t + 1.16e −37.3t (2) Section 9.5: Natural Response of the Critically Damped Unforced Parallel RLC Circuit 25 mH P 9.5-1 Find vc(t) for t > 0 for the circuit shown in Figure P 9.5-1. Answer: vc(t) = (3 + 6000t)e –2000t 30u(–t) mA 100 Ω V vc Figure P 9.5-1 Solution: t >0 di c dv + v c = 0, ic = 10−5 c dt dt 2 d vc dv ∴ + 4000 c + 4 × 106 vc = 0 2 dt dt 2 6 s + 4000 s + 4 × 10 = 0 ⇒ s = −2000, − 2000 ∴ vc ( t ) = A1e −2000t + A2te −2000t KVL a : 100ic + .025 t = 0− (Steady − State) iL = ic ( 0 − )=0 = ic ( 0 + vc ( 0− ) = 3 V = vc ( 0+ ) so vc ( 0+ ) = 3 = A1 dvc ( 0+ ) dt vc ( t ) = = 0 = −2000 A1 + A2 ⇒ A2 = 6000 ( 3 + 6000t ) e−2000t V ) ⇒ dvc ( 0+ ) dt = 0 + – 10 mF P 9.5-2 Find vc(t) for t > 0 for the circuit of Figure P 9.5-2. Assume steady-state 20 V conditions exist at t = 0–. t=0 10 Ω + – 1 Ω 1H + 1 4 Answer: vc(t) = – 8te–2t V F vc – Figure P 9.5-2 Solution: ( ) dvc t =0 KCL at vc : ∫−∞ vc dt + vc + 1 4 dt d 2 vc dv ⇒ + 4 c + 4vc = 0 dt dt t>0 s 2 + 4 s + 4 = 0, s = −2, − 2 t = 0− (Steady-State) ⇒ vc ( t ) = A1 e −2t + A2 t e −2t vc ( 0− ) = 0 = vc ( 0+ ) & iL ( 0− ) = 20 V = 2 A = iL ( 0+ ) 10 Ω Since vc ( 0+ ) = 0 then ic ( 0+ ) = −iL ( 0+ ) = −2 A ∴ So vc ( 0 ) dv ( 0 ) + = 0 = A1 + = − 8 = A2 dt ∴ vc ( t ) = − 8te −2t V c dvc ( 0+ ) dt ic ( 0+ ) = = −8 V S 1 4 10 mH P 9.5-3 Police often use stun guns to incapacitate potentially dangerous felons. The + t=0 hand-held device provides a series of high+ 4 v R = 106 Ω 10 V – voltage, low-current pulses. The power of the C – pulses is far below lethal levels, but it is enough to cause muscles to contract and put Figure P 9.5-3 the person out of action. The device provides a pulse of up to 50,000 V, and a current of 1 mA flows through an arc. A model of the circuit for one period is shown in Figure P 9.5-3. Find v(t) for 0 < t < 1 ms. The resistor R represents the spark gap. Select C so that the response is critically damped. Solution: Assume steady − state at t = 0− ∴ vc ( 0− ) = 10 4 V & iL ( 0− ) = 0 t>0 Also ∴ 0.01C diL + 106 iL = 0 (1) dt ⎡ d 2iL di ⎤ dvc : iL = −C = −C ⎢.01 2 + 106 L ⎥ dt dt dt ⎦ ⎣ KVL a : − vc + .01 ( 2) d 2iL di + 106 C L + iL = 0 2 dt dt Characteristic eq. ⇒ 0.01C s 2 + 106 s + 1 = 0 ⇒ s = −106 C ± (10 C ) 6 2 (.01C ) for critically damped: 1012 C2 − .04C = 0 ⇒ C = 0.04 pF ∴ s = −5 × 107 , −5 × 107 So iL ( t ) = A1e −5×10 t + A2te−5×10 7 7t diL + 0 ) = 100 ⎡⎣vc ( 0+ ) − 106 iL ( 0+ ) ⎤⎦ = 106 A ( s dt di ( 0 ) 7 So i L ( 0 ) = 0 = A1 and L = 106 = A2 ∴ iL ( t ) = 106 te−5×10 t A dt Now from (1) ⇒ Now v ( t ) = 106 iL ( t ) = 1012 te−5×10 t V 7 2 − 4 (.01C ) P 9.5-4 Reconsider Problem P 9.4-1 when L = 640 mH and the other parameters and conditions remain the same. vs(t) – 250t Answer: v(t) = (6 – 1500t)e V L + – + 80 Ω v(t) – Figure P 9.4-1 Solution: s2 + 1 1 1 1 s+ = 0 with = 500 and = 62.5 × 103 yields s = −250, −250 RC LC RC LC v ( t ) = Ae −250t + Bte −250t v ( 0) = 6 = A dv ( 0 ) = −3000 = − 250 A + B ⇒ B = − 1500 dt ∴ v ( t ) = 6e −250t − 1500te −250t 25μ F P 9.5-5 An automobile ignition uses an electromagnetic trigger. The RLC trigger circuit shown in Figure P 9.5-5 has a step input of 6 V, and v(0) = 2 V and i(0) = 0. The resistance R must be selected from 2 Ω < R < 7 Ω so that the current i(t) exceeds 0.6 A for greater than 0.5 s in order to activate the trigger. A critically damped response i(t) is required to avoid oscillations in the trigger current. Select R and determine and plot i(t). 1H Trigger di t + Ri + 2 + 4∫0 idt = 6 dt v (t ) Figure P 9.5-5 (1) taking the derivative with respect to t : d 2i di + R + 4i = 0 2 dt dt Characteristic equataion: s 2 + Rs + 4 = 0 Let R = 4 for critical damping ⇒ ( s + 2) 2 =0 So i ( t ) = Ate −2t + Be −2t i ( 0) = 0 ⇒ B = 0 di ( 0 ) = 4 − R (i ( 0)) = 4 − R ( 0) = 4 = A dt = 4te −2t A from (1) ∴ i (t ) F + – v(t) 6 u(t) V +– Solution: KVL : 1 4 i R Section 9-6: Natural Response of an Underdamped Unforced Parallel RLC Circuit P 9.6-1 A communication system from a space station uses short pulses to control a robot operating in space. The transmitter circuit is modeled in Figure P 9.6-1. Find the output voltage vc(t) for t > 0. Assume steadystate conditions at t = 0–. 0.8 H vc 250 Ω + – 5 × 10-6 F 250 Ω t=0 Answer: vc(t) = e– 400t [3 cos 300t + 4 sin 300t] V + – 6V Figure P 9.6-1 Solution: t>0 KCL at vc : also : Solving for i L vc 250 + iL + 5 ×10−6 vc = 0.8 dvc = 0 dt diL dt in (1) & plugging into ( 2 ) d 2 vc dv +800 c + 2.5×105 vc = 0 ⇒ s 2 +800s+ 250,000 = 0, s = −400± j 300 2 dt dt ∴ v (t ) = e c t = 0− −400t ⎡ ⎤ ⎣ A1 cos300t + A 2 sin 300t ⎦ (Steady − State) −6 V −6 = A = iL ( 0+ ) 500 500 Ω − + 6 = 3 V = vc ( 0+ ) vc ( 0 ) = 250 −6 500 iL ( 0− ) = ( Now from (1) : So vc ( 0 + ) dvc ( 0+ ) ) dvc ( 0+ ) dt = 3 = A1 = − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0 = 0 = − 400 A1 + 300 A2 ⇒ A2 = 4 dt ∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V ( 2) (1) P 9.6-2 The switch of the circuit shown in Figure P 9.6-2 is opened at t = 0. Determine and plot v(t) when C = 1/4 F. Assume steady state at t = 0–. Answer: v(t) = – 4e–2t sin 2t V 3Ω 6V t=0 + – 1Ω + 1 2 H C – v(t) Figure P 9.6-2 Solution: t = 0− i (0) = 2 A v (0) = 0 t = 0− KCL at node a: t v dv 1 + C + ∫ vdt + i ( 0 ) = 0 (1) dt L 0 1 in operator form have v + Csv + with s 2 + 4s + 8 = 0 1 v + i ( 0 ) = 0 or Ls 1 ⎞ ⎛ 2 1 ⎜s + s+ ⎟v = 0 C LC ⎠ ⎝ ⇒ s = −2 ± j 2 v ( t ) = e −2t [ B1 cos 2t + B2 sin 2t ] v ( 0 ) = 0 = B1 dv ( 0 ) 1 ⎡ −i ( 0 ) − v ( 0 ) ⎤⎦ = −4 [ 2] = −8 = 2 B2 or B2 = −4 dt C ⎣ So v ( t ) = −4e −2t sin 2 t V From (1) , = P 9.6-3 A 240-W power supply circuit is shown in Figure P 9.6-3a. This circuit employs a large inductor and a large capacitor. The model of the circuit is shown in Figure P 9.6-3b. Find iL(t) for t > 0 for the circuit of Figure P 9.6-3b. Assume steady-state conditions exist at t = 0–. Answer: iL(t) = e–2t(– 4 cos t + 2 sin t) A 4H iL t=0 14 2Ω F 8Ω 4Ω 7A (b) (a) Figure P 9.6-3 Solution: t>0 1 dvc vc + + iL = 0 4 dt 2 4diL KVL : vc = + 8 iL ( 2) dt KCL at vc : d 2iL di + 4 L + 5iL = 0 ⇒ s 2 + 4s + 5 = 0 ⇒ s = − 2 ± i 2 dt dt −2 t = e [ A1 cos t + A2 sin t ] ( 2 ) into (1) yields ∴ iL ( t ) t = 0− ( Steady − State ) vc ( 0− ) 2 ⎛ 48 ⎞ = 7 ⎜⎜ ⎟⎟ ⎝ 4 8+ 2⎠ ⇒ vc ( 0− ) = 8 V = vc ( 0+ ) −8 V = −4 A = iL ( 0+ ) 2Ω diL ( 0+ ) vc ( 0+ ) 8V A + = − 2iL ( 0 ) = − 2 ( −4 ) = 10 dt 4 4 s iL ( 0− ) = ∴ from ( 2 ) So i L ( 0 ∴ iL ( t ) + ) = -4 = A 1 and diL ( 0+ ) dt −2 t = e [ −4 cos t + 2 sin t = 10 = −2 A1 + A2 ⇒ A2 = 2 ]A (1) 600 P 9.6-4 The natural response of a parallel RLC circuit is measured and plotted as shown in Figure P 9.6-4. Using this chart, determine an expression for v(t). 500 400 300 200 Hint: Notice that v(t) = 260 mV at t = 5 ms and that v(t) = – 200 mV at t = 7.5 ms. Also, notice that the time between the first and third zerocrossings is 5 ms. v(t) (mV) 100 0 – 100 Answer: v(t) = 544e– 276t sin 1257t V – 200 – 300 – 400 0 5 10 15 20 25 Time (ms) Figure P 9.6-4 Solution: The response is underdamped so ∴ v ( t ) = e − α t [ k1 cos ωt + k2 sin ωt ] + k3 v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 ∴ v ( t ) = k2 e − α t sin ωt From Fig. P 9.6-4 t ≈ 5ms ↔ v ≈ 260mV (max) t ≈ 7.5ms ↔ v ≈ −200 mV (min) ∴ distance between adjacent maxima is ≈ ω = so 0.26 = k2 e −0.2 = k2 e −α (.005 ) −α 2π = 1257 rad s T sin (1257 (.005 ) ) (.0075) sin (1) ( 1257 (.0075) ) ( 2 ) Dividing (1) by (2) gives ⎛ sin ( 6.29 rad ) ⎞ − 1.3 = eα ( 0.0025) ⎜⎜ ⎟⎟ ⇒ ⎝ sin ( 9.43 rad ) ⎠ From (1) k 2 = 544 so v ( t ) = 544e−267t sin1257t e0.0025 α = 1.95 ⇒ α = 267 ( approx. answer ) 30 P 9.6-5 The photovoltaic cells of the proposed space station shown in Figure P 9.6-5a provide the voltage v(t) of the circuit shown in Figure P 9.6-5b. The space station passes behind the shadow of earth (at t = 0) with v(0) = 2 V and i(0) = 1/10 A. Determine and sketch v(t) for t > 0. Photocells (a) i 5Ω + v 1 10 F 2H – Space station electric motors The photovoltaic cells connected in parallel (b) Figure P 9.6-5 Solution: v ( 0 ) = 2 V and i ( 0 ) = 1 A 10 Char. eq. ⇒ s 2 + 1 1 s+ = 0 or s 2 + 2 s + 5 = 0 thus the roots are s = −1 ± j 2 RC LC So have v(t ) = e − t ⎡ B cos 2t + B2 sin 2t ⎤ now v(0+ ) = 2 = B1 ⎣ 1 ⎦ Need So dv ( 0+ ) dv ( 0 dt dt + v ( 0+ ) 1 1 V + + = ic ( 0 ) . KCL yields ic ( 0 ) = − − i ( 0+ ) = − C 5 2 s ) = 10 ⎛ − 1 ⎞ = − B + 2 B ⎜ ⎟ ⎝ 2⎠ 1 2 ⇒ B2 = − 3 2 3 Finally, we have v ( t ) = 2e − t cos 2t − e− t sin 2t V 2 t>0 Section 9-7: Forced Response of an RLC Circuit P 9.7-1 Determine the forced response for the inductor current if when (a) is = 1 A, (b) is = is u(t) A 0.5t A, and (c) is = 2e–250t A for the circuit of Figure P 9.7-1. 100 65 Ω 10 mH i Figure P 9.7-1 Solution: v dv + iL + C R dt diL KVL : v = L dt d 2iL L diL + iL + LC is = R dt dt 2 KCL : is = (a) is = l u (t ) ∴ assume i f = A d 2iL 1 diL 1 iL = is + + 2 dt RC dt LC 1 to get: 0 + 0 + A = 1 ⇒ A = 1× 10−5 = i f −3 .01 1 10 × ( )( ) Let iL = i f = A in (b) is = 0.5t u (t ) ∴ assume i f = At + B 0+ A 65 1 + ( At + B ) = 0.5 t (.01)(.001) (100 ) (.001) ⇒ 650 A + 100000 B = 0 and 100000 At = 0.5t A = 5 ×10−6 B = 3.25 × 10−8 i f = 5 × 10−6 t − 3.25 × 10−8 A (c) is = 2e −250t Assumming i f = Ae −250t does not work because i f cannot have the same form as is ∴ we choose i f = Bte−250t Be−250t −250 Bte−250t Bte−250t + + = 2 e −250t RC RC LC 150 B = 2 B = 0.0133 if = 0.0133 te −250t A 1 mF 7Ω P 9.7-2 Determine the forced response for the capacitor voltage, vf, for the circuit of Figure P 9.7-2 when vs u(t) V –+ 0.1 H + v – 833.3 μ F (a) vs = 2 V, (b) vs = 0.2t V, and (c) vs = 1e–30t V. Figure P 9.7-2 Solution: Represent the circuit by the differential equation: (a) vs = 2 ∴ assume v f = A Then 0 + 0 + 12000 A = 2 so A = 1 (b) d 2 v R dv 1 v = vs + + dt L dt LC 6000 = vf vs = 0.2t ∴ assume v f = At + B 70 A + 12000 At + 12000 B = 0.2t ⇒ 70 A + 12000 B = 0 and 12000 At = 0.2t 1 70 A , B= 60000 12000 t ∴v f = + 350 V 60000 A= (c) vs = e −30t ⇒ B = 350 ∴ assume v f = Ae−30t 900 A − 2100 Ae −30t + 12000 Ae−30t = e−30t vf = e −30t V 10800 ⇒ 10800 Ae−30t = e −30t ⇒ A = 1 10800 P 9.7-3 A circuit is described for t > 0 by the equation d 2v dv + 5 + 6v = vs 2 dt dt Find the forced response vf for t > 0 when (a) vs = 8 V, (b) vs = 3e–4t V, and (c) vs = 2e–2t V. 3 Answer: (a) vf = 8/6 V (b) vf = e– 4t V (c) vf = 2te–2t V 2 Solution: (a) (b) d 2v dv + 5 + 6v = 8 so we try v f = B. 2 dt dt Substituting v f = B into the differential equation gives 6 B = 8 ∴ vf = 8 / 6 V The differential equation is d 2v dv + 5 + 6v = 3e −4t so we try v f = Be −4t . 2 dt dt into the differential equation gives The differential equation is Substituting v f = Be −4t (−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = 3 / 2 ∴ vf = 3 / 2e −4t (c) d 2v dv + 5 + 6v = 2e −2t so we try vf = Bte−2t 2 dt dt (since − 2 is a natural frequency). Substituting vf = Bte −2t into the differential equation gives The differential equation is ⇒ (4t − 4) B + 5B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2 ∴ vf = 2te −2t Section 9-8: Complete Response of an RLC Circuit P 9.8-1 Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-1. iL i 2 kΩ + vc 1 kΩ – 11 mA t=0 6.25 H 1 μF + – 4V Figure P 9.8-1 Solution: First, find the steady state response for t < 0, when the switch is open. Both inputs are constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. After a source transformation at the left of the circuit: i L ( 0) = and 22 − 4 = 6 mA 3000 v C ( 0) = 4 V After the switch closes Apply KCL at node a: vC R +C d vC + iL = 0 dt Apply KVL to the right mesh: L d d i L + Vs − vC = 0 ⇒ vC = L i L + Vs dt dt After some algebra: Vs d2 1 d 1 d2 d 16 ⎛ 4 ⎞ i + i + i = − ⇒ i + (103 ) i L + ⎜ × 106 ⎟ i L = − × 103 L L 2 L 2 L dt R C dt LC R LC dt dt 25 ⎝ 25 ⎠ The characteristic equation is ⎛ 4 ⎞ s 2 + (103 ) s + ⎜ ×106 ⎟ = 0 ⇒ s1,2 = −200, − 800 rad/s ⎝ 25 ⎠ After the switch closes the steady-state inductor current is iL(∞) = -4 mA so i L ( t ) = −0.004 + A1 e −200 t + A2 e −800 t 4 ⎛ 4 ⎞d ⎡⎣( −200 ) A1 e −200 t + ( −800 ) A2 e−800 t ⎤⎦ + 4 vC (t ) = ⎜ ⎟ i L (t ) + 4 = 25 ⎝ 25 ⎠ dt = ( −32 ) A1 e −200 t + ( −128 ) A2 e −800 t + 4 Let t = 0 and use the initial conditions: 0.006 = −0.004 + A1 + A2 ⇒ 0.01 = A1 + A2 4 = ( −32 ) A1 + ( −128 ) A2 + 4 ⇒ A1 = ( −4 ) A2 So A1 = 8.01 and A2 = 2.00 and i L ( t ) = −0.004 + 8.01 e −200 t + 2.00 e −800 t A v C ( t ) = ( −104 ) e −200 t + (104 ) e −800 t + 4 V i (t ) = vC (t ) 1000 = ( −10 ) e −200 t + (10 ) e−800 t + 0.004 A Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-2. d2 d Hint: Show that 1 = 2 i (t ) + 5 i (t ) + 5i (t ) for t > 0 dt dt P 9.8-2 1Ω 2u(t) – 1 V + – 4Ω + v(t) – 0.25 F i(t) 4H Figure P 9.8-2 Answer: –3.62t i(t) = 0.2 + 0.246 e – 0.646 e–1.38t A for t > 0. Solution: First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = −1 = 0.2 A 1+ 4 and v (0) = 4 ( −1) = −0.8 V 1+ 4 For t > 0 Apply KCL at node a: v − Vs d +C v+i = 0 R1 dt Apply KVL to the right mesh: R2 i + L d d i − v = 0 ⇒ v = R2 i + L iL dt dt After some algebra: L + R1 R 2C d R1 + R 2 d2 Vs i + i + i= 2 dt R1 L C dt R1 L C R1 L C The forced response will be a constant, if = B so 1 = ⇒ d2 d i +5 i +5i =1 2 dt dt d2 d B + 5 B + 5B ⇒ B = 0.2 A . 2 dt dt To find the natural response, consider the characteristic equation: 0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 ) The natural response is in = A1 e−3.62 t + A2 e−1.38 t so i ( t ) = A1 e−3.62 t + A2 e−1.38 t + 0.2 Then d ⎛ ⎞ v ( t ) = ⎜ 4 i ( t ) + 4 i ( t ) ⎟ = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8 dt ⎝ ⎠ At t=0+ −0.2 = i ( 0 + ) = A1 + A2 + 0.2 −0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8 so A1 = 0.246 and A2 = -0.646. Finally i ( t ) = 0.2 + 0.246 e−3.62 t − 0.646 e−1.38 t A P 9.8-3 Answer: Determine v1(t) for t > 0 for the circuit shown in Figure P 9.8-3. 4 3 v1 (t ) = 10 + e−2.4 × 10 t − 6e−4 × 10 t V for t > 0 1 kΩ 10 V + – 1 kΩ + v1(t) – 1/6 μ F + v2(t) – t=0 1/16 μ F Figure P 9.8-3 Solution: First, find the steady state response for t < 0. The input is constant so the capacitors will act like an open circuits at steady state. v1 ( 0 ) = and 1000 (10 ) = 5 V 1000 + 1000 v2 ( 0 ) = 0 V For t > 0, Node equations: v1 − 10 ⎛ 1 v −v ⎞ d + ⎜ × 10−6 ⎟ v1 + 1 2 = 0 1000 ⎝ 6 1000 ⎠ dt ⎛1 ⎞ d ⇒ 2 v1 + ⎜ × 10−3 ⎟ v1 − 10 = v2 ⎝6 ⎠ dt v1 − v2 ⎛ 1 ⎞ d = ⎜ × 10−6 ⎟ v2 1000 ⎝ 16 ⎠ dt ⎛1 ⎞ d ⇒ v1 − v2 = ⎜ × 10−3 ⎟ v2 ⎝ 16 ⎠ dt After some algebra: d2 d v + ( 2.8 × 104 ) v1 + ( 9.6 × 107 ) v1 = 9.6 × 108 2 1 dt dt The forced response will be a constant, vf = B so d2 d B + ( 2.8 × 104 ) B + ( 9.6 ×107 ) B = 9.6 × 108 ⇒ B = 10 V . 2 dt dt To find the natural response, consider the characteristic equation: s 2 + ( 2.8 ×104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104 The natural response is 3 vn = A1 e −4×10 t + A2 e−2.4×10 4t so 3 v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10 4t + 10 At t = 0 5 = v1 ( 0 ) = A1 e −4×103 ( 0 ) + A2 e −2.4×104 ( 0 ) + 10 = A1 + A2 + 10 (1) Next ⎛1 ⎞ d 2 v1 + ⎜ ×10−3 ⎟ v1 − 10 = v2 ⎝6 ⎠ dt ⇒ d v1 = 12000v1 + 6000 v2 − 6 × 104 dt At t = 0 d v1 ( 0 ) = 12000v1 ( 0 ) + 6000 v2 ( 0 ) − 6 × 104 = 12000 ( 5 ) + 6000 ( 0 ) − 6 × 104 = 0 dt so 3 4 d v1 ( t ) = A1 −4 ×103 e−4×10 t + A2 −2.4 × 104 e−2.4×10 t dt ( ) ( ) At t = 0+ 0= d −4×103 ( 0 ) −2.4×104 ( 0 ) v1 ( 0 ) = A1 −4 × 103 e + A2 −2.4 × 104 e = A1 −4 × 103 + A2 −2.4 × 104 dt ( ) ( ) ( so A1 = -6 and A2 = 1. Finally v1 ( t ) = 10 + e −2.4 ×10 4t 3t − 6 e −4 ×10 V for t > 0 ) ( ) 1Ω P 9.8-4 Find v(t) for t > 0 for the circuit shown in Figure P 9.8-4 when v(0) = 1 V and iL(0) = 0. + 5 cos t V – Answer: v = 25e −3t − 1 ⎡ 429e−4t − 21cos t + 33 sin t ⎤⎦ V 17 ⎣ 0.5 H iL 1Ω 1 12 F + – v Figure P 9.8-4 Solution: t > 0 di ⎛ ⎞ KCL at top node : ⎜ 0.5 L − 5cos t ⎟ + iL + 1 dv = 0 12 dt dt ⎝ ⎠ di dv KVL at right loop : 0.5 L = 1 +v ( 2) 12 dt dt d d d 2iL diL 1 d 2 v + + = −5sin t (3) dt 2 dt 12 dt 2 d 2iL d 2 v dv 1 ⇒ 0.5 2 = + ( 4) 12 dt 2 dt dt dt of (1) ⇒ 0.5 dt of ( 2 ) Solving for d 2iL di in ( 4 ) and L in ( 2 ) & plugging into ( 3) 2 dt dt d 2v dv + 7 + 12v = − 30sin t 2 dt dt ⇒ s 2 + 7 s + 12 = 0 ⇒ s = − 3, − 4 so v(t ) = A1e−3t + A2 e−4t + v f Try v f = B1 cos t + B2 sin t & plug into D.E., equating like terms yields B1 = 21 , B2 = − 33 17 17 t = 0+ 5 −1 dv(0+ ) ic ( 0 ) = = 2A ∴ = 2 = 24 V 1 s 11 dt 12 ⎫ So v(0+ ) = 1 = A1 + A2 + 21 A1 = 25 17 ⎪ ⎬ 429 dv(0+ ) ⎪ A2 = − = 24 = − 3 A1 − 4 A2 − 33 17 17 dt ⎭ ∴ v(t ) = 25e −3t − 1 ( 429e −4t − 21cos t + 33sin t ) V 17 + (1) Find v(t) for t > 0 for the circuit of Figure P 9.8-5. P 9.8-5 Answer: v(t) = [ – 16e–t + 16e – 3t + 8]u(t) + [16e–(t–2) – 16e–3(t–2) – 8]u(t – 2) V 1 3 F + 2[u(t) – u(t – 2)] A 4Ω v 1H – Figure P 9.8-5 Solution: Use superposition − first consider the 2u (t ) source. KVL at right mesh : vc + siL + 4(iL − 2) = 0 also : iL = (1/ 3) svc ⇒ vc = (3 / s ) iL (1) (2) Plugging (2) into (1) yields ( s 2 + 4s + 3) iL = 0 , roots : s = −1 , −3 So iL (t ) = A1e − t + A2 e−3t − t = 0 ⇒ circuit is dead ∴ vc (0) = iL (0) = 0 Now from (1) diL (0+ ) = 8 − 4iL (0+ ) − vC (0+ ) = 8 A/s dt So iL (0) = 0 = A1 + A2 ⎫ ⎪ ⎬ A1 = 4 , A2 = −4 diL (0) = 8 = − A1 − 3 A2 ⎪ dt ⎭ −t −3t ∴ iL (t ) = 4e − 4e ∴ v1 (t ) = 8 − 4 iL (t ) = 8 − 16e − t + 16e −3t V Now for 2u (t − 2) source, just take above expression and replace t → t − 2 and flip signs ∴ v2 (t ) = −8 + 16e − (t − 2) − 16 e −3(t − 2) V ∴ v(t ) = v1 (t ) + v2 (t ) v(t ) = ⎡⎣8 − 16e − t + 16e −3t ⎤⎦ u (t ) + ⎡⎣ −8 + 16e− (t − 2) − 16 e−3( t − 2) ⎤⎦ u (t − 2) V P 9.8-6 An experimental space station power supply system is modeled by the circuit shown in Figure P 9.8-6. Find v(t) for t > 0. Assume steady-state conditions at t = 0–. t=0 (10 cos t)u(t) V – + 0.125 F 4Ω 4H + v(t) – + – 2Ω 5V i(t) Figure P 9.8-6 Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = − and 5 = −1.25 mA 4 v (0) = 5 V After the switch closes Apply KCL at node a: v d + 0.125 v = i dt 2 Apply KVL to the right mesh: −10 cos t + v + 4 After some algebra: d i+4i =0 dt d2 d v + 5 v + 6 v = 20 cos t 2 dt dt The characteristic equation is s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 rad/s Try vf = A cos t + B sin t d2 d A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t 2 ( dt dt ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t ( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t vf = 2 cos t + 2 sin t So A =2 and B =2. Then v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e−3 t v (t ) d + 0.125 v ( t ) = i ( t ) ⇒ dt 2 Next d v (t ) = 8 i (t ) − 4 v (t ) dt d V ⎛ 5⎞ v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8 ⎜ − ⎟ − 4 ( 5 ) = −30 dt s ⎝ 4⎠ Let t = 0 and use the initial conditions: 5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e−0 = 2 + A1 + A2 d v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t dt −30 = d v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2 dt So A1 = -23 and A2 = 26 and v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t 8Ω P 9.8-7 Find vc(t) for t > 0 in the circuit of Figure P 9.8-7 when (a) C = 1/18 F, (b) C = 1/10 F, and (c) C = 1/20 F. 4Ω 2u(t) A Answer: (a) vc(t) = 8e–3t + 24te – 3t – 8 V (b) vc(t) = 10e–t – 2e–5t – 8 V (c) vc(t) = e – 3t(8 cos t + 24 sin t) – 8 V C 2H a + v(t) – i(t) Figure P 9.8-7 Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = 0 A and v ( 0) = 0 V After the switch closes d v=i dt Apply KVL to the right mesh: d 8 i + v + 2 i + 4 (2 + i) = 0 dt d 12 i + v + 2 i = −8 dt ⎛ 1 ⎞ d2 d 4 After some algebra: v + ( 6) v + ⎜ ⎟v = − 2 dt dt C ⎝2C⎠ The forced response will be a constant, vf = B so Apply KCL at node a: C ⎛ 1 ⎞ d2 d 4 B + ( 6) B + ⎜ ⎟B = − 2 dt dt C ⎝2C⎠ (a) ⇒ B = −8 V d2 d v + ( 6 ) v + ( 9 ) v = −72 2 dt dt 2 The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3 When C = 1/18 F the differential equation is Then v ( t ) = ( A1 + A2 t ) e−3t − 8 . Using the initial conditions: 0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒ 0= So (b) d v ( 0 ) = −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0 dt A1 = 8 ⇒ A2 = 24 v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0 d2 d When C = 1/10 F the differential equation is 2 v + ( 6 ) v + ( 5 ) v = −40 dt dt 2 The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 Then v ( t ) = A1 e − t + A2 e−5 t − 8 . Using the initial conditions: 0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 ⎫ ⎪ ⎬ ⇒ A1 = 10 and A2 = −2 d 0 0 0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 ⎪ dt ⎭ So v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0 (c) d2 d v + ( 6 ) v + (10 ) v = −80 2 dt dt 2 The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j When C = 1/20 F the differential equation is Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 . Using the initial conditions: 0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒ 0= So A1 = 8 d v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒ dt v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0 A2 = 24 P 9.8-8 Find vc(t) for t > 0 for the circuit shown in Figure P 9.8-8. 4Ω 8Ω Hint: d2 d 2 = 2 vc (t ) + 6 vc (t ) + 2vc (t ) for t > 0 dt dt Answer: vc(t) = 0.123e – 5.65t + 0.877e – 0.35t + 1 V for t > 0. – 1 4 u(t) + 1 2 A + 1 4 2H F iL(t) Figure P 9.8-8 Solution: The circuit will be at steady state for t<0: so iL(0+) = iL(0-) = 0.5 A and vC(0+) = vC(0-) = 2 V. For t>0: Apply KCL at node b to get: 1 1 d 1 1 d = i L (t ) + v C (t ) ⇒ i L (t ) = − v C (t ) 4 4 dt 4 4 dt Apply KVL to the right-most mesh to get: 4 i L (t ) + 2 d ⎛1 d ⎞ i L (t ) = 8 ⎜ v C (t ) ⎟ + v C (t ) dt ⎝ 4 dt ⎠ Use the substitution method to get d ⎛1 1 d ⎛1 1 d ⎞ ⎞ ⎛1 d ⎞ 4⎜ − vC ( t ) ⎟ + 2 ⎜ − vC ( t ) ⎟ = 8 ⎜ vc ( t ) ⎟ + v ( t ) dt ⎝ 4 4 dt ⎝ 4 4 dt ⎠ ⎠ ⎝ 4 dt ⎠ c d2 d 2= v C (t ) + 6 v C (t ) + 2 v C (t ) or dt dt 2 d d2 The forced response will be a constant, vC= B so 2 = 2 B + 6 B + 2 B ⇒ B = 1 V . dt dt To find the natural response, consider the characteristic equation: 0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 ) vC(t) – The natural response is vn = A1 e −5.65t + A e −0.35t 2 vC ( t ) = A e so 1 −5.65 t +A e −0.35t 2 +1 Then i L (t ) = At t=0+ 1 1 d 1 + vC ( t ) = + 1.41A e −5.65t + 0.0875 A e−0.35t 1 2 4 4 dt 4 2 = vC ( 0 + ) = A + A + 1 1 2 1 1 = i L ( 0 + ) = + 1.41A1 + 0.0875 A 2 2 4 so A1 = 0.123 and A2 = 0.877. Finally vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t + 1 V P 9.8-9 In Figure P 9.8-9, determine the inductor current i(t) when is = 5u(t) A. Assume that i(0) = 0, vc(0) = 0. is 2Ω Answer: i(t) = 5 + e–2t [ – 5 cos 5t – 2 sin 5t] A 8 29 H 1 8F i Figure P 9.8-9 Solution: dv v +i+ =i 2 s dt d 2i ⎛ L ⎞ di LC + i + ⎜ ⎟ = 5 u (t ) dt ⎝ 2 ⎠ dt KCL : C 1 d 2i ⎛ 4 ⎞ di + i + ⎜ ⎟ = 5 u (t ) 29 dt ⎝ 29 ⎠ d t d 2i di + 4 + i = 145 u ( t ) dt dt Characteristic eqn: s 2 + 4s + 29 = 0 ⇒ roots : s = − 2 ± j5 ∴ in = e −2t [ A cos 5t + B sin 5t ] and i f = 145 = 5 29 −2t So i (t ) = 5 + e [ A cos 5t + B sin 5t ] Now i (0) = 0 = A + 5 ⇒ A = −5 di (0) = 0 = −2 A + 5 B ⇒ B = − 2 dt P 9.8-10 Railroads widely use automatic identification of railcars. When a train passes a tracking station, a wheel detector activates a radio-frequency module. The module’s antenna, as shown in Figure P 9.8-10a, transmits and receives a signal that bounces off a transponder on the locomotive. A trackside processor turns the received signal into useful information consisting of the train’s location, speed, and direction of travel. The railroad uses this information to schedule locomotives, trains, crews, and equipment more efficiently. One proposed transponder circuit is shown in Figure P 9.810b with a large transponder coil of L = 5 H. Determine i(t) and v(t). The received signal is is = 9 + 3e–2tu(t) A. (a) Vehicle-mounted transponder tag Wheel detector input Antenna + – (b) 0.5 F L 1Ω v 1.5 Ω i is 0.5 Ω Figure P 9.8-10 Solution: t = 0− t > 0 2 × 9 = 6 A = i (0+ ) 2 +1 1 & v (0− ) = × 9× 1.5 = 4.5 V = v(0+ ) 2 +1 i (0− ) = dv v + =i dt 1.5 s 5di dv v KVL : v + (0.5 + ) (0.5) = + i dt 1.5 dt KCL at middle node: i + 0.5 (1) (2) Solving for i in (1) and plugging into (2) yields d d 2 v ⎛ 49 ⎞ dv ⎛ 4 ⎞ ⎛2⎞ is where i = 9 + 3e−2t A 2 + + = + v i ⎜ ⎟ ⎜ ⎟ ⎜ ⎟s s 5 dt 2 ⎝ 30 ⎠ dt ⎝ 5 ⎠ dt ⎝ ⎠ 49 4 So the characteristic equation is s 2 + s + = 0 and its roots are s = − 0.817 ± j 0.365 30 5 −.817 t vn (t ) = e [ A1 cos(0.365 t ) + A2 sin(0.365 t )] Try vf (t ) = B0 + B1e −2t and substitute vf (t ) into the differential equation and equate like terms to get B0 = 4.5, B1 = −7.04 Then v(t ) = e −.817 t [ A1 cos(0.365t ) + A2 sin(0.365t ) ] + 4.5 − 7.04e−2t Now using the initial conditions gives v(0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 4 dv(0) 4 = 2is (0) − 2i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6 3 dt 3 ∴ 6 = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −22.82 and v(t ) dv(t ) − 0.5 so i (t ) = i (t ) − s dt 1.5 −0.817 t i (t ) =e [ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e−2t A i(t) P 9.8-11 Determine v(t) for t > 0 for the circuit shown in Figure P 9.8-11. 0.1 H + – Answer: – va(t) + 4Ω 6u(t) + 10 V 0.625 F 2va(t) vc(t) = 0.75 e–4t – 6.75 e–36t + 16 V for t > 0 + v(t) – Figure P 9.8-11 Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. va ( 0 ) = −4 i ( 0 ) i ( 0 ) = 2 ( −4 i ( 0 ) ) ⇒ i ( 0 ) = 0 A and v ( 0 ) = 10 V For t > 0 Apply KCL at node 2: va d + K va + C v = 0 R dt KCL at node 1 and Ohm’s Law: va = − R i so d 1+ K R v= i dt CR Apply KVL to the outside loop: L d i + R i + v − Vs = 0 dt After some algebra: d2 R d 1+ K R 1+ K R v+ v+ v= Vs 2 dt L dt LC LC ⇒ d2 d v + 40 v + 144 v = 2304 2 dt dt The forced response will be a constant, vf = B so d2 d B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V 2 dt dt The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 . v ( t ) = A1 e − 4 t + A2 e −36 t + 16 . Then Using the initial conditions: ⎫ 10 = v ( 0 ) = A1 e0 + A2 e0 + 16 ⇒ A1 + A2 = −6 ⎪ ⎬ ⇒ d 0 0 0 = v ( 0 ) = −4 A1 e − 36 A2 e ⇒ − 4 A1 − 36 A2 = 0 ⎪ dt ⎭ A1 = 0.75 and A2 = −6.75 So v ( t ) = 0.75 e−4 t − 6.75 e−36 t + 16 V for t > 0 P 9.8-12 The circuit shown in Figure P 9.8-12 is at steady state before the switch opens. The inductor current is given to be i(t) = 240 + 193e–6.25tcos(9.27t – 102°) mA for t ≥ 0. Determine the values of R1, R3, C, and L. t=0 R1 + – i(t) 24 V L + v(t) C 20 Ω – Figure P 9.8-12 Solution: Two steady state responses are of interest, before and after the switch opens. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. For t > 0, the switch is open. At steady state, inductor 24 current is i ( ∞ ) = . From the given equation, R1 + 20 i ( ∞ ) = lim i ( t ) = 0.24 . Thus, t →∞ 0.24 = 24 ⇒ R1 = 80 Ω . R1 + 20 For t < 0, the switch is closed and the circuit is at steady state. 24 = 0.24 + 0.193 cos ( −102° ) = 0.2 80 || R 3 + 20 ( ) Consequently, R 3 = 80 Ω After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠ Apply KVL to get v (t ) = L d i (t ) + R2 i (t ) dt Substituting v ( t ) into the first equation gives R3 d⎛ d d ⎛ ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ then R1 C L d2 dt ( i ( t ) + R1 C R 2 + L 2 ) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs Dividing by R1 C L : ⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs i t i t + + ⎜ ⎟ ⎜ ⎟ i (t ) = ( ) ( ) 2 ⎜ ⎟ ⎜ ⎟ R1 C L dt ⎝ R1 C L ⎠ dt ⎝ R1 C L ⎠ d2 Compare to d2 dt to get 2α = 2 i ( t ) + 2α R1 C R 2 + L R1 C L d i ( t ) + ω 0 2 i ( t ) = f (t ) dt , ω 02 = R1 + R 2 R1 C L and f (t ) = Vs R1 C L From the given equation, we have α = 6.25 and ω d = 9.27 rad/s . Consequently, ω 0 = ω d 2 + α 2 = 11.18 rad/s . Next 12.5 = R1 C R 2 + L R1 C L = 20 1 + L 80 C and 125 = R1 + R 2 R1 C L = 1.25 1 ⇒ 100 = CL CL So 12.5 = 20 1 + 1 80 C 100 C ⇒ 0 = 2000 C 2 − 12.5 C + 0.0125 ⇒ C = 1.25, 5 mF The corresponding values of the inductance are L = 8, 2 H . There are two solutions: R1 = 80 Ω, R 3 = 80 Ω, C = 1.25 mF and L = 8 H and R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H We have used the initial condition i ( 0 ) = 0.2 A but we have not yet used the initial condition v (t ) = L d i (t ) + R2 i (t ) ⇒ dt from the given equation, v (0) R2 i ( 0) 8 4 4 d i ( 0) = − = − = dt L L L L L i ( t ) = 0.24 + e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) ) A for t ≥ 0 d i ( t ) = ( −6.25 ) e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) ) dt + ( 9.27 ) e −6.25 t ( 0.04sin ( 9.27 t ) + 0.1888cos ( 9.27 t ) ) for t ≥ 0 d i ( 0 ) = ( −6.25 )( −0.04 ) + ( 9.27 )(1.888 ) = 2 dt Consequently, 2= d 4 i ( 0) = dt L ⇒ L=2H and we choose R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H P 9.8-13 The circuit shown in Figure P 9.8-13 is at steady state before the switch opens. Determine the inductor current, i(t), for t > 0. t=0 8Ω + – i(t) 18 V + v(t) 0.4 H 25 mF – Figure P 9.8-13 Solution: First, we find the initial conditions; For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v (0 −) = 12 ×18 = 12 V ( 8 || 24 ) + 12 and i (0 −) = 24 18 × = 0.75 A 8 + 24 ( 8 || 24 ) + 12 Next, represent the circuit by a differential equation. After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠ Apply KVL to get v (t ) = L 24 Ω d i (t ) + R2 i (t ) dt Substituting v ( t ) into the first equation gives d⎛ d d ⎛ ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ 12 Ω then R1 C L d2 dt 2 ( i ( t ) + R1 C R 2 + L ) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs Dividing by R1 C L : ⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs i t i t + + ⎜ ⎟ ⎜ ⎟ i (t ) = ( ) ( ) 2 ⎜ R1 C L ⎟ dt ⎜ R1 C L ⎟ R1 C L dt ⎝ ⎠ ⎝ ⎠ d2 Compare to d2 dt to get 2α = i t + 2α 2 ( ) R1 C R 2 + L R1 C L d i ( t ) + ω 02 i ( t ) = f (t ) dt , ω 02 = R1 + R 2 R1 C L and f (t ) = Vs R1 C L With the given element values, we have α = 17.5 and ω 0 2 = 250 . Consequently, the roots of the characteristic equation are s 1 = −α − α 2 − ω 0 2 = −25 and s 2 = −α + α 2 − ω 02 = −10 . The natural response is i n ( t ) = A1 e−10 t + A 2 e−25 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. if = 18 = 0.9 A 8 + 12 So i ( t ) = i n ( t ) + i f ( t ) = A1 e−10 t + A2 e−25 t + 0.9 It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0.75 = i ( 0 ) = A1 + A 2 + 0.9 The other initial condition comes from v (t ) R 2 d i (t ) = − i (t ) ⇒ dt L L then 12 12 d i ( 0) = − × 0.75 = 7.5 0.4 0.4 dt 7.5 = d i ( 0 ) = −10 A1 − 25 A 2 dt Solving these equations gives A1 = 0.25 and A2 = −0.4 so i ( t ) = 0.25 e−10 t − 0.4 e−25 t + 0.9 A for t > 0 (checked using LNAPTR 7/21/04) *P 9.8-14 The circuit shown in Figure P 9.8-14 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0. ia + – i(t) t=0 0.4 H 10 Ω 3ia 20 V 10 Ω 25 mF + v(t) – Figure P 9.8-14 Solution: First, we find the initial conditions; For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also i ( 0) d v ( 0) = =0 0.025 dt Next, represent the circuit after the switch closes by a differential equation. To do so, we find the Thevenin equivalent circuit for the part of the circuit to the left of the inductor. v s − v oc ⎫ ⎪ R1 ⎪ v s R 2 (1 + b ) ⎬ ⇒ v oc = v oc ⎪ R1 + R 2 (1 + b ) ia + bia = R 2 ⎪⎭ ia = i sc = i a (1 + b ) = v s R 2 (1 + b ) Rt = With the given values, v oc = 16 V and R t = 2 Ω . After the switch closes, apply KVL to get v oc i sc = vs R1 (1 + b ) R1 + R 2 (1 + b ) R1 R 2 = vs R1 + R 2 (1 + b ) (1 + b ) R1 R t i (t ) + L d i ( t ) + v ( t ) = voc dt d v (t ) dt Substituting i ( t ) into the first equation gives Apply KCL to get i (t ) = C d2 ⎛ 1 ⎞ voc ⎛R⎞ d v (t ) + ⎜ ⎟ v (t ) + ⎜ ⎟ v (t ) = CL dt ⎝ L ⎠ dt ⎝CL⎠ 2 d2 d v ( t ) + ω 02 v ( t ) = f (t ) dt dt Rt v 1 2α = , ω 02 = and f (t ) = oc L CL CL Compare to 2 to get v (t ) + 2α With the given element values, we have α = 2.5 and ω 0 2 = 100 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −2.5 ± j 9.682 and the circuit is underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 9.682 rad/s . The natural response is ( v n ( t ) = e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t ) Next, determine the forced response. The steady state response after the switch closes will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v f = v oc = 16 V ( v ( t ) = 16 + e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t so ) It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 16 + A1 ⇒ And 0= d v ( 0 ) = −2.5 A1 + 9.682 A 2 dt ⇒ A1 = −16 A2 = − 2.5 × 16 = −4.131 9.682 Finally, v ( t ) = 16 + e −2.5 t ( −16 cos 9.682 t − 4.131sin 9.682 t ) = 16 + 16.525 e −2.5 t cos ( 9.682 t + 165.5° ) V for t ≥ 0 (checked using LNAPTR 7/22/04) P 9.8-15 The circuit shown in Figure P 9.8-15 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0. t=0 50 Ω i(t) 2H + – + 50 Ω 20 V v(t) 5 mF – Figure P 9.8-15 Solution: First, we find the initial conditions; For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also i ( 0) v ( 0) d v ( 0) = − =0 0.005 50 × 0.005 dt Next, represent the circuit after the switch closes by a differential equation. After the switch closes, use KCL to get i (t ) = v (t ) d + C v (t ) R2 dt Use KVL to get v s = R1 i ( t ) + L d i (t ) + v (t ) dt Substitute to get vs = R1 R2 = CL Finally, v ( t ) + R1C d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt ⎛ R1 + R 2 d2 L ⎞d v t + R1C + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ ⎟ dt dt R R 2 2 ⎝ ⎠ ⎛ R1 R1 + R 2 d2 1 ⎞d = 2 v (t ) + ⎜ + v (t ) + v (t ) ⎟ ⎜ L R 2C ⎟ dt CL dt R 2CL ⎝ ⎠ vs Compare to d2 dt to get 2α = R1 L 2 + i ( t ) + 2α d i ( t ) + ω 02 i ( t ) = f (t ) dt R1 + R 2 1 , ω 02 = R 2C R 2CL f (t ) = and vs CL With the given element values, we have α = 14.5 and ω 0 2 = 200 . Consequently, the roots of the characteristic equation are s 1 = −11.3 and s 2 = −17.7 so the circuit is overdamped. The natural response is v n ( t ) = A1 e −11.3 t + A 2 e −17.7 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. vf = 1 v s = 10 V 2 So v n ( t ) = 10 + A1 e −11.3 t + A 2 e−17.7 t It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 10 + A1 + A 2 and 0= d v ( 0 ) = −11.3 A1 − 17.7 A 2 dt Solving these equations gives A1 = −27.6 and A 2 = 17.6 Finally, v ( t ) = 10 − 27.6 e−11.3 t + 17.6 e−17.7 t (checked using LNAPTR 7/26/04) t=0 P 9.8-16 The circuit shown in Figure P 9.8-16 is at steady state before the switch closes. Determine the inductor current, i(t), for t > 0. 16 Ω 9Ω + – + 25 mF 20 V 0.4 H i(t) Figure P 9.8-16 Solution: First, we find the initial conditions; For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v ( 0 − ) = 0 V and i ( 0 − ) = 0 A Also 9 i ( 0 ) + 0.4 d i ( 0) = v ( 0) ⇒ dt d i (0) = 0 dt Next, represent the circuit by a differential equation. After the switch closes use KVL to get R2 i (t ) + L d i (t ) = v (t ) dt Use KCL and KVL to get d ⎛ ⎞ v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) dt ⎝ ⎠ Substitute to get d d2 d v s = R1i ( t ) + R1CR 2 i ( t ) + R1C L 2 i ( t ) + R 2i ( t ) + L i ( t ) dt dt dt 2 d d = R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t ) dt dt then ⎛ R2 vs R1 + R 2 d2 1 ⎞d = 2 i (t ) + ⎜ + i (t ) + i (t ) ⎟ ⎜ L R1C ⎟ dt R1CL dt R CL 1 ⎝ ⎠ v(t) – Compare to d2 2 dt to get 2α = R2 L + i ( t ) + 2α d i ( t ) + ω 02 i ( t ) = f (t ) dt R1 + R 2 1 , ω 02 = R1C R1 C L and f (t ) = Vs R1 C L With the given element values, we have α = 12.5 and ω 0 2 = 156.25 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 0 2 = −12.5, − 12.5 so the circuit is critically damped. The natural response is ( ) i n ( t ) = A1 + A 2 t e−12.5 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. 20 if = = 0.8 A 16 + 9 So i ( t ) = i n ( t ) + i f ( t ) = A1 + A 2 t e−12.5 t + 0.8 ( ) It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have And 0= 0 = i ( 0 ) = A1 + 0.8 ⇒ A1 = −0.8 d i ( 0 ) = −12.5 A1 − A 2 dt ⇒ A 2 = 10 Thus i ( t ) = ( −0.8 + 10 t ) e−12.5 t + 0.8 for t > 0 (checked using LNAPTR 7/27/04) 75 Ω P 9.8-17 The circuit shown in Figure P 9.8-17 is at steady state before the switch opens. Determine the inductor current, i2(t), for t > 0. t=0 24 Ω + – 4H i1(t) 20 V 15 Ω Figure P 9.8-17 Solution: First, we find the initial conditions; For t < 0, the switch is closed and the circuit is at steady state. At steady state, the inductors act like short circuits. i1 ( 0 − ) = and 20 = 1.333 A 15 i 2 (0 −) = 0 A Next, represent the circuit by a differential equation. After the switch opens, KVL gives L1 d d i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t ) dt dt KVL and KCL give L1 d i 1 ( t ) + R1 ( i 1 ( t ) + i 2 ( t ) ) = 0 dt Use the operator method to get L1s i1 = R 2 i 2 + L 2 s i 2 L1s i1 + R1 ( i1 + i 2 ) = 0 1.6 H i2(t) L1s 2i1 + R1s i1 + R1s i 2 = 0 s ( R 2i 2 + L 2 s i 2 ) + R1 L1 (R i 2 2 + L 2 s i 2 ) + R1s i 2 = 0 ⎛ ⎞ L2 R1 R 2 + R1 ⎟ s i 2 + L 2 s 2 i 2 + ⎜ R 2 + R1 i2 = 0 ⎜ ⎟ L L 1 1 ⎝ ⎠ ⎛ R 2 R 1 R1 ⎞ R1 R 2 + + ⎟ s i2 + s 2i 2 + ⎜ i2 = 0 ⎜ L 2 L 2 L1 ⎟ L1 L 2 ⎝ ⎠ so ⎛ R 2 R1 R1 ⎞ d R1 R 2 d2 i t + + + ⎟ i 2 (t ) + i 2 (t ) = 0 2 2( ) ⎜ ⎜ L 2 L 2 L1 ⎟ dt dt L L 1 2 ⎝ ⎠ Compare to d2 dt to get 2α = R2 L2 2 + i ( t ) + 2α R1 L2 + R1 L1 d i ( t ) + ω 02 i ( t ) = f (t ) dt , ω 02 = R1R 2 L1L 2 and f (t ) = 0 With the given element values, we have α = 33.9 and ω 02 = 281.25 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −4.4, − 63.4 so the circuit is overdamped. The natural response is i n ( t ) = A1 e−4.4 t + A 2 e−63.4 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state the inductors act like short circuits. if = 0 A So i 2 ( t ) = i n ( t ) + i f ( t ) = A1 e−4.4 t + A2 e−63.4 t It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = i 2 ( 0 ) = A1 + A 2 L2 d i 2 ( 0 ) + R 2 i 2 ( 0 ) + R1 i 1 ( 0 ) + R 1 i 2 ( 0 ) ⇒ dt d i 2 ( 0 ) = −20 dt and −20 = d i ( 0 ) = −4.4 A1 − 63.4 A 2 dt Solving these equations gives A1 = −0.339 and A2 = 0.339 so i 2 ( t ) = −0.339 e−4.4 t + 0.339 e−63.4 t for t ≥ 0 (checked using LNAPTR 7/27/04) t=0 P 9.8-18 The circuit shown in Figure P 9.8-18 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0. 2H + – 20 V 50 Ω 5 mF 50 Ω i(t) + v(t) – Figure P 9.8-18 Solution: First, we find the initial conditions; For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also i (0) d v ( 0) = =0 0.005 dt Next, represent the circuit after the switch closes by a differential equation. After the switch closes i (t ) = C d v (t ) dt KCL and KVL give ⎛ 1 ⎛ d d ⎞⎞ vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t ) ⎜ R1 ⎝ dt dt ⎠ ⎟⎠ ⎝ Substituting gives ⎛ R2 ⎞ ⎛ R2 ⎞ ⎛ R2 ⎞ R2 d2 d d2 d vs = LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 + v t = 1 + LC v ( t ) + R 2C v ( t ) + ⎜1 + ( ) ⎟ ⎜ ⎟ ⎟⎟ v ( t ) 2 ⎜ ⎟ ⎜ ⎟ ⎜ R1 dt dt R R dt dt R 1 1 1 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ So the differential equation is R1v s LC ( R1 + R 2 ) = R1 R 2 d2 d 1 v (t ) + v (t ) + v (t ) dt LC L ( R1 + R 2 ) dt Compare to d2 dt to get 2α = ( 2 i ( t ) + 2α R1R 2 L R1 + R 2 ) d i ( t ) + ω 02 i ( t ) = f (t ) dt , ω 02 = 1 and CL f (t ) = ( R1v s LC R1 + R 2 ) With the given element values, we have α = 6.25 and ω 0 2 = 100 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −6.25 ± j 7.806 and the circuit is underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 7.806 rad/s . The natural response is ( v n ( t ) = e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t ) Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. vf = So 50 × 20 = 10 V 50 + 50 ( v ( t ) = 10 + e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t ) It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 10 + A1 ⇒ And 0= d v ( 0 ) = −6.25 A1 + 7.806 A 2 dt ⇒ A1 = −10 A2 = − 6.25 × 10 = −8.006 7.806 Finally, v ( t ) = 10 + e −6.25 t ( −10 cos 7.806 t − 8.006sin 7.806 t ) = 10 + 12.81 e −6.25 t cos ( 7.806 t + 141.3° ) V for t ≥ 0 (checked using LNAPTR 7/26/04) iL(t) P 9.8-19 Find the differential equation for vc(t) in the circuit of Figure P 9.8-19 using the direct method. Find vc(t) for time t > 0 for each of the following sets of component values: L R1 vs(t) = u(t) +– R2 C + vc(t) – Figure P 9.8-19 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Answer: (a) vc(t) = 1 2 – e–2t + –2t 1 2 e–4t V; (b) vc(t) = 1 4 –( 1 4 1 + 2 t)e–2t V; (c) vc(t) = 0.8 – e (0.8 cos 4t + 0.4 sin 4t) V Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L ⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠ ⎛ L ⎞d ⎛ R1 ⎞ d2 = LC 2 vC ( t ) + ⎜ + R1 C ⎟ vC ( t ) + ⎜1 + v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= R1 + R 2 2 The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 6s + 8 = ( s + 2 )( s + 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is vn = A1 e −2 t + A2 e−4 t V The complete response is vc ( t ) = iL ( t ) = vC ( t ) 1.309 At t = 0+ + 1 + A1 e −2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt 0 = vc ( 0 + ) = A1 + A2 + 0.5 0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819 Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 1= v f = vC ( ∞ ) = 4 R1 + R 2 The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 2 s2 + ⎜ + ⎟s+⎜ = s + 4s + 4 = ( s + 2 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = ( A1 + A2 t ) e −2 t V The complete response is vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+ 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 (( A 2 ) − A1 ) − A2 t e −2 t 0 = vc ( 0 + ) = A1 + 0 = iL ( 0 + ) = 1 4 1 + A2 − A1 4 Solving these equations gives A1 = -0.25 and A2 = -0.5, so 1 ⎛ 1 1 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠ (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Use the steady state response as the forced response: R2 4 v f = vC ( ∞ ) = 1= R1 + R 2 5 The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vc ( t ) = The complete response is iL ( t ) = At t = 0+ vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vC ( t ) 4 + A2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 0 = vc ( 0 + ) = 0.8 + A1 0 = iL ( 0 + ) = 0.2 + A2 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V P 9.8-20 Find the differential equation for vo(t) in the circuit of Figure P 9.8-20 using the direct method. Find vo(t) for time t > 0 for each of the following sets of component values: iL(t) R1 vs(t) = u(t) + – C L + + vc(t) – R2 vo(t) – Figure P 9.8-20 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Answer: 1 – e–2t + 1 (a) vo(t) = (b) vo(t) = (c) vo(t) = 0.2 – e–2t(0.2 cos 4t + 0.1 sin 4t) V 2 3 4 –( 3 4 + 2 3 2 e – 4t V t)e–2t V Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using iL ( t ) = vo ( t ) gives R2 d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt ⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 vs (t ) = LC 2 v o ( t ) + ⎜ + R1 C ⎟ v o ( t ) + ⎜ v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ o ( ) R2 dt ⎝ ⎠ ⎝ ⎠ R1 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= R1 + R 2 2 The characteristic equation is R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 s2 + ⎜ + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is vn = A1 e −2 t + A2 e −4 t V The complete response is 1 + A1 e −2 t + A2 e−4 t V 2 A1 −2 t A 2 −4 t vo ( t ) 1 iL ( t ) = e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + 1 d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2 At t = 0+ 0 = iL ( 0 + ) = 0 = vC ( 0 + ) = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6167 A1 + 0.2361 A2 2 Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω Use the steady state response as the forced response: R2 3 v f = vo ( ∞ ) = 1= R1 + R 2 4 The characteristic equation is R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 2 s2 + ⎜ + s+ = s + 4s + 4 = ( s + 2 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = ( A1 + A2 t ) e −2 t V The complete response is vo ( t ) = iL ( t ) = 3 + ( A1 + A2 t ) e −2 t V 4 vo ( t ) vC ( t ) = 3 iL ( t ) + 3 = 1 ⎛ A1 A2 ⎞ −2 t t⎟ e V +⎜ + 4 ⎝ 3 3 ⎠ 3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t d iL ( t ) = + ⎜⎜ ⎜ + t ⎟e ⎟+ 4 ⎝⎝ 3 3 ⎠ 3 ⎟⎠ dt At t = 0+ 0 = iL ( 0 + ) = A1 + 1 4 3 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) = 3 ⎛ 3 3 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠ (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 1= v f = vo ( ∞ ) = 5 R1 + R 2 The characteristic equation is R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 s2 + ⎜ + s+ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V The complete response is vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = vo ( t ) vC ( t ) = iL ( t ) + At t = 0+ 1 = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t 2 dt 0 = iL ( 0 + ) = 0.2 + A1 0 = vC ( 0 + ) = 0.2 + 2 A2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V Section 9-9: State Variable Approach to Circuit Analysis P 9.9-1 Find v(t) for t > 0 using the state variable method of Section 9.9 when C = 1/5 F in the circuit of Figure P 9.9-1. Sketch the response for v(t) for 0 < t < 10 s. Answer: v(t) = – 25e–t + e– 5t + 24 V 6Ω + C 4u(t) A 1H Figure P 9.9-1 Solution: t = 0− circuit is source free ∴ iL (0) = 0 & v(0) = 0 t>0 ⎛1⎞ = 4 KCL at top node: i L + ⎜ ⎟ dv (1) ⎝ 5 ⎠ dt di − 6i L = 0 KVL at right loop: ( v − 1) L dt Solving for i in (1) & plugging into (2) ⇒ 1 d 2v dv + 6 + 5v = 120 2 dt dt The characteristic equation is: s 2 + 6s + 5 = 0, The roots of the characteristic equation are s = −1, −5 ∴ Tthe natural response is: vn (t ) = A1 e −t + A2 e −5t Try vf = B & plug into D.E. ⇒ B = 24 = vf dv(0) = 20 − 5iL (0) = 20 V s dt So v(0) = 0 = A1 + A2 + 24 ⎫ A1 = −25, A2 = 1 ⎪ ⎬ ∴ v(t ) = −25 e −t + e−5t + 24 V dv(0) = 20 = − A1 − 5 A2 ⎪ dt ⎭ From (1) – v P 9.9-2 Repeat Problem P9.9-1when C = 1/10 F. Sketch the response for v(t) for 0 < t < 3 s. 6Ω + Answer: v(t) = e – 3t(– 24 cos t – 32 sin t) + 24 V C 4u(t) A – v 1H Figure P 9.9-1 Solution: At t = 0− the circuit is source free ∴ iL (0) = 0, & v(0) = 0 At t > 0 ( ) i = 4 − 1 dv dt 10 L KCL at top node : KVL at right node : v − (1) into (2) yields 2 ⇒ s + 6 s + 10 = 0, s = −3 ± j ∴ vn ( t ) = e −3t d 2v dt 2 diL +6 dt dv +10v = 240 dt ⎡⎣ A1 cos t + A2 sin t ⎤⎦ Try vf = B & plug into D.E. ⇒ vf = B = 24 dv (0) = 40−10 i L (0) = 40 V s dt dv (0) = 40 = −3 A1 + A2 So v (0) = 0 = A1 + 24 ⇒ A1 = −24 & dt From (1) ⇒ ∴ v (t ) = e −3 t [ −24cos t − 32sin t ] + 24 V − 6i L = 0 A2 = −32 (1) (2) Determine the current i(t) and the voltage v(t) for the circuit of Figure P 9.9-3. P 9.9-3 Answer: i(t) = (3.08e–2.57t – 0.08e–97.4t – 6) A v 0.2 H –3u(t) A i + – 20 mF 0.5 Ω 3A Figure P 9.9-3 Solution: i (0) = −3, v(0) = 0 t >0 dv v + +6=0 dt R di KVL: v = L dt KCL: i + C d 2i dt 2 + 100 di + 250i = −1500 dt s = − 2.57, − 97.4 i f (t ) = −1500 = −6 250 i (t ) = A1e −2.57 t + A2 e−97.4 t −6 ⎫ ⎪ A1 = 3 .081 ⎬ di (0) = 0 = − 2.57 A1 −97.4 A2 ⎪ A2 =−0.081 dt ⎭ i (0) = A1 + A2 − 6 = −3 i (t ) = 3.081 e v (t ) = 0.2 −2.57 t − 0.081e di = −1.58e dt −2.57 t −97.4 t −6 A + 1.58e−97.4 t V P 9.9-4 Clean-air laws are pushing the auto industry toward the development of electric cars. One proposed vehicle using an ac motor is shown in Figure P 9.9-4a. The motor-controller circuit is shown in Figure P 9.9-4b with L = 100 mH and C = 10 mF. Using the state equation approach, determine i(t) and v(t) where i(t) is the motor-control current. The initial conditions are v(0) = 10 V and i(0) = 0. + + – 2vx Transistorized dc to ac inverter + Integrated interior permanent magnet Sodium-sulfur battery ac motor and automatic transaxle 2Ω 2ix Electric power steering System controller – C – vx v L ix i 1Ω (a) (b) Figure P 9.9-4 Apply KCL to the supernode corresponding to the VCVS to get ix + vx 2 = 2ix + C dv dt ⇒ − ix + Apply KCL to the right node of the CCCS to get i + vx 2 = 0.01 dv dt vx = 2ix . 2 Apply KVL to the mesh consisting of the 2-Ω resistor, inductor and capacitor to get vx + v − L Apply KVL to the outside loop to get di di = 0 ⇒ v x + v = 0.1 dt dt v + ix + 2vx = 0 (1) (2) (3) (4) Combine equations (2) and (4) to get 4 1 2 4 i − v and v x = − i − v 9 9 9 9 Use (5) to eliminate ix and vx from (1) and (3) to get dv 5 1 di 2 5 0.01 = − i − v and 0.1 = − i + v dt 9 9 dt 9 9 Use operators to write 500 100 20 50 sv = − i− v and s i = − i + v 9 9 9 9 The characteristic equation is : s 2 + 13.33s + 333.33 = 0 ⇒ s1 , s2 = −6.67 ± j 17 ix = (5) (6) (7) The natural response is v ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and there is no forced because there is no forcing function. The constants A and B are evaluated using the initial conditions: dv(0) v(0) = 10 = A and = −111 = −6.67 A + 17 B ⇒ B = −2.6 dt Then v ( t ) = e −6.67 t ⎡⎣10 cos (17 t ) − 2.6 sin (17 t ) ⎤⎦ V Similarly, the natural response is i ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and again there is no forced because there is no forcing function. The constants A and B are evaluated using the initial conditions: di (0) i (0) = 0 = A and = 55.6 = −6.67 A + 17 B ⇒ B = −3.27 dt Then i ( t ) = e −6.67 t ⎡⎣3.27 sin (17 t ) ⎤⎦ V P 9.9-5 Studies of an artificial insect are being used to understand the nervous system of animals. A model neuron in the nervous system of the artificial insect is shown in Figure P 9.9-5. The input signal, vs, is used to generate a series of pulses, called synapses. The switch generates a pulse by opening at t = 0 and closing at t = 0.5 s. Assume that the circuit is at steady state and that v(0 –) = 10 V. Switch 6Ω 3Ω vs + – + v 1 6F 30 V 1 2H – Determine the voltage v(t) for 0 < t < 2 s. Figure P 9.9-5 Solution First consider t < 0 : v(0) = 10 V, i (0) = L 10 A 3 Next consider 0 < t < 0.5s α= R 3 1 and ω02 = LC = = L 2 12 s =−α ± α 2 − ω02 ⇒ s1 = −.028 and s2 = −2.97 The natural response is v(t ) = Ae −0.028t + Be−2.97 t and the forced response is vf = 0. The constants are evaluated using the initial conditions: v(0) = 10 = A + B ⎫ ⎪ A = 16. 89 dv(0) ⎬ = 20 = −0.028 A − 2.97 B ⎪ B = −6.89 dt ⎭ −0.028t −2.97 t so v(t ) = 16. 89 e − 6.89e Similarly i (t ) = − .079e −0.028t + 3.41e−2.97 t At t = 0.5 s, v(0.5) = 15.1 V and i (0.5) = 0.7 A For t > 0.55 s: v − 30 1 dv + iL + =0 6 6 dt di L KVL: v = 3 i L + 1 2 dt KCL: Characteristic equation: 0 = s 2 − 7 s − 18 ⇒ s = −1,9 vf = 10 V v(t ) = Ae9t + Be − t + 10 v(0.5) = 15.1 = 90 A + 0.61 B + 10 dv(0.5) = 10.7 = 810 A − 0.61 B dt t 0 → .5 .5 →2 ⎫ ⎪ ⎬ ⎪⎭ A = 17.6 × 10−3 B = 5.77 v(t) 16.89e−0.28τ − 6.89e−2.97 τ V 17.6 × 10-3e9t + 5.77e-t + 10 V Section 9-10: Roots in the Complex Plane 2 kΩ P 9.10-1 For the circuit of Figure P 9.10-1, determine the roots of the characteristic equation, and plot the roots on the s-plane. 12 – 6u(t) V 3 kΩ + – 2 mH 2 mH i1 i2 Figure P 9.10-1 Solution: s 2 + 3.5 × 106 s + 1.5 × 1012 = 0 s = −5 × 105 1 s2 = −3 × 106 P 9.10-2 For the circuit of Figure P 9.6-1, determine the roots of the characteristic equation and plot the roots on the s-plane. 0.8 H vc 250 Ω t=0 250 Ω + – 6V Figure P 9.6-1 Solution: s 2 + 800s + 250000 = 0 s = 400 ± j 300 + – 5 × 10-6 F 4H P 9.10-3 For the circuit of Figure P 9.10-3, determine the roots of the characteristic equation and plot the roots on the s-plane. vs + 1 4 – μF 4 kΩ Figure P 9.10-3 Solution: KCL: KVL: Characteristic equation: s 2 + 1×103 s + 1× 106 = 0 s = −500 ± j 866 dv v 1 × 10−6 + dt 4 4000 di vs = 4 + v dt i= P 9.10-4 An RLC circuit is shown in Figure P 9.10-4. (a) Obtain the two-node voltage equations using operators. (b) Obtain the characteristic equation for the circuit. (c) Show the location of the roots of the characteristic equation in the s-plane. (d) Determine v(t) for t > 0. 1H a 36u(t) V + – 12 Ω b 6 Ω 1 18 F + – v(t) Figure P 9.10-4 Solution: at t = 0 the initial conditions are v(0) = v (0) = 0, b dvb vb − va i (0) = 0 and C + + i =0 (1) dt 6 t=0 Node a: va (0) − 36 v (0) − vb (0) − i (0) + a = 0 then va (0) + 2va (0) = 36 so va (0) = 12 V 12 6 t ≥ 0 va − vs va − vb 1 + = 0 ∫ (va − vb ) dt + L 12 6 v −v dv 1 Node b : C b + b a + ∫ (va − vb )dt =0 dt 6 L v 1 1⎞ ⎛1 ⎛ 1 1⎞ Using operators ⎜ + + ⎟ va + ⎜ − − ⎟ vb = s 6 s⎠ 12 ⎝ 12 ⎝ 6 s⎠ 1 1 1 1 1 ( − − ) va + ( s + + ) vb = 0 6 s 18 6 s 2 Cramers rule: ( s + 5s + 6)vb = ( s + 6)vs Node a : Then vb = 36 + A1 e −2t + A2 e−3t vb (0) = 36 + A1 + A2 need (2) dvb (0) = −2 A1 − 3 A2 dt v ( 0 ) − vb (0) dvb (0) 1 12 = (−2 A1 − 3 A2 ) = a − i (0) = =2 dt 18 6 6 Use (2) and (3) to get Use 1 above: C A1 = −72 and A2 = 36 so v ( t ) = vb ( t ) = 36 − 72 e −2t + 36 e −3t , t ≥ 0 (3) Section 9-11 How Can We Check…? P 9.11-1 Figure P 9.11-1a shows an RLC circuit. The voltage, vs(t), of the voltage source is the square wave shown in Figure P 9.11-1a. Figure P 9.11-1c shows a plot of the inductor current, i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is correct. Answer: The plot is correct. 25 i(t) 100 Ω vs, V vs 0 4 8 12 + – 2 μF 12 mH 16 t, ms (a) (b) 400 mA (550.562u, 321.886m) (1.6405m, 256.950m) (3.6854m, 250.035m) 200 mA (1.0787m, 228.510m) I (L1) 0A –200 mA 0s I (L1) 2.0 ms 4.0 ms Time 6.0 ms 8.0 ms (c) Figure P 9.11-1a Solution: This problem is similar to the verification example in this chapter. First, check the steady-state inductor current v 25 i (t ) = s = = 250 mA 100 100 This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an underdamped response. That requires 12 ⋅10−3 = L < 4 R 2C = 4(100) 2 (2 ⋅10−6 ) = 8 ⋅10−2 This inequality is satisfied, which also agrees with the plot. The damped resonant frequency is given by 2 2 ⎛ ⎞ 1 1 1 ⎛ 1 ⎞ ⎟ = 5.95 ⋅103 ω = −⎜ − ⎜ ⎟ = d 6 − ⎜ ⎟ LC ⎝ 2RC ⎠ 2 ⋅10−6 12 ⋅10−3 ⎝ 2(100) (2 ⋅10 ) ⎠ ( )( ) The plot indicates a maxima at 550.6μs and a minima at 1078.7μs. The period of the damped oscillation is T = 2 (1078.7 μs − 550.6μs) = 1056.2μs d 2π 2π Finally, check that 5.95 ⋅103 = ω = = = 5.949 ⋅103 d T 1056.2 ⋅10−6 d The value of ωd determined from the plot agrees with the value obtained from the circuit. The plot is correct P 9.11-2 Figure P 9.11-2b shows an RLC circuit. The voltage, vs(t), of the voltage source is the square wave shown in Figure P 9.11-2a. Figure P 9.11-2c shows a plot of the inductor current, i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is correct. Answer: The plot is not correct. 15 i(t) 100 Ω vs, V vs 0 2 4 6 + – 0.2 μ F 8 mH 8 t, ms (b) (a) 300 mA (426.966u, 172.191m) 200 mA (1.7753m, 149.952m) 100 mA I (L1) 0A (831.461u, 146.570m) –100 mA 0s 2.0 ms 4.0 ms Time 6.0 ms 8.0 ms (c) Figure P 9.11-2 Solution: This problem is similar to the verification example in this chapter. First, check the steady-state inductor current. v 15 i (t ) = s = = 150 mA 100 100 This agrees with the value of 149.952 mA shown on the plot. Next, the plot shows an under damped response. This requires 8 ⋅10−3 = L < 4 R 2C = 4 (100) 2 (0.2 ⋅10−6 ) = 8 ⋅10−3 This inequality is not satisfied. The values in the circuit would produce a critically damped, not underdamped, response. This plot is not correct. PSpice Problems SP 9-1 The input to the circuit shown in Figure SP 9-1 is the voltage of the voltage source, vi(t). The output is the voltage across the capacitor, vo(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t for each of the following cases: vi(V) 5 0 (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Plot the output for these three cases on the same axis. Hint: Represent the voltage source using the PSpice part named VPULSE. 10 5 L + vi(t) – 15 t (s) R1 R2 C + vo(t) – Figure SP 9-1 Solution: Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom. SP 9-2 The input to the circuit shown in Figure SP 9-2 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across resistor R2. The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t for each of the following cases: (a) (b) (c) vi(V) 5 0 Plot the output for these three cases on the same axis. Hint: Represent the voltage source using the PSpice part named VPULSE. R1 + vi(t) – 15 t (s) 10 5 C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω L C R2 + vo(t) – Figure SP 9-2 Solution: Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.) V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom. 10 Ω SP 9-3 Determine and plot the capacitor voltage 50 Ω ig v(t) for 0 < t < 300 μs for the circuit shown in Figure SP 9-3a. The sources are pulses as shown in Figures SP 9-3b, c. 1 mH + v – 0.1 μ F + – vg (a) 0.2 A 5V vg ig 0 0 100 t ( μ s) 200 0 0 100 t ( μ s) (c) (b) Figures SP 9-3 Solution: 50 Ω 200 3 kΩ SP 9-4 Determine and plot v(t) for the circuit of Figure SP 9-4 when vs(t) = 5u(t) V. Plot v(t) for 0 < t < 0.25 s. vs(t) + – 6 kΩ + v(t) – 2 kΩ 3 kΩ 2 μF 3 μF Figure SP 9-4 Solution: Design Problems DP 9-1 Design the circuit shown in Figure DP 9-1 so that vc(t) = 1 2 + A1e–2t + A2e–4t V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec. iL(t) vs(t) = u(t) R1 L + – R2 C + vc(t) – Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L ⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠ = LC ⎛ L ⎞d ⎛ R1 ⎞ d2 v t + v t + R1 C ⎟ vC ( t ) + ⎜1 + 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎛ 1+ 1 ⎜ ⎛ 1 R1 ⎞ R2 s2 + ⎜ + ⎟s+⎜ ⎜ R 2 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s: ⎞ ⎟ ⎟ = s 2 + 6s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎟⎟ ⎠ R1 1 + = 6 and R2 C L 1+ R1 R2 LC =8 Using R1 = R 2 = R gives 1 R 1 + = 6 and =4 RC L LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then vc ( t ) = iL ( t ) = At t = 0+ vC ( t ) 1.309 + 1 + A1 e−2 t + A2 e−4 t V 2 d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt 0 = vc ( 0 + ) = A1 + A2 + 0.5 0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819 Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 DP 9-2 Design the circuit shown in Figure DP 9-1 so that vc(t) = 1 4 + (A1 + A2t)e–2t V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec. iL(t) vs(t) = u(t) L R1 + – R2 C + vc(t) – Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 4 R2 1 = ⇒ 3 R 2 = R1 4 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: vC ( t ) d KCL at the top node of R2 gives: + C vC ( t ) = iL ( t ) R2 dt d KVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L ⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠ ⎛ L ⎞d ⎛ R1 ⎞ d2 = LC 2 vC ( t ) + ⎜ + R1 C ⎟ vC ( t ) + ⎜1 + v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 2 s2 + ⎜ + ⎟s+⎜ = s + 4s + 4 = ( s + 2 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ Equating coefficients of like powers of s: 1 R2 C + R1 L 1+ = 4 and R1 R2 LC =4 Using R 2 = R and R1 = 3R gives 1 3R 1 + = 4 and =1 LC RC L These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω . vc ( t ) = iL ( t ) = vC ( t ) + 1 + ( A1 + A2 t ) e −2 t V 4 d 1 vC ( t ) = + dt 4 (( A 2 At t = 0+ 0 = vc ( 0 + ) = A1 + 0 = iL ( 0 + ) = 1 4 1 + A2 − A1 4 Solving these equations gives A1 = -0.25 and A2 = -0.5, so vc ( t ) = ) − A1 ) − A2 t e −2 t 1 ⎛ 1 1 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠ DP 9-3 Design the circuit shown in Figure DP 9-1 so that vc(t) = 0.8 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is underdamped and the damped resonant frequency is 4 rad/sec and the damping coefficient is 2. iL(t) vs(t) = u(t) R1 L + – R2 C + vc(t) – Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 1 vC ( ∞ ) = R1 + R 2 4 The specifications require that vC ( ∞ ) = so 5 R2 4 = ⇒ 4 R1 = R 2 5 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L ⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠ ⎛ L ⎞d ⎛ R1 ⎞ d2 + R1 C ⎟ vC ( t ) + ⎜1 + v t + v t 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ Equating coefficients of like powers of s: = LC R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R and R 2 = 4 R gives R 1 1 + = 4 and = 16 4R C L LC These equations do not have a unique solution. Try C = 1 1 F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 1 Ω and R 2 = 4 Ω . Next vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) = At t = 0+ vC ( t ) 4 + A2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 0 = vc ( 0 + ) = 0.8 + A1 0 = iL ( 0 + ) = 0.2 + A2 2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V DP 9-4 Show that the circuit shown in Figure DP 9-1 cannot be designed so that vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show that 1/RC + 10 RC = 4 would require the value of RC to be complex. iL(t) vs(t) = u(t) R1 L + – R2 C + vc(t) – Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives: vC ( t ) R2 vs ( t ) = L +C d vC ( t ) = iL ( t ) dt d iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get ⎞ ⎛ v (t ) ⎞ d ⎛ v (t ) d d vs ( t ) = L ⎜ C + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎟ ⎜ R2 ⎟ dt ⎜⎝ R 2 dt dt ⎠ ⎝ ⎠ ⎛ L ⎞d ⎛ R1 ⎞ d2 v t + + R1 C ⎟ vC ( t ) + ⎜1 + v t 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ Equating coefficients of like powers of s: = LC R1 1 + = 4 and R2 C L 1+ R1 R2 LC = 20 Using R1 = R 2 = R gives 1 R 1 + = 4 and = 10 RC L LC Substituting L = 1 into the first equation gives 10 C 0.4 ± 0.42 − 4 ( 0.1) 4 1 ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. ( RC ) 2 − DP 9-5 Design the circuit shown in Figure DP 9-5 so that vo(t) = 1 2 + A1 e–2t + A2 e– 4t V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec. iL(t) R1 L + vs(t) = u(t) + – C + R2 vc(t) – vo(t) – Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 1 so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using vo ( t ) = iL ( t ) gives R2 d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt ⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 vo ( t ) = LC 2 iL ( t ) + ⎜ + R1 C ⎟ iL ( t ) + ⎜ i t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) R2 dt ⎝ ⎠ ⎝ ⎠ R1 The characteristic equation is R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 s2 + ⎜ + s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s: ⎞ ⎟ ⎟ = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎟⎟ ⎠ R2 1 + = 6 and R1 C L 1+ R2 R1 LC =8 Using R1 = R 2 = R gives 1 R 1 + = 6 and =4 RC L LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 4 1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω 2 4 R Pick R = 1.309 Ω. Then 1 + A1 e −2 t + A2 e−4 t V 2 A1 −2 t A 2 −4 t v (t ) 1 iL ( t ) = o e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) = vC ( t ) = 1.309 iL ( t ) + 1 d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2 At t = 0+ 0 = iL ( 0 + ) = 0 = vC ( 0 + ) = A1 A2 1 + + 2.618 1.309 1.309 1 + 0.6167 A1 + 0.2361 A2 2 Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) = 1 −2 t 1 −4 t −e + e V 2 2 DP 9-6 Design the circuit shown in Figure DP 9-5 so that 3 vo(t) = 4 + (A1 + A2t)e–2t V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec. iL(t) R1 L + + – vs(t) = u(t) C + R2 vc(t) – vo(t) – Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions vC ( ∞ ) = R2 R1 + R 2 1, iL ( ∞ ) = The specifications require that vo ( ∞ ) = 1 R1 + R 2 and vo ( ∞ ) = R2 R1 + R 2 1 3 so 4 R2 3 = ⇒ 3R1 = R 2 4 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives: d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L Use the substitution method to get vs ( t ) = R1 C = R1 LC Using vo ( t ) = iL ( t ) gives R2 vo ( t ) = d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC The characteristic equation is ⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R1 C ⎟ iL ( t ) + ⎜ i t 2 L( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) dt ⎝ ⎠ ⎝ ⎠ R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 s2 + ⎜ + s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s: ⎞ ⎟ ⎟ = s 2 + 4s + 4 = ( s + 2 )2 ⎟ ⎟⎟ ⎠ R2 1 + = 4 and R1 C L 1+ R2 R1 LC =4 Using R1 = R and R 2 = 3R gives 1 3R 1 + = 4 and =1 RC L LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω 3 3 3 R Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω . vo ( t ) = iL ( t ) = 3 + ( A1 + A2 t ) e −2 t V 4 vo ( t ) vC ( t ) = 3 iL ( t ) + 3 = 1 ⎛ A1 A2 ⎞ −2 t t⎟ e V +⎜ + 4 ⎝ 3 3 ⎠ d 3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t iL ( t ) = + ⎜⎜ ⎜ + t ⎟e ⎟+ dt 4 ⎝⎝ 3 3 ⎠ 3 ⎟⎠ At t = 0+ 0 = iL ( 0 + ) = A1 + 1 4 3 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) = 3 ⎛ 3 3 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠ DP 9-7 Design the circuit shown in Figure DP 9-5 so that vc(t) = 0.2 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is underdamped, the damped resonant frequency is 4 rad/sec, and the damping coefficient is 2. iL(t) R1 L + vs(t) = u(t) + – C vc(t) – + R2 vo(t) – Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 R2 1 vC ( ∞ ) = and vo ( ∞ ) = 1, iL ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 1 The specifications require that vo ( ∞ ) = so 5 R2 1 = ⇒ R1 = 4 R 2 5 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: d KVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) KCL at the top node of the capacitor gives: R1 dt Use the substitution method to get vs ( t ) = R1 C = R1 LC Using vo ( t ) = iL ( t ) gives R2 vo ( t ) = d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC The characteristic equation is ⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R C i t + i t ( ) ( ) ⎜ ⎟ ⎜ L 1 ⎜ R2 ⎟ dt L ⎜ R 2 ⎟⎟ L ( ) dt 2 ⎝ ⎠ ⎝ ⎠ R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 + s2 + ⎜ s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s: ⎞ ⎟ ⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎟⎟ ⎠ R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R 2 = R and R1 = 4 R gives 1 R 1 + = 4 and = 16 4R C L LC 1 1 These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 4 Ω and R 2 = 1 Ω . Next vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vo ( t ) = 0.2 + e−2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 1 d vC ( t ) = iL ( t ) + iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t 2 dt iL ( t ) = At t = 0+ 0 = iL ( 0 + ) = 0.2 + A1 0 = vC ( 0 + ) = 0.2 + 2 A2 Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V DP 9-8 Show that the circuit shown in Figure DP 9-5 cannot be designed so that vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show that 1/RC + 10 RC = 4 would require the value of RC to be complex. iL(t) R1 L + vs(t) = u(t) + – C vc(t) – + R2 vo(t) – Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 R2 1 1, iL ( ∞ ) = 1 vC ( ∞ ) = and vo ( ∞ ) = R1 + R 2 R1 + R 2 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: d vC ( t ) = L iL ( t ) + R 2 iL ( t ) KVL around the right-hand mesh gives: dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) KCL at the top node of the capacitor gives: R1 dt Use the substitution method to get vs ( t ) = R1 C = R1 LC Using vo ( t ) = iL ( t ) gives R2 vo ( t ) = d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt R1 R2 LC The characteristic equation is ⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R1 C ⎟ iL ( t ) + ⎜ i t 2 L( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) dt ⎝ ⎠ ⎝ ⎠ R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 + s2 + ⎜ s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s: ⎞ ⎟ ⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎟⎟ ⎠ R2 1 + = 4 and R1 C L 1+ R2 R1 LC = 20 Using R1 = R 2 = R gives 1 R 1 + = 4 and = 10 LC RC L Substituting L = 1 into the first equation gives 10 C ( RC ) 2 0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied. L DP 9-9 A fluorescent light uses cathodes (coiled 1 2 tungsten filaments coated with an electron-emitting i t=0 substance) at each end that send current through mercury vapors sealed in the tube. Ultraviolet + 4 Ω radiation is produced as electrons from the cathodes 10 V – 1 3F knock mercury electrons out of their natural orbits. Some of the displaced electrons settle back into orbit, throwing off the excess energy absorbed in Figure DP 9-9 the collision. Almost all of this energy is in the form of ultraviolet radiation. The ultraviolet rays, which are invisible, strike a phosphor coating on the inside of the tube. The rays energize the electrons in the phosphor atoms, and the atoms emit white light. The conversion of one kind of light into another is known as fluorescence. One form of a fluorescent lamp is represented by the RLC circuit shown in Figure DP 99. Select L so that the current i(t) reaches a maximum at approximately t = 0.5 s. Determine the maximum value of i(t). Assume that the switch was in position 1 for a long time before switching to position 2 at t = 0. Hint: Use PSpice to plot the response for several values of L. Solution: R 1 s+ =0 L LC Select L so fast response and i achieve maximum at t = 0.5 s Characteristic equation: s 2 + 4 3 s + = 0 Try L = 1 H ⇒ s 2 + 4s + 3 = 0 or s = −3, −1 L L i (t ) = A1 e − t + A2 e −3t s2 + i (0) = 0 = A1 + A2 ⎫ ⎪ ⎬ A1 = 5, A2 = −5 di (0) v(0) 10 = = = − A1 − 3 A2 ⎪ dt L 1 ⎭ −t −3t i (t ) = 5e − 5e at t = 0.5 s i − 1.92 Section 10.2 Sinusoidal Sources P10.2‐1 Given the sinusoids v1 ( t ) = 8cos ( 250 t + 15° ) V and v 2 ( t ) = 6 cos ( 250 t − 45° ) V , determine the time by which v 2 ( t ) is advanced or delayed with respect to v1 ( t ) . Solution: T= The period of both sinusoids is 2π = 25.1327 ms 250 The difference in the phase angles is θ 2 − θ1 = −45° − 15° = −60° The delay time is td = −60 ( 25.1327 ) = −4.188 ms 2π (The minus sign indicates a delay.) The voltage v 2 ( t ) is delayed by 4.188 ms with respect to v1 ( t ) . P10.2‐2 Given the sinusoids v1 ( t ) = 8cos (100 t − 54° ) V and v 2 ( t ) = 8cos (100 t − 102° ) V , determine the time by which v 2 ( t ) is advanced or delayed with respect to v1 ( t ) . Solution: T= The period of both sinusoids is 2π = 62.8319 ms 100 The difference in the phase angles is θ 2 − θ1 = −102° − ( −54° ) = −48° −48° ( 62.8319 ) = −8.3776 ms 360° (The minus sign indicates a delay.) The voltage v 2 ( t ) is delayed by 48.3776 ms with respect to The delay time is v1 ( t ) . td = P10.2‐3 A sinusoidal current is given as i ( t ) = 125cos ( 5000 π t − 135° ) mA Determine the period, T, and the time, t1, at which the first positive peak occurs. Answer: T = 0.4 ms and t1 = 0.15 ms. Solution: The frequency is 5000 π rad/s or 2500 Hertz so the period is 1 T= = 0.0004 = 0.4 ms 2500 ⎛ π ⎞ Converting the angle from degrees to radians, we get −135° ⎜ ⎟ = −0.75 π radians . The ⎝ 180° ⎠ −0.75 π = −15 ms . The minus sign current sinusoid is shifted from 125cos ( 5000 π t ) mA by 5000 π indicates a delay. A positive peak occurs at t 1 = 0.15 ms . Since 15 ms is less than the period of i ( t 1 ) , the positive peak at t 1 = 0.15 ms is the first positive peak. P10.2‐4 Express the voltage shown in Figure P10.2‐7 in the general form v ( t ) = A cos (ω t + θ ) V where A ≥ 0 and −180° < θ ≤ 180° . Figure P10.2‐4 Solution: The amplitude is A = 45 mv and the period is given by period is T = 80 ms. The frequency is given by ω = T = 60 − 20 = 40 ms so the 2 2π = 78.54 rad/s . Noticing that v(t) is 0 80 ×10−3 at time 0 and is increasing at time 0, we can write v ( t ) = 45sin ( 78.54 t ) = 45cos ( 78.54 t − 90° ) mV 20 V P 10.2-5 Figure P 10.2-5 shows a sinusoidal voltage, v(t), plotted as a function of time, t. Represent v(t) by a function of the form A cos (ω + θ). v(t) 0V Answer: v(t) = 18 cos (393t–27°) –20 V 0s 20 ms t 40 ms Figure P 10.2-5 Solution: A = 18 V T = 18 − 2 = 16 ms ω= 2π 2π = = 393 rad/s 0.016 T ⎛ 16 ⎞ ⎟ = −27° ⎝ 18 ⎠ θ = − cos −1 ⎜ v ( t ) = 18 cos ( 393 t − 27° ) V 20 V P 10.2-6 Figure P 10.2-6 shows a sinusoidal voltage, v(t), plotted as a function of time, t. Represent v(t) by a function of the form A cos (ωt+θ). v(t) 0V –20 V 0s 20 ms 40 ms 60 ms t Figure P 10.2-6 Solution: A = 15 V T = 43 − 11 = 32 ms ω= 2π 2π = = 196 rad/s T 0.032 ⎛8⎞ ⎟ = −58° ⎝ 15 ⎠ θ = − cos −1 ⎜ v ( t ) = 15 cos (196 t − 58° ) V Section 10.3 Phasors and Sinusoids P10.3‐1 Express the current i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA In the general form i ( t ) = A cos (ω t + θ ) mA where A ≥ 0 and −180° < θ ≤ 180° . Solution: i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA = 2 cos ( 6 t + 120° ) + 2 cos ( 6 t − 60° − 90° ) mA Representing the sinusoids using phasors gives: I (ω ) = 2∠120° + 4∠ − 150° = ( −1 + j 1.732 ) + ( −3.464 − j 2 ) = −4.464 − j 0.268 = 4.472∠183.4° = 4.472∠ − 176.6° mA The corresponding sinusoid is: P10.3‐2 Express the voltage i ( t ) = 4.472 cos ( 6 t − 176.6° ) mA v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° ) V In the general form v ( t ) = A cos (ω t + θ ) V where A ≥ 0 and −180° < θ ≤ 180° . Solution: v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° ) = 5 2 cos ( 8 t ) + 2 cos ( 8 t + 45° − 90° ) = 5 2 cos ( 8 t ) + 2 cos ( 8 t − 45° ) V Representing the sinusoids using phasors gives: V (ω ) = 7.0711 + 10∠ − 45° = 7.0711 + ( 7.0711 − j 7.0711) = 14.1422 − j 7.0711 = 15.811∠ − 26.6° V The corresponding sinusoid is: v ( t ) = 15.811cos ( 8 t − 26.6° ) V P 10.3-3 Determine the polar form of the quantity (25∠36.9°) ( 80 ∠ − 53.1° ) (4 + j 8) + (6 − j 8) Answer: 200∠ − 16.2° Solution: (25∠36.9°) ( 80 ∠ − 53.1° ) 25 ⋅ 80∠ ( 36.9° − 53.1° ) 2000∠ − 16.2° = = = 200∠ − 16.2° (4 + j 8) + (6 − j 8) 10 ( 4 + 6 ) + j (8 − 8) P 10.3-4 Determine the polar and rectangular form of the expression ⎛ 3 2 ∠ − 45° ⎞ 5 ∠ + 81.87° ⎜⎜ 4 − j 3 + ⎟ 7 − j1 ⎟⎠ ⎝ Answer: 88.162 ∠30.127° = 76.2520 + j 44.2506 Solution: ⎛ ⎛ ⎞ 40 ∠ − 45° ⎞ 40 ∠ − 45° 8 ∠42° ⎜ 8 − j 3 + ⎟ = 8 ∠42° ⎜ 8 − j 3 + ⎟ 7 − j 12 ⎠ 13.892 ∠ − 59.744° ⎠ ⎝ ⎝ = 8 ∠42° ( 8 − j 3 + 2.8793 ∠14.744° ) = 8 ∠42° ( 8 − j 3 + 2.7845 + j 0.7328 ) = 8 ∠42° (10.7845 − j 2.2672 ) = 8 ∠42° (11.02∠ − 11.873° ) = 88.162 ∠30.127° = 76.2520 + j 44.2506 Determine the polar and rectangular form of the expression P 10.3-5 (60 ∠120°)(−16 + j12 + 20∠15°) 5∠ − 75° Solution: (60 ∠120°)(−16 + j12 + 20∠15°) (60 ∠120°)(−16 + j12 + 19.3185 + j 5.1764) = 5∠ − 75° 5∠ − 75° (60 ∠120°)(3.3185 + j 17.1764) = 5∠ − 75° (60 ∠120°)(17.494∠79.065°) = 5∠ − 75° 1049.6∠ − 160.93° = = 139.95∠109.07° = 45.714 + j 132.28 5∠ − 75° P10.3‐6 The circuit shown in Figure 10.3‐6 is at steady state. The input currents are i1 ( t ) = 10 cos ( 25 t ) mA and i 3 ( t ) = 10 cos ( 25 t + 135° ) mA Determine the voltage v2(t). Figure 10.3‐6 Solution: Using first Ohm’s law and then KCL v2 ( t ) = 250 i 2 ( t ) i 2 ( t ) = i1 ( t ) − i 3 ( t ) = 10 cos ( 25 t ) − 10 cos ( 25 t + 135° ) mA and Using phasors I 2 (ω ) = I 1 (ω ) − I 3 (ω ) = 10 − 10 ∠135 = 10 − ( −7.071 + j 7.071) mA = 17.071 − j 7.071 = 18.478∠ − 22.5° mA The corresponding sinusoid is i 2 ( t ) = 18.478cos ( 25 t − 22.5° ) mA Finally v2 ( t ) = 250 i 2 ( t ) = 4.6195cos ( 25 t − 22.5° ) V P10.3‐7 The circuit shown in Figure 10.3‐7 is at steady state. The inputs to this circuit are the current source current i1 ( t ) = 0.12 cos (100 t + 45° ) A and the voltage source voltage v 2 ( t ) = 24 cos (100 t − 60° ) V Determine the current i2(t). Figure P10.3‐7 Solution: Using Ohm’s and Kirchhoff’s laws v 2 (t ) 24 cos (100 t − 60° ) i 2 ( t ) = i1 ( t ) − = 0.12 cos (100 t + 45° ) − 96 96 = 0.12 cos (100 t + 45° ) − 0.25cos (100 t − 60° ) Using phasors I 2 (ω ) = 0.12 ∠45° − 0.25 ∠60° = ( 0.0849 + j 0.0849 ) − ( 0.1250 − j 0.2165 ) = −0.0401 + j 0.3014 = 0.3040∠97.6° A The corresponding sinusoid is i 2 ( t ) = 0.3040 cos (100 t + 97.6° ) A Checked using LNAPAC 8/16/11 P 10.3-8 i1 (t) Given that i2 (t) i1(t) = 30 cos (4t + 45°) mA and i2(t) = – 40 cos (4t) mA Determine v(t) for the circuit shown in 100 Ω 15 H + 120 Ω v(t) – Figure P 10.3-8. Figure P 10.3-8 Solution: V = j ( 4 )(15 ) ( I 1 + I 2 ) = j 60 ( 0.03∠45° − 0.04∠0° ) = j 60 ( 0.0212 + j 0.0212 − 0.04 ) = −1.273 − j1.127 So = 1.7∠ − 138.5° V v ( t ) = 1.7 cos ( 4t − 138.5° ) V (checked: LNAP 8/7/04) P 10.3-9 For the circuit shown in Figure P 10.3-9, find (a) the impedances Z1 and Z2 in polar form, (b) the total combined impedance in polar form, and (c) the steady-state current i(t). 3Ω Z1 i Answer: (a) Z1 = 5 ∠53.1°; Z 2 = 8 2 ∠ − 45° (b) Z1 + Z 2 = 11.7 ∠ − 20° (c) i (t ) = (8.55) cos (1250t + 20°) A 100 μF 3.2 mH 8Ω Z2 + – 100 cos (1250t) V Figure P 10.3-9 Solution: (a) (b) (c) Z1 =3+ j 4 = 5∠53.1° Ω and Z 2 =8− j8 = 8 2 ∠− 45° Ω Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω I= 100∠0° 100 100 = = ∠20.0° ⇒ Z1 +Z 2 11.7 ∠− 20° 11.7 i (t ) = 8.55 cos (1250 t + 20.0°) A P 10.3-10 The circuit shown in Figure P 10.3-10 is at steady state. The voltages vs(t) and v2(t) are given by + v1(t) – vs(t) = 7.68 cos (2t + 47°) V and vs(t) + – C R v2(t) = 1.59 cos (2t + 125°) V Find the steady-state voltage v1(t). + v2(t) – Figure P 10.3-10 Answer: v1(t) = 7.51 cos (2t + 35°) V Solution: V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125° = ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 ) = ( 5.23 + 0.91) + j ( 5.62 − 1.30 ) = 6.14 + j 4.32 = 7.51∠35° v1 ( t ) = 7.51 cos ( 2 t + 35° ) V P 10.3-11 The circuit shown in Figure P 10.3-11 is at steady state. The currents i1(t) and i2(t) are given by i1(t) = 744 cos (2t – 118°) mA and i2(t) = 540.5 cos (2t + 100°) mA Find the steady-state current i(t). Answer: i(t) = 460 cos (2t + 196°) mA 10 Ω v1(t) + – i1(t) i2(t) 6H 10 Ω 0.05 F + – 10 Ω v2(t) i(t) Figure P 10.3-11 Solution: I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠196° i ( t ) = 460 cos (2 t + 196°) mA i(t) P 10.3-12 Determine i(t) of the RLC circuit shown in Figure P 10.3-12 when 3H 6Ω vs vs = 2 cos (4t + 30°) V. + – 1 12 Answer: i(t) = 0.185 cos (4t – 26.3°) A Figure P 10.3-12 Solution: Vs = 2 ∠30° V and I = 2 ∠30° = 0.185 ∠ − 26.3° A 6+ j12+ 3 / j i (t ) = 0.185 cos (4 t − 26.3°) A F Section 10.4 Impedances P10.4-1 Figure P10.4-1a shows a circuit represented in the time domain. Figure P10.4-1b shows the same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.41a. Vs is the phasor corresponding to the voltage of the voltage source. Determine ZR, ZC, ZL1, ZL2, and Vs. + 8Ω + Z L1 ZR 2H Vs 5 sin 5t V vo +– 1 F 12 4H Vo +– Z L2 ZC – (a) – (b) Figure P10.4-1 Hint: 5 sin 5t = 5 cos (5t – 90°) Answer: Z R = 8Ω, Z C = 1 2.4 j 2.4 = = = − j 2.4Ω, Z L1 = j 5(2) = j10Ω, j j× j ⎛ 1⎞ j5 ⎜ ⎟ ⎝ 12 ⎠ ZL2 = j5(4) = j20 Ω, and Vs = 5 ∠–90° V Solution: ZR = 8 Ω, ZC = 1 = 2.4 j 2.4 = = − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω, j j× j 1 12 ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V. j5 1 P10.4-2 Figure P10.4-2a shows a circuit represented in the time domain. Figure P10.4-2b shows the same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.4- 2a. Is is the phasor corresponding to the current of the current source. Determine ZR, ZC, ZL1, ZL2, and Is. + 8Ω + ZR 2H vo 1 12 Z L1 Vo 4 cos (3t + 15°) A F 4H ZC Is Z L2 – – (a) (b) Figure P10.4-2 Answer: Z R = 8 Ω, Z C = 1 4 j4 = = = − j 4 Ω, Z L1 = j 3(2) = j 6 Ω, ⎛ 1 ⎞ j j× j j3 ⎜ ⎟ ⎝ 12 ⎠ ZL2 = j3(4) = j12 Ω, and Is = 4 ∠15° A Solution: ZR = 8 Ω, ZC = 1 = 4 j4 = = − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω, j j× j 1 12 ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° A. j3 2 P10.4-3 Represent the circuit shown in Figure P10.4-3 in the frequency domain using impedances and phasors. Figure P10.4-3 Solution: P10.4-4 Represent the circuit shown in Figure P10.4-4 in the frequency domain using impedances and phasors. Figure P10.4-4 Solution: 3 P10.4-5 Determine the current, i(t), for the circuit shown in Figure P10.4-5. Figure P10.4-5 Solution: Represent the circuit in the frequency domain using phasors and impedances: j 96 I − j 312.5 I + 250 I + 7.4∠ − 24° = 0 Using KVL: Solving: I= 7.4∠ − 24° 7.4∠ − 24° 7.4∠ − 24° = = = 0.0224∠16.9 A j 96 − j 312.5 + 250 − j 216.5 + 250 330.715∠ − 40.9° In the time domain: i ( t ) = 22.4 cos ( 200 t + 16.9 ) mA (Checked using LNAP) 4 P10.4-6 The input to the circuit shown in Figure P10.4-6 is the current i ( t ) = 120 cos ( 4000 t ) mA Determine the voltage, v(t), across the 40 Ω resistor. Figure P10.4-6 Solution: Represent the circuit in the frequency domain using phasors and impedances: Using KCL 0.120∠0° = I 1 + I 2 Using KVL j 60 I 2 + 40 I 2 − ( − j 12.5 ) I 1 = 0 ( ) j 60 I 2 + 40 I 2 − ( − j 12.5 ) 0.12∠0° − I 2 = 0 j 60 I 2 + 40 I 2 + ( − j 12.5 ) I 2 = ( − j 12.5 )( 0.12∠0° ) I2 = 1.5∠ − 90° 1.5∠ − 90° 1.5∠ − 90° = = = 0.024155∠ − 140° A j 60 + 40 + ( − j 12.5 ) 40 + j 47.5 62.1∠50° V = 40 I 2 = 0.9662∠ − 140° V In the time domain v ( t ) = 0.9662 cos ( 4000 t − 140° ) V (Checked using LNAP) 5 P10.4-7 The input to the circuit shown in Figure P10.4-7 is the current i ( t ) = 82 cos (10000 t ) μA Determine the voltage, v(t), across the 50 kΩ resistor. Figure P10.4-7 Solution: Represent the circuit in the frequency domain using phasors and impedances: Using KCL Using KVL In the time domain 82 ×10−6 ∠0° = 0 + I 1 ( 20 ×103 ) I 1 + ( − j 20 ×103 ) I 1 + V = 0 V = − ( 20 × 103 − j 20 ×103 )( 82 × 10−6 ) = 2.3193∠135° V v ( t ) = 2.3193cos (10000 t + 135° ) V (Checked using LNAP) 6 P 10.4-8 Each of the following pairs of element voltage and element current adheres to the passive convention. Indicate whether the element is capacitive, inductive, or resistive and find the element value. (a) v(t) = 15 cos (400t + 30°); i = 3 sin (400t + 30°) (b) v(t) = 8 sin (900t + 50°); i = 2 sin (900t + 140°) (c) v(t) = 20 cos (250t + 60°); i = 5 sin (250t + 150°) Answer: (a) L = 12.5 mH (b) C = 277.77 μF (c) R = 4 Ω Solution: (a) v = 15cos (400 t + 30°) V i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V v leads i by 90° ⇒ element is an inductor v 15 Z L = peak = = 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH ipeak 3 (b) i leads v by 90° ⇒ the element is a capacitor (c) 8 1 1 =4= = ⇒ C = 277.77 μ F ω C 900 C ipeak 2 v = 20 cos (250 t + 60°) V Zc = vpeak = i = 5sin (250 t +150°) =5cos (250 t + 60°) A Since v & i are in phase ⇒ element is a resistor v 20 ∴ R = peak = =4Ω ipeak 5 7 Figure P10.4-9 P10.4-9 This voltage and current for the circuit shown in Figure P10.4-9 are given by v ( t ) = 20 cos (20 t + 15°) V and i ( t ) = 1.49 cos (20 t + 63°) A Determine the values of the resistance, R, and capacitance, C. Solution: 1 V 20∠15° 20 =Z= = = ∠ (15° − 63° ) = 13.42∠ − 48° = 8.98 − j9.97 Ω 20 C I 1.49∠63° 1.49 1 Equating real and imaginary parts gives R = 9 Ω and C = = 5 mF . 20 × 9.97 R− j 8 P10.4-10 Figure P10.4-10 shows an ac circuit represented in both the time domain and the frequency domain. Determine the values of A, B, a and b. Figure P10.4-10 Solution: The impedance between nodes a and b is given by 18 + j (10 )( 2.5 ) = 18 + j 25 = 30.8∠54.2° To find the impedance between nodes b and c we first find the impedance of the capacitor: −j then 1 1 =−j = − j 25 0.04 (10 )( 0.004 ) 9 ( − j 25 ) − j 225 225∠ − 90° = = = 8.47∠ − 19.8° Ω 9 − j 25 26.57∠ − 70.2° 26.57∠ − 70.2° The impedance between nodes c and d is given by ( 5 ) ( j (10 )( 0.88) ) j 40 j 40 ⎛ 5 − j 8 ⎞ 1 1 −j = −j = − j 20 5 + j (10 )( 0.8 ) (10 )( 0.005 ) 5 + j 8 0.05 5 + j 8 ⎜⎝ 5 − j 8 ⎟⎠ 320 + j 200 − j 20 25 + 64 = 3.60 + j 2.25 − j 20 = 3.60 − j 17.75 Ω = So A = 30.8 V, B = 8.47 Ω, , a = 3.57 Ω and b = −17.75 Ω. 9 P 10.4-11 Represent the circuit shown in Figure P 10.4-11 in the frequency domain using impedances and phasors. 6Ω 15 cos 4t V + – 2H 5i(t) 0.125 F i(t) + v(t) – Figure P 10.4-11 Solution: P 10.4-12 Represent the circuit shown in Figure P 10.4-12 in the frequency domain using impedances and phasors. 12 cos (5t – 30°) V + – 2H 0.25 F 0.05 F 6Ω + v(t) – + – i(t) 15 cos (5t + 60°) V Figure P 10.4-12 Solution: 10 P 10.4-13 Find R and L of the circuit of Figure P 10.4-13 when v(t) = 10 cos (ωt + 40°) V; i(t) = 2 cos (ωt + 15°) mA, + v 6 and ω = 2 × 10 rad/s. – Answer: R = 4.532 kω, L = 1.057 mH R L i Figure P 10.4-13 Solution: Z= V 10 ∠40° = = − 5000∠205°Ω = 4532 + j 2113 = R + j ω L −I −2×10−3 ∠−165° so R = 4532 Ω and L = 2113 ω = 2113 = 1.057 mH 2×106 11 Section 10.5 Series and Parallel Impedances P10.5-1 Determine the steady state voltage v(t) in the circuit shown in Figure P10.5-1. Answer: v ( t ) = 32 cos ( 250 t − 57.9° ) V Figure P10.5-1 Solution: Represent the circuits in the frequency domain using phasors and impedances: Using voltage division: V = − In the time domain 16 16∠180° ( 40∠ − 15° ) = ( 40∠ − 15° ) = 32∠128.1° V 16 + j 12 20∠36.9° v ( t ) = 32 cos ( 250 t − 57.9° ) V 1 P10.5-2 Determine the voltage v(t) in the circuit shown in Figure P10.5-2. Answer: v ( t ) = 14.57 cos ( 800 t + 111.7° ) V Figure P10.5-2 Solution: Represent the circuit in the frequency domain: Replace the series impedances at the right of the circuit by an equivalent impedance Z S = j 112 + ( − j 250 ) + 80 = 80 − j 138 Ω Replace the parallel impedances at right of the circuit by an equivalent impedance ZP = (80 − j 138)120 = (80 − j 138 )120 80 − j 138 + 120 200 − j 138 = (159.51∠ − 59.9 )120 242.99∠ − 34.6 = 78.77∠25.3° Ω Using voltage division V=− 78.77∠ − 25.3° 78.77∠ − 25.3° 18∠0° = − 18∠0° = 14.57∠111.73° V j 100 + 78.77∠ − 25.3° 97.325∠42.97° In the time domain v ( t ) = 14.57 cos ( 800 t + 111.7° ) V (checked using LNAP) 2 P10.5-3 Determine the voltage v(t) in the circuit shown in Figure P10.5-3. Answer: v ( t ) = 14.1cos ( 2500 t − 35.2° ) V Figure P10.5-3 Solution: Represent the circuit in the frequency domain: Replace the parallel impedances at right of the circuit by an equivalent impedance: ( − j 100 )( j 250 ) = − j166.67 Ω − j 100 + j 250 Using voltage division V=− In the time domain − j 166.67 166.67∠ − 90° 22∠15° = − 22∠15° = 14.1∠ − 35.2° V 200 − j 166.67 260.3∠ − 39.8° v ( t ) = 14.1cos ( 2500 t − 35.2° ) V 3 P10.5-4 The input to the circuit shown in Figure P10.5-4 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i1(t). Answer: i1 ( t ) = 144 cos ( 25 t + 180° ) mA Figure P10.5-4 Solution: Represent the circuits in the frequency domain using phasors and impedances: Using current division I1 = In the time domain j 15 15 ( 48∠0° ) = ( 48∠0° ) = 144∠180° mA −5 j 15 − j 20 i1 ( t ) = 144 cos ( 25 t + 180° ) mA 4 P10.5-5 The input to the circuit shown in Figure P10.5-5 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i2(t). Figure P10.5-5 Solution: Represent the circuits in the frequency domain using phasors and impedances: Using current division I 2 ( ω) = In the time domain j15 15∠90° ( 48∠0° ) = ( 48∠0° ) = 28.8∠53.1° mA 20 + j15 25∠36.9° i 2 ( t ) = 28.8cos ( 25 t + 53.1° ) mA 5 P10.5-6 The input to the circuit shown in Figure P10.5-6 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i3(t). Answer: i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA Figure P10.5-6 Solution: Represent the circuits in the frequency domain using phasors and impedances: Using current division 7.5∠90° j15 || j15 j 7.5 I3 = ( 48∠0° ) = ( 48∠0° ) = ( 48∠0° ) = 16.8539∠69.44° V j 7.5 + 20 21.36∠69.56° ( j15 || j15 ) + 20 In the time domain i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA 6 P10.5-7 Figure P10.5-7 shows a circuit represented in the frequency domain. Determine the voltage phasor V1. Answer: V1 = 14.59∠ − 13.15° V Figure P10.5-7 Solution: V1 = j 80 || − j 50 − j 133.33 133.33∠ − 90° ( 20∠30° ) = ( 20∠30° ) = ( 20∠30° ) 125 + ( j 80 || − j 50 ) 125 − j 133.33 182.8∠ − 46.85° = 14.59∠ − 13.15° V 7 P10.5-8 Figure P10.5-8 shows a circuit represented in the frequency domain. Determine the current phasor I2. Answer: I 2 = 18.48∠ − 93.7° mA Solution: I2 = Figure P10.5-8 − j 50 − j 50 ( 20∠30° ) = ( 20∠30° ) − j 50 + 45 + j 80 45 + j 30 = 50∠ − 90 ( 20∠30° ) = 18.49∠ − 93.7° mA 54.08∠ 33.7 8 P10.5-9 Here’s an ac circuit represented both in the time domain and frequency domain: R1 + – 15 cos 20t V C Z1 + R2 v(t) + – L 15 0° V + Z2 – V(ω) – (b) (a) Z 1 = 15.3∠ − 24.1° Ω and Z 2 = 14.4∠53.1° Ω Suppose Determine the node voltage v(t) and the values of R1, R2, L and C. Solution: Consider Z1: 1 1 R1 − j = 15.3∠ − 24.1° = 14 − j 6.25 ⇒ R1 = 14 Ω and C = = 0.008 F = 8 mF 20 C 20 ( 6.25 ) Next consider Z2: 1 = 14.4∠53.1° ⇒ 1 1 + R 2 j 20 L 1 1 1 1 + = = ∠53.1 = 0.05556 − j 0.04167 R 2 j 20 L 14.4∠53.1° 14.4 Equating coefficients gives R2 = 1 1 = 18 Ω and L = = 1.2 H 0.05556 20 ( 0.04167 ) Next, consider the voltage divider: A∠31.5° = In the time domain, (15)(14.4 ) ∠36.9° 14.4∠36.9° (15∠0° ) = 15.3∠ − 24.1° + 14.4∠36.9° (14 − j 6.25)(11.52 + j 8.64 ) = 216∠36.9° 25.52 + j 2.39 = 216∠36.9° = 8.43∠31.5° V 25.63∠5.4° v ( t ) = 8.43cos ( 20 t + 31.5° ) V . 9 P 10.5-10 Find Z and Y for the circuit of 1 μF 160 μH 36 Ω Figure P 10.5-10 operating at 10 kHz. Z,Y Figure P 10.5-10 Solution: ZR ω = 2π f = 2π (10 ×103 ) = 62832 rad sec 1 1 = R = 36 Ω ⇔ YR = = = 0.0278 S 36 ZR Z L = jω L = j (62830)(160 ×10−6 ) = j10.053 ≈ j10 Ω ⇔ YL = ZC = 1 = − 0.1 j S ZL −j −j 1 = = − j15.915 ≈ − j 16 Ω ⇔ YC = = j 0.0625 S −6 ZC ω C (62830)(1×10 ) Yeq = YR + YL + YC = 0.0278 − j0.0375 = 0.0467 ∠ − 53.4° S Z eq = 1 = 21.43∠ 53.4° = 12.75 + j17.22 Ω Yeq 10 300 Ω P 10.5-11 For the circuit of Figure P 10.5-11, find the value of C required so that Z = 590.7Ω when f = 1 MHz. C Answer: C = 0.27 nF 47 μ H Z Figure P 10.5-11 Solution: Z L = j ω L = j (6.28 ×106 ) (47 × 10−6 ) = j 295 Ω ⎛ 1 ⎞ ⎜ ⎟( 300 + j 295 ) jω C ⎠ ⎝ Z eq = Z c || ( Z R +Z L ) = = 590.7 Ω 1 +300+ j 295 jω C 300+300 j 590.7 = ⇒ 590.7 − (590.7)(295 ω C ) + j (590.7)(300ω C ) = 300 + j 295 1+300 j ω C −300 ω C ( Equating imaginary terms ω =2π f = 6.28×106 rad sec ) (590.7) (300ω C ) = 295 ⇒ C = 0.27 nF 11 P 10.5-12 Determine the impedance Z for the circuit shown in Figure P 10.5-12. 2.5 H 2 mF 2 mF 100 Ω 2 mF Z 1.5 H Figure P 10.5-12 Solution: Replace series and parallel capacitors by an equivalent capacitor and series inductors by an equivalent inductor: Then 100 Z = jω 4 + 100 + 4 −j 1 ( jω 5 ×10−3 ( ) 1 jω 5 ×10 −3 ) ⎛ 200 ⎞ 200 100 ⎜ − j 1+ −j ⎟ ω ⎝ ⎠ = jω 4 + ω × = jω 4 + 2 ⎛ 200 ⎞ 1− j 1+ 100 + ⎜ − j ⎟ ω ω ⎠ ⎝ j j 2 ω 2 ω 2 ω = jω 4 + 100 4 − j 2 ω = 400 + j ⎛ 4 ω − 200 ω ⎞ Z = jω 4 + 100 ω ⎜ ⎟ 4 4 +ω2 4 +ω2 4 +ω2 ⎠ ⎝ 1+ 2 2 ω 12 P 10.5-13 The big toy from the hit movie Speaker Big is a child’s musical fantasy come true—a sidewalk-sized piano. Like a hopscotch grid, 20 Ω this once-hot Christmas toy invites anyone v = 12 cos ω t V + – C i 3 mH who passes to jump on, move about, and make music. The developer of the “toy” piano Figure P 10.5-13 used a tone synthesizer and stereo speakers as shown in Figure P 10.5-13 (Gardner, 1988). Determine the current i(t) for a tone at 796 Hz when C = 10 μF. Solution: j (2π ⋅ 796) (3 ⋅10 −3 ) = j15 Ω 12 = 0.48 ∠ − 37° A 20 + j15 i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A I= 13 P 10.5-14 Determine i(t), v(t), and L for the circuit shown in Figure P 10.5-14. Answer: i(t) = 1.34 cos (2t – 87°) A, v(t) = 7.29 cos (2t – 24°) V, and L = 4 H i(t) 2 cos(3t – 15°) A 8Ω L Figure P 10.5-14 Solution: Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A 8 I B ∠−51.87° Z1 = = = = 2 ∠−15° Is Z1 + Z 2 8+ j 3L 8 ∠0° ⎛ 3L ⎞ 82 + (3L) 2 ∠ tan −1 ⎜ ⎟ ⎝ 8 ⎠ Equate the magnitudes and the angles. ⎛ 3L ⎞ angles: + 36.87 = + tan −1 ⎜ ⎟ ⇒ L = 2 H ⎝ 8 ⎠ B 8 magnitudes: = ⇒ B =1.6 2 2 64+ 9 L 14 P 10.5-15 Spinal cord injuries result in paralysis of the lower body and can cause loss of bladder control. Numerous electrical devices have been proposed to replace the normal nerve pathway stimulus for bladder control. Figure P 10.5-15 shows the model of a bladder control system where vs = 20 cos ωt V and ω = 100 rad/s. Find the steady-state voltage across the 10-Ω load resistor. Answer: v(t) = 10 2 cos (100t + 45°) V 1 mF vs + – 50 Ω 100 μF 10 Ω Regular nerve pathway load Figure P 10.5-15 Solution: ⎛ 10 ⎞ V10 = Vs ⎜ ⎟ ⎝ 10− j10 ⎠ 10 ⎛ ⎞ = 20∠0° ⎜ ⎟ ⎝ 10 2∠− 45° ⎠ = 10 2∠45° v10 (t ) = 10 2 cos (100 t + 45°) V 15 P 10.5-16 There are 500 to 1000 deaths each year in Person's body the United States from electric shock. If a person makes a i good contact with his hands, the circuit can be represented by Figure P 10.5-16, where vs = 160 cos ωt V and ω = 2πf. Find the steady-state current i flowing Source vs +– 300 Ω 2 μF 100 mH through the body when (a) f = 60 Hz and (b) f = 400 Hz. Answer: (a) (b) i(t) = 0.53 cos (120πt + 5.9°) i(t) = 0.625 cos (800πt + 59.9°) A Figure P 10.5-16 Solution: (a) (b) 160 ∠0° 160 ∠0° = (− j1326) (300+ j 37.7) 303 ∠−5.9° − j1326 + 300+ j 37.7 = 0.53 ∠5.9° A i(t ) = 0.53cos (120π t +5.9°) A I= I= 160∠0° 160∠0° = (− j199)(300+ j 251) 256∠−59.9° − j199 +300+ j 251 = 0.625∠59.9° A i(t ) =0.625 cos (800π t +59.9°) A 16 P 10.5-17 Determine the steady-state voltage, v(t), and current, i(t), for each of the circuits shown in Figure P 10.5-17. – v(t) + 4Ω + 24 V – 40 Ω 10 Ω i(t) (a) – v(t) + 4H 24 cos (4t + 15°) V + – 40 Ω 10 mF i(t) (b) Figure P 10.5-17 Solution: (a) v (t ) = − i (t ) = 4 × 24 = −8 V 4 + ( 40 & 10 ) 40 24 8 × = = 1.6 A 40 + 10 4 + ( 40 & 10 ) 5 (b) Represent the circuit in the frequency domain using impedances and phasors. 17 V=− (16∠ − 90° )( 24∠15° ) = 33.66∠ − 65° V j16 × 24∠15° = 40 ( − j 25 ) j16 + ( 40 & − j 25 ) j16 + 40 − j 25 I= so and 40 × 40 − j 25 24∠15° = 1.78∠57° A 40 ( − j 25 ) j16 + 40 − j 25 v ( t ) = 33.66 cos ( 4t − 65° ) V i ( t ) = 1.78cos ( 4t + 57° ) A (checked: LNAP 8/1/04) 18 P 10.5-18 Determine the steady-state current, i(t), for the circuit shown in Figure P 10.5-18. 5H 5 mF 20 Ω 2 mF 30 Ω 4H i(t) +– 5 cos (10t + 30°) V Figure P 10.5-18 Solution: I= so 5∠30° 5∠30° 5∠30° + + = 0.100∠ − 23.1° + 0.0923∠98.2° + 0.1667∠ − 60° 30 + j 40 20 − j 50 j 50 − j 20 = 0.186∠ − 29.5° A i ( t ) = 0.186 cos (10t − 29.5° ) A (checked: LNAP 8/1/04) 19 P10.5-19 Determine the steady state voltage, v(t), for this circuit: 20 Ω 4H + v(t) 10 cos(5t + 45°) mA – 30 Ω 5 mF 2H 4 mF Solution: V = 0.01∠45° ⎡⎣( 20 & j 20 ) + ( 30 & ( − j 40 ) ) + ( j10 & ( − j 50 ) ) ⎤⎦ ⎡ 20 ( j 20 ) 30 ( − j 40 ) j10 ( − j 50 ) ⎤ = 0.01∠45° ⎢ + + ⎥ j10 − j 50 ⎦ ⎣ 20 + j 20 30 − j 40 = 0.01∠45° [14.14∠45° + 24∠ − 36.9° + 12.5∠90°] = 0.01∠45° [10 + j10 + 19.2 − j14.4 + j12.5] = 0.303∠60.5° V so v ( t ) = 0.303cos ( 5t + 60.5° ) V (checked: LNAP 8/1/04) 20 P10.5-20 Determine the steady state voltage, v(t), for this circuit: 2.5H 20 Ω 10 mF + – Solution: Let and 10 cos(4t + 60°) V 5H 5 mF + v(t) 40 Ω – ⎛ ( 20 − j 25) j10 = 250 − j 200 = 12.81∠75.5° Ω 1 ⎞ Z 1 = ⎜⎜ 20 − j ⎟⎟ & j10 = 4 ( 0.01) ⎠ 20 − j 25 + j10 20 − j15 ⎝ ⎛ ⎞ j 20 ( 40 − j 50 ) 1000 + j800 1 Z 2 = j 20 & ⎜⎜ − j + 40 ⎟⎟ = = = 25.61∠75.5° Ω 40 − j 30 ⎝ 4 ( 0.005 ) ⎠ j 20 + 40 − j 50 Then V= so Z2 Z1 + Z 2 × 10∠60° = 25.61∠75.5° × 10∠60° = 6.67∠60° V 12.81∠75.5° + 25.6∠75.5° v ( t ) = 6.67 cos ( 4t + 60° ) V (checked: LNAP 8/1/04) 21 P 10.5-21 The input to the circuit shown in Figure P 10.5-21 is the current source current is(t) = 25 cos (10t + 15°) mA The output is the current i1(t). Determine the steady-state response, i1(t). i1(t) 5H is(t) 5 mF 40 Ω 2H 2 mF 25 Ω Figure P 10.5-21 Solution: Represent the circuit in the frequency domain using impedances and phasors. Let 40 ( − j 50 ) ⎛ ⎞ 1 = j 50 + = 39.0∠51.3° Ω Z 1 = j 50 + ⎜ 40 & −3 ⎟ 40 − j 50 j10 × 2 × 10 ⎠ ⎝ and Z2 = − j j 20 ( 25 ) 1 + j 20 & 25 = − j 20 + = 12.5∠ − 38.7 Ω −3 25 + j 20 10 ( 5 ×10 ) Z1 and Z2 are connected in parallel. Current division gives I1 = so Z1 Z1 + Z 2 × 0.025∠15° = 0.024∠32.7° A i1 ( t ) = 0.024 cos (10t + 32.7° ) A (checked: LNAP 8/1/04) 22 P10.5-22 Determine the steady state voltage, v(t), and current i(t) for each of these circuits: – v(t) + 4Ω + 24 V – 40 Ω 10 Ω i(t) (a) – v(t) + 4H 24 cos (4t + 15°) V + – 40 Ω 10 mF i(t) (b) Solution: i (t ) = (a) v (t ) = 80 + 80 0.024 = 19.2 mA 40 + ( 80 + 80 ) 80 1 × ( 40 & ( 80 + 80 ) ) 0.024 = ( 32 )( 0.024 ) = 0.384 V 80 + 80 2 (b) Represent the circuit in the frequency domain using impedances and phasors. I= 80 + j80 × 0.024∠15° = 0.028∠25.5° A − j 25 + ( 80 + j80 ) V= 80 × ⎡ − j 25 & ( 80 + j80 ) ⎤⎦ × 0.024∠15° = 0.494∠ − 109.5° V 80 + j80 ⎣ So i ( t ) = 28cos (10t + 25.5° ) mA and v ( t ) = 0.494 cos (10t − 109.5° ) V (checked: LNAP 8/1/04) 23 P10.5-23 Determine the steady state current i(t) for this circuit: 25 Ω 2H 2 mF 20 Ω + – 40 Ω i(t) 16 cos(20t + 75°) V 5 mF 2H Solution: Represent the circuit in the frequency domain using impedances and phasors. Let Z 1 = 25 + j ( 20 ) 2 + Z 2 = 20 + 1 = 25 + j15 = 29.2∠31° Ω j ( 20 )( 0.002 ) 1 = 20 − j10 = 22.36∠ − 26.6° Ω j ( 20 )( 0.005 ) Z 3 = 40 + j ( 20 ) 2 = 40 + j 40 = 56.57∠45° Ω and let Z p = Z 2 & Z 3 = 18.86∠ − 8° = 18.67 − j 2.67 Ω Then I= so Z2 16∠75° × = 0.118∠6.1° A Z1 + Z p Z 2 + Z 3 i ( t ) = 0.118cos ( 20t + 6.1° ) A (checked: LNAP 8/2/04) 24 P 10.5-24 When the switch in the circuit shown in Figure P 10.5-24 is open and the circuit is at steady R2 R1 state, the capacitor voltage is v(t) = 14.14 cos (100t –45°) V When the switch is closed and the circuit is at steady + – 20 cos (100t) V state, the capacitor voltage is 0.5 μF + v(t) – Figure P 10.5-24 v(t) = 17.89 cos (100t –26.6°) V Determine the values of the resistances R1 and R2. Solution: Represent the circuit in the frequency domain using phasors and impedances. The impedance 1 capacitor is = − j 20, 000 . When the switch is closed j (100 ) ( 0.5 ×10−6 ) 17.89∠ − 26.6° = V = − j 20, 000 × 20∠0° R 2 − j 20, 000 Equating angels gives ⎛ −20, 000 ⎞ −26.6° = −90° − tan −1 ⎜ ⎜ R 2 ⎟⎟ ⎝ ⎠ When the switch is open 14.14∠ − 45° = V = ⇒ R2 = −20, 000 = 10015 Ω tan ( −63.4 ) − j 20, 000 × 20∠0° R1 + R 2 − j 20, 000 Equating angles gives ⎛ −20, 000 ⎞ −45° = −90° − tan −1 ⎜ ⎜ R1 + R 2 ⎟⎟ ⎝ ⎠ ⇒ R1 + R 2 = −20, 000 = 20, 000 tan ( −45° ) So R1 = 20, 000 − 10015 = 9985 Ω (checked: LNAP 8/2/04) 25 P10.5-25 Determine the steady state current i(t) for this circuit: 20 cos (5t + 30°) mA 4H 20 Ω 10 mF 15 Ω 2H 8H 40 Ω 8 mF i(t) Solution: Represent the circuit in the frequency domain using phasors and impedances. Let Z 1 = ( j 20 & 20 ) + 1 = 10 − j10 = 14.14∠ − 45° Ω j 0.05 ⎛ 1 ⎞ Z 2 = j 40 + 40 + ⎜ j10 & ⎟ + 15 = 55 + j 56.67 = 79∠46.3° Ω j 0.04 ⎠ ⎝ Z1 I=− × 20∠30° = 3.535∠129.3° mA Z1 + Z 2 so i ( t ) = 3.535cos ( 5t + 129.3° ) mA (checked: LNAP 8/2/04) 26 P10.5-26 Determine the steady state voltage, v(t), and current i(t) for each of these circuits: i(t) + – 50 Ω 4i(t) 20 V 40 Ω + v(t) – 2H 10 Ω (a) i(t) + – 5 mF 40 Ω 3H 20 cos (10t + 15°) V 4i(t) + v(t) – (b) Solution: (a) Using KCL and then KVL gives 20 = 80 mA 250 v ( t ) = 40 ( 5i ( t ) ) = 200 ( 0.08 ) = 16 V 20 = 50 i ( t ) + 40 ( 5 i ( t ) ) ⇒ i ( t ) = Then (b) Represent the circuit in the frequency domain using phasors and impedances. Where And Z 1 = 40 + j (10 ) 3 + 1 = 40 + j10 = 41.23∠26.6° Ω j (10 )( 0.005 ) Z 2 = j (10 ) 2 & 10 = 8 + j 4 = 8.944∠26.6° Ω Using KCL and then KVL gives 20∠15° = Z 1I + 5Z 2 I Then ⇒ I = 0.234∠ − 5.6° A V = Z 2 ( 5I ) = 10.47∠21° A so i ( t ) = 0.234 cos (10t − 5.6° ) A and v ( t ) = 10.47 cos (10t + 21° ) V (checked: 8/3/04) 27 P10.5-27 Determine the steady state voltage, v(t), for each of these circuits: 20 Ω 80 Ω 24 V + + – v(t) 40 Ω − 100 Ω (a) 3H 5 mF 25 Ω 20 Ω + + – v(t) – 24 cos (20t + 45°) V 4 mF 4H 20 Ω 2 mF 15 Ω (b) Solution: (a) Using voltage division twice v (t ) = 40 100 × 24 − × 24 = −12 V 40 + 80 20 + 100 28 (b) Represent the circuit in the frequency domain using phasors and impedances. Where Z 1 = 20 Ω ⎛ ⎞ 1 & 20 ⎟⎟ = 12.2 + j 70.2 = 71.30∠80.2° Ω Z 2 = j ( 20 ) 4 + ⎜⎜ ⎝ j ( 20 )( 0.002 ) ⎠ 1 + 25 = 25 + j 50 = 55.90∠63.4° Ω Z 3 = j ( 20 ) 3 + j ( 20 )( 0.005 ) Z4 = 1 + 15 = 15 − j12.5 = 19.53∠ − 39.8° Ω j ( 20 )( 0.004 ) Using voltage division twice V= so Z2 Z1 + Z 2 × 24∠45° − Z4 Z3 + Z4 × 24∠45° = 24.8∠80° V v ( t ) = 24.8cos ( 20t + 80° ) V (Checked using LNAP 10/5/04) 29 P 10.5-28 The input to the circuit shown in Figure P 10.5-28 is the voltage of the voltage source vs(t) = 5 cos (2t + 45°) V 0.1 F + – 4Ω vs(t) The output is the inductor voltage, v(t). Determine the steady-state output voltage. 3H + v(t) – Figure P 10.5-28 Solution: Represent the circuit in the frequency domain using phasors and impedances. 4 & j6 = 4 ( j 6 ) 24∠90° = = 3.33∠34° = 2.76 + j1.86 Ω 4 + j 6 7.2∠56° Using voltage division V= 3.33∠34° 3.33∠34° 3.33∠34° × 5∠45° = × 5∠45° = × 5∠45° = 3.98∠127° V − j 5 + 2.76 + j1.86 2.76 − j 3.14 4.18∠ − 48° The corresponding voltage in the time domain is v ( t ) = 3.98cos ( 2t + 127° ) V 30 P10.5-29 Determine the steady state voltage, v(t), for this circuit: + 8Ω 2H 5 sin 5t V v(t) +– 1 12 F 4H _ Solution: V1 (ω ) = V 2 (ω ) = j10 5 e − j 90 = 3.9 e − j 51 V 8 + j10 j 20 5 e − j 90 = 5.68 e − j 90 V j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V 31 P10.5-30 Determine the steady state voltage, v(t), for this circuit: + 8Ω 2H v(t) 1 12 4 cos (3t + 15°) A F 4H – Solution: V1 (ω ) = V 2 (ω ) = 8 ( j 6 ) j15 4 e = 19.2 e j 68 V 8 + j6 j12 ( − j 4 ) j15 4 e = 24 e − j 75 V j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 V 32 P10.5-31 The input to the circuit in Figure P10.5-31 is the voltage source voltage, v s ( t ) . The output is the voltage v o ( t ) . When the input is v s ( t ) = 8cos ( 40 t ) V , the output is v o ( t ) = 2.5cos ( 40 t + 14° ) V . Determine the values of the resistances R1 and R 2 . Figure P10.5-31 Solution: Using voltage division in the frequency domain: jω L R 2 L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L jω R1 R 2 where R p = R1 + R 2 Representing the given input and output in the frequency domain: 2.5∠14° Vo (ω ) = = 8∠0° Vi (ω ) In this case the angle of and the magnitude of ω L R1 ⎛ j ⎜ 90 − tan −1 ω ⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠ 2 e ⎜ ⎝ L ⎞⎟ R p ⎟⎠ Vo (ω ) L L ( R1 + R 2 ) tan ( 90° − 14° ) = = = 0.1 is specified to be 14° so 40 Rp R1 R 2 Vi (ω ) Vo (ω ) 2.5 is specified to be so 8 Vi (ω ) 40 L R1 = 2.5 ⇒ 8 1 + 16 values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω . L = 0.0322 . One set of R1 33 Section 10.6 Mesh and Node Equations P10.6‐1 The input to the circuit shown in Figure P10.6‐1 is the voltage v s ( t ) = 48cos ( 2500 t + 45° ) V Write and solve node equations to determine the steady state output voltage v o ( t ) . Figure P10.6‐1 Solution: Represent the circuit in the frequency domain as The node voltages are 48∠45° = V1 , V2, V3 and Vo. Express the dependent source voltage in terms of the node voltages: ⎛ V2 ⎞ V3 − V 2 = 25 I = 25 ⎜ ⎟ ⇒ V3 = 2.25 V 2 ⎝ 20 ⎠ Apply KCL to the supernode corresponding to the CCVS to get 48∠45° − V 2 j 50 = V2 20 + V3 − V o 30 + V3 − j 40 V3 48∠45° V 2 V 2 V3 − Vo = + + + j 50 j 50 20 30 − j 40 ⎛ 1 48∠45° ⎛ 1 1 ⎞ 1 ⎞ 1 =⎜ + ⎟ V2 + ⎜ + ⎟ V3 − V o j 50 30 ⎝ j 50 20 ⎠ ⎝ 30 − j 40 ⎠ ⎛ 1 48∠45° ⎛ 1 1 ⎞ 1 ⎞ 1 =⎜ + ⎟ V2 + ⎜ + ⎟ 2.25V 2 − Vo j 50 30 ⎝ j 50 20 ⎠ ⎝ 30 − j 40 ⎠ 48∠45° ⎛ 1 1 2.25 2.25 ⎞ 1 =⎜ + + + ⎟ V 2 − Vo j 50 30 ⎝ j 50 20 30 − j 40 ⎠ Apply KCL at the right node of the 30 Ω resistor to get V3 − V o 30 = ⎛ 1 1 ⎞ ⎛ 1 ⎞ ⇒ 0 = ⎜ − ⎟ 2.25 V 2 + ⎜ + ⎟ Vo j 25 ⎝ 30 ⎠ ⎝ 30 j 25 ⎠ Vo 1 2.25 2.25 ⎡ 1 ⎢ j 50 + 20 + 30 + − j 40 ⎢ In matrix form 2.25 ⎢ − ⎢ 30 ⎣ Solving, perhaps using MATLAB, 1 ⎤ ⎡ 48∠45° ⎤ 30 ⎥ ⎡ V 2 ⎤ ⎢ ⎥⎢ ⎥ = j 50 ⎥ ⎢ ⎥ 1 1 ⎥ ⎣ Vo ⎦ + 0 ⎢ ⎥⎦ ⎣ 30 j 25 ⎥⎦ − ⎡ V 2 ⎤ ⎡10.18∠ − 44.6°⎤ ⎢ ⎥=⎢ ⎥ V ⎣⎢ Vo ⎦⎥ ⎣ 14.67∠5.6° ⎦ P10.6‐2 Figure P10.6‐2 shows an ac circuit represented in the frequency domain. Determine the values of the phasor node voltages, Vb and Vc. Figure P10.6‐2 Solution: 12∠45° − V b Writing Node equations: = Vb + Vb − Vc 20 − j 25 15 − j 30 j 30 V b − Vc 12∠45° − Vc Vc + = 15 − j 30 40 + j 20 j 40 Rearranging: ⎞ ⎛ ⎞ 12∠45° ⎛ 1 1 1 1 =⎜ + + ⎟ Vb − ⎜ ⎟ Vc j 30 ⎝ j 30 20 − j 25 15 − j 30 ⎠ ⎝ 15 − j 30 ⎠ ⎛ ⎞ ⎛ 12∠45° 1 1 1 1 ⎞ = −⎜ + + ⎟ Vb + ⎜ ⎟ Vc 40 + j 20 ⎝ 15 − j 30 ⎠ ⎝ 15 − j 30 40 + j 20 j 40 ⎠ In matrix from: 1 1 1 ⎡ 1 ⎤ ⎡ 12∠45° ⎤ + + − ⎢ j 30 20 − j 25 15 − j 30 ⎥ ⎡ V b ⎤ ⎢ j 30 ⎥ 15 − j 30 ⎢ ⎥⎢ ⎥ = ⎢ ⎥ 1 1 1 1 ⎥ ⎣⎢ Vc ⎦⎥ ⎢ 12∠45° ⎥ ⎢ − + + ⎢ ⎢ 40 + j 20 ⎥ 15 − j 30 15 − j 30 40 + j 20 j 40 ⎥⎦ ⎣ ⎣ ⎦ Solving using MATLAB: V b = 7.69∠ − 19.8° and Vc = 10.18∠7.7° V Checked using LNAPAC P10.6‐3 Figure P10.6‐3 shows an ac circuit represented in the frequency domain. Determine the value of the phasor node voltage V. Answer: V = 71.0346∠ − 39.627° V Figure P10.6‐3 Solution: Writing a node equation: Rearranging Solving Finally 20∠30° − V −20∠30° − V + = 0.25∠15° j 40 25 − j 50 ⎛ 1 ⎞ 1 20∠30° −20∠30° + + − 0.25∠15° ⎜ ⎟V = 25 − j 50 j 40 ⎝ j 40 25 − j 50 ⎠ ( 0.012042∠ − 48.366° ) V = 0.85537∠87.993° V= 0.85537∠87.993° = 71.0346∠ − 39.627° V 0.012042∠ − 48.366° P10.6‐4 Figure P10.6‐4 shows an ac circuit represented in the frequency domain. Determine the values of the phasor mesh currents. Figure P10.6‐4 Solution: ( 40 + j 15) I 1 + ( 25 − j 50 ) ( I 1 − I 3 ) − 48∠75° = 0 ( 65 − j 35 ) I 1 − ( 25 − j 50 ) I 3 = 48∠75° Mesh 1: 48∠75° + ( − j 50 ) ( I 2 − I 3 ) + ( 32 + j 16 ) I 2 = 0 Mesh 2: ( 32 − j 34 ) I 2 + j 50I 3 = −48∠75° j 40I 3 − ( − j 50 ) ( I 2 − I 3 ) − ( 25 − j 50 ) ( I 1 − I 3 ) = 0 Mesh 3: In matrix form: ( −25 + j 50 ) I 1 + j 50 I 2 + ( 25 − j 160 ) I 3 = 0 0 −25 + j 50 ⎤ ⎡ I 1 ⎤ ⎡ 48∠75° ⎤ ⎡ 65 − j 35 ⎢ ⎥ ⎢ + j 50 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢ −48∠75° ⎥⎥ 0 32 − j 34 ⎢ ⎢⎣ −25 + j 50 ⎥⎦ + j 50 25 − j 60 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 Solving using MATLAB: I 1 = 0.794∠111°, I 2 = 0.790∠ − 61.7° and I 3 = 0.229∠176° A P 10.6-5 A commercial airliner has sensing devices to indicate to the cockpit crew that each R door and baggage hatch is closed. A device called vs a search coil magnetometer, also known as a b + – proximity sensor, provides a signal indicative of R a LR Door Ls the proximity of metal or other conducting material to an inductive sense coil. The inductance of the sense coil changes as the metal gets closer to Figure P 10.6-5 the sense coil. The sense coil inductance is compared to a reference coil inductance with a circuit called a balanced inductance bridge (see Figure P 10.6-5). In the inductance bridge, a signal indicative of proximity is observed between terminals a and b by subtracting the voltage at b, vb, from the voltage at a, va (Lenz, 1990). The bridge circuit is excited by a sinusoidal voltage source vs = sin (800 πt) V. The two resistors, R = 100 ω, are of equal resistance. When the door is open (no metal is present), the sense coil inductance, LS, is equal to the reference coil inductance, LR = 40 mH. In this case, what is the magnitude of the signal Va – Vb? When the airliner door is completely closed, LS = 60 mH. With the door closed, what is the phasor representation of the signal Va – Vb? Solution: vs = sin (2π ⋅ 400 t ) V R = 100 Ω LR = 40 mH ⎧ 40 mH LS = ⎨ ⎩ 60 mH door opened door closed With the door open VA − VB = 0 since the bridge circuit is balanced. With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω. The node equations are: KCL at node B: KCL at node A : Since VC = Vs =1 V VB − VC VB j100.5 + = 0 ⇒ VB = VC Z LR R j100.5+100 VA − VC VA + =0 R Z LS VB =0.709∠44.86° V and VA = 0.833∠33.55 V Therefore VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040 = 0.195∠ − 11.83° V P 10.6-6 Using a tiny diamond-studded burr operating at 190,000 rpm, cardiologists can remove life-threatening plaque deposits is coronary arteries. The procedure is fast, uncomplicated, and relatively painless (McCarty, 1991). The Rotablator, an angioplasty system, consists of an advancer/catheter, a guide wire, a console, and a power source. The advancer/catheter contains a tiny turbine that drives the flexible shaft that rotates the catheter burr. The model of the operational and control circuit is shown in Figure P 10.6-6. Determine v(t), the voltage that drives the tip, when vs = 2 cos (40t – 135°) V. Answer: v(t) = 1.414 cos (40t + 135°) V 1 20 vs + – 1 80 H F 1 80 F + 2 Ω 1 80 2i i F – v(t) Figure P 10.6-6 Solution: Represent the circuit in the frequency domain V1 − (−1+ j ) V1 V1 − V2 + + =0 −j2 j2 2 V2 − V1 V2 + − IC = 0 − j2 − j2 Also, expressing the controlling signal of the dependent source in terms of the node voltages ⎡ −1+ j ⎤ −1 + j yields ⇒ IC = 2 Ix = 2 ⎢ Ix = ⎥ = −1 − j A -2 j ⎣ -2 j ⎦ Solving these equations yields The node equations are: V2 = −3− j = 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V 1+ j 2 (checked: LNAP 7/19/04 Figure P 10.6-7, it is known that 1H v2(t) = 0.7571 cos (2t + 66.7°) V v3(t) = 0.6064 cos (2t–69.8°) V i1 5Ω P 10.6-7 For the circuit of A cos 2t V + – 1F 5H + v2 – 3i1 1 4 F + v3 – Determine i1(t). Figure P 10.6-7 Solution: V2 = 0.7571∠66.7° V V3 = 0.6064∠ − 69.8° V ⎫ I1 = I 2 + I 3 ⎪ ⎪ ⎧ I 3 =0.3032 ∠20.2° A V3 − V2 ⎪ ⎪ I2 = ⎬ yields ⎨ I 2 =0.1267∠−184° A j 10 ⎪ ⎪ I =0.195∠36° A ⎩ 1 ⎪ V3 I3 = ⎪ −j2 ⎭ therefore i1 (t ) =0.195cos (2 t +36°) A (checked: MATLAB 7/18/04) P10.6‐8 The input to the circuit shown in Figure P10.6‐8 is the voltage vs = 25cos(40t + 45°) V Determine the mesh currents i1 and i2 and the voltage vo. Figure P10.6‐8 Solution Represent the circuit in the frequency domain: Apply KVL to mesh 1: j 320 I 1 + 400 ( I 1 − I 2 ) − 25∠45° = 0 Apply KVL to mesh 2: 50 I 2 + j 240 I 2 + j 120 I 2 − 400 ( I 1 − I 2 ) = 0 In matrix form: −400 ⎤ ⎡ I 1 ⎤ ⎡ 25∠45° ⎤ ⎡ 400 + j 320 ⎢ ⎥= ⎢ −400 450 + j 360 ⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣ Solving using MATLAB: Using Ohm’s Law ⎡ I 1 ⎤ ⎡ 47.5∠ − 24.6°⎤ ⎢I ⎥ = ⎢ ⎥ mA ⎣ 2 ⎦ ⎣ 33.0∠ − 63.3° ⎦ Vo = 400 ( I 1 − I 2 ) = 12∠18.8° V In the time domain i1 ( t ) = 47.5cos ( 40 t − 24.6° ) mA , i 2 ( t ) = 33cos ( 40 t − 63.3° ) mA and v o ( t ) = 12 cos ( 40 t + 18.8° ) V P10.6‐9 The input to the circuit shown in Figure P10.6‐9 is the voltage vs(t) = 42cos(800t + 60°) mV Determine the output voltage vo(t). Answer: vo(t) = 823.5 cos(800t − 55.6°) mV Figure P10.6‐9 Solution Represent the circuit in the frequency domain: Apply KCL at the top node of the inductor, node a: 0.042∠60° − Va = Va + Va ⇒ Va = 4 ( 0.042∠60° ) 5− j 500 j 2000 2000 Apply KCL at the inverting input node of the op amp: Vb Va + = 0 ⇒ V b = −5 V a 2000 10, 000 Apply KCL at the top node of the capacitor, node b: Vb Vb V b − Vo + + = 0 ⇒ Vo = ( 3 + j 4 ) V b 10, 000 − j 5000 20, 000 Combining these results we get: ( 5∠53.1° )( 20∠ − 180° ) 0.042∠60° = 0.8235∠ − 55.6° 4 V o = ( 3 + j 4 )( −5 ) ( 0.042∠60° ) = ( ) 5− j 5.1∠ − 11.3° In the time domain v o ( t ) = 832.5cos ( 800 t − 55.6° ) mV P 10.6-10 The idea of using an induction coil in a C lamp isn’t new, but applying it in a commercially available product is. An induction coil in a bulb 1Ω 1Ω induces a high-frequency energy flow in mercury + 1Ω L vapor to produce light. The lamp uses about the same 2Ω vs +– amount of energy as a fluorescent bulb but lasts six times longer, with 60 times the life of a conventional incandescent bulb. The circuit model of the bulb and v – Induction bulb Figure P 10.6-10 its associated circuit are shown in Figure P 10.6-10. Determine the voltage v(t) across the 2-Ω resistor when C = 40 μF, L = 40 μH, vs = 10 cos (ω0t + 30°), and ω0 = 105 rad/s. Answer: v(t) = 6.45 cos (105t + 44°) V Solution: Represent the circuit in the frequency domain: The mesh equations are: −1 − j 4 ⎤ ⎡ I1 ⎤ ⎡10∠30°⎤ ⎡ (2 + j 4) ⎢ −1 −1 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ (2+1/ j 4) ⎢ ⎢⎣ − j 4 −1 (3+ j 4) ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ Using Cramer’s rule yields I3 = Then 2 + j8 (10∠30° ) = 3.225∠44° A 12+ j 22.5 V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45cos (105 t + 44° ) V (checked: LNAP 7/19/04) P 10.6-11 The development of coastal hotels in various parts of the world is a rapidly growing enterprise. The need for environmentally acceptable shark protection is manifest where these developments take place alongside shark-infested waters (Smith, 1991). One concept is to use an electrified line submerged in the water in order to deter the sharks, as shown in Figure P 10.611a. The circuit model of the electric fence is shown in Figure P 10.6-11b, where the shark is represented by an equivalent resistance of 100Ω. Determine the current flowing through the shark’s body, i(t), when vs = 375 cos 400t V. Electric fence (a) 100 μ F 25 μF i vs + – 100 Ω 250 mH Source Electric fence Shark (b) Figure P 10.6-11 Solution: Represent the circuit in the frequency domain: Mesh Equations: j 75 I1 − j 100 I 2 = 375 − j 100 I1 + (100+ j 100) I 2 = 0 Solving for I 2 yields I 2 = 4.5 + j 1.5 = 3 ∠53.1° A ⇒ i 2 ( t ) 3cos ( 400t + 53.1° ) A (checked: LNAP 7/19/04) P 10.6‐12 Determine the node voltage at nodes a and b in each of these circuits: Solution (a) 24 − v a 40 24 − v b The node equations are 25 or 1 1 ⎡1 + ⎢ 40 20 + 15 ⎢ 1 ⎢ − ⎢⎣ 20 Solving using MATLAB gives = + va − vb 20 va − vb 20 + = va 15 vb 50 1 ⎤ ⎡ v ⎤ ⎡ 24 ⎤ ⎥ ⎢ a ⎥ ⎢ 40 ⎥ 20 =⎢ ⎥ ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 24 ⎥ ⎢v ⎥ + + 25 20 50 ⎥⎦ ⎣ b ⎦ ⎢⎣ 25 ⎥⎦ − v a = 8.713 V and v b = 12.69 V (b) Use phasors and impedances to represent the circuit in the frequency domain as where Z 1 = 25 + j ( 20 ) 4 = 25 + j80 = 83.82∠72.7° Ω ⎛ ⎞ 1 Z 2 = ⎜⎜ 40 & ⎟ + j ( 20 ) 5 = 3.56 + j88.6 = 88.68∠87.7° Ω j ( 20 )( 0.004 ) ⎟⎠ ⎝ Z 3 = 20 Ω Z 4 = 15 + j ( 20 ) 2 = 15 + j 40 = 42.72∠69.4° Z 5 = j ( 20 ) 3 + 1 = j 50 = 50∠90° Ω j ( 20 )( 0.005 ) 24∠45° − Va The node equations are Z2 24∠45° − V b Z1 1 1 ⎡ 1 ⎢Z + Z + Z 3 4 ⎢ 2 ⎢ 1 − ⎢ Z3 ⎢⎣ Solving using MATLAB gives So = + Va Z4 + Va − V b Va − Vb Z3 Z3 = Vb Z5 ⎤ ⎡ 24∠45° ⎤ ⎥ ⎡ Va ⎤ ⎢ Z ⎥ 2 ⎥⎢ ⎥ = ⎢ ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 24∠45° ⎥ + + ⎥ ⎢V ⎥ ⎢ ⎥ Z 1 Z 3 Z 5 ⎥⎦ ⎣ b ⎦ ⎢⎣ Z 1 ⎥⎦ − 1 Z3 Va = 7.89∠44.0° V b = 8.45∠45.1° v a ( t ) = 7.89 cos ( 20t + 44° ) V v a ( t ) = 8.45cos ( 20t + 45.1) V (checked: LNAP 8/3/04) P 10.6‐13 Determine the voltage v(t): Solution: Represent the circuit in the frequency domain using impedances and phasors The mesh currents are I and 0.05∠ − 30° A . Apply KVL to the top mesh to get 15∠45° + ( − j 25 ) I + (15 + j 32 )( I − 0.05∠ − 30° ) + 25I = 0 So Then So I= −15∠45° + (15 + j 32 )( 0.05∠ − 30° ) = 0.3266∠ − 143.6° = −0.2629 − j 0.1939 A 25 − j 25 + 15 + j 32 V = ( − j 25 ) I = 8.166∠126.4° = −4.8475 + j 6.5715 V v ( t ) = 8.166 cos ( 8t + 126.4° ) V (checked: LNAP 8/3/04) P 10.6‐14 Determine the voltage vo(t) when vs(t) = 25 cos (100t‐15°) V. Solution: Represent the circuit in the frequency domain using impedances and phasors. The mesh currents are I and 10I. Apply KVL to the supermesh corresponding to the dependant current source to get ( j500 ) I + ( − j5)(10I ) + 40 (10I ) − 25∠ − 15° = 0 So I= 25∠ − 15° = 0.04152∠ − 63.37° A 400 + j 450 The output voltage is V = 40 (10I ) = 16.61∠ − 63.37° V So v ( t ) = 16.61cos (100t − 63.37° ) V (checked: LNAP 8/3/04) P 10.6‐15 Determine the mesh current is when i(t) = 0.8394 cos (10t‐138.5°) A. Determine the values of L and R. Solution: Represent the circuit in the frequency domain using phasors and impedances. Apply KVL to the center mesh to get 0.8394∠138.5° = I = So 8∠210° − 30∠ − 15° R + j10 L R = 35 Ω ⇒ R + j10 L = 35 + j 25 = 35 + j (10 ) 2.5 and L = 2.5 H (checked: LNAP 8/3/04) P 10.6-16 The circuit shown in Figure P 10.6-16 has two inputs: 50 Ω 5 mF v1(t) = 50 cos (20t – 75°) V 6H 40 Ω 2H v2(t) = 35 cos (20t + 110°) V When the circuit is at steady state, the node voltage is + – v1(t) R + v(t) – L + v2(t) – v(t) = 21.25 cos (20t – 168.8°) V Determine the values of R and L. Figure P 10.6-16 Solution: Represent the circuit in the frequency domain using phasors and impedances. Apply KCL at the top node of R and L to get ( 50∠ − 75° ) − V + 35∠100° − V = j 40 40 ⇒ V R & jω L 50∠ − 75° 35∠110° ⎛ 1 1 1 1 ⎞ + =⎜ + + −j ⎟V 40∠90° 40 20 L ⎠ ⎝ j 40 40 R Using the given equation for v(t) we get 21.25∠ − 168.8° = V = Then Finally 1.587∠161.7° 1 1 0.025 (1 − j ) + − j R 20 L 1 1 1.587∠161.7° −j = − 0.025 (1 − j ) = 0.04 − j 0.01176 R 20 L 21.25∠ − 168.8° R= 1 1 = 25 Ω and L = = 4.25 H 0.04 20 ( 0.01176 ) (checked: LNAP 8/3/04) P 10.6‐17 Determine the steady state current i(t): Solution: Represent the circuit in the frequency domain using phasors and impedances. The node equations are 50∠0° − Va 15 50∠0° − V b − j 20 or 1 1 ⎡1 ⎢15 + j100 + 25 ⎢ 1 ⎢ − ⎢ j100 ⎣ + = Vb − Va j100 V b − Va j100 = + Va 25 Vb j 50 1 ⎤ ⎡ Va ⎤ ⎡ 50∠0° ⎤ ⎥ ⎢ ⎥ j100 ⎥ ⎢ ⎥ = ⎢ 15 ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 50∠0° ⎥ ⎢V ⎥ + + j 50 j100 − j 20 ⎥⎦ ⎣ b ⎦ ⎢⎣ − j 20 ⎥⎦ − ⎡ 0.1067 − j 0.010 ⎢ j 0.010 ⎣ j 0.010 ⎤ ⎡ Va ⎤ ⎡3.333⎤ ⎢ ⎥= j 0.020 ⎥⎦ ⎣ V b ⎦ ⎢⎣ j 2.5 ⎥⎦ Solving, e.g. using MATLAB, gives Va = 33.05∠ − 12.6° V and V b = 108.9∠1.9° V Then So I= Va = 1.322∠ − 12.6° A 25 i ( t ) = 1.322 cos ( 25t − 12.6° ) A (checked: LNAP 8/3/04) P 10.6‐18 Determine the steady state current i(t): Solution: Represent the circuit in the frequency domain using phasors and impedances. Label the node voltages. The node equations are 24∠15° − Va 25 24∠15° − V b − j 6.25 or Then So Va + Va − Vb j 40 10 Va − Vb Vb + = 10 45 1 ⎤ ⎡ V ⎤ ⎡ 24∠15° ⎤ ⎥⎢ a⎥ ⎢ ⎥ 10 25 ⎥⎢ ⎥ = ⎢ ⎥ 1 1 1⎥ 24∠15° ⎥ ⎢ ⎢V ⎥ + + j 6.25 45 10 ⎥⎦ ⎣ b ⎦ ⎢⎣ 6.25∠ − 90° ⎥⎦ 1 1 ⎡1 ⎢ 25 − j 40 + 10 ⎢ 1 ⎢ − ⎢⎣ 10 − ⎡ 0.140 − j 0.025 ⎢ −0.10 ⎣ Solving gives = − 0.10 ⎤ ⎡ Va ⎤ ⎡ 0.960∠15° ⎤ ⎢ ⎥= 0.1222 + j 0.160 ⎥⎦ ⎣ V b ⎦ ⎢⎣3.840∠105°⎥⎦ Va = 24.67∠32.6° V and V b = 25.59∠25.2° V I= Va − Vb 10 = 0.3347∠134.9° A i ( t ) = 0.3347 cos (10t + 134.9° ) A (checked: LANP 8/4/04) P 10.6‐19 Determine the steady state voltage vo(t): Solution: Represent the circuit in the frequency domain using phasors and impedances. 20∠0° − V V V − 5V = + j 40 25 − j 20 5V − Vo Vo = 10 − j10 The node equations are ⎡1 ⎢ 25 − ⎢ ⎢ ⎢⎣ 1 1 j −j 5 40 1 − 2 ⎡ 0.04 − j 0.225 ⎢ −0.50 ⎣ Solving gives So ⎤ ⎡ V ⎤ ⎡ − j 0.5⎤ ⎥⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎥ 1 1 + j ⎥ ⎢⎣ Vo ⎥⎦ ⎢⎣ 0 ⎥⎦ 10 10 ⎥⎦ 0 ⎤ ⎡ V ⎤ ⎡ − j 0.5⎤ ⎢ ⎥= 0.10 + j 0.10 ⎥⎦ ⎣ Vo ⎦ ⎢⎣ 0 ⎥⎦ 0 V = 2.188∠ − 10.1° V and Vo = 7.736∠ − 55.1° V v o ( t ) = 7.736 cos ( 5t − 55.1° ) V (checked: LNAP 8/4/04) P 10.6‐20 Determine the steady state current i(t) in each of these circuits: 4 i(t) + – 20 Ω 36 mA i(t) 8Ω (a) 4 i(t) + – 20 Ω 2H i(t) 36 cos (25t) mA 2 mF 15 Ω 4 mF (b) Solution: (a) Use KVL to see that the voltage across the 8 Ω resistor is 20i ( t ) − 4i ( t ) = 16i ( t ) . Apply KCL to the supernode corresponding to the dependent voltage source to get 16i ( t ) 0.036 = i ( t ) + = 3i ( t ) 8 so i ( t ) = 12 mA (b) Represent the circuit in the frequency domain using phasors and impedances. Z 1 = 20 + where 1 = 20 − j 20 Ω j ( 25 )( 0.002 ) ⎛ ⎞ 1 Z 2 = j 50 + ⎜⎜ 15 & ⎟⎟ = 43.3∠83.9° Ω j 25 0.004 ( )( ) ⎝ ⎠ V = Z 1I − 4I = ( Z 1 − 4 ) I Use KVL to get Then apply KCL to the supernode corresponding to the dependent source to get 0.036∠0° = I + so so I= (Z 1 − 4) I Z2 Z 2 ( 0.036∠0° ) Z1 + Z 2 − 4 ⎛ Z1 + Z 2 − 4 ⎞ =⎜ ⎟⎟ I ⎜ Z 2 ⎝ ⎠ = 50.4∠35.7° mA i ( t ) = 50.4 cos ( 25t + 35.7° ) mA (checked: LNAP 8/4/04) 10.6‐21 The input to the circuit show in Figure 10.9‐24 is the current i s ( t ) = 50 cos ( 200 t ) mA Determine the steady‐state mesh current i 2 . Figure P10.6‐21 Solution: Represent the circuit in the frequency domain using phasors and impedances: Apply KVL to the right mesh to get: ( j 20 − j 62.5) I 2 + 60 ( I 2 − 0.050∠0° ) = 0 In the time domain ⇒ I2 = 3∠0° = 0.0408∠35.3° A 60 − j 42.5 i 2 ( t ) = 40.8cos ( 200 t + 35.3° ) mA MATLAB, 11/20/09 10.6‐22 The input to the circuit show in Figure 10.9‐24 is the current i s ( t ) = 80 cos ( 250 t ) mA The steady‐state mesh current in the right mesh is i s ( t ) = 66.56 cos ( 250 t + 33.7° ) mA Figure P10.6‐22 Determine the value of the resistance R. Solution: Represent the circuit in the frequency domain using phasors and impedances. The mesh currents are I 1 = 0.080∠0° A and I 2 = 0.06656∠33.7° A Apply KVL to the right to get ( j 150 − j 200 )( 0.06656∠33.7° ) + R ( 0.06656∠33.7° − 0.080∠0° ) = 0 ( − j 50 )( 0.06656∠33.7° ) + R ( 0.044376∠123.7° ) = 0 R= ( 50∠90° )( 0.06656∠33.7° ) = 74.9955 75 Ω 0.044376∠123.7° MATLAB, 11/20/09 P10.6-23 This circuit shown in Figure P10.6-23 is at steady state. The voltage source voltages are given by v1(t) = 12 cos (2t – 90°) V and v2(t) = 5 cos (2t + 90°) V The currents are given by i1(t) = 744 cos (2t – 118°) mA , i2(t) = 540.5 cos (2t + 100°) mA Determine the values of R1 , R 2 , L and C. Figure P10.6-23 Solution: Represent the circuit in the frequency domain using impedances and phasors: I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠ − 164° In the time domain i ( t ) = 460 cos (2 t − 164°) mA Replacing series impedances by equivalent impedances gives Z 1 = R1 + j ω L and Z 2 = R2 − j From KVL 1 ωC Z 1 I 1 + 10 I − V1 = 0 ⇒ Z 1 = V1 − 10 I I1 = 12∠ − 90° − 10 ( 0.460∠ − 164° ) 0.744∠ − 118° − j 12 − 10 ( −0.443 − j 0.125 ) 0.744∠ − 118° 4.43 − j 10.75 11.63∠ − 67.6° = = 0.744∠ − 118° 0.744∠ − 118° = 15.63∠50.4° = = 10 + j 12 Ω and − Z 2 I 2 + V 2 − 10 I = 0 ⇒ Z 2 = V 2 − 10 I I2 = = 5∠90° − 10 ( 0.460∠ − 164° ) 0.5405∠100° j 5 − 10 ( −0.443 − j 0.125 ) 0.5405∠100° 4.43 + j 6.25 7.66∠54.7° = = 0.5405∠100° 0.5405∠100° = 14.14∠ − 55.3° = 10 − j 10 Ω Next and 12 =6H 2 1 ⇒ R 2 = 10 Ω and C = = 0.05 F 2 (10 ) 10 + j 12 = R1 + j ω L = R1 + j 2 L ⇒ R1 = 10 Ω and L = 10 − j 10 = R 2 − j 1 1 = R2 − j 2C ωC 10.7 Thevenin and Norton Equivalent Circuits P10.7‐1 Determine the Thevenin equivalent circuit of the circuit shown in Figure P10.7‐1 when (a) ω = 1000 rad/s, (b) ω = 2000 rad/s and (c) ω = 4000 rad/s. Figure P10.7‐1 Solution: Represent the circuit in the frequency domain as Determine the open circuit voltage and Thevenin impedance: Voc = ZC 2500 + Z C (12∠0° ) Zt = 2500 Z C 2500 + Z C + ZL a. Voc = 2.3534∠ − 78.69° V and Z t = 775.22∠82.875° Ω b. Voc = 1.194∠ − 84.29° V and Z t = 2252.6∠89.37° Ω c. Voc = 0.59925∠ − 87.14° V and Z t = 4875.3∠89.93° Ω The Thevenin Equivalent Circuit changes whenever the input frequency changes. P10.7‐2 Determine the Thevenin equivalent of this circuit when vs(t) = 5 cos (4000t‐30°) V. 1 80 mF 20 mH vs + – 80 Ω Solution: Find Voc : ⎛ 80 + j80 ⎞ Voc = ( 5 ∠−30° ) ⎜ ⎟ ⎝ 80 + j80− j 20 ⎠ ⎛ 80 2∠− 45° ⎞ = ( 5 ∠−30° ) ⎜ ⎟ ⎝ 100∠36.90° ⎠ = 4 2∠ − 21.9° V Find Z t : Zt = The Thevenin equivalent is ( − j 20 )( 80 + j80 ) = 23 ∠ − 81.9° − j 20 + 80 + j80 Ω P10.7‐3 Determine the Thevenin equivalent of this circuit 2v 600 Ω + 9 cos 500t – v + – + – 1 150 mF 300 Ω Solution: First, determine Voc : The mesh equations are 600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0° −2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒ j 3 I1 + (1 − j 3) I 2 = 0 I 2 = 0.0124∠ − 16° A Using Cramer’s rule: Voc = 300 I 2 = 3.71∠ − 16° V Then Next, determine I sc : −2 V − V = 0 ⇒ V = 0 ⇒ I sc = The Thevenin impedance is ZT = The Thevenin equivalent is 9∠0° = 0.015∠0° A 600 Voc 3.545∠−16° = = 247∠ − 16° Ω I sc 0.015∠0° P10.7‐4 Determine the Thevenin equivalent of this circuit when vs(t) = 10 cos (10,000t−53.1°) V. 3i / 2 2Ω 200 μH vs + – a i 25 μ F b Solution: First, determine Voc : The node equation is: Voc Voc − (6 + j8) 3 ⎛ Voc − (6+ j8) ⎞ + − ⎜ ⎟=0 − j4 j2 2⎝ j2 ⎠ Voc =3+ j 4 =5∠53.1° V Vs = 10∠53° = 6 + j 8 V Next, determine I sc : The node equation is: V V V − (6 + j8) 3 ⎡ V − (6 + j8) ⎤ + + − ⎢ ⎥=0 2 − j4 j2 2⎣ j2 ⎦ V= I sc = 3 + j4 1− j V 3+ j 4 = 2 2− j 2 The Thevenin impedance is ZT = Vs = 10∠53° = 6 + j 8 V ⎛ 2− j 2 ⎞ Voc = 3 + j4 ⎜ ⎟ = 2 − j2 Ω I sc ⎝ 3+ j 4 ⎠ The Thevenin equivalent is (checked: LNAP 7/18/04) P10.7‐5 Determine the frequency at which Y is a pure conductance 20 k Ω Y Solution: Y = G + YL + YC 1 Y = G when YL + YC = 0 or + jω C = 0 jω L 1 1 1 , fO = ωO = = 2π LC 2π 39.6×10−15 LC = 0.07998×107 Hz =800 kHz (80 on the dial of the radio) 1 nF 39.6 μH P 10.7-6 Consider the circuit of Figure P 10.7-6, where we wish to determine the current I. Use a series of source transformations to reduce the part of the circuit connected to the 2-Ω resistor to a Norton equivalent circuit, and then find the current in the 2-Ω resistor by current division. j4 Ω 3 30° A –j3 Ω 4Ω –j2 Ω 2Ω I Figure P 10.7-6 Solution: Z1 = (− j 3)(4) = 2.4∠ − 53.1° Ω − j 3+ 4 =1.44 − j1.92 Ω Z 2 = Z1 + j 4 = 1.44 + j 2.08 = 2.53∠55.3° Ω Z3 = 3.51∠ − 37.9° Ω = 2.77 − j 2.16 Ω ⎛ 3.51∠−37.9° ⎞ ( 3.51∠−37.9° ) = 1.9∠ − 92° A I = ( 2.85∠− 78.4° ) ⎜ ⎟ = ( 2.85∠− 78.4° ) ( 5.24∠− 24.4° ) ⎝ 2.77 − j 2.16 + 2 ⎠ (checked: LNAP 7/18/04) P 10.7-7 For the circuit of Figure P 10.7-7, 200 Ω determine the current I using a series of source transformations. 20 45° V + – 100 Ω 160 μ H 10 μ F The source has ω = 25 × 103 rad/s. I Answer: i(t) = 4 cos (25,000t – 44°) mA Figure P 10.7-7 Solution: Z2 = I= (200)(− j 4) = 4∠ − 88.8° Ω 200− j 4 0.4∠− 44° = 4∠ − 44° mA −4 j +100+ j 4 i (t ) = 4 cos (25000 t − 44°) mA P10.7‐8 Determine the Thevenin equivalent of the circuit in (a): a 8Ω Zt j10 Ω –j5 a + – +– Vt b j20 Ω –j2.4 Ω b (a) (b) Solution: V1 = V2 = j10 5 e − j 90 = 3.9 e− j 51 8 + j10 j 20 5 e − j 90 = 5.68 e− j 90 j 20 − j 2.4 Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 Zt = 8 ( j10 ) − j 2.4 ( j 20 ) + = 4.9 + j 1.2 8 + j10 − j 2.4 + j 20 P10.7‐9 Determine the voltage v(t) for this circuit: + 8Ω 2H 5 sin 5t V v(t) +– 1 12 F 4 cos (3t + 15°) A 4H – Solution: V1 (ω ) = V 2 (ω ) = j10 5 e − j 90 = 3.9 e − j 51 8 + j10 j 20 5 e − j 90 = 5.68 e − j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° ) V. 10.8 Superposition P10.8‐1 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐1 when the voltage source voltages are vs1 (t) = 12 cos (2500 t) V and vs2 (t) = 12 cos (4000 t) V Figure 10.8‐1 Solution: Use superposition in the time domain. These circuits can be used to find the part of io caused by vs1 and the part of io caused by vs2. In the frequency domain: 12∠0° = 0.03578∠26.6° V 300 + j 250 − j 400 12∠0° I2 = = 0.03578∠ − 26.6° V 300 − j 250 + j 400 I1 = In the time domain i o1 ( t ) = 35.78cos ( 2500 t + 26.6° ) mA and i o2 ( t ) = 35.78cos ( 4000 t − 26.6° ) mA and i o ( t ) = i o1 ( t ) + i o2 ( t ) = 35.78cos ( 2500 t + 26.6° ) + 35.78cos ( 4000 t − 26.6° ) mA V P10.8‐2 Determine the steady state voltage v(t) in the circuit shown in Figure P10.8‐2 when the current source current is (a) 400 rad/s and (b) 200 rad/s. Figure 10.8‐2 Solution: (a) Represent the circuit in the frequency domain as Use superposition in the frequency domain to write Vo = − = 100 j 150 (12∠0° ) + 100 ( 0.1∠0° ) 100 + j 150 100 + j 150 −1200 + j 1500 = 10.66∠72.35° 100 + j 150 In the time domain v o ( t ) = 10.66cos ( 400 t + 72.35 ) V (b) Use superposition in the time domain. These circuits can be used to find the part of vo caused by the current source and the part of vo caused by the voltage source. In the frequency domain: Vo1 = 100 Vo2 = − j 75 ( 0.1∠0° ) = 6∠53.1° V 100 + j 75 100 (12∠0° ) = 6.656∠123.7° V 100 + j 150 In the time domain v o1 ( t ) = 6cos ( 200 t + 53.1° ) V and v o2 ( t ) = 6.656cos ( 400 t + 123.7° ) V and v o ( t ) = v o1 ( t ) + v o2 ( t ) = 6cos ( 200 t + 53.1° ) + 6.656cos ( 400 t + 123.7° ) V P10.8‐3 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐3 when the voltage source voltage is vs (t) = 8 + 8 cos (400t − 135° ) V Figure 10.8‐3 Solution: Use superposition in the time domain: An inductor in a dc circuit acts like a short circuit so: i1 ( t ) = 8 = 0.533 A 15 Represent the right circuit the frequency domain: 8∠ − 135° = 0.32∠ − 188° A 15 + j 20 i 2 ( t ) = 0.32cos ( 400 t − 188° ) A I2 = In the time domain and i ( t ) = i1 ( t ) + i 2 ( t ) = 0.533 + 0.32cos ( 400 t − 188° ) A P10.8‐4 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐4 when the voltage source voltages are vs1 (t) = 10 cos (800t + 30° ) V and vs2 (t) = 15 sin (200t − 30° ) V Figure 10.8‐4 Solution: Use superposition in the time domain: Represent these circuits the frequency domain: I1 = 10∠30° 15∠ − 120° = 0.1342∠33.4° A = 0.0447∠ − 33.4° A and I 2 = − 100 + j 200 100 + j 50 In the time domain i ( t ) = i1 ( t ) + i 2 ( t ) = 44.7 cos ( 800 t − 33.4° ) + 134.2cos ( 200 t + 33.4° ) mA 4i(t) P 10.8-5 The input to the circuit shown in + – Figure P 10.8-5 is the current source current 20 Ω is(t) = 36 cos (25t) + 48 cos (50t + 45°) mA Determine the steady-state current, i(t). 2H i(t) is(t) 2 mF 15 Ω 4 mF Figure P 10.8-5 Solution: Use superposition in the time domain. Let i s1 ( t ) = 36 cos ( 25 t ) mA and i s 2 ( t ) = 48cos ( 50 t + 45° ) mA We will find the response to each of these inputs separately. Let ii(t) denote the response to isi(t) for i = 1,2. The sum of the two responses will be i(t), i.e. i ( t ) = i1 ( t ) + i 2 ( t ) Represent the circuit in the frequency domain as Use KVL to get Vi = Z i I i − 4I i Apply KCL to the supernode corresponding to the dependent voltage source. I si = I i + Vi Z2 = Z1 + Z 2 − 4 Z2 or Ii = Z 2I s i Z1 + Z 2 − 4 Ii Consider the case i = 1 : is1(t) = 26cos(25t) mA. Here ω = 25 rad/s and I si = 36∠0° mA Z 1 = 20 + 1 = 20 − j 20 Ω j ( 25 )( 0.002 ) ⎛ ⎞ 1 Z 2 = j 50 + ⎜⎜15 & ⎟⎟ = 43.3∠83.9° Ω 25 0.004 j ( )( ) ⎝ ⎠ and I 1 = 50.4∠35.7° mA so i ( t ) = 50.4 cos ( 25t + 35.7° ) mA Next consider i = 2 : is2 = 48cos(50t + 45°) mA. Here ω = 50 rad/s and I s2 = 48∠45° mA Z 1 = 20 + 1 = 20 − j10 Ω j ( 50 )( 0.002 ) ⎛ ⎞ 1 Z 2 = j100 + ⎜⎜15 & ⎟ = 95.5∠89.1° Ω j ( 50 )( 0.004 ) ⎟⎠ ⎝ (Notice that Z1 and Z2 change when ω changes.) I 2 = 52.5∠55.7° mA so i 2 ( t ) = 52.5cos ( 50t + 55.7° ) mA Finally, using superposition in the time domain gives i ( t ) = 50.4 cos ( 25t + 35.7° ) + 52.5cos ( 50t + 55.7° ) mA (checked: LNAP 8/7/04) 20 Ω 20 Ω 4H 5 mF P 10.8-6 The inputs to the circuit shown in 10 Ω Figure P 10.8-6 are i(t) vs1(t) = 30 cos (20t + 70°) V and vs2(t) = 18 cos (10t – 15°) V + – vs1(t) + 15 Ω vs2(t) – 2H Figure P 10.8-6 Solution: Use superposition in the time domain. Let i1(t) be the part of i(t) due to vs1(t) and i2(t) be the part of i(t) due to vs2(t). To determine i1(t), set vs2(t) = 0. Represent the resulting circuit in the frequency domain to get where Z 1 = 20 + j80 = 82.46∠76° Ω Z 2 = 10 + ( j 40 & 15 ) = 23.15 + j 4.93 = 23.67∠12° Ω Z 3 = 20 + 1 = 20 − j10 = 22.36∠ − 26.6° Ω j ( 20 )( 0.005 ) Next, using Ohm’s law and current division gives I1 = so Z3 Z 3 ( 30∠70° ) 30∠70° × = = 0.182∠ − 17.6° A Z 1 + ( Z 2 & Z 3 ) Z 2 + Z 3 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3 i ( t ) = 0.182 cos ( 20t − 17.6° ) A To determine i2(t), set vs1(t) = 0. Represent the resulting circuit in the frequency domain to get where Z 4 = 20 + j 40 = 44.72∠63.4° Ω Z 5 = 10 + ( j 20 & 15 ) = 19.6 + j 7.2 = 20.88∠20.2° Ω Z 6 = 20 + 1 = 20 − j 20 = 28.28∠ − 45° Ω j (10 )( 0.005 ) Next, using Ohm’s law and current division gives I2 = Z4 Z 1 (18∠ − 15° ) 18∠ − 15° × = = 0.377∠18° A Z 6 + ( Z 4 & Z 5 ) Z 4 + Z 5 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3 so i 2 ( t ) = 0.377 cos (10t + 18° ) A Using superposition, i ( t ) = i1 ( t ) + i 2 ( t ) = 0.182 cos ( 20t − 17.6° ) + 0.377 cos (10t + 18° ) A (checked: LNAP 8/8/04) P 10.8-7 The input to the circuit shown in Figure 5Ω 25 Ω P 10.8-7 is the voltage source voltage vs(t) = 5 + 30 cos (100t) V + – i(t) vs(t) Determine the steady-state current, i(t). 20 μF 50 mH Figure P 10.8-7 Solution: Use superposition in the time domain. Let vs1(t) = 5 V and vs2(t) = 30cos(100t) V. Find the steady state response to vs1(t). When the input is constant and the circuit is at steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. So 5 i1 ( t ) = = 1 A 5 Find the steady state response to vs2(t). Represent the circuit in the frequency domain using impedances and phasors. I2 = So 30∠0° = 4.243∠ − 45 A 5 + j5 i 2 ( t ) = 4.243cos (100t − 45° ) A Using superposition i ( t ) = i1 ( t ) + i 2 ( t ) = 1 + 4.243cos (100t − 45° ) P10.8‐8 Determine the voltage v(t) for the circuit + 8Ω 2H 5 sin 5t V v(t) +– 1 12 F 4 cos (3t + 15°) A 4H – Solution: V1 (ω ) = V 2 (ω ) = j10 5 e − j 90 = 3.9 e − j 51 8 + j10 j 20 5 e − j 90 = 5.68 e − j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° ) V. P10.8‐9 Determine the current i(t) for this circuit when v1(t)=10cos(10t) V 5Ω v1 + – i(t) 1.5 H 10 mF 10 Ω 3A Solution: Use superposition. First, find the response to the voltage source acting alone: Z eq = − j10⋅10 = 5(1 − j ) Ω 10− j10 Replacing the parallel elements by the equivalent impedance. The write a mesh equation : −10 + 5 I1 + j15 I1 + 5(1 − j ) I1 = 0 ⇒ I1 = 10 = 0.707∠ − 45° A 10+ j10 Therefore: i1 (t ) = 0.707 cos(10 t − 45° ) A Next, find the response to the dc current source acting alone: Current division: I2 = − Using superposition: i (t ) = 0.707 cos(10 t − 45°) − 2 A 10 × 3 = −2 A 15 Section 10-9: Phasor Diagrams P 10.9-1 Using a phasor diagram, determine V when V = V1 – V2 + V3* and V1 = 3 + j3, V2 = 4 + j2, and V3 = –3 – j2. (Units are volts.) Answer: V = 5 ∠143.1° V . Solution: V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( −3− j 2 ) = −4 + j 3 * * P 10.9-2 Consider the series RLC circuit of Figure P 10.9-2 when R = 10 Ω, L = 1 mH, C = 100 μF, and ω = 103 rad/s. Find I and plot the phasor diagram j ωL R 10 0° V + – I 1 j ωC Figure P 10.9-2 Solution: I= VR = R I = 7.4∠42° V VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V VS = 10∠0° V 10∠0° = 0.74∠42° A 10+ j1− j10 10.10 Op Amps in AC Circuits P10.10‐1 The input to the circuit shown in Figure P10.10‐1 is the voltage v s ( t ) = 2.4 cos ( 500 t ) V . Determine the output voltage vo(t). Answer: v o ( t ) = 6.788cos ( 500 t + 135° ) V Figure P10.10‐1 Solution: Represent the circuit in the frequency domain as Recognizing this circuit as an inverting amplifier, we can write 14.14∠ − 45° ⎞ ⎛ 20 || − j 20 ⎞ ⎛ Vo = ⎜ − ⎟ ( 2.4∠0 ) = ⎜ (1∠180° ) ⎟ ( 2.4∠0 ) = 6.788∠135° V 5 5 ⎝ ⎠ ⎝ ⎠ In the time domain v o ( t ) = 6.788cos ( 500 t + 135° ) V (Checked using LNAPAC 3/15/12) P10.10‐2 The input of the circuit shown in Figure P10.10‐2 is the voltage v s ( t ) = 1.2 cos ( 400 t + 20° ) V . Determine the output voltage vo(t). Figure P10.10‐2 Solution: Represent the circuit in the frequency domain as Recognizing this circuit as a noninverting amplifier, we can write ⎛ 48 ⎞ V0 = ⎜1 + ⎟ (1.2∠20° ) = (1 + j 4.8 )(1.2∠20° ) = 5.88∠98° V ⎝ − j 10 ⎠ In the time domain v o ( t ) = 5.88cos ( 400 t + 98° ) V (Checked using LNAPAC 3/15/12) P10.10‐3 The input of the circuit shown in Figure P10.10‐3 is the voltage v s ( t ) = 3.2 cos ( 200 t ) V . Determine the output voltage vo(t). Figure P10.10‐3 Solution: Represent the circuit in the frequency domain as Recognizing this circuit as a voltage divider followed by a noninverting amplifier, we can write ⎞ 20 20 ⎛ 40 ⎞ ⎛ ⎛ ⎞ Vo = ⎜1 + ⎟ ⎜ ⎟ ( 3.2∠0° ) = ⎜ ⎟ ( 9.6∠0° ) = 4.293∠63.4° V ⎝ 20 ⎠ ⎝ − j 40 + 20 ⎠ ⎝ 44.72∠ − 63.4° ⎠ In the time domain v o ( t ) = 4.293cos ( 200 t + 63.4° ) V (Checked using LNAPAC 3/15/12) P10.10‐4 The input of the circuit shown in Figure P10.10‐4 is the voltage v s ( t ) = 1.2 cos ( 2000 t ) V . Determine the output voltage vo(t). Figure P10.10‐4 Solution: Represent the circuit in the frequency domain as Recognizing this circuit as a voltage divider followed by a voltage follower, we can write ⎛ − j 2.5 ⎞ ⎛ 2.5∠ − 90° ⎞ Vo = ⎜ ⎟ (1.2∠0° ) = ⎜ ⎟ (1.2∠0° ) = 0.1974∠ − 80.54° V ⎝ 15.2∠ − 9.46° ⎠ ⎝ 15 − j 2.5 ⎠ v o ( t ) = 0.1974 cos ( 400 t − 80.54° ) V In the time domain (Checked using LNAPAC 3/15/12) P10.10‐5 The input of the circuit shown in Figure P10.10‐5 is the voltage v s ( t ) = 1.2 cos ( 2000 t ) V . Determine the output voltage vo(t). Figure P10.10‐5 Solution: Represent the circuit in the frequency domain as Recognizing this circuit as a inverting amplifier, we can write In the time domain ⎛ − j 50 ⎞ Vo = ⎜ − ⎟ (1.2∠0 ) = 12∠90° V 5 ⎠ ⎝ v o ( t ) = 12 cos ( 400 t + 90° ) V (Checked using LNAPAC 3/15/12) P 10.10-6 Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10-6. Z2 Z1 + – Vs Z4 Z3 – – + + + Vo – Figure P 10.10-6 Solution: Label the nodes: The ideal op amps force Va = 0 and Vc = 0. Apply KCL at node a to get Vb = Apply KCL at node c to get Vo = Z2 Z1 + Z 2 Z4 Z3 + Z4 Vs Vb Therefore Z4 Z2 Vo = × Vs Z 3 + Z 4 Z 1 + Z 2 P 10.10‐7 Determine the ratio Vo/Vs for both of the circuits shown in Figure P 10.10‐7. Z1 – Z1 Z3 Z3 + + + + – Vs Z4 Z2 + – Vo Vs Z4 Z2 – – (a) (b) Figure P 10.10-7 Solution: Label a node voltage as Va in each of the circuits. In both circuits, we can apply KCL at the node between Z3 and Z4 to get Vo = In (a) Va = = ( Z4 Z3 + Z4 Z 2 || Z 3 + Z 4 ( ) Z 1 + Z 2 || Z 3 + Z 4 ( ) Va Vs Z2 Z3 +Z4 ( ) ) ( Z1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4 ) Vs so Z2 Z4 Va = Vs Z 1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4 ( ) ( In (b) Va = Z2 Z1 + Z 2 Vs so Z4 Z2 Vo = × Vs Z 3 + Z 4 Z 1 + Z 2 Vo ) P 10.10‐8 Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10‐8. Z4 Z3 – Z1 Z5 + + – Vs + Z6 Z2 Vo – Figure P 10.10‐8 Solution: Label the node voltages Va and Vb as shown: Apply KCL at the node between Z1 and Z2 to get Z2 Va = Vs Z1 + Z 2 Apply KCL at the node between Z1 and Z2 to get Z3 + Z4 Vb = Va Z3 Apply KCL at the node between Z5 and Z6 to get Z6 Vo = Vb Z5 + Z6 so Z6 Z3 + Z4 Z2 Vo = × × Vs Z 5 + Z 6 Z3 Z1 + Z 2 + P 10.10-9 When the input to the circuit shown in Figure P 10.10-9 is the voltage source voltage R1 vs(t) = 2 cos (1000t) V + – the output is the voltage – + R2 vs(t) vo(t) + vo(t) = 5 cos (1000t – 71.6°) V C=1 mF Determine the values of the resistances R1 and R2. v(t) 1 kΩ – Figure P 10.10-9 Solution: The network function of the circuit is 1 1+ R2 1+ R2 R2 ⎞ jω C ⎛ 1000 = 1000 = ⎜1 + = ⎟ 1 V s ⎝ 1000 ⎠ R + 1 + j ω C R1 1 + j 10−3 R1 1 jω C Vo Converting the given input and output sinusoids to phasors gives Vo Vs = 5 ∠71.6° 2 Consequently 1+ R2 5 ∠71.6° 1000 = 2 1 + j 10−3 R1 Equating angles gives ( 71.6° = − tan −1 10−3 R1 ) ⇒ R1 = tan ( 71.6° ) × 10 3 = 3006 Ω Equating magnitudes gives R2 R2 1+ 1+ 5 1000 1000 = = 2 2 1 + 10−3 R1 1 + 10−3 × 3006 ( ) ( ) 2 ⎛5 ⎞ ⇒ R 2 = ⎜ 10 − 1⎟ × 10 3 = 6906 Ω ⎝2 ⎠ – P 10.10-10 When the input to the circuit 10 kΩ source voltage vs(t) + – C R 10 kΩ shown in Figure P 10.10-10 is the voltage + – v(t) + – vs(t) = 4 cos (100t) V + 10 kΩ – the output is the voltage vo(t) = 8 cos (100t + 135°) V Figure P 10.10‐10 Determine the values of C and R. Solution: Represent the circuit in the frequency domain as Apply KCL at the top node of the impedance of the capacitor to get Vs − V 1 V V = + 4 ⇒ Vs = 1 + j ( 5 × 105 ) C V 4 1 10 10 2 j100C ( ) Apply KCL at the inverting node of the op amp to get V Vo + =0 104 R ⇒ Vo = − R V 104 R 2 ×104 = Vs 1 + j ( 5 ×105 ) C Vo so − Converting the input and output sinusoids to phasors gives Vo Vs = 8∠135° = 2∠135° 4∠0° R R 4 2 ×10 2 ×104 2∠135° = = ∠180° − tan −1 ( 5 × 105 ) C 5 2 1 + j ( 5 ×10 ) C 1 + ⎡⎣( 5 × 105 ) C ⎤⎦ − so vo(t) ( ) Equating angles gives ( 135° = 180° − tan −1 ( 5 × 105 ) C ) ⇒ C= tan ( 45° ) = 2 × 10−6 = 2 μ F 5 ×105 Next, equating magnitudes gives R R 4 4 2 ×10 2= = 2 ×10 2 1 + ( 5 × 105 )( 2 × 10−6 ) ⇒ R = 104 = 10 kΩ P10.10‐11 The input to the circuit shown in Figure P10.10‐11 is the voltage source voltage, v s ( t ) . The output is the voltage v o ( t ) . The input v s ( t ) = 2.5cos (1000 t ) V causes the output to be v o ( t ) = 8cos (1000 t + 104° ) V . Determine the values of the resistances R1 and R2 . Answers: R1 = 1515 Ω and R 2 = 20 kΩ. Figure P10.10-11 Solution: R2 1 jω C = R2 1 + jω CR 2 R2 R2 R1 1 + jω CR 2 Vo (ω ) =− =− R1 Vi (ω ) 1 + jω CR 2 R2 R1 Vo (ω ) j (180− tan −1 ω CR 2 ) e = 2 Vi (ω ) 1 + (ω CR 2 ) In this case the angle of the magnitude of Vo (ω ) tan (180° − 104° ) is specified to be 104° so CR 2 = = 0.004 and Vi (ω ) 1000 Vo (ω ) 8 is specified to be so 2.5 Vi (ω ) R2 R1 1 + 16 = 8 ⇒ 2.5 R2 R1 = 13.2 . One set of values that satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ . Section 10.11 The Complete Response Figure P10.11‐1 P10.11‐1 The input to the circuit shown in Figure P10.11‐1 is the voltage v s = 12 cos ( 4000 t ) V . The output is the capacitor voltage, vo. Determine vo. Solution: Before the switch closes the circuit is represented in the frequency domain as Express the dependent source voltage in terms of the node voltages: 150 Ι a = Va − Vo Using Ohm’s law 150 so Va 200 = Va − Vo Vo = 1 Va 4 Apply KCL to the supernode corresponding to the dependent source to get Vs − V a 100 = Va 200 + Vo − j 12.5 ⇒ Vo = − j 12.5 ⎛ − j 12.5 − j 12.5 ⎞ Vs − ⎜ + ⎟ Va 100 200 ⎠ ⎝ 100 Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo ) − j 0.125 (12∠0° ) = 1.2∠ − 53.1° V 1 − j 0.75 The corresponding sinusoid is 1.2 cos ( 4000 t − 53.1° ) V . The initial capacitor voltage is Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo = v o ( 0 ) = 1.2 cos ( −53.1° ) = 0.7205 V . The steady state response after the switch closes is the forced response. The circuit is represented in the frequency domain as Express the dependent source voltage in terms of the node voltages: 150 Ι a = Va − Vo Using Ohm’s law Va 150 = Va − Vo 200 1 Vo = Va so 4 Apply KCL to the supernode corresponding to the dependent source to get Vs − V a 100 = Va 200 + Vo − j 12.5 + ⎛ 1 1 ⎞ 1 1 ⎞ ⎛ 1 ⇒ Vo ⎜ + ⎟= + Vs − ⎜ ⎟ Va 25 ⎝ 100 200 ⎠ ⎝ − j 12.5 25 ⎠ 100 Vo Multiply by 200 to get Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo = 24 = 0.937∠ − 38.7° V 20 + j 16 The corresponding sinusoid is the forced response: v f ( t ) = 0.937 cos ( 4000 t − 38.7° ) V The natural response is v n ( t ) = k e− t τ V To determine the time constant τ we need find to find the Thevenin resistance of the part of the circuit connected to the capacitor after the switch closes. Here’s the circuit: The terminals separate the capacitor from the part of the circuit connected to the capacitor. Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input to zero and (3) connect a current source to the terminals to get The Thevenin resistance is given by Rt = vt it Express the dependent source voltage in terms of the node voltages to get v a − v t = 150 i a = 150 va ⇒ va = 4 vt 200 Apply KCL to the supernode corresponding to the dependent source to get it = va 100 The time constant is The natural response is + va 200 + vt 25 = 4vt 100 + 4vt 200 + vt 25 = vt 10 ⇒ Rt = vt it = 10 Ω τ = R t C = 10 ( 20 × 10−6 ) = 0.2 × 10−3 = 0.2 ms v n ( t ) = k e − t τ = k e − 5000 t V The complete response is v o ( t ) = 0.937 cos ( 4000 t − 38.7° ) + k e −5000 t V for t ≥ 0 Using the initial condition we calculate 0.7205 = v o ( 0 ) = 0.937 cos ( −38.7° ) + k ⇒ k = −0.0108 Finally v o ( t ) = 0.937 cos ( 4000 t − 38.8° ) − 0.0108 e −5000t V for t ≥ 0 Figure P10.11‐2 P10.11‐2 The input to the circuit shown in Figure P10.11‐2 is the voltage v s = 12 cos ( 4000 t ) V . The output is the capacitor voltage, vo. Determine vo. Solution: Before the switch opens the circuit is represented in the frequency domain as Express the dependent source voltage in terms of the node voltages: 150 Ι a = Va − Vo Using Ohm’s law Va 150 = Va − Vo 200 1 so Vo = Va 4 Apply KCL to the supernode corresponding to the dependent source to get Vs − V a 100 = Va 200 + Vo − j 12.5 + ⎛ 1 1 ⎞ 1 1 ⎞ ⎛ 1 Vs − ⎜ ⇒ Vo ⎜ + ⎟= + ⎟ Va 25 ⎝ 100 200 ⎠ ⎝ − j 12.5 25 ⎠ 100 Vo Multiply by 200 to get Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo = The corresponding sinusoid is The initial condition is 24 = 0.937∠ − 38.7° V 20 + j 16 0.937 cos ( 4000 t − 38.7° ) V v o ( 0 ) = 0.937 cos ( −38.7° ) = 0.7313 V The steady state response after the switch closes is the forced response. The circuit is represented in the frequency domain as Express the dependent source voltage in terms of the node voltages: 150 Ι a = Va − Vo Using Ohm’s law 150 Va 200 = Va − Vo Vo = so 1 Va 4 Apply KCL to the supernode corresponding to the dependent source to get Vs − V a 100 = Va 200 + Vo − j 12.5 ⇒ Vo = − j 12.5 ⎛ − j 12.5 − j 12.5 ⎞ + Vs − ⎜ ⎟ Va 100 200 ⎠ ⎝ 100 Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo ) Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo = − j 0.125 (12∠0° ) = 1.2∠ − 53.1° V 1 − j 0.75 The corresponding sinusoid is the forced response: v f ( t ) = 1.2 cos ( 4000 t − 53.1° ) V The natural response is v n ( t ) = k e− t τ V To determine the time constant τ we need find to find the Thevenin resistance of the part of the circuit connected to the capacitor after the switch closes. Here’s the circuit: The terminals separate the capacitor from the part of the circuit connected to the capacitor. Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input to zero and (3) connect a current source to the terminals to get The Thevenin resistance is given by Rt = vt it Express the dependent source voltage in terms of the node voltages to get v a − v t = 150 i a = 150 va 200 ⇒ va = 4 vt Apply KCL to the supernode corresponding to the dependent source to get it = va 100 + va 200 = 4vt 100 + 4vt 200 = 0.06 v t ⇒ Rt = vt it = 16.67 Ω The time constant is τ = R t C = 16.67 ( 20 ×10−6 ) = 0.333 × 10−3 = 0.333 ms The natural response is v n ( t ) = k e − t τ = k e − 3000 t V The complete response is v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + k e −3000 t V for t ≥ 0 Using the initial condition we calculate 0.7313 = v o ( 0 ) = 1.2 cos ( −53.1° ) + k ⇒ k = 0.0108 Finally v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + 0.0108 e −3000 t V for t ≥ 0 Section 10.12 Using MATLAB to Analyze Electric Circuits 10.12-1 Determine the mesh currents for the circuit shown in Figure P10.12-1 when vs(t) = 12 cos(2500t + 60°) V and is(t) = 2 cos(2500t − 15°) mA Figure P10.12‐1 Solution: Represent the circuit in the frequency domain: Represent the source current in terms of the mesh currents: I 2 − I 1 = I s = 0.002∠ − 15° A Apply KVL to the supermesh corresponding to the current source: ( 2000 − j 2000 ) I 1 + j 1200 ( I 1 − I 3 ) + 1500 ( I 2 − I 3 ) + ( 4000 − j 1000 ) I 2 = 0 Apply KVL to mesh 3: ( 2400 + j 1875) I 3 + 1500 ( I 3 − I 2 ) + j 1200 ( I 3 − I 1 ) = −Vs = −12∠60° −1 1 0 ⎡ ⎤ ⎡ I 1 ⎤ ⎡0.002∠ − 15° ⎤ ⎢ ⎥ ⎢ ⎥ In matrix form: ⎢ 2000 − j 800 5500 − j 1000 −1500 − j 1200 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢ 0 ⎥ ⎢⎣ − j 1200 3900 + j 3075 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ −12∠60° ⎥⎦ −1500 Solving, using MATLAB, gives In the time domain: ⎡ I 1 ⎤ ⎡1.549∠ − 164° ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ I 2 ⎥ = ⎢ 1.039∠ − 65° ⎥ mA ⎢ I 3 ⎥ ⎢⎣ 2.904∠ − 148° ⎥⎦ ⎣ ⎦ ⎡ i1 ⎤ ⎡1.549 cos ( 2500 t − 164° ) ⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢ 1.039 cos ( 2500 t − 65° ) ⎥ mA ⎢ i 3 ⎥ ⎢ 2.904 cos ( 2500 t − 148° ) ⎥ ⎦ ⎣ ⎦ ⎣ 10.12-2 Determine the node voltages for the circuit shown in Figure P10.12-2 when vs(t) = 12 cos(400t + 45°) V. Figure P10.12‐2 Solution: Represent the circuit in the frequency domain: Represent the dependent source voltage in terms of the node voltages currents: V2 V3 − V 2 = 10000 ⇒ V3 = 3.5 V 2 4000 Apply KCL to the supernode corresponding to the dependent voltage source: Vs − V 2 j 2000 = V2 4000 + V3 j 3200 + V3 − V 4 4000 ⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ ⎛ 1 ⎞ =⎜ + + ⎟ V2 + ⎜ ⎟ V3 − ⎜ ⎟ V4 j 2000 ⎝ 4000 j 2000 ⎠ ⎝ 4000 ⎠ ⎝ 4000 j 3200 ⎠ Vs − V 4 V3 − V 4 V4 Apply KCL at node 3: + = 10, 000 4000 − j 20, 000 Rearranging: Vs Rearranging: Vs 10, 000 =− ⎡ ⎢ 3.5 ⎢ ⎢ 1 1 + In matrix form: ⎢ ⎢ 4000 j 2000 ⎢ 0 ⎢ ⎣ Solving, using MATLAB, gives In the time domain: ⎛ 1 ⎞ 1 1 +⎜ + + ⎟ V4 4000 ⎝ 4000 10, 000 − j 20, 000 ⎠ V3 ⎡ ⎤ ⎤ ⎢ 0 ⎥ ⎥ −1 0 ⎥ ⎥ ⎡ V2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ Vs ⎥ 1 1 1 + − ⎥ ⎥ ⎢ V3 ⎥ = ⎢ 4000 j 3200 4000 ⎥ ⎢ V ⎥ ⎢ j 2000 ⎥ ⎥ ⎣ 4 ⎦ ⎢ Vs ⎥ 1 1 1 1 − + + ⎢ ⎥ ⎥ 4000 4000 10, 000 − j 20, 000 ⎦ ⎣10, 000 ⎦ ⎡ V 2 ⎤ ⎡ 3.236∠34° ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ V3 ⎥ = ⎢11.324∠34° ⎥ V ⎢ V 4 ⎥ ⎢⎣10.798∠29°⎥⎦ ⎣ ⎦ ⎡ v 2 ⎤ ⎡ 3.236 cos ( 400 t + 34° ) ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ v 3 ⎥ = ⎢11.324 cos ( 400 t + 34° ) ⎥ V ⎢ v 4 ⎥ ⎢10.798cos ( 400 t + 29° ) ⎥ ⎦ ⎣ ⎦ ⎣ 10.12-3 Determine the mesh currents for the circuit shown in Figure P10.12-3 Figure P10.12‐3 Solution: Represent the source current in terms of the mesh currents: I 1 − I 2 = 4.2∠30° A Apply KVL to the supermesh corresponding to the current source: j 8 ( I 1 − I o ) + 5 ⎡⎣ j 8 ( I 1 − I o ) ⎤⎦ + ( 4 + j 5 ) I 2 + ( 3 − j 8 ) I 1 = 0 5 I o + 6 ( j 8) ( I 1 − I o ) = 0 Apply KVL to mesh 3: In matrix form: −1 0 ⎤ ⎡ I 1 ⎤ ⎡ 4.2∠30°⎤ ⎡ 1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢3 + j 40 4 + j 5 − j 48 ⎥ ⎢ I 2 ⎥ = ⎢ 0 ⎥ ⎢⎣ − j 48 0 5 + j 48⎥⎦ ⎢⎣I o ⎥⎦ ⎢⎣ 0 ⎥⎦ Solving, using MATLAB, gives ⎡ I 1 ⎤ ⎡ 2.204∠93.1°⎤ ⎢ ⎥ ⎢ ⎥ ⎢ I 2 ⎥ = ⎢ 3.758∠178° ⎥ A ⎢ I o ⎥ ⎢⎣ 2.192∠99° ⎥⎦ ⎣ ⎦ 10.12-4 Determine the node voltages for the circuit shown in Figure P10.12-4. Figure P10.12‐4 Solution: ⎛ 24∠45° − V 2 ⎞ V 2 V 2 − Vo + 10 ⎜ + ⎟= 2+ j4 2+ j4 j6 ⎝ ⎠ 4 24∠45° − V3 V 2 − Vo Vo + = Apply KCL at node 3: j6 5− j2 4 11 1 ⎡1 1 ⎤ ⎡ 11 ⎤ − 24∠45° ) ⎥ ( ⎢4 + j 6 + 2+ j 4 ⎥ ⎢ j6 ⎡V ⎤ 2+ j4 ⎥⎢ 2⎥ = ⎢ ⎥ In matrix form: ⎢ 1 1 1 1 ⎥ ⎣ Vo ⎦ ⎢ 1 ⎢ ⎥ − + + ⎥ ( 24∠45° ) ⎥ ⎢ ⎢ j6 5− j 2 j 6 4⎦ ⎣ ⎣5− j 2 ⎦ V ⎡ 2 ⎤ ⎡ 22.13∠40.8°⎤ Solving, using MATLAB, gives ⎢V ⎥ = ⎢ ⎥ V ⎣ o ⎦ ⎣10.07∠30.4° ⎦ Apply KCL at node 2: 24∠45° − V 2 10.12-5 The input to the circuit shown in Figure P10.12-5 is the voltage source voltage vs(t) = 12 cos(20,000t + 60°) V . and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output sinusoids. Figure P10.12‐5 Solution: Represent the circuit in the frequency domain: Write a node equation: 12∠60° − Vo 200, 000 + 12∠60° − Vo − j 25, 000 + 12∠60° =0 20, 000 ⎛ ⎛ ⎞ 1 1 1 ⎞ 1 1 Rearrange: ⎜ + + + ⎟12∠60° = ⎜ ⎟ Vo ⎝ 200, 000 − j 25, 000 20, 000 ⎠ ⎝ 200, 000 − j 25, 000 ⎠ Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for Introduction to Electric Circuits): %----------------------------------------------% Describe the input voltage source. %----------------------------------------------w = 20000; A = 12; theta = (pi/180)*60; Vs = A*exp(j*theta); %-----------------------------------------------% Describe the impedances. %-----------------------------------------------R1=50e3; R2=200e3; ZC=-j*25e3; %-----------------------------------------------% Calculate the phasor corresponding to the % output voltage. %-----------------------------------------------Vo=(1/R2 + 1/ZC + 1/R1)*Vs/(1/R2 + 1/ZC); B = abs(Vo) phi = angle(Vo) %-----------------------------------------------% %-----------------------------------------------T = 2*pi/w; tf = 2*T; N = 100; dt = tf/N; t = 0 : dt : tf; %-----------------------------------------------% Plot the input and output voltages. %-----------------------------------------------for k = 1 : 101 vs(k) = A * cos(w * t(k) + theta); vo(k) = B * cos(w * t(k) + phi); end plot (t, vs, t, vo) to get the plot: 10.12-6 The input to the circuit shown in Figure P10.12-6 is the voltage source voltage vs(t) = 3 cos(4000t + 30°) V . and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output sinusoids. Figure P10.12‐6 Solution: Represent the circuit in the frequency domain: Recognize this circuit as an inverting amplifier to write: 20 || − j 12.5 ⎛ 20 || − j 12.5 ⎞ Vo = ⎜ − (12∠60° ) ⎟ Vs = − 5 5 ⎝ ⎠ Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for Introduction to Electric Circuits): %----------------------------------------------% Describe the input voltage source. %----------------------------------------------w = 4000; A = 3; theta = (pi/180)*30; Vs = A*exp(j*theta); %-----------------------------------------------% Describe the impedances. %-----------------------------------------------R1=5e3; R2=20e3; ZC=-j*12.5e3; Zp=R2*ZC/(R2+ZC); %-----------------------------------------------% Calculate the phasor corresponding to the % output voltage. %-----------------------------------------------Vo=(-Zp/R1)*Vs; B = abs(Vo) phi = angle(Vo) %-----------------------------------------------% %-----------------------------------------------T = 2*pi/w; tf = 2*T; N = 100; dt = tf/N; t = 0 : dt : tf; %-----------------------------------------------% Plot the input and output voltages. %-----------------------------------------------for k = 1 : 101 vs(k) = A * cos(w * t(k) + theta); vo(k) = B * cos(w * t(k) + phi); end plot (t, vs, t, vo) to get the plot: Section 10.14 How Can We Check…? P 10.14-1 Computer analysis of the circuit in Figure P 10.14-1 indicates that the values of the node voltages are V1 = 20 ∠–90° and V2 = 44.7 ∠–63.4°. Are the values correct? Hint: Calculate the current in each circuit element using the values of V1 and V2. Check to see whether KCL is satisfied at each node of the circuit. j10 Ω V1 10 Ω 2A V2 10 Ω 3Ix Ix Figure P 10.14-1 Solution: Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the division in rectangular form. Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40 KCL at node 1: 2− V1 10 − V1 − V 2 j 10 = 2− − j 20 − j 20 − ( 20 − j 40 ) − = 2+ j2−2− j2 = 0 10 j 10 KCL at node 2: V1 − V 2 j 10 − ⎛ V1 ⎞ − j 20 − ( 20 − j 40 ) 20 − j 40 ⎛ − j 20 ⎞ + 3⎜ ⎟ = − + 3⎜ ⎟ = ( 2 + j 2) − ( 2 − j4) − j 6 = 0 j 10 10 10 ⎝ 10 ⎠ ⎝ 10 ⎠ V2 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. P 10.14-2 Computer analysis of the circuit in Figure P 10.14-2 indicates that the mesh currents are i1(t) = 0.39 cos (5t + 39°) A and i2(t) = 0.28 cos (5t + 180°) A. i1 8Ω 2H 5 sin 5t V Is this analysis correct? +– Hint: Represent the circuit in the frequency domain using impedances and phasors. Calculate the voltage across each circuit element using the values of I1 and I2. Check to see whether KVL is satisfied for each mesh of the circuit. 1 12 F i2 4H Figure P 10.14-2 Solution: I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180° Generally, it is more convenient to multiply complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the multiplication in rectangular form. Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284 KVL for mesh 1: 8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10 ≠ 0 Since KVL is not satisfied for mesh 1, the mesh currents are not correct. Here is a MATLAB file for this problem: Vs = -j*5; Z1 = 8; Z2 = j*10; Z3 = -j*2.4; Z4 = j*20; % Mesh equations in matrix form Z = [ Z1+Z2 0; 0 Z3+Z4 ]; V = [ Vs; -Vs ]; I = Z\V abs(I) angle(I)*180/3.14159 % Verify solution by obtaining the algebraic sum of voltages for % each mesh. KVL requires that both M1 and M2 be zero. M1 = -Vs + Z1*I(1) +Z2*I(1) M2 = Vs + Z3*I(2) + Z4*I(2) P 10.14-3 Computer analysis of the circuit in Figure P 10.14-3 indicates that the values of the node voltages are v1(t) = 19.2 cos (3t + 68°) V v1 8Ω 2H and 4 cos (3t + 15°) A v2(t) = 2.4 cos (3t + 105°) V. Is this analysis correct? 1 12 F 4H Hint: Represent the circuit in the frequency domain using impedances and phasors. Calculate the current in each circuit element using the values of V1 and V2. Check to see whether KCL is satisfied at each node of the circuit. v2 Figure P 10.14-3 Solution: V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V KCL at node 1 : 19.2 ∠ 68° 19.2 ∠ 68° + − 4∠15 = 0 8 j6 KCL at node 2: 24 ∠105° 24 ∠105° + + 4∠15 = 0 j12 − j4 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. Here is a MATLAB file for this problem: Is = 4*exp(j*15*3.14159/180); Z1 = 8; Z2 = j*6; Z3 = -j*4; Z4 = j*12; Y = [ 1/Z1 + 1/Z2 0; 0 1/Z3 + 1/Z4 ]; I = [ Is; -Is ]; V = Y\I abs(V) angle(V)*180/3.14159 % Verify solution by obtaining the algebraic sum of currents for % each node. KCL requires that both M1 and M2 be zero. M1 = -Is + V(1)/Z1 + V(1)/Z2 M2 = Is + V(2)/Z3 + V(2)/Z4 P 10.14-4 A computer program reports that the currents of the circuit of Figure P 10.14-4 are I = 0.2 ∠53.1° A, I1 = 632 ∠–18.4° mA, and I2 = 190 ∠71.6° mA. Verify this result. j500 Ω I 100 0° V + – 3000 Ω –j1000 Ω I1 I2 Figure P 10.14-4 Solution: First, replace the parallel resistor and capacitor by an equivalent impedance ZP = (3000)(− j 1000) = 949 ∠ − 72° = 300 − j 900 Ω 3000− j 1000 The current is given by VS 100 ∠0° = = 0.2∠53° A I= j 500+ Z P j 500+ 300− j 900 Current division yields ⎛ − j 1000 ⎞ I1 = ⎜ ⎟ ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA ⎝ 3000 − j 1000 ⎠ ⎛ ⎞ 3000 I2 = ⎜ ⎟ ( 0.2 ∠53° ) = 190∠71.4° mA ⎝ 3000 − j 1000 ⎠ The reported value of I1 is off by an order of magnitude. P 10.14-5 The circuit shown in Figure P 10.14-5 was built using a 2 percent resistor having a nominal resistance of 500 Ω and a 10 percent capacitor with a nominal capacitance of 5 μF. The steady-state capacitor voltage was measured to be v(t) = 18.3 cos (200t – 24°) V The voltage source represents a “signal generator.” Suppose that the signal generator was adjusted so carefully that errors in the amplitude, frequency, and angle of the voltage source voltage are all negligible. Is the measured response explained by the component tolerances? That is, could the measured v(t) have been produced by this circuit with a resistance R that is within 2 percent of 500 Ω and a capacitance C that is within 5 percent of 5 μF? R + 20 cos (200t) V + – C v(t) – Figure P 10.14-5 Solution: Represent the circuit in the frequency domain using phasors and impedances. Use voltage division to get 1 j 200C 18.3∠ − 24° = × 20∠0° 1 R+ j 200C 1 1 0.915∠ − 24° = = ∠ − tan −1 ( 200CR ) So 2 1 + j 200CR 1 + ( 200CR ) Equating angles gives −24° = − tan −1 ( 200CR ) ⇒ 200CR = tan ( 24° ) = 0.4452 The nominal component values cause 200CR = 0.5. So we expect that the actual component values are smaller than the nominal values. Try C = 5 (1 − 0.10 ) × 10−6 = 4.5 μ F Then R= 0.4452 = 494.67 Ω 200 × 4.5 × 10−6 500 − 494.67 = 0.01066 = 1.066% this resistance is within 2% of 500 Ω. We conclude 500 that the measured angle could have been caused by a capacitance that is within 10% of 5 μF and the resistance is within 2% of 500 Ω. Let’s check the amplitude. We require Since 1 1 + ( 0.4452 ) 2 = 0.9136 0.915 So the measured amplitude could also have been caused by the given circuit with C = 4.5 μF and R = 494.67 Ω. We conclude that he measured capacitor voltage could indeed have been produced by the given circuit with a resistance that is within 2 % of 500 Ω and a capacitance that is within 10% of 5 μF. PSpice Problems SP 10-1 The circuit shown in Figure SP 10.1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and is(t) = 2 sin (6t + 60°) A 1H vs(t) + – 3Ω 1 12 F + v(t) – is(t) Figure SP 10.1 Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) is sinusoidal and has the same frequency as vs(t) and is(t). Solution: SP 10-2 The circuit shown in Figure SP 10.1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and is(t) = 2 sin (18t + 60°) A 1H vs(t) + – 3Ω 1 12 F + v(t) – is(t) Figure SP 10.1 Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) is not sinusoidal. Solution: SP 10-3 The circuit shown in Figure SP 10-1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and is(t) = 0.8 A 1H vs(t) + – 3Ω 1 12 F + v(t) – is(t) Figure SP 10.1 Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice part VSIN and IDC for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) looks sinusoidal, but it’s not sinusoidal because of the dc offset. Solution: SP 10-4 The circuit shown in Figure SP 10- 1H 3Ω 1 has two inputs, vs(t) and is(t), and one output, v(t). When inputs are given by vs(t) + – 1 12 F vs(t) = Vm sin 6t V and is(t) = Im A + v(t) – is(t) Figure SP 10.1 the output will be vo(t) = A sin (6t + θ) + B V Linearity requires that A be proportional to Vm and that B be proportional to Im. Consequently, we can write A = k1Vm and B = k2Im, where k1 and k2 are constants yet to be determined. (a) Use PSpice to determine the value of k1 by simulating the circuit using Vm = 1 V and Im = 0. (b) Use PSpice to determine the value of k2 by simulating the circuit using Vm = 0 V and Im = 1. (c) Knowing k1 and k2, specify the values of Vm and Im that are required to cause vo(t) = 5 sin (6t + θ) + 5 V Simulate the circuit using PSpice to verify the specified values of Vm and Im. Solution: The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of Vm and Im are Vm = 12.5 V and Im = -1.667 A. Design Problems DP 10-1 Design the circuit shown in Figure DP 10-1 to produce the specified output voltage vo(t) = 8cos(1000t + 104°) V when provided with the input voltage vi(t) = 2.5cos(1000t) V. Figure DP 10-1 Solution: R2 1 jω C = R2 1 + jω CR 2 R2 R2 1 + jω CR 2 R1 Vo (ω ) =− =− Vi (ω ) R1 1 + jω CR 2 R2 R1 Vo (ω ) j (180 − tan −1 ω CR 2 ) e = 2 Vi (ω ) 1 + (ω CR 2 ) In this case the angle of the magnitude of Vo (ω ) tan (180° − 104° ) is specified to be 104° so CR 2 = = 0.004 and 1000 Vi (ω ) Vo (ω ) 8 is specified to be so Vi (ω ) 2.5 R2 R1 1 + 16 = 8 ⇒ 2.5 R2 R1 = 13.2 . One set of values that satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ . DP 10-2 Design the circuit shown in Figure DP 10-2 to produce the specified output voltage vo(t) = 2.5cos(1000t − 76°) V when provided with the input voltage vi(t) = 12cos(1000t) V. Figure DP 10-2 Solution: R2 Vo (ω ) = Vi (ω ) where K = 1 jω C = R2 1 + jω CR 2 R2 1 + jω CR 2 K = R2 1 + jω CR p R1 + 1 + jω CR 2 R1 R1 + R 2 and R p = R1 R 2 R1 + R 2 Vo (ω ) K − j tan −1 ω CR p = e 2 Vi (ω ) 1 + (ω CR p ) In this case the angle of C Rp = C R1 R 2 R1 + R 2 =− Vo (ω ) is specified to be -76° so Vi (ω ) tan ( −76 ) V (ω ) 2.5 is specified to be = 0.004 and the magnitude of o so 1000 Vi (ω ) 12 R2 K 2.5 = ⇒ 0.859 = K = . One set of values that satisfies these two equations is R1 + R 2 1 + 16 12 C = 0.2 μ F, R1 = 23.3 kΩ, R 2 = 142 kΩ . DP 10-3 Design the circuit shown in Figure DP 10-3 to produce the specified output voltage vo(t) = 2.5cos(40t + 14°) V when provided with the input voltage vi(t) = 8cos(40t) V. Figure DP 10-3 Solution: jω L R 2 L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) In this case the angle of jω ω = L R1 ⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠ 2 e ⎛ L ⎞ j ⎜ 90 − tan −1 ω ⎟ ⎜ ⎟ R p ⎠ ⎝ Vo (ω ) is specified to be 14° so Vi (ω ) V (ω ) L L ( R1 + R 2 ) tan ( 90° − 14° ) 2.5 is specified to be so = = = 0.1 and the magnitude of o 40 Rp R1 R 2 Vi (ω ) 8 40 L R1 = 2.5 ⇒ 8 L = 0.0322 . One set of values that satisfies these two equations is R1 1 + 16 L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω . DP 10-4 Show that it is not possible to design the circuit shown in Figure DP 10-4 to produce the specified output voltage vo(t) = 2.5cos(40t − 14°) when provided with the input voltage vi(t) = 8cos(40t) V. Figure DP 10-4 Solution: jω L R 2 L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) In this case the angle of jω ω = L R1 ⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠ Vo (ω ) is specified to be −14°. This requires Vi (ω ) L L ( R1 + R 2 ) tan ( 90 + 14 ) = = = −0.1 Rp R1 R 2 40 This condition cannot be satisfied with positive element waves. 2 e ⎛ L ⎞ j ⎜ 90 − tan −1 ω ⎟ ⎜ R p ⎟⎠ ⎝ v DP 10-5 A circuit with an unspecified R, L, and C is shown in Figure DP 10-5. The input source is is = 10 cos 1000t A, R is 10 Ω C L and the goal is to select the R, L, and C so that the node voltage is v = 80 cos 1000t V. Figure DP 10-5 Solution: Z1 =10 Ω 1 Z2 = jω C Z3 = R + jω L 1 S 10 Y2 = jω C 1 Y3 = R + jω L Y1 = v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 80∠ − θ V iS (t ) = 10 cos 1000 t A ⇒ I s =10∠0° A try θ = 0° . Then ⎡1 ⎤ 1 + jωC ⎥ = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L (80∠−∞ ) ⎢ + ⎣10 R + jω L ⎦ 2 Equate real part: 40 − 40ω LC = R where ω = 1000 rad sec Equate imaginary part: 40 RC = L Solving yields R = 40(1− 4×107 RC 2 ) Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 ) which yields C = 2.5×10−5 F= 25 μ F so L = 40 RC = 0.02 H=20 mH Now check the angle of the voltage. First Y1 = 1/10 = 0.1 S Y2 = j 0.25 S Y3 = 1/(20+ j 20) = .025− j.025 S then Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V So the angle of the voltage is θ =0° , which satisfies the specifications. R DP 10-6 The input to the circuit shown in Figure DP 10-6 is the voltage source voltage + + – vs(t) = 10 cos (1000t) V vs(t) C vo(t) – The output is the steady-state capacitor voltage vo(t) = A cos (1000t + θ) V Figure DP 10-6 (a) Specify values for R and C such that θ = –30°. Determine the resulting value of A. (b) Specify values for R and C such that A = 5 V. Determine the resulting values of θ. (c) Is it possible to specify values for R and C such that A = 4 and θ = –60°? (If not, justify your answer. If so, specify R and C.) (d) Is it possible to specify values of R and C such that A = 7.07 V and θ = –45°? (If not, justify your answer. If so, specify R and C.) Solution: Represent the circuit in the frequency domain using phasors and impedances. Using voltage division gives 1 10 j1000C ×10∠0° = A∠θ = 1 1 + j103 RC R+ j1000C Equating magnitudes and angles gives 2 A= 10 ⇒ 1 + 106 R 2C 2 ⎛ 10 ⎞ ⎜ ⎟ −1 ⎝ A⎠ RC = 1 + j103 RC and θ = − tan −1 (103 RC ) (a) Pick C = 1 μF, then R = θ = −30° ⇒ ⇒ RC = RC = tan ( 30° ) 0.577 = . 103 103 0.577 = 577 Ω and A = 8.66 V. 106 ×103 2 (b) tan ( −θ ) 103 A=5 V ⇒ ⎛ 10 ⎞ ⎜ ⎟ −1 3 ⎝5⎠ RC = = 3. 3 10 10 Pick C = 1 μF, then R = 3 = 1732 Ω and θ = −60° . 10 × 103 −6 2 (c) A=4 θ = −60° ⎛ 10 ⎞ ⎜ ⎟ −1 2.29 ⎝ 4⎠ RC = = 3 3 10 10 ⇒ ⇒ RC = tan ( 60° ) 1.73 = 3 103 10 Since RC cannot be both 0.00229 and 0.00173 simultaneously, the specifications cannot be satisfied using this circuit. 2 (d) A = 7.07 θ = −45° ⇒ ⇒ ⎛ 10 ⎞ ⎜ ⎟ −1 ⎝ 7.07 ⎠ RC = = 10−3 3 10 RC = tan ( 45° ) = 10−3 103 Both specifications can be satisfied by taking R = 1000 Ω and C = 1 μF. Chapter 11: AC Steady State Power Exercises Exercise 11.3-1 Determine the instantaneous power delivered to an element and sketch p(t) when the element is (a) a resistance R and (b) an inductor L. The voltage across the element is v(t) = Vm cos (ωt + θ) V. V2 V2 Answer: (a) PR = m [1 + cos (2ωt + 2θ)] W (b) PL = m cos (2ωt + 2θ – 90°) W 2R 2ω L Solution: (a) When the element is a resistor, the current has the same phase angle as the voltage: i (t ) = v(t ) Vm cos (ω t + θ ) A = R R The instantaneous power delivered to the resistor is given by 2 2 2 V V Vm V pR (t ) = v(t ) ⋅ i (t ) = Vm cos (ω t + θ ) ⋅ cos (ω t + θ ) = m cos 2 (ω t + θ ) = m + m cos (2ω t + θ ) R R 2R 2R (b) When the element is an inductor, the current will lag the voltage by 90°. Z L = jω L = ω L∠90° Ω ⇒ I= V V ∠θ V = m = m ∠ (θ −90° ) Z ω L∠90° ω L The instantaneous power delivered to the inductor is given by 2 V V pL (t ) = i (t ) ⋅ v(t ) = m cos (ω t + θ − 90° ) ⋅Vm cos (ω t + θ ) = m cos ( 2ω t + 2θ −90° ) W ωL 2ω L Exercise 11.4-1 Find the effective value of the following currents: (a) cos 3t + cos 3t; (b) sin 3t + cos(3t + 60°); (c) 2 cos 3t + 3 cos 5t Answer: (a) 2 (b) 0.366 (c) 2.55 Ex. 11.4-1 (a) I 2 i (t ) = 2 cos 3 t A ⇒ I eff max = = 2A 2 2 (b) i (t ) = cos (3 t − 90° ) + cos (3 t + 60° ) A 1 3 +j = 0.518∠ − 15° A 2 2 0.518 i (t ) = 0.518 cos (3t − 15°) A ⇒ I eff = = 0.366 A 2 I = (1∠−90° ) + (1∠60° ) = − j + (c) 2 I eff 2 ⎛ 2 ⎞ ⎛ 3 ⎞ =⎜ ⎟ +⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ 2 ⇒ I eff = 2.55 A Exercise 11.5-1 Determine the average power delivered to each element of the circuit shown in Figure E 11.5-1. Verify that average power is conserved. Answer: 4.39 + 0 = 4.39 W Figure E 11.5-1 Solution: Analysis using Mathcad (ex11_5_1.mcd): A := 12 Enter the parameters of the voltage source: Enter the values of R and L R := 10 L := 4 The impedance seen by the voltage source is: The mesh current is: I := ω := 2 Z := R + j ⋅ω ⋅L A Z ⎯ I ⋅( I ⋅Z) The complex power delivered by the source is: Sv := Sv = 4.39 + 3.512i 2 ⎯ I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := Sr = 4.39 2 ⎯ I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := Sl = 3.512i 2 Verify Sv = Sr + Sl : Sr + Sl = 4.39 + 3.512i Sv = 4.39 + 3.512i Exercise 11.5-2 Determine the complex power delivered to each element of the circuit shown in Figure E 11.5-2. Verify that complex power is conserved. Answer: 6.606 + j5.248 – j3.303 + 6.606 = j1.982 VA Figure E 11.5-2 Solution: Analysis using Mathcad (ex11_5_2.mcd): Enter the parameters of the voltage source: A := 12 R := 10 Enter the values of R, L an dC ω := 2 L := 4 C := 0.1 The impedance seen by the voltage source is: Z := R + j ⋅ω ⋅L + The mesh current is: I := A Z The complex power delivered by the sourceis: Sv := 1 j ⋅ω ⋅C ⎯ I ⋅( I ⋅Z) 2 ⎯ I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := 2 ⎯ I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := 2 ⎯⎛ 1 ⎞ I ⋅⎜ I ⋅ ⎟ j ⋅ω ⋅C ⎠ ⎝ The complex power delivered to the capacitor is: Sc := 2 Verify Sv = Sr + Sl + Sc : Sr + Sl + Sc = 6.606 + 1.982i Sv = 6.606 + 1.982i Sr = 6.606 Sl = 5.284i Sc = −3.303i Sv = 6.606 + 1.982i Exercise 11.6-1 A circuit has a large motor connected to the ac power lines [ω = (2π)60 = 377 rad/s]. The model of the motor is a resistor of 100 ω in series with an inductor of 5 H. Find the power factor of the motor. Answer: pf = 0.053 lagging Solution: ( ) (377) (5) ⎤ ⎡ pf = cos (∠ Z) = cos ⎡ tan −1 ω L ⎤ = cos ⎢ tan −1 = 0.053 lagging R ⎥⎦ 100 ⎥⎦ ⎣⎢ ⎣ Exercise 11.6-2 A circuit has a load impedance Z = 50 + j80 Ω, as shown in Figure 11.6-5. Determine the power factor of the uncorrected circuit. Determine the impedance ZC required to obtain a corrected power factor of 1.0. Answer : ZC = –j111.25 Ω Figure 11.6-5 Solution: ( ) ( ) pf = cos (∠ Z) = cos ⎡ tan −1 X ⎤ = cos ⎡ tan −1 80 ⎤ = 0.53 lagging ⎢⎣ ⎢⎣ 50 ⎥⎦ R ⎥⎦ (50)2 + (80)2 XC = = −111.25 Ω ⇒ ZC = − j 111.25 Ω 50 tan (cos −1 1) −80 Exercise 11.6-3 Determine the power factor for the total plant of Example 11.6-1 when the resistive heating load is decreased to 30 kW. The motor load and the supply voltage remain as described in Example 11.6-1. Answer: pf = 0.915 Figure from Example 11.6-1 Solution: PT = 30 + 86 = 116 kW and QT = 51 kVAR S T = PT + j QT = 116 + j 51 = 126.7 ∠23.7° kVA pf plant = cos 23.7° = 0.915 Exercise 11.6-4 A 4-kW, 110-Vrms load, as shown in Figure 11.6-5, has a power factor of 0.82 lagging. Find the value of the parallel capacitor that will correct the power factor to 0.95 lagging when ω = 377 rad/s. Answer: C = 0.324 mF Figure 11.6-5 Solution: P = V rms I rms cos θ Z= V rms I rms ⇒ I rms = P 4000 = = 44.3 A V rms cos θ (110)(.82) ∠ cos −1 (0.82) = 2.48 ∠34.9° = 2.03+ j 1.42 = R + j X To correct power factor to 0.95 requires (2.03) 2 + (1.42) 2 R2 + X 2 = = − 8.16 Ω (2.03) tan (18.19° ) −1.42 R tan (cos −1 pfc) − X −1 C= =325 μ F ω X1 X1 = Exercise 11.7-1 Determine the average power absorbed by the resistor in Figure 11.7-2a for these two cases: (a) vA(t) = 12 cos 3t V and vB(t) = 4 cos 3t V; (b) vA(t) = 12 cos 4t V and vB(t) = 4 cos 3t V Answer: (a) 2.66 W (b) 4.99 W Figure 11.7-2a Solution: (a) I = I1 + I 2 = (1.414∠ − 45° ) + ( 0.4714∠135° ) = 0.9428∠ − 45° A 0.94282 ( 6 ) = 2.66 W 2 (b) 1.22 I1 = 1.2∠53° A ⇒ p1 = ( 6 ) = 4.32 W 2 0.47142 I 2 = 0.4714∠135° A ⇒ p2 = ( 6 ) = 0.666 W 2 ∴ p = p1 + p2 = 4.99 W ⇒ p= Exercise 11.8-1 For the circuit of Figure 11.8-1, find ZL to obtain the maximum power transferred when the Thévenin equivalent circuit has Vt = 100 ∠0°V and Zt = 10 + j14 Ω. Also determine the maximum power transferred to the load. Answer: ZL = 10–j14 Ω and P = 125 W Figure 11.8-1 Solution: For maximum power transfer Z L = Z*t = 10 − j14 Ω 100 I= =5 A (10+ j14) + (10− j14) 2 ⎛ 5 ⎞ PL = ⎜ ⎟ Re {10− j14} = 125 W ⎝ 2⎠ Exercise 11.8-2 A television receiver uses a cable to connect the antenna to the TV, as shown in Figure E 11.8-2, with vs = 4 cos ωt mV. The TV station is received at 52 MHz. Determine the average power delivered to each TV set if (a) the load impedance is Z = 300 Ω; (b) two identical TV sets are connected in parallel with Z = 300 Ω for each set; (c) two identical sets are connected in parallel and Z is to be selected so that maximum power is delivered at each set. Answer: (a) 9.6 nW (b) 4.9 nW (c) 5 nW Figure E 11.8-2 Solution: If the station transmits a signal at 52 MHz then ω = 2π f = 104π ×106 rad/sec so the received signal is vs (t ) = 4 cos (104π × 106 t ) mV (a) If the receiver has an input impedance of Zin =300 Ω then Zin 300 1 ⎛ 1 ⎞ ( 2.4 ×10 Vin = Vs = × 4 × 10−3 = 2.4 mV ⇒ P = Vin2 ⎜ ⎟= R + Zin 200+300 2 ⎝ Zin ⎠ 2(300) ) −3 2 (b) If two receivers are connected in parallel then Z in = 300||300 = 150 Ω and Vin = total P = Zin 150 VS = (4 ×10−3 ) = 1.71× 10−3 V R + Zin 200+150 2 Vin ⎛ 1 ⎞ (1.71×10−3 ) 2 = 9.7 nW or 4.85 nW to each set ⎜ ⎟= 2 ⎝ Zin ⎠ 2(150) (c) In this case, we need Zin = R || R = 200 Ω ⇒ R = 400 Ω , where R is the input impedance of each television receiver. Then Ptotal = Vin 2 (2×10−3 ) 2 = = 10 nW ⇒ 5 nW to each set 2 Zin 2(200) = 9.6 nW Exercise 11.9-1 Determine the voltage vo for the circuit of Figure E 11.9-1. Hint: Write a single mesh equation. The currents in the two coils are equal to each other and equal to the mesh current. Answer: vo = 14 cos 4t V Figure E 11.9-1 Solution: Coil voltages: V1 = j 24 I 1 + j 16 I 2 = j 40 I V 2 = j 16 I 1 + j 40 I 2 = j 56 I Mesh equation: 24 = V1 + V 2 = j 40 I + j 56 I = j 96 I 24 1 =−j 4 j 96 ⎛ 1⎞ Vo = V2 = ( j 56 ) ⎜ − j ⎟ = 14 ⎝ 4⎠ vo = 14 cos 4t V I= Exercise 11.9-2 Determine the voltage vo for the circuit of Figure E 11.9-2. Hint: This exercise is the same as Exercise 11.9-1, except for the position of the dot on the vertical coil. Answer: vo = 18 cos 4t V Figure E 11.9-2 Solution: Coil voltages: V1 = j 24 I 1 − j 16 I 2 = j 8 I V 2 = − j 16 I 1 + j 40 I 2 = j 24 I Mesh equation: 24 = V1 + V 2 = j 8 I + j 24 I = j 32 I 24 3 =−j 4 j 32 ⎛ 3⎞ Vo = V 2 = ( j 24 ) ⎜ − j ⎟ = 18 ⎝ 4⎠ vo = 18 cos 4t V I= Exercise 11.9-3 Determine the current io for the circuit of Figure E 11.9-3. Hint: The voltage across the vertical coil is zero because of the short circuit. The voltage across the horizontal coil induces a current in the vertical coil. Consequently, the current in the vertical coil is not zero. Answer: io = 1.909 cos (4t – 90°) A Solution: Figure E 11.9-3 0 = V 2 = j 16 I 1 + j 40 I 2 40 I 2 = −2.5 I 2 16 V s = V1 = j 24 I 1 + j 16 I 2 ⇒ I1 = − = j (24(−2.5) + 16) I 2 = − j 44 I 2 24 6 = j 11 − j 44 I o = I 1 − I 2 = (−2.5 − 1) I 2 I2 = = −3.5 I 2 ⎛ 6⎞ = −3.5 ⎜ j ⎟ = − j 1.909 ⎝ 11 ⎠ io = 1.909 cos ( 4t - 90° ) A Exercise 11.9-4 Determine the current io for the circuit of Figure E 11.9-4. Hint: This exercise is the same as Exercise 11.9-3, except for the position of the dot on the vertical coil. Answer: io = 0.818 cos (4t – 90°) A Figure E 11.9-4 Solution: 0 = V 2 = − j 16 I 1 + j 40 I 2 40 I 2 = 2.5 I 2 16 V s = V1 = j 24 I 1 − j 16 I 2 ⇒ I1 = = j (24(2.5) − 16) I 2 = j 44 I 2 24 6 =−j j 44 11 I o = I 1 − I 2 = (2.5 − 1) I 2 I2 = = 1.5 I 2 6⎞ ⎛ = 1.5 ⎜ − j ⎟ = − j 0.818 ⎝ 11 ⎠ io = 0.818 cos ( 4t - 90° ) A Exercise 11.10-1 Determine the impedance Zab for the circuit of Figure E 11.10-1. All the transformers are ideal. Answer: Zab = 4.063Z Figure E 11.10-1 Ex. 11.10-1 Z1 = 1 ⎛ Z⎞ Z⎞ 1 Z Z ⎛ = , = = 9 ⎜ Z + ⎟ and Z 3 = 2 (Z + Z 2 ) Z Z + ⎜ 2 2 2 ⎜ 2 ⎟ ⎟ n3 4 n1 n2 ⎝ n3 ⎠ 4⎠ ⎝ then 1⎛ Z ⎞⎞ ⎛ Z ab = Zin = Z + Z3 = Z + ⎜ Z +9⎜ Z+ ⎟ ⎟ = 4.0625 Z 4⎝ 4 ⎠⎠ ⎝ PROBLEMS Section 11.3 Instantaneous Power and Average Power P 11.3-1 An RLC circuit is shown in Figure P 11.3-1. Find the instantaneous power delivered to the inductor when is = 1 cos ωt A and ω = 6283 rad/s. Figure P 11.3-1 Solution: 1∠0° = V V V + + ⇒ V = 14.6∠ − 43° V 20 j 63 − j16 V = 0.23∠ − 133° A j 63 p(t ) = i (t )v(t ) = 0.23cos (2π ⋅103 t − 133° ) × 14.6 cos (2π ⋅103 t − 43° ) I= = 3.36 cos (2π ⋅103 t −133° ) cos (2π ⋅103 t − 43° ) = 1.68 (cos (90° ) + cos (4π ⋅103 t −176° )) =1.68 cos (4π ⋅103 t −176° ) P 11.3-2 Find the average power absorbed by the 0.6-kΩ resistor and the average power supplied by the current source for the circuit of Figure P 11.3-2. Figure P 11.3-2 Solution: Current division: ⎡ 1800 − j 2400 ⎤ I=4 5⎢ ⎥ ⎣1800− j 2400+ 600 ⎦ =5 5 ∠ − 8.1° mA 2 2 P600Ω I 600 ⎛5⎞ = = 300(25) ⎜ ⎟ = 1.875 ×104 μW = 18.75 mW 2 ⎝ 2⎠ Psource = ⎛ 5⎞ V I cos θ 1 = (600) ⎜ 5 ⎟ 4 5 cos(−8.1° ) = 2.1× 104 μW = 21 mW 2 2 ⎝ 2⎠ ( ) P 11.3-3 Use nodal analysis to find the average power absorbed by the 20-Ω resistor in the circuit of Figure P 11.3-3. Answer: P = 200 W Figure P 11.3-3 Solution: Node equations: 20I X −100 20I X − V + IX + = 0 ⇒ I X (20 − j15) − V = − j 50 10 − j5 V − 20I X V − 3I X + = 0 ⇒ I X (−40 + j 30) + V (2 − j ) = 0 10 − j5 Solving the node equations using Cramer’s rule yields IX = j 50(2 − j ) 50 5∠63.4° = = 2 5∠10.3° A (40 − j 30) − (20 − j15)(2 − j ) 25∠53.1° Then 2 PAVE ( I = X (20) = 10 2 5 2 ) 2 = 200 W P 11.3-4 Nuclear power stations have become very complex to operate, as illustrated by the training simulator for the operating room of the Pilgrim Power Station shown in Figure P 11.34a. One control circuit has the model shown in Figure P 11.3-4b. Find the average power delivered to each element. Answer: Psource current = −12.8 W P8 Ω = 6.4 W PL = 0 W Pvoltage source = 6.4 W P 11.3-4 Solution: A node equation: (V − 16) V + − (2 2∠45°) = 0 j4 8 ⎛ 2⎞ ⇒ V = ⎜⎜16 ⎟ ∠18.4° V 5 ⎟⎠ ⎝ Then I= 16 − V = 3.2 ∠ − 116.6° A j4 2 PAVE 8Ω PAVE current source = − 1 V = × 2 8 2 ⎛ 2⎞ ⎜16 ⎟ 5⎠ 1 ⎝ = × = 6.4 W absorbed 2 8 1 1⎛ 2⎞ V 2 2 cos θ = − ⎜ 16 ⎟ 2 2 cos ( 26.6° ) = −12.8 W absorbed 2 2⎝ 5⎠ ( ) ( ) PAVE inductor = 0 1 1 PAVE voltage source = − (16) I cosθ = − (16)( 3.2)cos( − 116.6°) = 6.4 W absorbed 2 2 P 11.3-5 Find the average power delivered to each element for the circuit of Figure P 11.3-5. Figure P 11.3-5 Solution: A node equation: −20 + V1 V1 + (3 / 2)V1 + = 0 ⇒ V1 = 50 5 ∠ − 26.6° V 10 15 − j 20 Then I= V1 + ( 3/2 ) V1 ( 5 / 2 ) V1 = 5 5 ∠26.6° A = 15 − j 20 25∠ − 53.1° Now the various powers can be calculated: 2 2 PAVE 10Ω PAVE current source = − 1 1 V (20) cosθ = − (50 5)(20) cos (−26.6°) = −1000 W absorbed 2 2 PAVE 15Ω PAVE voltage source = − 1 V1 1 (50 5) = = = 625 W absorbed 2 10 2 10 ( ) ( )( 2 5 5 I = (15 ) = − 2 2 2 (15 ) = 937.5 W absorbed ) 1 3 1 I V1 cos θ = − 5 5 75 5 cos ( −53.1° ) = −562.5 W absorbed 2 2 2 PAVE capacitor = 0 W P 11.3-6 A student experimenter in the laboratory encounters all types of electrical equipment. Some pieces of test equipment are battery-operated or operate at low voltage so that any hazard is minimal. Other types of equipment are isolated from electrical ground so that there is no problem if a grounded object makes contact with the circuit. Some types of test equipment, however, are supplied by voltages that can be hazardous or have dangerous voltage outputs. The standard power supply used in the United States for power and lighting in laboratories is the 120, grounded, 60Hz sinusoidal supply. This supply provides power for much of the laboratory equipment, so an understanding of its operation is essential in its safe use (Bernstein, Figure P 11.3-6 1991). Consider the case where the experimenter has one hand on a piece of electrical equipment and the other hand on a ground connection, as shown in the circuit diagram of Figure P 11.3-6a. The hand-to-hand resistance is 200 Ω. Shocks with an energy of 30 J are hazardous to humans. Consider the model shown in Figure P 11.3-6b, which represents the human with R. Determine the energy delivered to the human in 1 s. Solution: Z= I = 200 ( j 200 ) 200 ∠90° 200 = = ∠45° Ω 200 (1 + j ) 2 ∠45° 2 ⎛ 120∠0D j 200 ⎞ = 0.85 ∠ − 45° A, I R = ⎜ ⎟ I = 0.6∠0° A 200 200 + j 200 D ⎝ ⎠ ∠45 2 P = I R = ( 0.6 ) ( 200 ) = 72 W and w = ( 72 )(1) = 72 J 2 2 P 11.3-7 An RLC circuit is shown in Figure P 11.3-7 with a voltage source vs = 7 cos 10t V. (a) Determine the instantaneous power delivered to the circuit by the voltage source. (b) Find the instantaneous power delivered to the inductor. Answer: (a) p = 7.54 + 15.2 cos (20t – 60.3°) W (b) p = 28.3 cos (20t – 30.6°) W Figure P 11.3-7 Solution: Z = j3 + 4(− j 2) = 0.8 + j1.4 4− j 2 = 1.6 ∠60.3° Ω ∴I = 7∠0° V = = 4.38 ∠ − 60.3° A Z 1.6∠60.3° i (t ) = 4.38cos (10 t − 60.3° ) A The instantaneous power delivered by the source is given by (7)(4.38) [cos (60.3°) + cos (20 t −60.3°)] 2 = 7.6 + 15.3cos (20 t − 60.3°) W p(t ) = v(t ) ⋅ i (t ) = (7 cos 10 t )(4.38cos (10 t − 60.3°)) = The inductor voltage is calculated as VL = I ⋅ Z L = (4.38 ∠ − 60.3°)(j 3) = 13.12 ∠29.69° V vL (t ) = 13.12 cos (10 t + 29.69°) V The instantaneous power delivered to the inductor is given by pL (t ) = vL (t ) ⋅ i (t ) = ⎡⎣ (13.12 cos (10t + 29.69°)(4.38cos (l0t − 60.3° ) ⎤⎦ 57.47 [ cos (29.69°+ 60.3°)+ cos (20 t + 29.69°−60.3°)] 2 = 28.7 cos (20t − 30.6°) W = P 11.3-8 (a) Find the average power delivered by the source to the circuit shown in Figure P 11.3-8. (b) Find the power absorbed by resistor R1. Answer: (a) 30 W (b) 20 W Figure P 11.3-8 Solution: The equivalent impedance of the parallel resistor (1)( j ) = 1 1+ j Ω . Then and inductor is Z = ( ) 1+ j 2 I= (a) Psource = I V cos θ = 2 10∠0° 20 20 = = ∠ − 18.4° A 1 j 3 + 10 1+ (1+ j ) 2 ⎛ 20 ⎞ ⎟ ⎝ 10 ⎠ cos −18.4° = 30.0 W ( ) 2 (10 ) ⎜ 2 ⎛ 20 ⎞ 2 (1) I R1 ⎜⎝ 10 ⎟⎠ (b) PR 1 = = = 20 W 2 2 Section 11.4 Effective Value of a Periodic Waveform P 11.4-1 Find the rms value of the current i for (a) i = 2 – 4 cos 2t A, (b) i = 3 sin π t + π t A, and (c)i = 2 cos 2t + 4 2 cos (2t + 45°) + 12 sin 2t A. Answer: (a) 2 3 (b) 2.35 A (c) 5 2 A Solution: ( Treat i as two sources of different frequencies.) i = 2 − 4 cos 2t = i1 + i 2 (a) 2A source: I eff = lim T →∞ 1 T ∫ T o 2 (2) dt = 2 A and 4 cos 2t source: Ieff = 4 A 2 The total is calculated as 2 I eff (b) 2 ⎛ 4 ⎞ = ( 2) + ⎜ ⎟ = 12 A ⇒ I rms = I eff = 12 = 2 3 A ⎝ 2⎠ 2 i ( t ) = 3cos (π t − 90° ) + 2 cos π t ⇒ I = ( 3∠ − 90° ) + ( 2 ∠0° ) = 2 − j 3 = 3.32∠ − 64.8° A I rms = (c) 3.32 = 2.35 A 2 i ( t ) = 2 cos 2t + 4 2 cos ( 2t + 45° ) + 12 cos ( 2 t − 90° ) ( ) I = ( 2∠0° ) + 4 2 ∠45° + (12∠ − 90° ) = ( 2 + 4 ) + ( j 4 − j12 ) =10∠ − 53.1° A I rms = 10 =5 2 A 2 2 cos P 11.4-2 Determine the rms value for each of the waveforms shown in Figure P 11.4-2. Answer: (a) 4.10 V (b) 4.81 V (c) 4.10 v (V) 6 v (V) v (V) 6 6 2 2 2 2 5 7 10 t (s) 2 (a) 5 7 10 t (s) 3 5 8 10 t (s) (c) (b) Figure P 11.4-2 Solution: (a) (b) (c) ) 1 5 (∫ Vrms = 1 5 ( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) = 1 116 = 4.81 V (8 + 108) = 5 5 Vrms = 1 5 ( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) = 1 84 = 4.10 V (12 + 72 ) = 5 5 2 0 2 5 62 dt + ∫ 22 dt = 2 2 0 3 0 5 2 2 2 5 3 2 1 5 (∫ ) Vrms = 2 0 5 36 dt + ∫ 4 dt = 2 2 5 0 2 3 5 0 3 1 84 = 4.10 V ( 72 + 12 ) = 5 5 P 11.4-3 Determine the rms value for each of the waveforms shown in Figure P 11.4-3. Answer: (a) 4.16 V (b) 4.16 V (c) 4.16 Solution: (a) 2 2 Vrms 1 4⎛ 4 2 ⎞ = ⎜ t + ⎟ dt 3 ∫1 ⎝ 3 3 ⎠ = 1 4 4 2 2 t + 1) dt ( ∫ 27 1 4 1 4t 2 + 2 4 1 4 7 10 t (s) (a) 4 4 2 = ( 4 t + 4 t + 1) dt 27 ∫1 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝ v (V) 6 ⎞ 4 +t 1 ⎟ ⎟ ⎠ = 4 ( (85.33 − 1.33) + ( 2 )(16 − 1) + 3) 27 = 4 (117 ) = 4.16 V 27 (b) v (V) 6 2 1 4 7 10 t (s) (b) v (V) 6 2 2 Vrms = 1 4 ⎛ 4 22 ⎞ ⎜ − t + ⎟ dt 3 ∫1 ⎝ 3 3 ⎠ = 4 4 2 ( −2 t + 11) dt ∫ 1 27 4 4 2 = ( 4 t − 44 t + 121) dt 27 ∫1 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝ 4 1 44t 2 − 2 4 1 ⎞ 4 + 121 t 1 ⎟ ⎟ ⎠ = 4 (84 + ( −22 )15 + (121) 3) 27 = 4 (117 ) = 4.16 V 27 3 6 9 (c) Figure P 11.4-3 t (s) 2 (c) Vrms 1 3⎛ 4 4 3 4 3 2 2 ⎞ = t + 2 ⎟ dt = 2 t + 3) dt = ( ( 4 t + 12 t + 9 ) dt ⎜ ∫ ∫ 3 0⎝3 27 0 27 ∫ 0 ⎠ 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝ 3 12t 2 + 2 0 ⎞ 3 + 9t 0 ⎟ ⎟ 0 ⎠ 3 = 4 ( 36 + 54 + 27 ) 27 = 4 (117 ) = 4.16 V 27 P 11.4-4 Find the rms value for each of the waveforms of Figure P 11.4-4. Answer: Vrms = 1.225 V, Irms = 5 mA v(t) V 2 Sinusoid + Constant –T/ 2 0 T/ 2 T t (a) i(t) mA 10 Sinusoidal –5 0 5 15 t ( ms) 10 (b) Figure P 11.4-4 Solution: (a) ⎛ 2π v ( t ) = 1 + cos ⎜ ⎝ T vdc eff 2 t ⎛1 T ⎞ = ⎜ ∫ 0 1dt ⎟ = ⎝T ⎠ T T 0 ⎞ t ⎟ = vdc + vac ⎠ ⎛T ⎞ 2 = ⎜ − 0 ⎟ = 1 V and vac eff = T ⎝ ⎠ 1 V 2 2 veff = vdc eff + vac eff 2 (b) 2 2 = ⎛ 1 ⎞ 1 +⎜ ⎟ = 1.225 V ⎝ 2⎠ 2 The period of the sinusoid is half the period of i(t). Let T be the period of i(t) and T/2 be the period of the sinusoid. Then I RMS 4π T ⎧ 0<t < ⎪ 10sin T t 4 ⎪ 4π 3T T ⎪ <t < i ( t ) = ⎨−10sin t 2 4 T ⎪ 0 otherwise ⎪ ⎪ ⎩ 3T / 4 100 ⎡ T / 4 2 4π 4π ⎤ = t dt + ∫ t dt ⎥ sin sin 2 ∫ ⎢ T /4 T ⎣ 0 T T ⎦ ⎡⎛ 8π sin 100 ⎢⎜ t T ⎢⎜ − = 18 π T ⎢⎜ 2 ⎢⎣⎝ T I rms = 25 T /4 8π ⎞ ⎛ t⎟ ⎜ t sin T ⎟ + ⎜ − 18π ⎟ ⎜2 T ⎠0 ⎝ ⇒ I rms = 5 mA 2T / 4 ⎞ t⎟ ⎟ ⎟ ⎠T /2 ⎤ ⎥ ⎥ = 25 ⎥ ⎥⎦ P 11.4-5 Find the rms value of the voltage v(t) shown in Figure P 11.4-5. Answer: Vrms = 4.24 V v(t) (V) 9 –0.3 –0.2 –0.1 0 0.1 0.2 0.3 0.4 0.5 t (s) Figure P 11.4-5 Solution: ⎧90 t ⎪ v ( t ) = ⎨90 ( 0.2−t ) ⎪0 ⎩ 2 = Vrms 0 ≤ t ≤ 0.1 0.1 ≤ t ≤ 0.2 0.2 ≤ t ≤ 0.3 2 0.2 2 1 ⎡ 0.1 2 ⎤ = 90 + ⎡ − ⎤ 90 t dt 90 0.2 t dt ( ) ( ) ⎦ ⎥⎦ .3 ∫0.1 ⎣ .3 ⎢⎣ ∫ 0 902 = .3 V rms ⎡ 0.1 t 2 dt + 0.2 0.2−t 2 dt ⎤ ) ⎥ ∫ ∫0.1 ( ⎣⎢ 0 ⎦ ⎡ .001 .001 ⎤ ⎢⎣ 3 + 3 ⎥⎦ = 18 V = 18 = 4.24 V P 11.4-6 Find the effective value of the current waveform shown in Figure P 11.4-6. Answer: Ieff = 8.66 10 5 –1 0 1 2 3 4 5 t (s) Figure P 11.4-6 Solution: I eff = 3 1 t 2 1⎡ 2 2 ( ) (10) (5) 2 dt ⎤ = 8.66 i t dt = dt + ∫ ∫ ∫ 0 0 2 ⎢ ⎥⎦ 3⎣ T P 11.4-7 Calculate the effective value of the voltage across the resistance R of the circuit shown in Figure P 11.4-7 when ω = 100 rad/s. Hint: Use superposition. Answer: Veff = 4.82 V 10 cos ω t A 5A 6 sin ω t A R 12 Ω Figure P 11.4-7 Solution: Use superposition: V1 = 5∠0° V V2 = 2.5 V (dc) V3 = 3∠−90° V V1 and V3 are phasors having the same frequency, so we can add them: V1 + V3 = ( 5∠0° ) + ( 3∠ − 90° ) = 5 − j3 = 5.83∠ − 31.0° V Then vR (t ) = v1 (t ) + ( v2 (t ) + v3 (t ) ) = 2.5 + 5.83cos (100 t − 31.0°) V Finally 2 V 2 R eff ⎛ 5.83 ⎞ = (2.5) + ⎜ ⎟ = 23.24 V ⇒ VR eff = 4.82 V ⎝ 2 ⎠ 2 Section 11.5 Complex Power i(t) P 11.5-1 The complex power delivered by the voltage source in Figure P 11.5-1 is S = 3.6 + j7.2 V A. Determine the values of the resistance, R, and inductance, L. + – R 12 cos 4t V L Answer: R = 4 Ω and L = 2 H Figure P 11.5-1 Solution: I* = R+ j4 L = 2 ( 3.6 + j 7.2 ) 2S = = 0.6 + j 1.2 = 1.342∠63.43° A 12∠0° 12∠0° 12∠0° = 8.94∠63.43 = 4 + j 8 ⇒ R = 4 Ω and L = 2 H 1.342∠ − 63.43 i(t) P 11.5-2 The complex power delivered by the voltage source in Figure P 11.5-2 is S = 18 + j9 VA. Determine the values of the resistance, R, and inductance, L. + – 12 cos 4t V R Answer: R = 4 Ω and L = 2 H Figure P 11.5-2 Solution: I* = 2 (18 + j 9 ) 2S = = 3 + j 1.5 = 3.35∠26.56° A 12∠0° 12∠0° 1 1 1 1 3.35∠ − 26.56° + = −j = = 0.2791∠ − 26.56 = 0.250 − j 0.125 R j4 L R 4L 12∠0° ⇒ R = 4 Ω and L = 2 H L i(t) P 11.5-3 Determine the complex power delivered by the voltage source in the circuit shown in Figure P 11.5-3. Answer: S = 7.2 + j3.6 VA + – 4Ω 12 cos 4t V 8Ω 2H Figure P 11.5-3 Solution: Let Zp = Next I= Finally S= 8 ( j 8) j 8 1− j 8 + j 8 = × = = 4+ j4 V 8 + j 8 1+ j 1− j 2 12∠0° 12∠0° = = 1.342∠ − 26.6° A 4 + Z p 4 + (8 + j 8) (12∠0° )(1.342∠ − 26.6° ) * = 7.2 + j 3.6 2 VA P 11.5-4 Many engineers are working to develop photovoltaic power plants that provide ac power. An example of an experimental photovoltaic system is shown in Figure P 11.5-4a. A model of one portion of the energy conversion circuit is shown in Figure P 11.5-4b. Find the average, reactive, and complex power delivered by the dependent source. Answer: S = + j8/9 VA (a) 5 mH 1Ω + 5 cos 400t A 2Ω v1 1 800 F + – –( 1 8)v1 – (b) Figure P 11.5-4 Soution: Represent the circuit in the frequency domain and label the node voltages: The node equations are: V1 V1 − V2 + = 0 ⇒ V1 (1 − j ) + j V2 = 10 2 j2 ⎛1 ⎞ V2 − ⎜ V1 ⎟ V2 − V1 V2 ⎝ 8 ⎠ = 0 ⇒ V 0.25 + j + V 2 = 0 = + ) 2( ) 1( j2 − j2 1 5∠0 = Using MATLAB: V1 = 5.33∠36.9° V j ⎤ ⎡ V1 ⎤ ⎡10 ⎤ ⎡ 1− j ⎢ 0.25 + j 2 ⎥ ⎢ V ⎥ = ⎢ 0 ⎥ ⇒ V = 2.75∠ − 67.2° V ⎣ ⎦⎣ 2⎦ ⎣ ⎦ 2 then 1 − V1 − V2 I = 8 = 2.66∠126.9° A 1 Now the complex power can be calculated as ( ) * ⎛ 5.33 ⎛ ⎛1⎞ ⎞ ⎞ I* ⎜ −⎜ ⎟ V1 ⎟ 2.667∠−126.9° ⎜ − ∠36.9° ⎟ ⎝ ⎝8⎠ ⎠ = ⎝ 8 ⎠ = j 0.8889 = j 8 VA S= 2 9 2 Finally S = P + jQ = j 8 8 ⇒ P = 0, Q = VAR 9 9 (Checked using MATLAB and LNAP) 12 Ω P 11.5-5 For the circuit shown in Figure P 11.5-5, determine I and the complex power S delivered by the source when V = 50 ∠120° V rms. Answer: S = 100 + j75 VA + V – I j20 Ω –j10 Ω 20 Ω Figure P 11.5-5 Solution: I= S = VI* = 50∠120° 50∠120° = = 2.5∠83.13° A 16+ j12 20∠36.87° ( 50∠120° )( 2.5∠−83.13° ) = 125 ∠36.87° = 100 + j 75 VA 50 mF P 11.5-6 For the circuit of Figure P 11.5-6, determine the complex power of the R, L, and C elements and show that the complex power delivered by the sources is equal to the complex power absorbed by the R, L, and C elements. 10 Ω 6 cos 10t A + – 5 cos 10t V 2H Figure P 11.5-6 Solution: KVL: (10+ j 20 ) I1 = 5∠0° − j 2 I 2 ⇒ (10 + j 20 ) I1 + j 2 I 2 = 5∠0° KCL: I1 + I 2 = 6∠0° Solving these equations using Cramer’s rule: Δ= I1 = 10 + j 20 1 j2 = 10 + j18 1 1 5 j2 5 − j12 = = 0.63∠232° A = −0.39 − j 0.5 A Δ 6 1 10 + j18 I 2 = 6 − I1 = 6 + 0.39 + j.5 = 6.39 + j.5 = 6.41∠4.47° A Now we are ready to calculate the powers. First, the powers delivered: 1 ( 5∠0° )( −I 2* ) = 2.5 ( 6.41∠(180− 4.47) ) = −16.0 + j1.25 VA 2 1 S 6∠ 0° = [5− j 2 I 2 ]( 6∠0° ) = ⎡⎣5− j 2( 6.39+ j.5 ) ⎤⎦ 3 = 18.0− j 38.3 VA 2 S Total = S5∠ 0° + S 6∠ 0° = 2.0− j 37.2 VA S5∠ 0° = delivered Next, the powers absorbed: 1 10 2 2 10 I1 = (.63) = 2.0 VA 2 2 j 20 2 S j20Ω = I1 = j 4.0 VA 2 1 2 2 S − j2Ω = ( − j 2 ) I 2 = − j ( 6.41) = − j 41.1 VA 2 S Total = 2.0 − j 37.1 VA S10Ω = absorbed To our numerical accuracy, the total complex power delivered is equal to the total complex power absorbed. P 11.5-7 A circuit is shown in Figure P 11.5-7 with an unknown impedance Z. However, it is known that v(t) = 100 cos (100t + 20°) V and i(t) = 25 cos (100t–10°) A. (a) Find Z. (b) Find the power absorbed by the impedance. (c) Determine the type of element and its magnitude that should be placed across the impedance Z (connected to terminals a–b) so that the voltage v(t) and the current entering the parallel elements are in phase. Answer: (a) 4 ∠30° Ω (b) 1082.5 W (c) 1.25 mF a i(t) + v(t) Z – b Figure P 11.5-7 Solution: (a) (b) Z= V 100∠20° = = 4∠30° Ω I 25∠ − 10° P= I V cos θ (100 )( 25 ) cos 30° = = 1082.5 W 2 2 1 = 0.25∠ − 30° = 0.2165 − j 0.125 S . To cancel the phase angle we add a capacitor Z having an admittance of YC = j 0.125 S . That requires ω C =0.125 ⇒ C = 1.25 mF . (c) Y = P 11.5-8 Find the complex power delivered by the voltage source and the power factor seen by the voltage source for the circuit of Figure P 11.5-8. + v1 – 1Ω 4Ω 10 cos 2t V + – (3 4) v1 13F Figure P 11.5-8. Solution: Apply KCL at the top node to get 10 − V1 V1 3 − V1 + = 0 ⇒ V1 = 4∠36.9° V 3 4 4 1− j 2 V I1 = 1 = 1∠36.9° A Then 4 The complex power delivered by the source is calculated as − S= Finally (1∠36.9° )* (10∠0° ) = 2 5∠ − 36.9° VA pf = cos ( −36.9° ) = .8 leading R P 11.5-9 The circuit in Figure P 11.5-9 consists of a source connected to a load. (a) Suppose R = 9 Ω and L = 5 H. Determine the + L – 24 cos (3t +75°) V average, complex, and reactive powers delivered by the source to the load. Source Load (b) Suppose R = 15 Ω and L = 3 H. Determine the average, complex, and reactive powers delivered Figure P 11.5-9 by the source to the load. (c) Suppose the source delivers 8.47 + j14.12 VA to the load. Determine the values of the resistance, R, and the inductance, L. (d) Suppose the source delivers 14.12 + j8.47 VA to the load. Determine the values of the resistance, R, and the inductance, L. Solution: Represent the circuit in the frequency domain as (a) 24∠75° 24∠75° = = 1.37∠16° A 9 + j15 17.5∠59° 24 (1.37 ) 1 ∠ ( 75 − 16 ) ° = 16.44∠59° = 8.47 + j14.1 VA S = ( 24∠75° )(1.37∠16° )∗ = 2 2 I= so P = 8.47 W and Q = 14.1 VAR (b) (c) 24∠75° 24∠75° = = 1.37∠44° A 15 + j 9 17.5∠31° 1 S = ( 24∠75° )(1.37∠44° )∗ = 16.44∠31° = 14.1 + j8.47 VA 2 I= ∗ ∗ ⎛ 2 ( 8.47 + j14.12 ) ⎞ ⎛ 2 (16.44∠59° ) ⎞ ∗ I=⎜ ⎟ =⎜ ⎟ = (1.37∠ − 16° ) = 1.37∠16° A 24∠75° ⎝ ⎠ ⎝ 24∠75° ⎠ R + j 3L = 24∠75° = 17.5∠59° = 9 + j15 Ω 1.37∠16° R = 9 Ω and L = (d) 15 =5 H 3 ∗ ⎛ 2 (14.12 + j8.47 ) ⎞ ∗ I=⎜ ⎟ = (1.37∠ − 44° ) = 1.37∠44° A 24∠75° ⎝ ⎠ R + j 3L = 24∠75° = 17.5∠31° = 15 + j 9 Ω 1.37∠44° R = 15 Ω and L = 9 =3H 3 P 11.5-10 The circuit in Figure P 11.5-10 consists of a source connected to a load. Suppose the amplitude of the source voltage is doubled so that vi (t) = 48 cos (3t + 75°)V. How will each of the following change? (a) The impedance of the load (b) The complex power delivered to the load (c) The load current Solution: Represent the circuit in the frequency domain as Doubling the amplitude of vi(t): (c) doubles the amplitude of the load current. (a) does not change the impedance. (b) multiplies the complex power by 22 = 4. R + – 24 cos (3t +75°) V Source Load Figure P 11.5-10 L P 11.5-11 The circuit in Figure P 11.5-11 consists of a source connected to a load. Suppose the phase angle of the source voltage is doubled so that vi (t) = 24 cos (3t + 150°) V. How will the following change? (a) The impedance of the load (b) The complex power delivered to the load (c) The load current Solution: Represent the circuit in the frequency domain as Doubling the angles of vi(t) increases the angle of vi(t) by 75° (c) increases the angle of the load current by 75°. (a) does not change the impedance. (b) does not change the complex power. R + – 24 cos (3t +75°) V Source Load Figure P 11.5-11 L R P 11.5-12 The circuit in Figure P 11.5-12 consists of a source connected to a load. The complex power delivered by the source to the load is S = 6.61 + j1.98 VA. Determine the values of R and C. + – C 12 cos 2t V Source 4H Load Figure P 11.5-12 Solution: Represent the circuit in the frequency domain as 1 S = VI∗ 2 so ⇒ ∗ ⎛ 2S ⎞ I=⎜ ⎟ ⎝V⎠ ∗ ⎛ 2 ( 6.61 + j1.98 ) ⎞ ∗ − j1.67 A I (ω ) = ⎜ ⎟ = (1.10 + j 0.33) = (1.10 − j 0.33) = 1.15e j0 12 e ⎝ ⎠ Using KVL gives 1 ⎛ ⎞ + j8 ⎟ (1.15e − j16.7 ) = 12e j 0 ⎜R− j 2C ⎝ ⎠ 1 ⎞ 12e j 0 ⎛ = 10.4e j16.7 = 10 + j 3 Ω R + j ⎜8 − ⎟= − j16.7 2C ⎠ 1.15e ⎝ R = 10 Ω So And 8− 1 =3 ⇒ 2C 1 1 =5 ⇒ C = F 2C 10 R P 11.5-13 Design the circuit shown in Figure P 11.5-13, that is, specify values for R and L so that the complex power delivered to the RL circuit is 8 + j6 VA. Answer: R = 5.76 Ω and L = 2.16 H 12 cos 2t + – L Figure P 11.5-13 Solution: Analysis using Mathcad (ex11_5_3.mcd): Enter the parameters of the voltage source: A := 12 ω := 2 Enter the Average and Reactive Power delivered to the RL circuit: P := 8 The complex power delivered to the RL circuit is: Q := 6 S := P + j ⋅Q 2 The impedance seen by the voltage source is: Z := Calculate the required values of R and L R := Re( Z) The mesh current is: I := A Z A ⎯ 2 ⋅S L := Im( Z) ω ⎯ I ⋅( I ⋅Z) The complex power delivered by the source is: Sv := 2 ⎯ I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := 2 ⎯ I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := 2 Verify Sv = Sr + Sl : Sr + Sl = 8 + 6i Sv = 8 + 6i R = 5.76 Sv = 8 + 6i Sr = 8 Sl = 6i L = 2.16 P11.5-14 The source voltage in the circuit shown in Figure P11.5-14 is V s = 24 ∠ 30° V. Consequently I 1 = 3.13 ∠ 25.4° A, I 2 = 1.99 ∠ 52.9° A and V 4 = 8.88 ∠ −10.6° V Determine (a) the average power absorbed by Z 4, (b) the average power absorbed by Z 1, and (c) the complex power delivered by the voltage source. (All phasors are given using peak, not RMS, values.) Figure P11.5-14 Solution: a. The average power absorbed by Z 4 is P4 = ( 8.88)(1.99 ) cos 2 ( −10.6° − 52.9° ) = 3.942 W . ⎛ 3.13 2 ⎞ b. The average power absorbed by Z 1 is P1 = ⎜ ⎟ 4 = 19.59 W . ⎝ 2 ⎠ c. The complex power delivered by the voltage source is S= ( 24∠30° )( 3.13∠25.4° ) * = 37.56∠4.6° = 37.44 + j 3.01 VA 2 Section 11.6 Power Factor P 11.6-1 An industrial firm has two electrical loads connected in parallel across the power source. Power is supplied to the firm at 4000 V rms. One load is 30 kW of heating use, and the other load is a set of motors that together operate as a load at 0.6 lagging power factor and at 150 kVA. Determine the total current and the plant power factor. Answer: I = 42.5 A rms and pf = 1/ 2 Solution: Heating: P = 30 kW Motor: θ = cos −1 ( 0.6 ) = 53.1° ⎪⎫ S = 150 kVA Total (plant): ⎧ P = 150 cos 53.1° = 90 kW ⎬ ⇒ ⎨ ⎩Q = 150sin 53.1° = 120 kVAR ⎭⎪ P = 30 + 90 = 120 kW ⎫ ⎬ ⇒ S = 120 + j120 = 170∠45° VA Q = 0 + 120 = 120 kVAR ⎭ The power factor is pf = cos 45° = 0.707 lagging. The rms current required by the plant is I = S 170 kVA = = 42.5 A rms . V 4 kV P 11.6-2 Two electrical loads are connected in parallel to a 400-V rms, 60-Hz supply. The first load is 12 kVA at 0.7 lagging power factor; the second load is 10 kVA at 0.8 lagging power factor. Find the average power, the apparent power, and the power factor of the two combined loads. Answer: Total power factor = 0.75 lagging Solution: Load 1: P1 = S cosθ = (12 kVA )( 0.7 ) = 8.4 kW Q1 = S sin ( cos −1 (.7 ) ) = (12 kVA ) sin ( 45.6° ) = 8.57 kVAR Load 2: P2 = (10 kVA )( 0.8 ) = 8 kW Q2 = 10sin ( cos −1 ( 0.8 ) ) = 10sin ( 36.9° ) = 6.0 kVAR Total: S = P + jQ = 8.4 + 8 + j ( 8.57 + 6.0 ) = 16.4 + j14.57 = 21.9∠41.6° kVA The power factor is pf = cos ( 41.6° ) = 0.75 . The average power is P = 16.4 kW. The apparent power is |S| = 21.9 kVA. P 11.6-3 The source of Figure P 11.6-3 delivers 50 VA with a power factor of 0.8 lagging. Find the unknown impedance Z. 6Ω 20 0° V Answer: Z = 6.39 ∠26.6° Ω + – Z j8 Ω Figure P 11.6-3 Solution: The source current can be calculated from the apparent power: 2( 50∠ cos −1 0.8 ) Vs I s∗ 2S ∗ ⇒I = = = 5∠36.9° A S= s 20∠0° Vs 2 I s = 5∠ − 36.9° = 4 − j3 A Next I1 = Vs 20∠0° = = 2∠ − 53.1° = 1.2 − j1.6 A 6 + j8 10∠53.1° I 2 = I s − I1 = 4 − j 3 − 1.2 + j1.6 = 2.8 − j1.4 = 3.13∠ − 26.6° A Finally, Z= Vs 20∠0° = = 6.39∠26.6° Ω I 2 3.13∠ − 26.6° P 11.6-4 Manned space stations require several continuously available ac power sources. Also, it is desired to keep the power factor close to 1. Consider the model of one communication circuit shown in Figure P 11.6-4. If an average power of 500 W is dissipated in the 20Ω resistor, find (a) Vrms, (b) Is rms, (c) the power factor seen by the source, and (d) |Vs|. Is Vs + – I –j20 Ω + j20 Ω 20 Ω V = ⎜V ⎜ 0° V – IL Figure P 11.6-4 Solution: (Using all rms values.) 2 (a) V 2 2 P = I R= ⇒ V = P ⋅ R = ( 500 )( 20 ) ⇒ R (b) Is = I + I L = V = 100 Vrms V V 100∠0° 100∠0° + = + = 5 − j 5 = 5 2∠ − 45° A 20 j 20 20 j 20 (c) Zs = − j 20 + ( 20 )( j 20 ) = 10 20 + j 20 pf = cos ( −45° ) = (d) 2∠ − 45° Ω 1 leading 2 No average power is dissipated in the capacitor or inductor. Therefore, PAVE = PAVE = 500 W ⇒ source 20Ω Vs I s cos θ = 500 ⇒ Vs = 500 = I s cos θ ( 500 = 100 V ⎛ 1 ⎞ 5 2 ⎜ ⎟ ⎝ 2⎠ ) I P 11.6-5 Two impedances are supplied by V = 100 ∠160° Vrms, as shown in Figure P 11.6-5, where I = 2 ∠190° A rms. The first load draws P1 = 23.2 W and Q1 = 50 VAR. Calculate I1, I2, the power factor of each impedance, and the total power factor of the circuit. I1 + V – Z1 I2 Z2 Figure P 11.6-5 Solution: Load 1: V = 100∠160° Vrms I = 2∠190° Arms = −1.97 − j 0.348 Arms P1 = 23.2 W, Q1 = 50 VAR S1 = P1 + jQ1 = 23.2 + j50 = 55.12∠65.1° VA pf1 = cos 65.1° = 0.422 lagging I1∗ = Load 2: S1 55.12∠65.1° = = 0.551∠ − 94.9°, so I1 = 0.551∠94.9° Arms Vs 100∠160° I 2 = I − I 1 = −1.97 − j 0.348 + 0.047 − j 0.549 = 2.12∠ − 155° Arms S 2 = V I 2 * = (100∠160° )( 2.12∠155° ) * = 212∠315° = 212∠ − 45° = 150 − j 150 VA pf 2 = cos ( −45° ) = 0.707 leading Total: S = S1 + S 2 = ( 23.2 + j50 ) + (150 − j150 ) = 173.2 − j100 = 200∠ − 30° VA pf = cos ( −30° ) = 0.866 leading P 11.6-6 A residential electric supply three-wire circuit from a transformer is shown in Figure P 11.6-6a. The circuit model is shown in Figure P 11.6-6b. From its 7.2 kV nameplate, the refrigerator motor is known to have a rated current of 8.5 A rms. It is reasonable to assume an inductive impedance angle of 45° for a small motor at rated load. Lamp and range loads are 100 W and 12 kW, respectively. (a) (b) Calculate the currents in line 1, line 2, and the neutral wire. Calculate: (i) Prefrig, Qrefrig, (ii) Plamp, Qlamp, and (iii) Ptotal, Qtotal, Stotal, and overall power factor. Line 1 120 Vrms Refrig Neutral Kitchen range Lamp 120 Vrms Line 2 + – (a) Refrig 120 0° Vrms Range + – Lamp 120 0° Vrms (b) (c) The neutral connection resistance increases, because of corrosion and looseness, to 20 Ω. (This must be included as part of the neutral wire.) Use mesh analysis and calculate the voltage across the lamp. Solution: (Using rms values) Z refrig = 120 = 14.12 Ω 8.5 Z refrig = 14.12∠45° = 10 + j10 Ω V 2 (120 ) = = = 144 Ω 100 P 2 R lamp R range (a) Ι refrig = and ( 240 ) = 2 12, 000 = 4.8 Ω 120∠0° 120∠0° = 8.5∠ − 45° Arms , I lamp = = 0.83∠0° Arms 10 + j10 144 I range = From KCL: 240∠0° = 50∠0° Arms 4.8 I1 = I refrig + I range = 56 − j 6 = 56.3∠ − 6.1° Arms I 2 = −I lamp − I range = 50.83∠180° Arms I N = −I1 − I 2 = 7.92∠131° Arms (b) Prefrig = I refrig 2 2 Rrefrig = 722.5 W and Qrefrig = I refrig X refrig = 722.5 VAR Plamp = 100 W and Q lamp = 0 Ptotal = 722 + 100 + 12, 000 = 12.82 kW ⎫ ⎬ ⇒ S = 12,822 + j 722 = 12.84∠3.2° kVA Qtotal = 722 + 0 + 0 = 722 VAR ⎭ The overall power factor is pf = cos ( 3.2° ) = 0.998 , (c) Mesh equations: − 20 ⎡ 30 + j10 ⎢ − 20 164 ⎢ ⎢⎣ −10 − j10 − 144 Solve to get: − 10 − j10 ⎤ ⎡ I A ⎤ ⎡120∠0° ⎤ − 144 ⎥⎥ ⎢⎢ I B ⎥⎥ = ⎢⎢120∠0° ⎥⎥ ⎥⎦ 158.8 + j10 ⎥⎦ ⎢⎣ I C ⎥⎦ ⎢⎣ 0 I A = 64.3 − j1.57 = 64.3∠ − 1.7° Arms I B = 61.3 − j 0.19 = 61.3∠ − 0.5°Arms I C = 60 + j 0 = 60∠0° Arms The voltage across the lamp is Vlamp = Rlamp I B − IC = 144 1.27∠ − 8.6° = 183.2 Vrms (checked: MATLAB 7/20/04) P 11.6-7 A motor connected to a 220-V supply line from the power company has a current of 7.6 A. Both the current and the voltage are rms values. The average power delivered to the motor is 1317 W. (a) Find the apparent power, the reactive power, and the power factor when ω = 377 rad/s. (b) Find the capacitance of a parallel capacitor that will result in a unity power factor of the combination. (c) Find the current in the utility lines after the capacitor is installed. Answer: (a) pf = 0.788 (b) C = 56.5 μF (c) I = 6.0 A rms Solution: (Using all rms values) (a) VI = 220 ( 7.6 ) = 1672 VA P 1317 = = .788 VI 1672 θ =cos −1pf = 38.0° ⇒ Q = VIsinθ =1030 VAR pf = (b) To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing –Q, then V2 (220)2 ⇒ X c = 47Ω 1030 = = Xc Xc ∴C= 1 (c) 1 ω X = (377)(47) = 56.5μ F P = VI cosθ where θ = 0° then 1317 = 220I ∴ I = 6.0Arms for corrected pf * Note I = 7.6Arms for uncorrected pf . P 11.6-8 Two loads are connected in parallel across a 1000-V rms, 60-Hz source. One load absorbs 500 kW at 0.6 power factor lagging, and the second load absorbs 400 kW and 600 kVAR. Determine the value of the capacitor that should be added in parallel with the two loads to improve the overall power factor to 0.9 lagging. Answer: C = 2.2 μF Solution: First load: ( ) S 1 = P + jQ = P 1 + j tan ( cos −1 (.6 ) ) = 500 (1 + j tan 53.1° ) = 500 + j 667 kVA S 2 = 400 + j 600 kVA Second load: Total: S = S1 + S 2 = 900 + j1267 kVA S desired = P + jP tan (cos −1 (.90)) = 900 + j 436 VA From the vector diagram: S desired = S + jQ . Therefore 900 + j 436 = 900 + j1277 + j Q ⇒ Q = −841 VAR 2 2 V V (1000) 2 j * 841 = − ⇒ = = = j1189 ⇒ Z = − j1189 = − j Z * 377 C − j841 − j841 Z Finally, C= 1 = 2.20 μ F (1189)(377) 6.4 Ω P 11.6-9 A voltage source with a complex internal impedance is connected to a load, as shown in Figure V P 11.6-9. The load absorbs 1 kW of average power at 1 100 V rms with a power factor of 0.80 lagging. The source frequency is 200 rad/s. 24 mH + + – ZL VL – Figure P 11.6-9 (a) Determine the source voltage V1. (b) Find the type and value of the element to be placed in parallel with the load so that maximum power is transferred to the load. Solution: (a) S = P + jQ = P + jP tan (cos −1 pf ) = 1000 + j1000 tan (cos −1 0.8) = 1000 + j 750 VA S 1000+ j750 = = 10 + j7.5 ⇒ I = 10 − j 7.5 Arms Let VL = 100∠0° Vrms. Then I* = VL 100∠0° ZL = (b) VL 100∠0° = = 8∠36.9° = 6.4 + j 4.8 Vrms I 12.5∠−36.9° V1 =[6.4 + j (200)(.024) + Z L )(I ) = (12.8 + j 9.6)(10 − j 7.5) = 200∠0° Vrms 1 For maximum power transfer, we require ( 6.4 + j 4.8 )* = Z L || Z new = . YL +Ynew 1 1 1 = YL + Ynew ⇒ Ynew = − = j 0.15 S (6.4 − j4.8) 6.4 − j 4.8 6.4 + j 4.8 Then Z new = − j 6.67 Ω so we need a capacitor given by 1 1 = 6.67 ⇒ C = = 750 μ F (6.67)(200) ωC P 11.6-10 The circuit shown in Figure P 11.6-10a can be represented in the frequency domain as shown in Figure P 11.6-10b. In the frequency domain, the value of the mesh current is I = 1.076 ∠–38.3° A. (a) Determine the complex power supplied by the voltage source. (b) Given that the complex power received by Z1, is 6.945 + j 13.89 VA, determine the values of R1 and L1. (c) Given that the real power received by Z3 is 4.63 W at a power factor of 0.56 lagging, determine the values of R3 and L3. Z1 = R1 + jωL1 R1 L1 R2 + – 48 cos (6t) V R3 48 0° V + – I Z2 = R2 − j C2 L3 Z3 = R3 + jωL3 (a) (b) Figure P 11.6-10 Solution: ( 48∠0° )(1.076∠ − 38.3° ) (a) S = * 2 (b) V1 = Z1 = 2 S1 I * V1 I = = 2 ( 6.945 + j 13.89 ) (1.076∠ − 38.3° ) * = 25.82∠38.3° = 20.27 + j 16 VA = 2 (15.53∠63.4° ) = 28.87∠25.1° V (1.076∠38.3° ) 28.87∠25.1° = 26.83∠63.4° = 12 + j 24 = 12 + j 6 ( 4 ) Ω 1.076∠ − 38.3° R1 = 12 Ω and L1 = 4 H. (c) θ = cos −1 ( 0.56 ) = 56° , S = V3 = Z3 = 2 ( 8.268∠56° ) (1.076∠ − 38.3° ) * V3 I = P 4.63 = = 8.268 VA pf 0.56 = 15.37∠17.7° V 15.37∠17.7° = 14.28∠56° = 8 + j 11.83 = 8 + j 6 (1.97 ) Ω 1.076∠ − 38.3° R3 = 8 Ω and L3 = 2 H. 1 ωC2 R P 11.6-11 The circuit in Figure P 11.6-11 consists of a source connected to a load. The source delivers 14.12 W to the load at a power factor of 0.857 lagging. What are the values of the resistance, R, and the inductance, L? + – 24 cos (3t +75°) V Source L Load Figure P 11.6-11 Solution: Represent the circuit in the frequency domain as ⎧0.857 = cos (θ ) ⎪ pf = 0.857 lagging ⇒ ⎨ and ⎪ θ >0 ⎩ so θ = 31°. Next 14.12 = P = S cos θ = S ( 0.857 ) so S = Then and so 14.12 = 16.48 VA 0.857 S = 16.48∠31° = 14.12 + j8.49 ∗ ∗ ⎡ 2 (16.48∠31° ) ⎤ ⎛ 2S ⎞ I=⎜ ⎟ =⎢ ⎥ = 1.37∠44° 24 75 ∠ ° ⎝V⎠ ⎣ ⎦ 24∠75° R + j 3L = = 17.5∠31° = 15 + j 9 Ω 1.37∠44° R = 15 Ω L=3H P 11.6-12 The circuit in Figure P 11.6-12 consists of a source connected to a load. Determine the impedance of the load and the complex power delivered by the source to the load under each of the following conditions: + – A 24 cos (3t + 75°) V Source B Load Figure P 11.6-12 (a) (b) (c) (d) The source delivers 14.12 + j8.47 VA to load A and 8.47 + j14.12 VA to load B. The source delivers 8.47 + j14.12 VA to load A, and the impedance of load B is 15 + j9 Ω. The source delivers 14.12 W to load A at a power factor of 0.857 lagging, and the impedance of load B is 9 + j15 Ω. The impedance of load A is 15 + j9 Ω, and the impedance of load B is 9 + j15 Ω. Solution: Represent the circuit in the frequency domain as (a) ∗ ⎛ 2 (14.12 + j8.47 ) ⎞ IA = ⎜ ⎟ = 1.37∠44° A 24∠75° ⎝ ⎠ ∗ ⎛ 2 ( 8.47 + j14.12 ) ⎞ IB = ⎜ ⎟ = 1.37∠16° A 24∠75° ⎝ ⎠ I = I A + I B = (1.37∠44° ) + (1.37∠16° ) = ( 0.986 + j 0.954 ) + (1.319 + j 0.377 ) = 2.305 + j1.331 = 2.662∠30° A Z= S= 24∠75° = 9.016∠45° 2.662∠30° 1 ∗ ( 24∠75° )( 2.662∠30° ) = 31.9∠45° = 22.59 + j 22.59 VA 2 (b) ∗ ⎛ 2 ( 8.47 + j14.12 ) ⎞ IA = ⎜ ⎟ = 1.37∠16° A 24∠75° ⎝ ⎠ IB = 24∠75° = 1.37∠44° A 15 + j 9 I = I A + I B = 2.662∠30° A Z= 24∠75° = 9.016∠45° Ω 2.662∠30° S = 22.59 + j 22.59 VA (c) P = 14.12 W = 0.857 = cos ( 75 − θ A ) ⎫⎪ ⎬ 75 − θ A > 0 ⎪⎭ Then IA = 24 I A 2 ⇒ cos ( 75 − θ A ) θ A = 75° − 31° = 44° 2 (14.12 ) = 1.37 24 cos ( 31° ) so I A = 1.37∠44° A Also 24∠75° = 137∠16° A 9 + j15 I = I A + I B = 2.662∠30° A IB = (d) 24∠75° = 1.37∠44° 15 + j 9 24∠75° IB = = 1.37∠16° 9 + j15 I = I A + I B = 2.662∠30° A IA = Z= 24∠75° = 9.016∠45° Ω 2.662∠30° S = 22.59 + j 22.59 VA P 11.6-13 Figure P 11.6-13 shows two possible representations of an electrical load. One of these representations is used when the power factor of the load is lagging, and the other is used when the power factor is leading. Consider two cases: (a) At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 lagging. (b) At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 leading. R R L C Figure P 11.6-13 In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the value of the capacitance, C, or the inductance, L. Solution: ⎛X⎞ Let Z = R + j X = R 2 + X 2 ∠ tan −1 ⎜ ⎟ be the impedance of the load. Further, let V = A∠θ ⎝R⎠ and I = B∠φ be the voltage across and current through the load. Then A∠θ A ⎛X⎞ = ∠ (θ − φ ) R 2 + X 2 ∠ tan −1 ⎜ ⎟ = R + j X = Z = B∠φ B ⎝R⎠ ⎛X⎞ Equating angles gives tan −1 ⎜ ⎟ = θ − φ ⎝R⎠ Also, the complex power delivered to the load is S = ( A∠θ )( B∠φ ) * = So AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2 pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf ) (Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a) pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9° X = tan ( 36.9 ) = 0.75 ⇒ X = 6 ( 0.75 ) = 4.5 Ω R Choose the inductor to implement the positive reactance. Then We require 4.5 = ω L = 4 L ⇒ L = 1.125 H b) pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9° X = tan ( −36.9 ) = −0.75 ⇒ X = 6 ( −0.75 ) = −4.5 Ω R Choose the capacitor to implement the negative reactance. Then We require −4.5 = − 1 1 =− 4C ωC ⇒ C= 8 = 0.889 F 9 P 11.6-14 Figure P 11.6-14 shows two possible representations of an electrical load. One of these representations is used when the power factor of the load L R C R is lagging, and the other is used when the power factor is leading. Consider two cases: (a) At the frequency ω = 4 rad/s, the load has the Figure P 11.6-14 power factor pf = 0.8 lagging. (b) At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 leading. In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the value of the capacitance, C, or the inductance, L. Solution: 1 1 1 1 1 1 ⎛ R⎞ = −j = + 2 ∠ tan −1 ⎜ − ⎟ be the admittance of the load. Further, Let Y = + 2 R jX R X R X ⎝ X⎠ let V = A∠θ and I = B∠φ be the voltage across and current through the load. Then 1 1 B∠φ B ⎛ R⎞ + 2 ∠ tan −1 ⎜ − ⎟ = Y = = ∠ (φ − θ ) 2 R X A∠θ A ⎝ X⎠ ⎛ R ⎞ tan −1 ⎜ − ⎟ = (φ − θ ) ⎝ ω L⎠ Also, the complex power delivered to the load is Equating angles gives S = ( A∠θ )( B∠φ ) * = AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2 pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf ) So (Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) Therefore R ⎛ R⎞ tan −1 ⎜ − ⎟ = (φ − θ ) = − cos −1 ( pf ) ⇒ − = tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) ) X ⎝ X⎠ R ⇒ = tan ( cos −1 ( pf ) ) X a) pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9° R 6 = tan ( 36.9 ) = 0.75 ⇒ X = =8 Ω 0.75 X Choose the inductor to implement the positive reactance. Then We require 8=ω L = 4L ⇒ L = 2 H pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9° b) R 6 = tan ( −36.9 ) = −0.75 ⇒ X = = −8 Ω −0.75 X Choose the capacitor to implement the negative reactance. Then We require −8 = − 1 1 =− 4C ωC ⇒ C= 1 = 31.25 mF 32 R P 11.6-15 Figure P 11.6-15 shows two electrical loads. Express the power factor of each load in terms of ω, R, and L. R L L (a) (b) Figure P 11.6-15 Solution: Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex power delivered to the load is S = ( A∠θ )( B∠φ ) * = So AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2 pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf ) (Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a.) Let Also Equating angles gives ⎛ω L ⎞ Z = R + j ω L = R 2 + ω 2 L2 ∠ tan −1 ⎜ ⎟ ⎝ R ⎠ A∠θ A Z= = ∠ (θ − φ ) B∠φ B ⎛ω L ⎞ tan −1 ⎜ ⎟ = θ −φ ⎝ R ⎠ ⎛ −1 ⎛ ω L ⎞ ⎞ ⎛ω L⎞ −1 tan −1 ⎜ ⎟ = θ − φ = cos ( pf ) ⇒ pf = cos ⎜ tan ⎜ ⎟ ⎟ lagging ⎝ R ⎠ ⎝ R ⎠⎠ ⎝ ⎛ω L ⎞ (The power factor is lagging because tan −1 ⎜ ⎟ is an angle in the first quadrant.) ⎝ R ⎠ so b.) Let Y= 1 1 1 1 + = −j = R jω L R ωL Also ⎛ R ⎞ 1 1 + 2 2 ∠ tan −1 ⎜ − ⎟ 2 R ω L ⎝ ω L⎠ B∠φ B = ∠ (φ − θ ) A∠θ A ⎛ R ⎞ tan −1 ⎜ − ⎟ = (φ − θ ) ⎝ ω L⎠ Y= Equating angles gives ⎛ R ⎞ R −1 = tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) ) tan −1 ⎜ − ⎟ = (φ − θ ) = − cos ( pf ) ⇒ − ωL ⎝ ω L⎠ ⎛ ⎛ R ⎞⎞ so pf = cos ⎜ tan −1 ⎜ ⎟ ⎟ lagging ⎝ ω L ⎠⎠ ⎝ ⎛ R ⎞ (The power factor is lagging because tan −1 ⎜ ⎟ is an angle in the first quadrant.) ⎝ω L ⎠ R P 11.6-16 Figure P 11.6-16 shows two electrical loads. Express the power factor of each load in terms of ω, R, and C. R C C (b) (a) Figure P 11.6-16 Solution: Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex power delivered to the load is AB AB S = ( A∠θ )( B∠φ ) * = cos (θ − φ ) + j sin (θ − φ ) 2 2 So pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf ) (Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a.) Let Also Z = R+ 1 jω C = R− j ⎛ 1 1 1 ⎞ = R 2 + 2 2 ∠ tan −1 ⎜ − ⎟ ωC ω C ⎝ ωCR⎠ Z= A∠θ A = ∠ (θ − φ ) B∠φ B ⎛ 1 ⎞ tan −1 ⎜ − ⎟ =θ −φ ⎝ ωCR⎠ Equating angles gives So ⎛ 1 ⎞ −1 tan −1 ⎜ − ⎟ = θ − φ = cos ( pf ) ⇒ ⎝ ωCR⎠ ⎛ ⎛ 1 ⎞⎞ pf = cos ⎜ tan −1 ⎜ − ⎟ ⎟ leading ⎝ ω C R ⎠⎠ ⎝ ⎛ −1 ⎞ (The power factor is leading because tan −1 ⎜ ⎟ is an angle in the third quadrant.) ω C R ⎝ ⎠ Y= b.) Let Also Equating angles gives So 1 1 + jω C = + ω 2 C 2 ∠ tan −1 (ω C R ) 2 R R B∠φ B Y= = ∠ (φ − θ ) A∠θ A tan −1 (ω C R ) = (φ − θ ) tan −1 (ω C R ) = − cos −1 ( pf ) ⇒ pf = cos ( − tan −1 (ω C R ) ) leading (The power factor is leading because − tan −1 (ω C R ) is an angle in the third quadrant.) P11.6-17 The source voltage in the circuit shown in Figure P11.6-17 is V s = 24 ∠ 30° V. Consequently I 1 = 3.13 ∠ 25.4° A, I 2 = 1.99 ∠ 52.9° A and V 4 = 8.88 ∠ −10.6° V Determine (a) the average power absorbed by Z 4, (b) the average power absorbed by Z 1, and (c) the complex power delivered by the voltage source. (All phasors are given using peak, not RMS, values.) Figure P11.6-17 Solution: ⎛ ⎛ −2 ⎞ ⎞ a. The power factor of Z 1 is pf1 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( −26.6° ) = 0.894 leading . ⎝ 4 ⎠⎠ ⎝ ⎛ ⎛ 8 ⎞⎞ b. The power factor of Z 3 is pf 3 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( 69.4° ) = 0.352 lagging . ⎝ 3 ⎠⎠ ⎝ c. The power factor of Z 4 is pf 4 = cos ( −10.6° − 52.9° ) = cos ( −63.5° ) = 0.45 leading . P11.6-18 The current source the circuit shown in Figure P11.6-18 supplies 131.16 – j36.048 VA and the voltage source supplies 64.2275 – 87.8481VA. Determine the values of the impedances Z1 and Z2. Figure P11.6-18 Solution: The current source this circuit supplies 131.16 – j36.048 VA and the voltage source supplies 64.2275 – 87.8481VA. Determine the values of the impedances Z1 and Z2. The voltage V1 is calculated from the current source current and complex power: 2S 2 (131.16 − j 36.048 ) = = 136.02∠ − 0.368° V I* ( 2∠15° ) * The current I2 is calculated from the voltage source voltage and complex power: V1 = * * ⎛ 2S ⎞ ⎛ 2 ( 64.2275 + j87.8481) ⎞ I2 = ⎜ ⎟ = ⎜ ⎟ = 2.1765∠ − 83.829° A 100∠ − 30° ⎝V⎠ ⎝ ⎠ The impedances are calculated as Z1 = Z2 = V1 2∠15° + I 2 = V1 − 100∠ − 30° −I 2 136.02∠ − 0.368° = 40 + j 30 Ω 2∠15° + ( 2.1765∠ − 83.829° ) = 136.02∠ − 0.368° − 100∠ − 30° = 20 − j 25 Ω −2.1765∠ − 83.829° Section 11.7 The Power Superposition Principle 14 A P 11.7-1 Find the average power absorbed by the 2-Ω resistor in the circuit of Figure P 11.7-1. Answer: P = 413 W + 110 cos 20t V 2Ω – 12 Ω 1 100 F Figure P 11.7-1 Solution: Use superposition since we have two different frequency sources. First consider the dc source (ω = 0): ⎛ 12 ⎞ I1 = 14 ⎜ ⎟ = 12 A ⎝ 12 + 2 ⎠ P1 = I12 R = (12) 2 (2) = 288 W Next, consider the ac source (ω = 20 rad/s): After a source transformation, current division gives − j 60 ⎡ ⎤ ⎢ ⎥ 25 (12 − j 5) I 2 = −9.167 ⎢ ∠116.6° A ⎥ = 5 ⎢ − j 60 + 2+ j 4 ⎥ ⎢⎣ (12 − j 5) ⎥⎦ 2 Then I (125)(2) P2 = 2 (2) = =125 W 2 2 Now using power superposition P = P1 + P2 = 288 + 125 = 413 W 0.2 H + P 11.7-2 Find the average power absorbed by the 8-Ω resistor in the circuit of Figure P 11.7-2. Answer: P = 22 W 5 cos 2000t A 18 – 40 cos 8000t V mF 1 mH 8Ω Figure P 11.7-2 Solution: Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s source: Current division yields ⎡ 8 ⎤ ⎢ −j2 ⎥ 5 I1 = 5 ⎢ ∠63.4° A ⎥= 5 ⎢ 8 +8 ⎥ ⎢⎣ − j 2 ⎥⎦ 2 Then I 8 P1 = 1 = 20 W 2 Next consider ω = 8000 rad/s source. Current division yields ⎡ 8 ⎤ ⎢ j7 ⎥ 5 I 2 = − j5 ⎢ ∠ − 171.9° A ⎥= 8 50 ⎢ +8 ⎥ ⎢⎣ j 7 ⎥⎦ 2 Then I 8 P2 = 2 =2W 2 Now using power superposition P = P1 + P2 = 22 W P 11.7-3 For the circuit shown in Figure P 11.7-3, determine the average power absorbed by each resistor, R1 and R2. The voltage source is vs = 10 + 10 cos (5t + 40°) V, and the current source is is = 4 cos (5t – 30°) A. 2H 10 Ω 6i2 R1 R2 5 Ω + – is 1 10 vs F i2 Figure P 11.7-3 P11.7-3 Use superposition since we have two different frequencies. First consider the dc source (ω = 0): ⎛1⎞ i 2 (t ) = 0 and i1 (t ) = 10 ⎜ ⎟ = 1 A ⎝ 10 ⎠ PR1 = i12 R1 = 12 (10) = 10 W PR 2 = 0 W Next consider ω = 5 rad/s sources. Apply KCL at the top node to get (10I1 −10∠40°) =0 j10 −10 I1 + (5− j 2) I 2 = 0 −6 I 2 + I1 + I 2 − ( 4 ∠−30° ) + Apply KVL to get Solving these equations gives I1 = −0.56 ∠ − 64.3° A and I 2 = −1.04∠ − 42.5° A Then PR1 = I12 R1 2 ( 0.56 ) = 2 2 (10) = 1.57 W and PR 2 = I 22 R 2 2 (1.04 ) = 2 Now using power superposition PR = 10 + 1.57 = 11.57 W and PR 2 = 0 + 2.7 = 2.7 W 1 2 (5) = 2.7 W P 11.7-4 For the circuit shown in Figure P 11.7-4, determine the effective value of the resistor voltage vR and the capacitor voltage vC. + 4 cos 10t V + – 2 Ω 6 sin 5t V + – 20 mF Figure P 11.7-4 Solution: Use superposition since we have two different frequencies. First consider the ω = 10 rad/s source: V1 4∠0° = = 0.28 + j 0.7 A Z 2− j5 V = 2 I1 = 2(0.28 + j.7) = 0.56 + j1.4 = 1.51 ∠68.2° V I1 = R1 V = − j 5 I1 = 3.77 ∠ − 21.8° V C1 Next consider ω = 5 rad/s source. V2 6∠−90° = = 0.577 − j 0.12 A Z 2 − j10 VR = 2 I 2 = 2(.577 − j 0.12) =1.15− j 0.24 I2 = 2 V C2 =1.17∠−11.8° V = − j10I 2 =5.9∠258.3° V Now using superposition vR (t) = 1.51cos (10 t + 68.2°) + 1.17 cos (5 t − 11.8°) V vC (t) = 3.77 cos (10 t − 21.8°) + 5.9 cos (5 t − 258.3°) V Then 2 2 2 Reff ⎛ 1.51 ⎞ ⎛ 1.17 ⎞ =⎜ ⎟ +⎜ ⎟ = 1.82 ⇒ VReff =1.35 V ⎝ 2⎠ ⎝ 2 ⎠ 2 Ceff ⎛ 3.77 ⎞ ⎛ 5.9 ⎞ =⎜ ⎟ +⎜ ⎟ = 24.52 ⇒ VCeff = 4.95 V ⎝ 2 ⎠ ⎝ 2⎠ V 2 V 2 vR – + vC – Section 11.8 The Maximum Power Transfer Theorem P 11.8-1 Determine values of R and L for the circuit shown in Figure P 11.8-1 that cause maximum power transfer to the load. Answer: R = 800 Ω and L = 1.6 H 4000 Ω R 5 cos (1000t + 60°) V + – 0.5 μF L Source Load Figure P 11.8-1 Solution: Z t = 4000 || − j 2000 = 800 − j1600 Ω Z L = Z*t =800+ j1600 Ω ⎧ R =800 Ω R + j1000 L = 800 + j1600 ⇒ ⎨ ⎩ L =1.6 H P 11.8-2 Is it possible to choose R and L for the circuit shown in Figure P 11.8-2 so that the average power delivered to the load is 12 mW? Answer: Yes R 2.5 cos 100t mA 0.2 μF 25 kΩ Source L Load Figure P 11.8-2 Solution: Z t = 25, 000 || − j 50, 000 = 20, 000 − j10, 000 Ω Z L =Z*t = 20,000+ j10,000 Ω ⎧ R = 20 kΩ ⎪ R + jω L = 20, 000 + j10, 000 ⇒ ⎨100 L =10,000 ⎪ L =100 H ⎩ After selecting these values of R and L, 2 Since Pmax ⎛ 0.14×10−2 ⎞ 3 I = 1.4 mA and Pmax = ⎜ ⎟ ( 20×10 ) = 19.5 mW 2 ⎝ ⎠ > 12 mW , yes, we can deliver 12 mW to the load. P 11.8-3 The capacitor has been added to the load in the circuit shown in Figure P 11.8-3 in order to maximize the power absorbed by the 4000-Ω resistor. What value of capacitance should be used to accomplish that objective? Answer: 0.1 μF 800 Ω + – 0.32 H 5 cos (5000t + 45°) V Source 4000 Ω C Load Figure P 11.8-3 Solution: ⎛ −j ⎞ R⎜ ω C ⎟⎠ R − jω R 2C Z t = 800 + j1600 Ω and Z L = ⎝ = j 1 + (ω RC ) 2 R− ωC ⎛ −j ⎞ R⎜ ω C ⎟⎠ R − jω R 2C * ⎝ ZL = Zt ⇒ = = 800 − j1600 Ω j 1+ (ω RC ) 2 R− ωC Equating the real parts gives 800 = R 4000 = 2 1+ (ω RC ) 1+[(5000)(4000)C ]2 ⇒ C = 0.1 μ F P 11.8-4 What is the value of the average power delivered to the 2000-Ω resistor in the circuit shown in Figure P 11.8-4? Can the average power delivered to the 2000-Ω resistor be increased by adjusting the value of the capacitance? Answer: 8 mW. No. 400 Ω 5 cos 1000t V 0.8 H + – 2000 Ω Source 1 μF Load Figure P 11.8-4 Solution: Z t = 400 + j 800 Ω and Z L = 2000 || − j1000 = 400 − j 800 Ω Since Z L = Z*t the average power delivered to the load is maximum and cannot be increased by adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is VR = 5 ZL = 2.5 − j 5 = 5.59e − j 63.4 V Zt +ZL 2 ⎛ 5.59 ⎞ 1 P=⎜ = 7.8 mW ⎟ ⎝ 2 ⎠ 2000 So is the average power delivered to the 2000 Ω resistor. P 11.8-5 What is the value of the resistance R in Figure P 11.8-5 that maximizes the average power delivered to the load? R 5 cos 1000t V 0.8 H + – 2000 Ω Source 1 μF Load Figure P 11.8-5 Solution: Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power provided by source is delivered to the load. Section 11.9 Coupled Inductors M a P 11.9-1 Two magnetically coupled coils are connected as shown in Figure P 11.9-1. Show that an equivalent inductance at terminals a–b is Lab = L1 + L2 – 2M. L1 L2 b Figure P 11.9-1 Solution: Vs = I jω L1 − I jω M + I jω L2 − I jω M ⇒ jω ( L1 + L2 − 2 M ) = Vs I Therefore Lab = L1 + L2 − 2M a P 11.9-2 Two magnetically coupled coils are shown connected in Figure P 11.9-2. Find the equivalent inductance Lab. M L1 b Figure P 11.9-2 Solution: The coil voltages are given by: V1 = I1 jω L1 + I 2 jω M V2 = I 2 jω L2 + I1 jω M From KVL the coil voltages are equal so I1 jω L1 + I 2 jω M = I 2 jω L2 + I1 jω M ⇒ I1 jω ( L1 − M ) = I 2 jω ( L2 − M ) ⎛ L −M ⇒ I1 = ⎜ 2 ⎝ L1 − M ⎞ ⎟ I2 ⎠ From KCL ⎛ L −M ⎞ ⎛ L + L −2 M ⎞ I s = I1 + I 2 = ⎜ 2 + 1⎟ I 2 = ⎜ 2 1 ⎟ I2 ⎝ L1 − M ⎠ ⎝ L1 − M ⎠ Next ⎛ L1 − M I2 = ⎜ ⎜ L2 + L1 − 2 M ⎝ ⎞ ⎛ L −M ⎟⎟ I s and I1 = ⎜ 2 ⎝ L1 − M ⎠ ⎞ ⎛ L2 − M ⎟ I2 = ⎜ ⎠ ⎝ L1 − M ⎞ ⎛ L1 − M ⎟ ⎜⎜ ⎠ ⎝ L2 + L1 − 2 M ⎞ ⎟⎟ I s ⎠ Substituting gives ⎛ L2 − M ⎞ ⎛ L1 − M ⎞ V1 = jω L1 I1 + jω M I 2 = jω M ⎜ I + jω M ⎜ I ⎜ L2 + L1 − 2 M ⎟⎟ s ⎜ L2 + L1 − 2 M ⎟⎟ s ⎝ ⎠ ⎝ ⎠ ⎛ M ( L2 − M ) + M ( L1 − M ) ⎞ ⎛ L1 L 2 − M 2 j = jω ⎜ I = ω ⎟⎟ s ⎜⎜ ⎜ L2 + L1 − 2 M ⎝ ⎠ ⎝ L1 + L 2 − 2 M ⎛ L1 L 2 − M 2 ⎞ V1 Z= = jω ⎜ Finally ⎜ L1 + L 2 − 2 M ⎟⎟ Is ⎝ ⎠ ⎞ ⎟⎟ I s ⎠ L2 P 11.9-3 The source voltage of the circuit shown in Figure P 11.9-3 is vs = 141.4 cos 100t V. Determine i1(t) and i2(t). 2Ω vs + – M = 0.6 H i1 0.4 H i2 1.6 H 200 Ω Figure P 11.9-3 Solution: Mesh equations: −141.4∠0° + 2 I1 + j 40 I1 − j 60 I 2 = 0 200 I 2 + j160 I 2 − j 60 I1 = 0 ⇒ I 2 = (0.23∠51°) I1 Solving yields I1 = 4.17∠ − 68° A and I 2 = 0.96∠ − 17°A Finally i1 ( t ) = 4.17 cos(100t −68°) A and i 2 ( t ) = 0.96 cos(100t −17°) A (checked: LNAPAC 7/21/04) P 11.9-4 A circuit with a mutual inductance is shown in Figure P 11.9-4. Find the voltage V2 when ω = 5000. 10 Ω V1 = 10 0° V + – M = 10 mH 1 mH 100 mH 400 Ω + V2 – Figure P 11.9-4 Solution: Mesh equations: (10+ j5 ) I 1 − j50 I 2 = 10 − j 50 I1 + ( 400+ j 500 ) I 2 = 0 Solving the mesh equations using Cramer’s rule: I2 = Then (10+ j5)( 0 )−( − j50 )(10 ) 2 (10+ j5) ( 400+ j500 ) − ( − j50 ) = 0.062 ∠29.7° A V2 = 400 I 2 = 400 ( 0.062 ∠29.7° ) = 24.8 ∠29.7° 1 150 P 11.9-5 Determine v(t) for the circuit of Figure P 11.9-5 when vs = 10 cos 30t V. Answer: v(t) = 23 cos (30t + 9°) V 28 Ω F + vs + – v(t) 0.1 H 0.3 H 0.2 H – Figure P 11.9-5 Solution: Mesh equations: 10 = − j 5I 1 + ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦ 0 = 28I 2 + ⎡⎣ j 6I 2 + j 3 ( I 1 − I 2 ) ⎤⎦ − ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦ or − j 6 ⎤ ⎡ I 1 ⎤ ⎡10 ⎤ ⎡ j4 ⎢ − j 6 28 + j 9 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ Solving the mesh equations, e.g. using MATLAB, yields I1 = 2.62∠ − 72° A and I 2 = 0.53∠0° A then V = j 9(I1 − I 2 ) + j 3I 2 = j 9I1 − j 6I 2 = 23 ∠10° V Finally v (t ) = 23cos (30t + 10°) V (checked: LNAPAC 7/21/04) P 11.9-6 Find the total energy stored in the circuit shown in Figure P 11.9-6 at t = 0 if the secondary winding is (a) open-circuited, (b) short-circuited, (c) connected to the terminals of a 7-Ω resistor. Answer: (a) 15 J 12 Ω M = 0.6 H 0.3 H 10 cos 5t A (b) 0 J (c) 5 J 1.2 H Figure P 11.9-6 Solution: (a) I 2 = 0 ⇒ I1 = 10 ∠0° A ⇒ i1 (0) = 10 A 2 w= (b) L1 i1 (0) 2 = (0.3) (10) 2 = 15 J 2 Mesh equations: j 6 I 2 - j 3 I1 = 0 ⇒ I1 = 2 I 2 I1 = 10∠0° A ⇒ I 2 = 5∠0° A Then w= 1 1 1 1 L1i 2 2 (0) + L 2i12 (0) − M i1 (0) i 2 (0) = (0.3)(10) 2 + (1.2)(5) 2 − (0.6) (10)(5) = 0 2 2 2 2 (c) (7 + j 6) I 2 − j 3 I1 = 0 I 2 = 3.25 ∠49.4° A i 2 (t ) = 3.25cos(5t + 49.4°) A i 2 (0) = 2.12 A Finally 1 1 w = (0.3) (10) 2 + (1.2) (2.12) 2 − (0.6) (10) (2.12) = 5.0 J 2 2 8 mH P 11.9-7 Find the input impedance, Z, of the circuit of Figure P 11.9-7 when ω = 1000 rad/s. Answer: Z = 8.4 ∠14° Ω 1 6 mF 5 mH 6 mH 3Ω Z Figure P 11.9-7 Solution: Mesh equations: − VT + j8 I1 + j 5(I1 − I 2 ) − j 6 I1 + j 6 (I1 − I 2 ) + j 5 I1 = 0 3 I 2 + j 6 (I 2 − I1 ) − j 5 I1 = 0 Solving yields I2 = j 11 I1 3+ j 6 ⎛ j 11 ⎞ VT = I1 ( j 18) + I 2 (− j 11) = ⎜ j 18+ ⎟I 3+ j 6 ⎠ 1 ⎝ Then Z = j 11 VT = j 18 + = 8.28 ∠13° = 8.0667 + j 1.8667 Ω I1 3+ j 6 (checked: LNAPAC 7/21/04) P 11.9-8 A circuit with three mutual inductances is shown in Figure P 11.9-8. When vs = 10 cos 2t V, M1 = 2 H, and M2 = M3 = 1 H, determine the capacitor voltage v(t). M1 5Ω 4H 3H M2 vs 2Ω 6Ω M3 + – + 1 10 v F 2H – Figure P 11.9-8 Solution: The coil voltages are given by V1 = j 6 I1 − j 2 (I1 − I 2 ) − j 4 I 2 = j 4 I1 − j 2 I 2 V2 = j 4 ( I1 − I 2 ) − j 2 I1 + j 2 I 2 = j 2 I1 − j 2 I 2 V3 = j8 I 2 − j 4 I1 + j 2 ( I1 − I 2 ) = − j 2 I1 + j 6 I 2 The mesh equations are 5 I1 + V1 + 6 (I1 − I 2 ) + V2 = 10∠0° − V2 + 6 (I 2 − I1 ) + 2 I 2 + V3 − j 5 I 2 = 0 Combining and solving yields 11 + j 6 10 −6 − j 4 0 60 + j 40 I2 = = = 1.2 ∠0.28° A 11 + j 6 −6 − j 4 50 + j 33 −6 − j 4 8 + j 3 Finally V = − j 5 I 2 = 6.0 ∠ − 89.72° A ⇒ v(t ) = 6 cos(2 t − 89.7°) V 0.25 mF P 11.9-9 The currents i1(t) and i2(t) in Figure, P 11.9-9 are mesh currents. Represent the circuit in the frequency domain and write the mesh equations. 6H 5H 40 Ω i1(t) i2(t) 8H + – 15 cos (25t + 30°) V 80 Ω P 11.9-9 Solution: The equations describing the coupled coils give: V1 = j 200 ( I 1 − I 2 ) + j125 I 2 = j 200I 1 − j 75 I 2 V 2 = j150 I 2 + j125 ( I 1 − I 2 ) = j125 I 1 + j 25 I 2 The mesh equation for the left mesh is − j 160 I 1 + V1 − 15∠30° + 40 I 1 = 0 − j 160 I 1 + j 200I 1 − j 75 I 2 − 15∠30° + 40 I 1 = 0 ( 40 + j 40 ) I 1 − j 75 I 2 = 15∠30° The mesh equation for the right mesh is V 2 + 80 I 2 − V1 = 0 j125 I 1 + j 25 I 2 + 80 I 2 − ( j 200I 1 − j 75 I 2 ) = 0 − j 75 I 1 + ( 80 + j100 ) I 2 = 0 P 11.9-10 Determine the mesh currents for the circuit shown in Figure P 11.9-10. 3H 10 Ω + – – + 4H v1 12 cos 5t V 6H + – i1 v2 20 Ω 50 Ω i2 P 11.9-10 Solution: In the frequency domain, the coil voltages are given by V1 = j ω L1 I 1 − j ω M I 2 V2 = j ω L 2 I 2 − j ω M I 1 The mesh equations are R1 I 1 + V1 + R 2 ( I 1 − I 2 ) = Vs V2 + R3 I 2 − R 2 ( I 1 − I 2 ) = 0 Substituting for the coil voltages gives ⎡ R1 + R 2 + j ω L1 ⎢ ⎢ ⎣ − ( R2 + j ω M ) − ( R 2 + j ω M ) ⎤ ⎡ I 1 ⎤ ⎡ Vs ⎤ ⎥⎢ ⎥ = ⎢ ⎥ R 2 + R 3 + j ω L 2 ⎦⎥ ⎣I 2 ⎦ ⎣ 0 ⎦ Using the given values ⎡ 30 + j 20 − ( 20 + j 15 ) ⎤ ⎡ I 1 ⎤ ⎡12∠0 ⎤ ⎢ ⎥⎢ ⎥ = 70 + j 30 ⎦ ⎣I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣ − ( 20 + j 15 ) Solving, for example using MATLAB, gives ⎡ I 1 ⎤ ⎡0.4240∠ − 28.9° ⎤ ⎢ ⎥=⎢ ⎥ ⎣ I 2 ⎦ ⎣ 0.1392∠ − 15.2° ⎦ Back in the time domain, the mesh currents are ⎡ i1 ⎤ ⎡0.4240 cos ( 5t − 28.9° ) ⎤ ⎥ A ⎢i ⎥ = ⎢ − ° t 0.1392 cos 5 15.2 ( ) 2 ⎣ ⎦ ⎣ ⎦ (checked using LNAP, 9/9/04) P 11.9-11 Determine the coil voltages, v1, v2, v3, and v4, for the circuit shown in Figure P 11.911. v1 – 2 cos (5t + 45°) A + 4H 10 Ω 3H + 10 Ω v2 v3 + 6H – 5H 20 Ω 4H – – 5H v4 + 1.25 cos (5t − 45°) A 2.75 cos (5t) A Figure P 11.9-11 Solution: In the frequency domain, the coil voltages are given by V1 = − j ω L1 I a + j ω M 1 ( I a − I b ) = j ω ( M 1 − L1 ) I a − j ω M 1 I b = ( − j 5 )( 2∠45° ) + ( − j 15 )(1.25∠ − 45° ) = 21.25∠ − 106.9° V V2 = j ω L 2 ( I b − I a ) + j ω M 1 I a = j ω ( M 1 − L2 ) I a + j ω L2 I b = ( − j 15 )( 2∠45° ) + ( j 30 )(1.25∠ − 45° ) = 48.02∠6.4° V V3 = j ω L 3 ( I c − I a ) − j ω M 2 I c = − j ω L 2 I a + j ω ( L 3 − M 2 ) I c = ( − j 25 )( 2∠45° ) + ( j 5 )( 2.75∠0° ) = 41.43∠ − 31.4° V V4 = − j ω L 4 I c + j ω M 2 ( I c − I a ) = − j ω M 2 I a + j ω ( M 2 − L 4 ) I c = ( − j 20 )( 2∠45° ) + ( − j 5 )( 2.75∠0° ) = 50.66∠ − 56.1° V Back in the time domain, the coil voltages are v1 = 21.25 cos ( 5t − 106.9° ) V , v 2 = 48.02 cos ( 5t + 6.3° ) V , v 3 = 41.43 cos ( 5t − 31.4° ) V and v 4 = 50.66 cos ( 5t − 56.1° ) V (checked using LNAP, 9/9/04) 4H Solution: (a) In (b) and (c) the coils are coupled by mutual inductance, but not in (a). This circuit can be represented in the frequency domain, using phasors and impedances. + 5Ω 5H vo(t ) – + P 11.9-12 Figure P 11.9-12 shows three similar circuits. In each, the input to the circuit is the voltage of the voltage source, vs(t). The output is the voltage across the right-hand coil, vo(t). Determine the steady-state output voltage, vo(t), for each of the three circuits. – vs(t ) = 5.94 cos (3t + 140°) V (a) 4H + 2H 5Ω 5H vo(t ) + – – vs(t ) = 5.94 cos (3t + 140°) V Voltage division gives Vo = (b) j 15 5.94∠140° 5 + j 12 + j 15 4H = ( 0.5463∠10.5° )( 5.94∠140° ) = 3.245∠150.5° V vo ( t ) = 3.245 cos ( 3 t + 150.5° ) V (b) The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. 5Ω 5H vo(t ) – + In the time domain, the output voltage is given by + 2H – vs(t ) = 5.94 cos (3t + 140°) V (c) Figure P 11.9-12 This circuit of a single mesh. Notice that the mesh current, I(ω), enters the undotted ends of both coils. Apply KVL to the mesh to get 5 I (ω ) + ( j12 I (ω ) + j 6 I (ω ) ) + ( j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0 5 I (ω ) + ( j12 + j 6 + j 6 + j15 ) I (ω ) − 5.94∠140° = 0 I (ω ) = 5.94∠140° 5.94∠140° 5.94∠140° = = = 0.151∠57° A 5 + j (12 + 6 + 6 + 15 ) 5 + j 39 39.3∠83 Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by Vo (ω ) = j 15 I (ω ) + j 6 I (ω ) = j 21 I (ω ) = j 21 ( 0.151∠57° ) = (21∠90°) ( 0.151∠57° ) = 3.17∠147° V In the time domain, the output voltage is given by vo ( t ) = 3.17 cos ( 3 t + 147° ) V (c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the lefthand coil is located at the right of the coil in (c) and at the left of the coil in (b). As before, our first step is to represent the circuit in the frequency domain, using phasors and impedances. This circuit consists of a single mesh. Notice that the mesh current, I(ω), enters the dotted end of the left-hand coil and the undotted end of the right-hand coil. Apply KVL to the mesh to get 5 I (ω ) + ( j12 I (ω ) − j 6 I (ω ) ) + ( − j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0 5 I (ω ) + ( j12 − j 6 − j 6 + j15 ) I (ω ) − 5.94∠140° = 0 I (ω ) = 5.94∠140° 5.94∠140° 5.94∠140° = = = 0.376∠68.4° A 5 + j (12 − 6 − 6 + 15 ) 5 + j 15 15.8∠71.6 Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by Vo (ω ) = j 15 I (ω ) − j 6 I (ω ) = j 9 I (ω ) = j 9 ( 0.376∠68.4° ) = (9∠90°) ( 0.376∠68.4° ) = 3.38∠158.4° V In the time domain, the output voltage is given by vo ( t ) = 3.38 cos ( 3 t + 158.4° ) V P 11.9-13 Figure P 11.9-13 shows three similar circuits. In each, the input to the circuit is the voltage of the voltage source, vs(t) = 5.7 cos (4t + 158°) V The output in each circuit is the voltage across the right-hand coil, vo(t). Determine the steadystate output voltage, vo(t), for each of the three circuits. + 4Ω + – vs(t) 5H 4H 2H 4Ω vo(t) + – vs(t) + 5H 4H – vo(t) – (a) (b) 2H 4Ω + – vs(t) + 5H 4H vo(t) – (c) Figure P 11.9-13 Solution: (a) In (b) and (c) the coils are coupled by mutual inductance, but not in (a). This circuit can be represented in the frequency domain, using phasors and impedances. Voltage division gives j 16 || j 20 5.7∠158° = ( 0.9119∠24° )( 5.7∠158° ) = 5.198∠182° V 4 + ( j 16 || j 20 ) In the time domain, the output voltage is given by Vo = vo ( t ) = 5.2 cos ( 4 t + 182° ) V (b) The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that both coil currents, I1(ω) and I2(ω), enter the undotted ends of their respective coils. The device equations for coupled coils are: and V1 (ω ) = j 16 I1 (ω ) + j 8 I 2 (ω ) (1) V2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω ) (2) The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for V1(ω) and V2(ω) gives j 16 I1 (ω ) + j 8 I 2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω ) j 8 I1 (ω ) = j 12 I 2 (ω ) I 1 (ω ) = 3 I 2 (ω ) 2 Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get I ( ω ) = I 1 ( ω ) + I 2 (ω ) = I 1 (ω ) = Therefore 3 5 I 2 ( ω ) + I 2 (ω ) = I 2 (ω ) 2 2 3 2 I (ω ) and I 2 (ω ) = I (ω ) 5 5 (3) Substituting the expressions for I1(ω) and I2(ω) from Equation 3 into Equation 1 gives 16 ( 3) + 8 ( 2 ) ⎛3 ⎞ ⎛2 ⎞ V1 (ω ) = j 16 ⎜ I (ω ) ⎟ + j 8 ⎜ I (ω ) ⎟ = j I (ω ) = j 12.8 I (ω ) 5 ⎝5 ⎠ ⎝5 ⎠ Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get 4 I (ω ) + V1 (ω ) − 5.7∠158° = 0 Using Equation 4 gives Solving for I (ω) gives 4 I (ω ) + j 12.8 I (ω ) − 5.7∠158° = 0 I (ω ) = 5.7∠158° 5.7∠158° = = 0.425∠85° A 4 + j 12.8 13.41∠73° Now the output voltage can be calculated using Equation 4: (4) Vo (ω ) = V1 (ω ) = j 12.8 I (ω ) = j 12.8 ( 0.425∠85° ) = (12.8∠90° )( 0.425∠85° ) = 5.44∠175° V In the time domain, the output voltage is given by vo ( t ) = 5.44 cos ( 4 t + 175° ) V (c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the righthand coil is located at the bottom of the coil in (b) and at the top of the coil in (c). Here is the circuit from (c) represented in the frequency domain, using impedances and phasors. The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that one of coil currents, I1(ω), enters the undotted end of the coil while the other coil current, I2(ω), enters the dotted end of the coil. The device equations for coupled coils are: and V1 (ω ) = j 16 I1 (ω ) − j 8 I 2 (ω ) (5) V2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω ) (6) The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for V1(ω) and V2(ω)gives j 16 I1 (ω ) − j 8 I 2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω ) j 24 I1 (ω ) = j 28 I 2 (ω ) I 1 (ω ) = 28 7 I 2 (ω ) = I 2 (ω ) 24 6 Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get 7 13 I 2 (ω ) + I 2 (ω ) = I 2 (ω ) 6 6 7 6 I1 (ω ) = I (ω ) and I 2 (ω ) = I (ω ) Therefore 13 13 Substituting the expressions for I1(ω) and I2(ω) from Equation 7 into Equation 5 gives I (ω ) = I 1 (ω ) + I 2 (ω ) = (7) 16 ( 7 ) − 8 ( 6 ) ⎛7 ⎞ ⎛6 ⎞ V1 (ω ) = j 16 ⎜ I (ω ) ⎟ − j 8 ⎜ I (ω ) ⎟ = j I (ω ) = j 4.9 I (ω ) 13 ⎝ 13 ⎠ ⎝ 13 ⎠ Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get 4 I (ω ) + V1 (ω ) − 5.7∠158° = 0 Using Equation 8 gives 4 I (ω ) + j 4.9 I (ω ) − 5.7∠158° = 0 Solving for I (ω) gives I (ω ) = 5.7∠158° 5.7∠158° = = 0.901∠107° A 4 + j 4.9 6.325∠51° Now the output voltage can be calculated using Equation 8: Vo (ω ) = V1 (ω ) = j 4.9 I (ω ) = j 4.9 ( 0.901∠107° ) = ( 4.9∠90° )( 0.901∠107° ) = 4.41∠197° V In the time domain, the output voltage is given by vo ( t ) = 4.41 cos ( 4 t + 197° ) V (8) P11.9-14 The circuit shown in Figure 11.9-14 is represented in the time domain. Determine coil voltages v1 and v2. Answer: v1 = 104.0 cos (6t + 46.17 °) V and v2 = 100.6 cos (6t + 63.43 °) V. Figure 11.9-14 Solution: Represent the circuit in the frequency domain as shown. Then V1 = j 48 (1.5∠ − 90° ) + j 30 ( 2.5∠0° ) = j 48 ( − j 1.5 ) + j 30 ( 2.5 ) = 72 + j 75 = 103.97∠46.17° V and V 2 = j 36 ( 2.5∠0° ) + j 30 (1.5∠ − 90° ) = j 36 ( 2.5 ) + j 30 ( − j 1.5 ) = 45 + j 90 = 100.62∠63.43° V P11.9-15 The circuit shown in Figure P11.9-15 is represented in the frequency domain (e.g. –j30 Ω is the impedance due to the mutual inductance of the coupled coils). Suppose V ( ω ) = 70∠0° V. Then I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A. Determine the values of B and θ. Answer: B = 1.75 A and θ = −90 ° Figure 11.9-15 Solution: Suppose V ( ω ) = 70∠0° V. Then I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A. V = − j 25 I 1 + j 50 I 1 + j 30 I 2 = j 25 I 1 + j 30 I 2 = j 25 I 1 + j 30 ( − j 0.875 ) I1 = V − j 30 ( − j 0.875 ) V − 26.25 70 − 26.25 43.75 = = = = − j 1.75 j 25 j 25 j 25 j 25 P11.9-16 Determine the values of the inductances L1 and L2 in the circuit shown in Figure P11.9-16, given that i ( t ) = 0.319 cos ( 4 t − 82.23° ) A and v ( t ) = 0.9285cos ( 4 t − 62.20° ) V . Figure P11.9-16 Solution: Represent the circuit in the frequency domain as Apply KVL to the left mesh to get 5∠0° + j 3 ( 0.9285∠ − 62.20° ) ⎛ 0.9285∠ − 62.20° ⎞ 5∠0° = j 4 L1 ( 0.319∠ − 82.23° ) + j 12 ⎜ − ⎟ ⇒ L1 = 4 j 4 ( 0.319∠ − 82.23° ) ⎝ ⎠ = 6.0004 − j 0.0005 ≅6H Apply KVL to the left mesh to get ⎛ 0.9285∠ − 62.20° ⎞ 0.9285∠ − 62.20° = j 4 L 2 ⎜ − ⎟ + j 12 ( 0.319∠ − 82.23° ) 4 ⎝ ⎠ L2 = ( 0.9285∠ − 62.20° ) − j 12 ( 0.319∠ − 82.23° ) ⎛ 0.9285∠ − 62.20° ⎞ j 4⎜ − ⎟ 4 ⎝ ⎠ = 3.9998 + j 0.0005 ≅4H P11.9-17 Determine the complex power supplied by the source in the circuit shown in Figure P11.9-17. Figure P11.9-17 Solution: Represent the circuit in the frequency domain as The coil voltages are given by V1 = j 100 I 1 − j 80 I 2 and V 2 = j 120 I 2 − j 80 I 1 V1 + 40 ( I 1 − I 2 ) + 20 I 1 − 180∠0° = 0 Using KVL V 2 + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0 Substituting the coil voltages: ( j 100 I 1 − j 80 I 2 ) + 40 ( I 1 − I 2 ) + 20 I 1 = 180∠0° ( j 120 I 2 − j 80 I 1 ) + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0 ( 60 + j 100 ) I 1 − ( 40 + j 80 ) I 2 = 180∠0° − ( 40 + j 80 ) I 1 + ( 90 + j 70 I 1 ) I 2 = 0 Simplifying I 1 = 2.731∠ − 26.9° A and I 2 = 2.142∠ − 1.36° A Solving gives The complex power delivered by the source is S= VI * (180∠0° )( 2.731∠ − 26.9° ) * = = 219.19 + j111.20 VA 2 2 P11.9-18 The input to the circuit shown in Figure P11.9-18 is v s ( t ) = 12 cos ( 5 t ) V The impedance of the load is 20 + j 15 Ω. Determine the complex power (a) supplied by the source, (b) received by the 20 Ω resistor, (c) received by the coupled inductors and (d) received by the load. . Figure P11.9-18 Solution: Represent the circuit in the frequency domain as The coil voltages are given by V1 = j 20 I 1 − j 25 I 2 and V 2 = j 40 I 2 − j 25 I 1 Using KVL 20 I 1 + V1 − 12∠45° = 0 and 20 I 2 + j15 I 2 + V 2 = 0 Substituting the coil voltages: 20 I 1 + j 20 I 1 − j 25 I 2 = 12∠0° 20 I 2 + j15 I 2 + j 40 I 2 − j 25 I 1 = 0 Writhing these equations in matrix form − j 25 ⎤ ⎡ I 1 ⎤ ⎡12∠0° ⎤ ⎡ 20 + j 20 ⎢ ⎥= ⎢ − j 25 20 + j 55⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣ I 1 = 0.4676∠ − 22.8° A and I 2 = 0.1998∠ − 2.86° A Solving gives (a) The complex power delivered by the source is S= (12∠0° ) I1 * = (12∠0° )( 0.4676∠ − 22.8° ) * = 2.5855 + j1.0893 VA 2 2 (b) The complex power received by the 20 Ω resistor is I S= 1 2 2 ( 20 ) = ( 0.4676 ) 2 2 ( 20 ) = 2.1865 + j 0 VA (c) The complex power received by the coupled inductors is S= V1I1 * V2 I 2 * + = 0 + j 0.79 VA 2 2 (d) The complex power received by the load is I S= 2 2 2 ( 20 + j 15) ( 0.1998) = 2 2 ( 20 + j 15) = 0.399 + j 0.299 VA (checked using LNAP 2/6/09) P11.9-19 Figure P11.9-18a shows a source connected to a 160 Ω load. In Figure P11.9-18a an ideal transformer and capacitor have been inserted between the source and the load. (a) Determine the value of the average power delivered to the 160 Ω load in Figure P11.918a. (b) Determine the values of n and C Figure P11.9-18b that maximize the average power in the load. (c) Using the values of n and C from part (b), determine the value of the average power delivered to the 160 Ω load in Figure P11.9-18b. (a) (b) Figure P11.9-19 Solution: (a) Represent the circuit in the frequency domain. Using KVL I= and 120∠0° = 0.7058∠ − 0.674° A 170 + j 2 P= 0.7058 2 (160 ) = 39.85 W 2 (b) For maximum power transfer to the 160 Ω load 1 1 ⎛ 1 ⎞ 160 ) = 10 ⇒ n = 4 and − ( j 2 ) = ⎜ 2 ⎟ − j 2 ( n 250 C ⎝4 ⎠ ⇒ C = 0.125 mF (c) Represent the circuit in the frequency domain I1 = and 120∠0° 2 ⎛1⎞ 10 + j 2 + ⎜ ⎟ ( − j 32 + 160 ) ⎝4⎠ = 120∠0° 1 = 6∠0° A and I 2 = I1 = 1.5∠0° A 20 4 P= 1.5 2 (160 ) = 180 W 2 Section 11.10 The Ideal Transformer P 11.10-1 Find V1, V2, I1, and I2 for the circuit of Figure P 11.10-1, when n = 5. I1 2 + 3j 12 0° V + – 1:n I2 + + V1 V2 – 100 – j75 – Ideal Figure P 11.10-1 Solution: Z = (2 + j 3) + I1 = 12∠0° 12∠0° = =2A Z 6 ⎛ 100− j 75 ⎞ ⎛ 100− j 75 ⎞ V1 = I1 ⎜ ⎟ = (2) ⎜ ⎟ = 10∠ − 36.9° V 2 n ⎝ ⎠ ⎝ 25 ⎠ V2 = nV1 = 5 (10∠−36.9°) = 50 ∠−36.9° V I2 = I1 2 = A n 5 (100− j 75) =6Ω 52 P 11.10-2 A circuit with a transformer is shown in Figure P 11.10-2. (a) Determine the turns ratio. (b) Determine the value of Rab. (c) Determine the current supplied by the voltage source. Answer: (a) n = 5 (b) Rab = 400 Ω i a 5 mA rms 1:n + 10 V rms + – V0 – Rab b Ideal Figure P 11.10-2 Solution: (a) V0 = (5 ×10−3 )(10, 000) = 50 V n= (b) (c) V N2 50 = 0 = = 5 N1 V1 10 Rab = Is = 1 1 R2 = (10 × 103 ) = 400 Ω 2 25 n 10 10 = = 0.025 A = 25 mA Rab 400 10 kΩ P 11.10-3 Find the voltage Vc in the circuit shown in Figure P 11.10-3. Assume an ideal transformer. The turns ratio is n = 1/3. Answer: Vc = 21.0 ∠ − 105.3° 30 Ω j20 Ω 1:n + 5Ω –j8 Ω 80 –50° V – + – Vc Ideal Figure P 11.10-3 Solution: 1 Z 2 = 9 Z 2 = 9 (5 − j8) = 45 − j 72 Ω n2 Using voltage division, the voltage across Z 1 is Z1 = ⎛ ⎞ 45− j 72 V1 = ( 80∠ − 50° ) ⎜ ⎟ = 74.4 ∠ − 73.3° V ⎝ 45− j 72+ 30 + j 20 ⎠ Then V2 = nV1 = 74.4 ∠ − 73.3° = 24.8 ∠ − 73.3° V 3 Using voltage division again yields ⎛ − j8 ⎞ ⎛ 8∠−90° ⎞ Vc = V2 ⎜ = ( 24.8∠−73.3° ) ⎜ ⎟ ⎟ = 21.0∠ − 105.3° V ⎝ 89∠−58° ⎠ ⎝ 5− j 8 ⎠ P 11.10-4 An ideal transformer is connected in the circuit shown in Figure P 11.10-4, where vs = 50 cos 1000t V and n = N2/N1 = 5. Calculate V1 and V2. 2Ω vs + – 1:5 + + v1 v2 – 200 Ω – Ideal Figure P 11.10-4 Solution: n = 5, Z1 = 200 ( 5) 2 = 8 Ω ⇒ V1 = 8 ( 50∠0° ) = 40∠0° V ⇒ V2 = n V1 = 200∠0° V 8+ 2 P11.10-5 Figure P11.10-5 shows a load connected to a source through an ideal transformer. The input to the circuit is v s ( t ) = 12 cos ( 5 t ) V Determine (a) The values of the turns ration, n, and load inductance, L, required for maximum power transfer to the load. (b) The complex power delivered to the transformer by the source. (c) The complex power delivered to the load by the transformer. Figure P11.10-5 Solution: * (a) For maximum power transfer: Equating real parts gives n = 1 1 ⎛ ⎞ 288 + j 20 L ) = ⎜ 18 − j ⎟ = 18 + j 5 2 ( 20 × 0.01 ⎠ n ⎝ ( ) 5 42 288 = 4 . Equating imaginary parts gives L = =4 H. 20 18 (b) Represent the circuit in the frequency domain as Replace the transformer and load by an equivalent impedance Z equiv = 1 ( 288 + j 80 ) = 18 + j 5 Ω 42 I1 = and 12∠0° 12∠0° 1 = = ∠0° A (18 − j 5) + (18 + j 5) 36 3 ⎛1 ⎞ V1 = (18 + j 5 ) I 1 = (18 + j 5 ) ⎜ ∠0° ⎟ = 6.227∠15.5° V ⎝3 ⎠ The secondary coil current and voltages and 1 1⎛1 1 ⎞ I 2 = − I 1 = − ⎜ ∠0° ⎟ = − ∠0° = −0.0833∠0° A 4 4⎝3 12 ⎠ 4 V 2 = V1 = 24.91∠15.5° V 1 The complex power delivered to the transformer by the source. 2 I V1I1 * = 1 + j 0.277 VA = 1 (18 + j 5 ) 2 2 (c) The complex power delivered to the load by the transformer is I VI * − 2 2 = 1 + j 0.277 VA = 2 2 2 2 ( 288 + j80 ) (checked using LNAP and MATLAB 2/6/09) a Ideal P 11.10-6 Find the Thévenin equivalent at terminals a–b for the circuit of Figure P 11.10-6 v when v = 16 cos 3t V. Answer: Voc = 12 and Zt = 3.75 Ω 2Ω 2Ω + – 6Ω b 1:2 Figure P 11.10-6 Solution: Z= 1 ( 2 + 6) = 2 Ω 22 ⎞ ⎛ 6 ⎞ ⎛⎛ 2 ⎞ Voc = ⎜ ⎟ ( 2) ⎜ ⎜ ⎟16∠0° ⎟ = 12∠0° V ⎝ 6 + 2 ⎠ ⎝⎝ 2 + 2 ⎠ ⎠ Z= 1 1 2 = Ω 2 ( ) 2 2 ⎛ ⎞ 1 1 ⎜ 16∠0° ⎟ I sc = −I 2 = I1 = ⎜ ⎟ = 3.25∠0° A 2 2⎜ 2+ 1 ⎟ ⎝ 2 ⎠ Then Zt = 12∠0° = 3.75∠0° Ω 3.2∠0° 2Ω P 11.10-7 Find the input impedance Z for the circuit of Figure P 11.10-7. Answer: Z = 6 Ω 2:1 6Ω Z Ideal Figure P 11.10-7 Solution: 1 V1 2 V − V2 V1 = I3 = 1 2 4 V V I 2 = I3 − 2 = 1 6 6 V 1 I1 = − I 2 = − 1 2 12 V I T = I 3 + I1 = 1 6 V Z= 1 =6 IT V2 = P 11.10-8 In less developed regions in mountainous areas, small hydroelectric generators are used to serve several residences (Mackay, 1990). Assume each house uses an electric range and an electric refrigerator, as shown in Figure P 11.10-8. The generator is represented as Vs operating at 60 Hz and V2 = 230 ∠0° V. Calculate the power consumed by each home connected to the hydroelectric generator when n = 5. Source and line 1:n 2 + j3 Vs + + – 10 Ω 20 Ω V2 20 mH – Ideal Range Refrigerator Figure P 11.10-8 Solution: ZL = 1 ⎛ 20 (10+ j 7.54) ⎞ 8.1∠23° = = 0.3 + j 0.13 Ω 52 ⎜⎝ 20+10+ j 7.54 ⎟⎠ 25 V V ( 230 ) = 88 kW/home PL = L = 2 = 2 R 2 2 R L 2( 0.3) 2 2 Therefore, 529 kW are required for six homes. 2 P 11.10-9 Three similar circuits are shown in Figure P 11.10-9. In each of these circuits 8Ω vs(t) vs(t) = 5 cos (4t + 45°) V. + – i2(t) + + v1(t) 4H 3H – Determine v2 (t) for each of the three circuits. Answer: v2(t) = 0 V (a) v2(t) = 1.656 cos (4t + 39°) V (b) (c) v2(t) = 2.88 cos (4t + 45°) V i1(t) v2(t) 12 Ω – (a) 8Ω i1(t) i2(t) 2H + vs(t) + – v1(t) + 4H 3H – v2(t) 12 Ω – (b) 8Ω vs(t) + – i1(t) i2(t) 10:8.66 + + v1(t) v2(t) – – Ideal (c) Figure P 11.10-9 Solution: (a) Coil voltages: V1 = j16 I1 V2 = j12 I 2 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 12 Ω Substitute the coil voltages into the mesh equations and do some algebra: 8 I1 + j16 I1 = 5∠45° ⇒ I1 = 0.28∠ − 18.4° 12 I 2 + j12 I 2 = 0 ⇒ I 2 = 0 V2 = −12 I 2 = 0 (b) Coil voltages: V1 = j16 I1 + j8 I 2 V2 = j12 I 2 + j8 I1 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute the coil voltages into the mesh equations and do some algebra: 8 I1 + ( j16 I1 + j8 I 2 ) = 5∠45° 12 I 2 + ( j12 I 2 + j8 I1 ) = 0 I1 = − 12 + j12 3 I 2 = ( j − 1) I 2 j8 2 ⎡ ⎤ ⎛3⎞ ⎢( 8 + j16 ) ⎜ 2 ⎟ ( j − 1) + j8⎥ I 2 = 5∠45° ⇒ I 2 = 0.138∠ − 141° ⎝ ⎠ ⎣ ⎦ V2 = −12 I 2 = 1.656∠39° (c ) Coil voltages and currents: 10 V2 8.66 8.66 I1 = − I2 10 V1 = Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute into the second mesh equation and do some algebra: 2 ⎛ 10 ⎞ 8.66 ⎛ 10 ⎞ −12 ⎜ − I1 ⎟ = V1 ⇒ V1 = 12 ⎜ ⎟ I1 ⎝ 8.66 ⎠ 10 ⎝ 8.66 ⎠ 2 ⎛ 10 ⎞ 8 I1 + 12 ⎜ ⎟ I1 = 5∠45° ⇒ I1 = 0.208∠45° ⎝ 8.66 ⎠ ⎛ 10 ⎞ 12 (10 ) 0.208∠45° = 2.88∠45° V2 = −12 I 2 = −12 ⎜ − I1 ⎟ = 8.66 ⎝ 8.66 ⎠ P 11.10-10 Find V1 and I1 for the circuit of Figure P 11.10-10 when n = 5. I1 1 + j3 10 0° V + – 1:n I2 + V1 100 – j75 – Ideal Figure P 11.10-10 Solution: I1 = 10∠0° 10∠0° = = 2∠0° A 100 − j 75 (1 + j 3) + ( 4 − j 3) 1 j 3 + + ( ) 52 V1 = ( 4 − j 3) 2∠0° = 10∠ − 36.9° V P 11.10-11 Determine v2 and i2 for the circuit shown in Figure P 11.10-11 when n = 2. Note that i2 does not enter the dotted terminal. Answer: v2 = 0.68 cos (10t + 47.7°) V i2 = 0.34 cos (10t + 42°) A 20 mF 5 cos 10t V + – i1 5Ω 2Ω 1:n + v1 – v2 – + 20 mH i2 Ideal Figure P 11.10-11 Solution: I1 = 5∠0° 5∠0° 5∠0° = = = 0.68∠42° A 2 + j 0.2 5.5 + j 4.95 7.4∠− 42° − + 5 j 5 ( ) 22 1 I 2 = I1 = 0.34∠42° A 2 V2 = ( 2 + j 0.2 ) I 2 = ( 2.01∠5.7° )( 0.34∠42° ) = 0.68∠47.7° V so v2 (t ) = 0.68cos (10 t + 47.7°) V and i 2 (t ) = 0.34 cos (10 t + 42°) A P11.10-12 The circuit shown in Figure P11.10-12 is represented in the frequency domain. Given the line current is I Line = 0.5761∠ − 75.88° A , determine PSource , the average power supplied by the source, PLine , the average power delivered to the line and PLoad , the average power delivered to the load. Hint: Use conservation of (average) power to check your answers. Answer: PSource = 42.15 W, PLine = 0.6638 W and PLoad = 41.49 W. Figure P11.10-12 Solution: ⎛5⎞ ⎜ ⎟120∠0° ⎝1⎠ = 0.5761∠ − 75.88° A I Line = 2 ⎛5⎞ 4 + j 10 + ⎜ ⎟ (10 + j 40 ) ⎝1⎠ ⎛5⎞ I Source = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A ⎝1⎠ ⎛5⎞ I Load = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A ⎝1⎠ (120 )( 2.8805) cos 0 − −75.88° = 42.16 W PSource = )) ( ( 2 0.5761 2 PLine = ( 4 ) = 0.6638 W 2 2.8805 2 PLoad = (10 ) = 41.49 W 2 P11.10-13 The circuit shown in Figure P11.10-13 is represented in the frequency domain. Determine R and X, the real and imaginary parts of the equivalent impedance, Z eq. Answer: R = 180 Ω and X =110 Ω Figure P11.10-13 Solution: 2 Z eq ⎛6⎞ = − j 250 + ⎜ ⎟ ( 5 + j 10 ) = 180 + j 110 Ω ⎝1⎠ P11.10-14 Figure P11.10-14 shows a load connected to a source through an ideal transformer. Determine the complex power delivered to the transformer by the source. Answer: S = 698.3 + j1745.7 VA Figure P11.10-14 Solution: Represent the circuit in the frequency domain. Using the element equation of the ideal transformer: 15 V 2 = (120∠15° ) = 450∠15° V 4 Using Ohm’s law: V2 450∠15° = = 8.3563∠ − 53.2° I2 = 20 + j 50 53.852∠68.2° The power supplied by the source is equal to the power received by the load. The power received by the load is calculated as 2 8.35632 S= ( 20 + j 50 ) = ( 20 + j 50 ) = 698.3 + j1745.7 VA 2 2 I2 (checked using LNAP and MATLAB 2/17/12) Section 11.11 How Can We Check …? P 11.11-1 Computer analysis of the circuit shown in Figure P 11.11-1 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 2.327 cos (4t – 25.22°) A and i2(t) = 1.229 cos (4t – 11.19°) A Check the results of this analysis by checking that the average power supplied by the voltage source is equal to the sum of the average powers received by the other circuit elements. 4Ω v1(t) + – i1(t) 2Ω 2H i2(t) 2H Figure P 11.11-1 P11.11-1 The average power supplied by the source is Ps = (12 )( 2.327 ) cos 2 ( 30° − ( −25.22° ) ) = 7.96 W Capacitors and inductors receive zero average power, so the average power supplied by the voltage source should be equal to the sum of the average powers received by the resistors: PR = 2.327 2 1.1292 4 + ( ) ( 2 ) = 10.83 + 1.27 = 12.10 W 2 2 The average power supplied by the voltage source is not equal to the sum of the average powers received by the other circuit elements. The mesh currents cannot be correct. (What went wrong? It appears that the resistances of the two resistors were interchanged when the data was entered for the computer analysis. Notice that PR = 2.327 2 1.1292 ( 2) + ( 4 ) = 5.41 + 2.55 = 7.96 W 2 2 The mesh currents would be correct if the resistances of the two resistors were interchanged. The computer was used to analyze the wrong circuit.) P 11.11-2 Computer analysis of the circuit shown in Figure P 11.11-2 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 1.647 cos (4t – 17.92°) A and i2(t) = 1.094 cos (4t – 13.15°) A Check the results of this analysis by checking that the complex power supplied by the voltage source is equal to the sum of the complex powers received by the other circuit elements. 4Ω vs(t) + – i1(t) 2Ω 2H i2(t) 4H Figure P 11.11-2 Solution: The average complex supplied by the source is Ss = (12∠30° )(1.647∠ − 17.92° ) * = (12∠30° )(1.647∠17.92° ) = 9.88∠47.92° = 6.62 + j 7.33 2 2 The complex power received by the 4 Ω resistor is S4Ω = ( 4 ×1.647∠ − 17.92° )(1.647∠ − 17.92° ) * = 5.43 + j 0 2 VA The complex power received by the 2 Ω resistor is S2Ω = ( 2 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 1.20 + j 0 2 VA The current in the 2 H inductor is (1.647∠ − 17.92° ) − (1.094∠ − 13.15° ) = 0.5640∠ − 27.19° The complex power received by the 2 H inductor is S 2H = ( j 8 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 1.27 2 VA The complex power received by the 4 H inductor is S 4H = ( j 16 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 9.57 2 VA W S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 1.27 ) + ( 0 + j 9.57 ) = 6.63 + j 10.84 ≠ Ss The complex power supplied by the voltage source is not equal to the sum of the complex powers received by the other circuit elements. The mesh currents cannot be correct. (Suppose the inductances of the inductors were interchanged. Then the complex power received by the 4 H inductor would be S 4H = ( j 16 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 2.54 2 VA The complex power received by the 2 H inductor would be S 2H = ( j 8 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 4.79 2 VA S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 2.54 ) + ( 0 + j 4.79 ) = 6.63 + j 7.33 ≈ S s The mesh currents would be correct if the inductances of the two inductors were interchanged. The computer was used to analyze the wrong circuit.) P 11.11-3 Computer analysis of the circuit shown in Figure P 11.11-3 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 1.001 cos (4t – 47.01°) A and i2(t) = 0.4243 cos (4t – 15.00°) A Check the results of this analysis by checking that the equations describing currents and voltages of coupled coils are satisfied. vs(t) + – i1(t) 4H 3H 6H i2(t) 15 Ω Figure P 11.11-3 Solution: The voltage across the left coil must be equal to the voltage source voltage. Notice that the mesh currents both enter the undotted ends of the coils. In the frequency domain, the voltage across the left coil is ( j 16 )(1.001∠ − 47.01° ) + ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° + 5.092∠75° = (11.715 + j 10.923) + (1.318 + j 4.918 ) = 13.033 + j 15.841 = 20.513∠50.55° The voltage across the left coil isn’t equal to the voltage source voltage so the computer analysis isn’t correct. What happened? A data entry error was made while doing the computer analysis. Both coils were described as having the dotted end at the top. If both coils had the dot at the top, the equation for the voltage across the right coil would be ( j 16 )(1.001∠ − 47.01° ) − ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° − 5.092∠75° = (11.715 + j 10.923) − (1.318 + j 4.918 ) = 10.397 + j 6.005 = 12.007∠30.01° This is equal to the voltage source voltage. The computer was used to analyze the wrong circuit. Computer analysis of the circuit shown in Figure P 11.11-4 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 25.6 cos (4t + 30°) mA and i2(t) = 64 cos (4t + 30°) mA Check the results of this analysis by checking that the equations describing currents and voltages of ideal transformers are satisfied. P 11.11-4 2:5 vs(t) + – i1(t) i2(t) 75 Figure P 11.11-4 Solution: First check the ratio of the voltages across the coils. n1 2 12∠30° = 2.5 ≠ = n2 5 ( 75)( 0.064∠30° ) The transformer voltages don’t satisfy the equations describing the ideal transformer. The given mesh currents are not correct. That’ enough but let’s also check the ratio of coil currents. (Notice that the reference direction of the i2(t) is different from the reference direction that we used when discussing transformers.) n1 2 0.064∠30° = 2.5 ≠ = 0.0256∠30° n2 5 The transformer currents don’t satisfy the equations describing the ideal transformer. n1 1 to be 2.5 instead of 0.4 = In both case, we calculated . This suggests that a data entry 2.5 n2 error was made while doing the computer analysis. The numbers of turns for the two coils was interchanged. PSpice Problems The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source, vs(t) = 7.5 sin (5t + 15°) V The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to plot the input and output voltages. Hint: Represent the voltage source using the PSpice part called VSIN. SP 11-1 8Ω 3H vs(t) + – 5H 2H + vo(t) – Figure SP 11-1 Solution: The coupling coefficient is k = 3 = 0.94868 . 2×5 4Ω 8Ω 3H vs(t) + – 5H 2H + vo(t) 4Ω – Figure SP 11-1 The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source, vs(t) = 7.5 sin (5t + 15°) = 7.5 cos (5t – 75°) V The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to determine the average power delivered to the coupled inductors. Hint: Represent the voltage source using the PSpice part called VAC. Use printers (PSpice parts called IPRINT and VPRINT) to measure the ac current and voltage of each coil. SP 11-2 Solution: Here is the circuit with printers inserted to measure the coil voltages and currents: Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the current of coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the voltage of coil 1 as 4.292∠-58.74°: FREQ 7.958E-01 VM(N00984) 2.498E+00 VP(N00984) 1.072E+02 FREQ IM(V_PRINT1)IP(V_PRINT1) 7.958E-01 4.484E-01 -9.457E+01 FREQ IM(V_PRINT2)IP(V_PRINT2) 7.958E-01 6.245E-01 -7.277E+01 FREQ 7.958E-01 VM(N00959) VP(N00959) 4.292E+00 -5.874E+01 The power received by the coupled inductors is p= ( 4.292 )( 0.4484 ) cos 2 = 0.78016 − .78000 ≈ 0 ( −58.74 − ( −94.57 ) ) + ( 2.498)( 0.6245) cos 2 (107.2 − ( −72.77 ) ) The input to the circuit shown in Figure SP 11-3 is the voltage of the voltage source, vs(t) = 48 cos (4t + 114°) V The output is the voltage across the 9-Ω resistor, vo(t). Use PSpice to determine the average power delivered to the transformer. Hint: Represent the voltage source using the PSpice part called VAC. SP 11-3 i(t) 8Ω 2:3 + – + 9Ω vs(t) vo(t) – Figure SP 11-3 Solution: The inductance are selected so that L2 L1 = N2 N1 = 3 and the impedance of these inductors are 2 much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit. Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the current of coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage of coil 1 as 16.00∠114.1°: FREQ 6.366E-01 VM(N00984) 2.400E+01 VP(N00984) 1.141E+02 FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 4.000E+00 1.140E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 6.366E-01 2.667E+00 -6.590E+01 FREQ 6.366E-01 VM(N00959) 1.600E+01 VP(N00959) 1.141E+02 The power received by the transformer is p= (16 )( 4 ) cos (114 − 114 ) + 2 = 32 − 32.004 ≈ 0 ( 24 )( 2.667 ) cos 2 (114 − ( −66 ) ) SP 11-4 Determine the value of the input impedance, Zt, of the circuit shown in Figure SP 11-4 at the frequency ω = 4 rad/s. Hint: Connect a current source across the terminals of the circuit. Measure the voltage across the current source. The value of impedance will be equal to the ratio of the voltage to the current. 8Ω 2Ω 5:2 4H Zt Figure SP 11-4 Solution: The inductance are selected so that L2 L1 = N2 N1 = 2 and the impedance of these inductors are 5 much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit. FREQ 6.366E-01 VM(N00921) 1.011E+02 VP(N00921) 7.844E+01 VR(N00921) 2.025E+01 VI(N00921) 9.903E+01 The printer output gives the voltage across the current source as 20.25 + j 99.03 = 101.1∠78.44° V The input impedance is Zt = 20.25 + j 99.03 = 20.25 + j 99.03 Ω = 101.1∠78.44° Ω 1 ⎛ 52 ⎞ (We expected Z t = 8 + ⎜ 2 ⎟ ( 2 + j ( 4 )( 4 ) ) = 20.5 + j 100 Ω . That’s about 1% error.) ⎝2 ⎠ Design Problems DP 11-1 A 100-kW induction motor, shown in Figure DP 11-1, is receiving 100 kW at 0.8 power factor lagging. Determine the additional apparent power in kVA that is made available by improving the power factor to (a) 0.95 lagging and (b) 1.0. (c) Find the required reactive power in kVAR provided by a set of parallel capacitors for parts (a) and (b). (d) Determine the ratio of kVA released to the kVAR of capacitors required for parts (a) and (b) alone. Set up a table recording the results of this problem for the two values of power factor attained. Distribution line Capacitor Induction motor Figure DP 11-1 Solution: (a) P 100 ⎧ = = 125 kVA | S |= P = 100 kW ⎫ ⎪ pf 0.8 ⇒ ⎬ ⎨ pf=0.8 ⎭ ⎪Q =| S |sin ( cos −1 0.8 ) = 125sin ( 36.9° ) = 75 kVAR ⎩ Now pf = 0.95 so P 100 S= = =105.3 kVA pf 0.95 Q = S sin (cos −1 0.95) =105.3sin (18.2°) =32.9 kVAR so an additional 125 − 105.3 = 19.7 kVA is available. (b) Now pf = 1 so P 100 = =100 kVA pf 1 Q = S sin (cos −1 1) = 0 S= and an additional 125−100= 25 kVA is available. (c) In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In part (b), the capacitors are required to reduce Q by 75 – 0 = 75 kVAR. (d) Corrected power factor Additional available apparent power Reduction in reactive power 0.95 19.7 kVA 1.0 25 kVA 42.1 kVAR 75 kVAR DP 11-2 Two loads are connected in parallel and supplied from a 7.2-kV rms 60-Hz source. The first load is 50-kVA at 0.9 lagging power factor, and the second load is 45 kW at 0.91 lagging power factor. Determine the kVAR rating and capacitance required to correct the overall power factor to 0.97 lagging. Answer: C = 1.01 μF Solution: This example demonstrates that loads can be specified either by kW or kVA. The procedure is as follows: First load: Second load: Total load: ⎧⎪ P1 = S1 pf =( 50 )( 0.9 ) = 45 W S1 =50 VA ⎫ ⎬ ⇒ ⎨ −1 pf =0.9 ⎭ ⎪⎩Q1 = S1 sin (cos 0.9) =50sin (25.8°) = 21.8 kVAR P 45 ⎧ S2 = 2 = = 49.45 kVA P2 = 45 W ⎫ ⎪ pf 0.91 ⎬ ⇒ ⎨ pf = 0.91 ⎭ ⎪Q = S sin (cos −1 0.91) = 49.45sin (24.5°) = 20.5 kVAR ⎩ 2 2 S L = S1 + S 2 = (45 + 45) + j(21.8+20.5) = 90 + j 42.3 kVA Specified load: P 90 ⎧ Ss = s = =92.8 kVA Ps =90 W ⎫ ⎪ pf 0.97 ⎬ ⇒ ⎨ pf = 0.97 ⎭ ⎪Q = S sin (cos −1 0.97) =92.8sin (14.1°) = 22.6 kVAR ⎩ s s The compensating capacitive load is Qc = 42.3 − 22.6 = 19.7 kVAR . The required capacitor is calculated as 2 V (7.2 × 103 ) 2 1 Xc = c = = 2626 Ω ⇒ C = = 1.01 μ F 3 Qc 19.7 × 10 377 (2626) DP 11-3 (a) Determine the load impedance Zab that will absorb maximum power if it is connected to terminals a–b of the circuit shown in Figure DP 11-3. (b) Determine the maximum power absorbed by this load. (c) Determine a model of the load and indicate the element values. 5Ω a 100 mH + 10 cos 100 t V + – – + vab 0.5 vab – b Figure DP 11-3 Solution: Find the open circuit voltage: −10 + 5I + j10I − 0.5 Voc = 0 and 10 − Voc 5 Voc = 8∠36.9° = 6.4 + j 4.8 V I= so Find the short circuit current: I sc = 10∠0° = 2∠0° A 5 The the Thevenin impedance is: Zt = Voc = 3.2 + j 2.4 Ω I sc The short circuit forces the controlling voltage to be zero. Then the controlled voltage is also zero. Consequently the dependent source has been replaced by a short circuit. (a) Maximum power transfer requires Z L = Z t * = 3.2 − j 2.4 Ω . (c) ZL can be implemented as the series combination of a resistor and a capacitor with 1 R = 3.2 Ω and C = = 4.17 mF . (100 ) (2.4) (b) Pmax = | Voc |2 64 = = 2.5 W 8R 8 ( 3.2 ) DP 11-4 Select the turns ratio n necessary to provide maximum power to the resistor R of the circuit shown in Figure DP 11-4. Assume an ideal transformer. Select n when R = 4 and 8 Ω. 3 Ω j4 Ω 1:n R Vs + – j3 Ω Ideal Figure DP 11-4 Solution: When n is selected to deliver maximum power to Z3, the value of the maximum power is given as 2 ⎛ R ⎞ Vs ⎜ 2⎟ ⎝n ⎠ 2 P= 2 2 R⎞ ⎛ 3 ⎛ ⎞ 3 + + + 4 ⎜ ⎟ ⎜ ⎟ n2 ⎠ ⎝ n2 ⎝ ⎠ When R = 4 Ω, 2 n 2 R Vs P= 25n 4 + 48n 2 + 25 4 2 2 3 dP 2 ⎡ 2 n(25n + 48n + 25) − n (100n + 96n ) ⎤ = R Vs ⎢ ⎥ dn (25n 4 + 48n 2 + 25) 2 ⎣ ⎦ 5 4 ⇒ − 50n + 50n = 0 ⇒ n = 1 ⇒ n = 1 0= When R = 8 Ω, a similar calculation gives n = 1.31. DP 11-5 An amplifier in a short-wave radio operates at 100 kHz. The load Z2 is connected to a source through an ideal transformer, as shown in Figure DP 11-5. The load is a series connection of a 10-Ω resistance and 10-μH inductance. The Zs consists of a 1-Ω resistance and a 1-μH inductance. (a) Select an integer n in order to maximize the energy delivered to the load. Calculate I2 and the energy to the load. (b) Add a capacitance C in series with Z2 in order to improve the energy delivered to the load. Zs Vs I2 1:n + – Z2 Ideal Figure DP 11-5 Solution: Maximum power transfer requires Equating real parts gives 10 + j 6.28 = Z1 = (1 + j 0.628 ) * n2 10 = 1 ⇒ n = 3.16 n2 Equating imaginary parts requires jX = − j 0.628 ⇒ 3.162 X = −6.28 This reactance can be realized by adding a capacitance C in series with the resistor and inductor that comprise Z2. Then −6.28 = X = − 1 1 + 6.28 ⇒ C = = 0.1267 μ F 5 5 ( 2π ×10 ) C ( 2π ×10 ) (12.56 ) DP 11-6 A new electronic lamp (e-lamp) has been developed that uses a radio-frequency sinusoidal oscillator and a coil to transmit energy to a surrounding cloud of mercury gas as shown in Figure DP 11-6a. The mercury gas emits ultraviolet light that is transmitted to the phosphor coating, which, in turn, emits visible light. A circuit model of the e-lamp is shown in Figure DP 11.6b. The capacitance C and the resistance R are dependent on the lamp’s spacing design and the type of phosphor. Select R and C so that maximum power is delivered to R, which relates to the phosphor coating (Adler, 1992). The circuit operates at ω0 = 107 rad/s. Phosphorus coating 100 Ω Mercury vapor 1 mH Coil Coil V0 sin ω 0 t V + – C R (b) (a) Figure DP 11-6 Solution: Maximum power transfer requires 1 || R = (100 + j107 ×10−6 ) * 7 j10 C R = 100 − j10 1 + j107 R C R = (100 − j10) (1 + j107 R C ) = 100 + 108 R C + j (109 R C − 10 ) Equating real and imaginary parts yields R = 100 + 108 R C and 109 R C − 10 = 0 then RC = 10−8 ⎛ 10−8 ⎞ 10−8 ⇒ R = 100 + 108 R ⎜ = 0.101 nF ⎟ = 99 Ω ⇒ C = 99 ⎝ R ⎠ Chapter 12: Three-Phase Circuits Exercises Exercise 12.2-1 The Y-connected three-phase voltage source has Vc = 120∠ − 240° V rms . Find the line-to-line voltage Vbc. Answer: 207.8 ∠90° V rms Solution: VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120° Vbc = 3 (120 ) ∠ − 90° 1 Exercise 12.3-1 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω. Answer: SA = 129 + j81 VA, SB = 90 + j90 VA, SC = 136 – j34 VA, and S = 355 + j137 VA Ia A B b a A Ib B – – + + Va ZB Vb ZA n N IN n ZC + – Vc c C Ic C Figure 12.3-1 Solution: Mathcad analysis (12x4_1.mcd): Vp := 120 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.079 − 0.674i IaA = 1.272 180 π ⋅ arg( IaA ) = −32.005 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 80 + j⋅ 50 Va IbB := ZA j⋅ Vb ZB IbB = 1.061 π ⋅ 120 ZB := 80 + j⋅ 80 IbB = −1.025 − 0.275i 180 π 180 ⋅ arg( IbB) = −165 IcC := ZC := 100 − j⋅ 25 Vc ZC IcC = −0.809 + 0.837i IcC = 1.164 180 π ⋅ arg( IcC) = 134.036 2 3 Exercise 12.3-2 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = ZB = ZC = 40 + j30 Ω. Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA Ia A B b a A Ib B – – + + Va ZB Vb ZA n N IN n ZC + – Vc c C Ic C Figure 12.3-1 Solution: Mathcad analysis (12x4_2.mcd): Vp := 120 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.92 − 1.44i IaA = 2.4 180 π ⋅ arg( IaA ) = −36.87 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 40 + j⋅ 30 Va IbB := ZA j⋅ Vb ZB IbB = 2.4 π 180 ⋅ 120 ZB := ZA IbB = −2.207 − 0.943i 180 π ⋅ arg( IbB) = −156.87 IcC := ZC := ZA Vc ZC IcC = 0.287 + 2.383i IcC = 2.4 180 π ⋅ arg( IcC) = 83.13 4 5 Exercise 12.3-3 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω. Intermediate Answer: VnN = 28.89 ∠–150.5 V rms Answer: S = 392 + j142 VA IaA b B a A IbB – – + + ZB Va ZA Vb n – N + VNn ZC Vc + – c C IcC Figure 12.3-2 Solution: Mathcad analysis (12x4_3.mcd): Vp := 120 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: π 180 ⋅ − 120 ZA := 80 + j⋅ 50 j⋅ Vc := Va⋅ e π 180 ⋅ 120 ZB := 80 + j⋅ 80 ZC := 100 − j⋅ 25 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = −25.137 − 14.236i VnN = 28.888 ⋅ Vp 180 π ⋅ arg( VnN) = −150.475 6 Calculate the line currents: IaA := IaA = 1.385 − 0.687i IaA = 1.546 180 π Check: ⋅ arg( IaA ) = −26.403 Va − VnN ZA IbB := Vb − VnN ZB IbB = −0.778 − 0.343i IbB = 0.851 180 π IcC := ⋅ arg( IbB) = −156.242 Vc − VnN ZC IcC = −0.606 + 1.03i IcC = 1.195 180 π ⋅ arg( IcC) = 120.475 IaA + IbB + IcC = 0 Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 191.168 + 119.48i Total power delivered to the load: SB = 57.87 + 57.87i ⎯ SC := IcC⋅ IcC⋅ ZC SC = 142.843 − 35.711i SA + SB + SC = 391.88 + 141.639i 7 Exercise 12.3-4 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = ZB = ZC = 40 + j30 Ω. Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA IaA b B a A IbB ZB – – + + Va ZA Vb n – N + VNn ZC Vc + – c C IcC Figure 12.3-2 Solution: Mathcad analysis (12x4_4.mcd): Vp := 120 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 ⋅0 Describe the three-phase load: j⋅ Vb := Va⋅ e π 180 ⋅ − 120 ZA := 40 + j⋅ 30 j⋅ Vc := Va⋅ e ZB := ZA π 180 ⋅ 120 ZC := ZA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC ⋅ Vp 8 9 Exercise 12.5-1 Consider the three-phase circuit shown in Figure 12.5-1. The voltages of the Y-connected source are 360 360 360 ∠–30° V rms, Vb = ∠–150° V rms, and Vc = ∠90° V rms Va = 3 3 3 The Δ-connected load is balanced. The impedance of each phase is ZΔ = 180 ∠45° Ω. Determine the phase and line currents when the line-to-line voltage is 360 V rms. Partial Answer: IAB = 2 ∠45° A rms and IaA = 3.46 ∠15° A rms I aA b a – I bB + + B Z3 – – Va Vb n I AB + – + VAB Z1 VBC Vc + A I CA VCA I BC – c – Z2 + C I cC Figure 12.5-1 Solution: Balanced delta load: phase currents: line currents: (See Table 12.5-1) Z Δ = 180∠ − 45° I AB VAB 360∠0° = = = 2∠45° A D Z Δ 180∠− 45 I BC = VBC 360∠−120° = = 2∠− 75° A Z Δ 18∠− 45° I CA = VCA 360∠120° = = 2∠165° A ° Z Δ 180∠− 45 I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A I B = 2 3∠−105° A I C = 2 3∠135° A 10 Exercise 12.6-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc = 110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j270 Ω. Determine the phase currents in the Δ-connected load. Answer: IAB = 0.49 ∠–32.5° A rms, IBC = 0.49 ∠–152.5° A rms, ICA = 0.49 ∠87.5° A rms Figure 12.6-1a Solution: Three-wire Y-to-Delta Circuit with line impedances Mathcad analysis (12x4_4.mcd): Vp := 110 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ − 120 Z1 := 150 + j⋅ 270 Describe the delta connected load: j⋅ Vc := Va⋅ e π 180 Z2 := Z1 ⋅ 120 Z3 := Z1 Convert the delta connected load to the equivalent Y connected load: ZA := Z1⋅ Z3 ZB := Z1 + Z2 + Z3 ZA = 50 + 90i Z2⋅ Z3 Z1 + Z2 + Z3 ZB = 50 + 90i Z1 + Z2 + Z3 ZC = 50 + 90i ZaA := 10 + j⋅ 25 Describe the three-phase line: Z1⋅ Z2 ZC := ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 14 VnN = −1.172 × 10 − 14 + 1.784i× 10 − 14 VnN = 2.135 × 10 180 π ⋅ Vp ⋅ arg( VnN) = 123.304 11 Calculate the line currents: IaA := IaA = 0.392 − 0.752i IaA = 0.848 180 π Check: ⋅ arg( IaA ) = −62.447 Va − VnN IbB := ZA + ZaA Vb − VnN ZB + ZbB IbB = −0.847 + 0.036i IbB = 0.848 180 π IcC := ⋅ arg( IbB) = 177.553 Vc − VnN ZC + ZcC IcC = 0.455 + 0.716i IcC = 0.848 180 π ⋅ arg( IcC) = 57.553 IaA + IbB + IcC = 0 Calculate the phase voltages of the Y-connected load: VAN := IaA ⋅ ZA VAN = 87.311 180 π ⋅ arg( VAN) = −1.502 VBN := IbB⋅ ZB VBN = 87.311 180 π ⋅ arg( VBN) = −121.502 VCN := IcC⋅ ZC VCN = 87.311 180 π ⋅ arg( VCN) = 118.498 Calculate the line-to-line voltages at the load: VAB := VAN − VBN VAB = 151.227 180 π ⋅ arg( VAB) = 28.498 VBC:= VBN − VCN VBC = 151.227 180 π ⋅ arg( VBC) = −91.502 VCA := VCN − VAN VCA = 151.227 180 π ⋅ arg( VCA) = 148.498 Calculate the phase currents of the Δ -connected load: IAB := VAB Z3 IAB = 0.49 180 π ⋅ arg( IAB) = −32.447 IBC := VBC Z1 IBC = 0.49 180 π ⋅ arg( IBC) = −152.447 ICA := VCA Z2 ICA = 0.49 180 π ⋅ arg( ICA) = 87.553 12 Exercise 12.7-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc = 110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j 270 Ω. Determine the average power delivered to the Δconnected load. Intermediate Answer: IaA = 0.848 ∠–62.5° A rms and VAN = 87.3 ∠–1.5° V rms Answer: P = 107.9 W Figure 12.6-1a Solution: Continuing Ex. 12.6-1: Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 35.958 + 64.725i Total power delivered to the load: SB = 35.958 + 64.725i ⎯ SC := IcC⋅ IcC⋅ ZC SC = 35.958 + 64.725i SA + SB + SC = 107.875 + 194.175i 13 Exercise 12.8-1 The line current to a balanced three-phase load is 24 A rms. The line-to-line voltage is 450 V rms, and the power factor of the load is 0.47 lagging. If two wattmeters are connected as shown in Figure 12.8-2, determine the reading of each meter and the total power to the load. Answer: P1 = –371 W, P2 = 9162 W, and P = 8791 W IB W2 B IA A + W1 B A Z Z VAC Z C – C FIGURE 12.8-2 Solution: P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2 1 pf = .4 lagging ⇒ θ = 61.97 ° So P T = 450(24) ⎡⎣ cos 91.97° + cos 31.97° ⎤⎦ = 8791 W ∴ P 1 = − 371 W P2 = 9162 W 14 Exercise 12.8-2 The two wattmeters are connected as shown in Figure 12.8-2 with P1 = 60 kW and P2 = 40 kW, respectively. Determine (a) the total power and (b) the power factor. Answers: (a) 100 kW (b) 0.945 leading IB W2 B IA A W1 + A B Z Z VAC Z C – C Figure 12.8-2 Ex. 12.8-2 Consider Fig. 12.8-1 with P1 = 60 kW P2 = 40 kW . (a.) P = P1 + P2 = 100 kW (b.) use equation 12.9-7 to get tan θ = 3 P2 − P1 40−60 = 3 = − .346 ⇒ θ = − 19.11° PL + P2 100 then pf = cos ( − 19.110°) = 0.945 leading 15 Problems Section 12-2: Three Phase Voltages P 12.2-1 A balanced three-phase Y-connected load has one phase voltage: Vc = 277 ∠45° V rms The phase sequence is abc. Find the line-to-line voltages VAB, VBC, and VCA. Draw a phasor diagram showing the phase and line voltages. Solution: Given VC = 277 ∠45° and an abc phase sequence: VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75° VB = 277 ∠( 45° +120 )° = 277 ∠165° VAB = VA − VB =( 277 ∠− 75° )−( 277 ∠165° ) =( 71.69 − j 267.56 ) −( −267.56+ j 71.69 ) =339.25− j 339.25 = 479.77 ∠− 45° 480 ∠− 45° Similarly: VBC = 480 ∠ − 165° and VCA = 480 ∠75° P 12.2-2 A three-phase system has a line-to-line voltage VBA = 12,470 ∠–35° V rms with a Y load. Find the phase voltages when the phase sequence is abc. Solution: VAB 3∠30° = 12470 ∠145° V VAB = VA × 3∠30° ⇒ VA = In our case: So VAB = −VBA = − (12470 ∠−35° ) VA = 12470 ∠145° = 7200∠115° 3∠30° Then, for an abc phase sequence: VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125° VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V 1 P 12.2-3 A three-phase system has a line-to-line voltage Vab = 1500 ∠30° V rms with a Y load. Determine the phase voltage. Solution: Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the line-to-line voltage is So the phase voltage is Vab = 1500 ∠30° V 1500 ∠30° Va = = 866∠0° V 3∠30° 2 Section 12.3 P 12.3-1 The Y-to-Y Circuit Consider a three-wire Y-to-Y circuit. The voltages of the Y-connected source are Va = (208/ 3 ) ∠0° V rms, Vb = (208/ 3 ) ∠–120° V rms, and Vc = (208/ 3 )∠120° V rms. The Y-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω. (a) (b) (c) (d) Find the phase voltages. Find the line currents and phase currents. Show the line currents and phase currents on a phasor diagram. Determine the power dissipated in the load. Solution: Balanced, three-wire, Y-Y circuit: where Z A = Z B = Z C = 12∠30 = 10.4 + j 6 MathCAD analysis (12p4_1.mcd): Vp := Describe the three-phase source: 208 3 π j⋅ Va := Vp⋅ e ⋅0 180 j⋅ ⋅ − 120 j⋅ 180 Vb := Va⋅ e Describe the balanced three-phase load: π Vc := Va⋅ e ZA := 10.4 + j⋅ 6 π ⋅ 120 180 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := Va − VnN ZA IaA = 8.663 − 4.998i Check: − 14 ⋅ Vp IbB := VnN = 2.762 × 10 Vb − VnN IbB = −8.66 − 5.004i IaA = 10.002 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 IbB = 10.002 180 ⋅ arg( IaA ) = −29.982 π − 15 IaA + IbB + IcC = 4.696 × 10 IcC := ZB Vc − VnN ZC −3 IcC = −3.205 × 10 + 10.002i IcC = 10.002 ⋅ arg( IbB) = −149.982 180 π ⋅ arg( IcC) = 90.018 − 14 − 1.066i× 10 1 Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB 3 SA = 1.04 × 10 + 600.222i Total power delivered to the load: ⎯ SC := IcC⋅ IcC⋅ ZC 3 SB = 1.04 × 10 + 600.222i 3 3 SC = 1.04 × 10 + 600.222i 3 SA + SB + SC = 3.121 × 10 + 1.801i× 10 Consequently: (a) The phase voltages are 208 ∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms 3 (b) The currents are equal the line currents (c) Va = I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms and I c = I cC = 10∠90° A rms (d) The power delivered to the load is S = 3.121 + j1.801 kVA . 2 P 12.3-2 A balanced three-phase Y-connected supply delivers power through a three-wire plus neutral-wire circuit in a large office building to a three-phase Y-connected load. The circuit operates at 60 Hz. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. Each transmission wire, including the neutral wire, has a 2-Ω resistance, and the balanced Y load has a 10-Ω resistance in series with 100 mH. Find the line voltage and the phase current at the load. Solution: Balanced, three-wire, Y-Y circuit: where Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms and Vc = 120∠120° Vrms Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 ) and Z aA = 10 + j 37.7 Ω = Z bB = Z cC = 2 Ω Mathcad Analysis (12p4_2.mcd): 3 Consequently, the line-to-line voltages at the source are: Vab = Va × 3∠30° = 120∠0°× 3∠30° = 208∠30° Vrms, Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms The line-to-line voltages at the load are: VAB = VA × 3∠30° = 118.3∠3°× 3∠30° = 205∠33° Vrms, Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms and the phase currents are I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms 4 P 12.3-3 A Y-connected source and load are shown in Figure P 12.3-3. (a) Determine the rms value of the current iA (t). (b) Determine the average power delivered to the load. ia(t) + – – + 1H 1H 12 Ω 12 Ω 10 cos 16t V 10 cos (16t – 120°) V Source Load 12 Ω – + 10 cos (16t + 120°) V 1H Figure P 12.3-3 Solution: Balanced, three-wire, Y-Y circuit: where Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms and Z A = Z B = Z C = 12 + j (16 )(1) = 12 + j16 Ω MathCAD analysis (12p4_3.mcd): 5 Vp := Describe the three-phase source: 10 2 π j⋅ Va := Vp⋅ e ⋅0 j⋅ 180 π ⋅ − 120 j⋅ 180 Vb := Va⋅ e Vc := Va⋅ e ZA := 12 + j⋅ 16 Describe the balanced three-phase load: π ⋅ 120 180 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := IaA = 0.212 − 0.283i IaA = 0.354 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 ⋅ arg( IaA ) = −53.13 Va − VnN ZA − 15 ⋅ Vp IbB := VnN = 1.675 × 10 Vb − VnN IbB = −0.351 − 0.042i IbB = 0.354 180 π ⋅ arg( IbB) = −173.13 Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 1.5 + 2i Total power delivered to the load: SB = 1.5 + 2i IcC := ZB Vc − VnN ZC IcC = 0.139 + 0.325i IcC = 0.354 180 π ⋅ arg( IcC) = 66.87 ⎯ SC := IcC⋅ IcC⋅ ZC SC = 1.5 + 2i SA + SB + SC = 4.5 + 6i Consequently (a) The rms value of ia(t) is 0.354 A rms. (b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W 6 P 12.3-4 An unbalanced Y-Y circuit is shown in Figure P 12.3-4. Find the average power delivered to the load. Hint: VNn(ω) = 27.4 ∠–63.6 V Answer: 436.4 W a 10 Ω 5 mH A 20 Ω 60 mH b 10 Ω 5 mH B 40 Ω 40 mH C 60 Ω + – 100 cos (377t) – N + n 100 cos (377t +120°) c 10 Ω 5 mH 20 mH + – 100 cos (377t + 240°) Source Line Load Figure P 12.3-4 Solution: Unbalanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω Z C = 60 + j ( 377 ) ( 20 × 10−3 ) = 60 + j 7.54 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω Mathcad Analysis (12p4_4.mcd): 7 Vp := 100 Describe the three-phase source: π j⋅ Va := Vp⋅ e ⋅0 j⋅ 180 π ⋅ 120 180 Vb := Va⋅ e π j⋅ ⋅ − 120 180 Vc := Va⋅ e Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 ZB := 40 + j⋅ ω⋅ 0.04 Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZC := 60 + j⋅ ω⋅ 0.02 ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) VnN = 12.209 − 24.552i Calculate the line currents: IaA := IaA = 2.156 − 0.943i IaA = 2.353 180 π 180 VnN = 27.42 Va − VnN ZA + ZaA IbB := π IcC := ZB + ZbB 180 π ⋅ arg( IbB) = 100.492 SA = 55.382 + 62.637i Total power delivered to the load: ( ZC + ZcC IcC = 1.244 Calculate the power delivered to the load: ⎯ ⎯ IbB⋅ IbB IaA ⋅ IaA ⋅ ZA SB := ⋅ ZB SA := 2 2 ) Vc − VnN IcC = −0.99 − 0.753i IbB = 2.412 ⋅ arg( IaA ) = −23.619 ( ⋅ arg( VnN) = −63.561 Vb − VnN IbB = −0.439 + 2.372i ⋅ Vp ) SB = 116.402 + 43.884i 180 π ⋅ arg( IcC) = −142.741 SC := ⎯ (IcC ⋅ IcC) 2 ⋅ ZC SC = 46.425 + 5.834i SA + SB + SC = 218.209 + 112.355i The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W 8 P 12.3-5 A balanced Y-Y circuit is shown in Figure P 12.3-5. Find the average power delivered to the load. a 10 Ω 5 mH A 20 Ω 60 mH b 10 Ω 5 mH B 20 Ω 60 mH C 20 Ω + – 100 cos (377t) – N + n 100 cos (377t +120°) c 10 Ω 5 mH 60 mH + – 100 cos (377t + 240°) Source Line Load Figure P 12.3-5 Solution: Balanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = Z B = Z C = 20 + j ( 377 ) ( 60 × 10−3 ) = 20 + j 22.6 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω Mathcad Analysis (12p4_5.mcd): 9 Vp := 100 Describe the three-phase source: π j⋅ Va := Vp⋅ e ⋅0 j⋅ 180 π ⋅ 120 180 Vb := Va⋅ e π j⋅ ⋅ − 120 180 Vc := Va⋅ e Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 ZB := ZA ZC := ZA Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15 VnN = −8.982 × 10 − 14 Calculate the line currents: IaA := IaA = 1.999 − 1.633i IaA = 2.582 180 π − 14 + 1.879i× 10 ⋅ arg( IaA ) = −39.243 VnN = 2.083 × 10 Va − VnN ZA + ZaA IbB := 180 π ⋅ arg( VnN) = 115.55 Vb − VnN IcC := ZB + ZbB IbB = 0.415 + 2.548i π ⋅ arg( IbB) = 80.757 Vc − VnN ZC + ZcC IcC = −2.414 − 0.915i IbB = 2.582 180 ⋅ Vp IcC = 2.582 180 π ⋅ arg( IcC) = −159.243 Calculate the power delivered to the load: SA := ⎯ (IaA ⋅ IaA ) ⋅ ZA 2 SA = 66.645 + 75.375i Total power delivered to the load: SB := ⎯ (IbB ⋅ IbB) ⋅ ZB 2 SB = 66.645 + 75.375i SC := ⎯ (IcC ⋅ IcC) ⋅ ZC 2 SC = 66.645 + 75.375i SA + SB + SC = 199.934 + 226.125i The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W 10 P 12.3-6 An unbalanced Y-Y circuit is shown in Figure P 12.3-6 Find the average power delivered to the load. Hint: VNn (ω) = 1.755∠–29.5 V Answer: 436.4 W a A 4Ω 1H b B 2Ω 2H c C 4Ω + – 10 cos (4t – 90°) – N + n 10 cos (4t + 150°) 2H + – 10 cos (4t + 30°) Source Line Load Figure P 12.3-6 Solution: Unbalanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = 4 + j ( 4 )(1) = 4 + j 4 Ω, Z B = 2 + j ( 4 )( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_6.mcd): 11 Vp := 10 Describe the three-phase source: π j⋅ Va := Vp⋅ e ⋅ − 90 j⋅ 180 π ⋅ 150 180 Vb := Vp⋅ e π j⋅ Vc := Vp⋅ e ⋅ 30 180 Enter the frequency of the 3-phase source: ω := 4 ZA := 4 + j⋅ ω⋅ 1 Describe the three-phase load: ZB := 2 + j⋅ ω⋅ 2 ZC := 4 + j⋅ ω⋅ 2 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = 1.528 − 0.863i Calculate the line currents: IaA := IaA = −1.333 − 0.951i IaA = 1.638 180 π 180 VnN = 1.755 Va − VnN ZA IbB := π Vb − VnN IbB = 0.39 + 1.371i 180 π ) ⋅ arg( IbB) = 74.116 SA = 5.363 + 5.363i Total power delivered to the load: ( Vc − VnN ZC IcC = 0.943 − 0.42i IcC = 1.032 Calculate the power delivered to the load: ⎯ ⎯ IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ( IcC := ZB IbB = 1.426 ⋅ arg( IaA ) = −144.495 ⋅ arg( VnN) = −29.466 ) SB = 2.032 + 8.128i 180 π ⋅ arg( IcC) = −24.011 SC := ⎯ (IcC ⋅ IcC) 2 ⋅ ZC SC = 2.131 + 4.262i SA + SB + SC = 9.527 + 17.754i The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W 12 P 12.3-7 A balanced Y-Y circuit is shown in Figure P 12.3-7. Find the average power delivered to the load. a A 4Ω 2H b B 4Ω 2H c C 4Ω + – 10 cos (4t – 90°) – N + n 10 cos (4t + 150°) 2H + – 10 cos (4t + 30°) Source Line Load Figure P 12.3-7 Solution: Balanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = Z B = Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_7.mcd): 13 Vp := 10 Describe the three-phase source: j⋅ Va := Vp⋅ e π ⋅ − 90 180 π j⋅ ⋅ 150 j⋅ 180 Vb := Vp⋅ e Vc := Vp⋅ e π ⋅ 30 180 Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load: ZA := 4 + j⋅ ω⋅ 2 ZB := ZA ZC := ZA The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va Calculate the line currents: IaA := IaA = −1 − 0.5i π Va − VnN ZA IbB := Vb − VnN IbB = 1.118 ⋅ arg( IaA ) = −153.435 180 π ) ⋅ arg( IbB) = 86.565 SA = 2.5 + 5i Total power delivered to the load: ( Vc − VnN ZC IcC = 0.933 − 0.616i IcC = 1.118 Calculate the power delivered to the load: ⎯ ⎯ IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ( IcC := ZB IbB = 0.067 + 1.116i IaA = 1.118 180 − 15 VnN = 1.517 × 10 ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC ) SB = 2.5 + 5i 180 π ⋅ arg( IcC) = −33.435 SC := ⎯ (IcC ⋅ IcC) 2 ⋅ ZC SC = 2.5 + 5i SA + SB + SC = 7.5 + 15i The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W 14 Section 12-4: The Δ- Connected Source and Load A balanced three-phase Δ-connected load has one line current: P 12.4-1 IB = 50 ∠–40° A rms Find the phase currents IBC, IAB, and ICA. Draw the phasor diagram showing the line and phase currents. The source uses the abc phase sequence. Solution: Given I B = 50∠ − 40° A rms and assuming the abc phase sequence we have I A = 50∠80° A rms and I C = 50∠200° A rms From Eqn 12.6-4 I A = I AB × 3∠ − 30° ⇒ I AB = so IA 3∠ − 30° 50∠80° = 28.9∠110° A rms 3∠−30° = 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms I AB = I BC P 12.4-2 A three-phase circuit has two parallel balanced Δ loads, one of 5-Ω resistors and one of 20-Ω resistors. Find the magnitude of the total line current when the line-to-line voltage is 480 V rms. Solution: The two delta loads connected in parallel are equivalent to a single delta load with Z Δ = 5 || 20 = 4 Ω The magnitude of phase current is 480 Ip = = 120 A rms 4 The magnitude of line current is I L = 3 I p = 208 A rms 1 Section 12-5: The Y- to Δ- Circuit P 12.5-1 Consider a three-wire Y-to-Δ circuit. The voltages of the Y-connected source are Va = (208/ 3 ) ∠–30°V rms, Vb = (208/ 3 ) ∠–150° V rms, and Vc = (208/ 3 ) ∠90°V rms. The Δ-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω. Determine the line currents and calculate the power dissipated in the load. Answer: P = 9360 W Solution: We have a delta load with Z = 12∠30° . One phase current is I AB ⎛ 208 ⎞ ⎛ 208 ⎞ ∠−30° ⎟ −⎜ ∠−150° ⎟ ⎜ V V −V 3 ⎠ ⎝ 3 ⎠ = 208∠0° = 17.31∠ − 30° A rms = AB = A B = ⎝ 12∠30° 12∠30° Z Z The other phase currents are I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms One line currents is I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) × ( ) 3∠ − 30° = 30∠ − 60° A rms The other line currents are I B = 30∠ − 180° A rms and I C = 30∠60° A rms The power delivered to the load is P = 3( 208 ) (30) cos ( 60° − 30° ) = 9360 W 3 1 P 12.5-2 A balanced Δ-connected load is connected by three wires, each with a 4-Ω resistance, to a Y source with Va = (480/ 3 )∠–30° V rms, Vb = (480/ 3 ) ∠–150° V rms, and Vc = (480/ 3 )∠–90° V rms. Find the line current IA when ZΔ = 39 ∠–40° Ω. Answer: IA = 17 ∠0.9 A Solution: The balanced delta load with Z Δ = 39∠− 40° Ω is equivalent to a balanced Y load with ZY = ZΔ = 13∠ − 40° = 9.96 − j 8.36 Ω 3 Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω 480 ∠−30° 3 then I A = = 17∠0.9° A rms ° 16.3 ∠−30.9 2 P 12.5-3 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠30°V rms. Determine the phase currents in the load when Z = 3 + j4 Ω. Sketch a phasor diagram. a A Z n + – – + Vab Z c b – + Z B C Figure P 12.5-3 Solution: Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠30° V rms 380 ∠30° So one phase voltage is Va = = 200∠0° V rms 3∠30° So VAB = 380∠30° V rms VA = 220∠0° V rms VBC = 380∠-90° V rms VB = 220∠−120° V rms VCA = 380∠150° V rms VC = 220∠120° V rms One phase current is IA = VA 220∠0° = 44∠ − 53.1° A rms 3+ j4 Z The other phase currents are I B = 44∠−173.1° A rms amd I C = 44∠66.9° A rms 3 P 12.5-4 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠0° V rms. Determine the line and phase currents in the load when Z = 9 + j12 Ω. Solution: Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠0° V rms Va = So one phase voltage is So 380 ∠0° = 200∠ − 30° V rms 3∠30° Vab = 380∠0° V rms Va = 220∠ − 30° V rms Vbc = 380∠-120° V rms Vb = 220∠−150° V rms Vca = 380∠120° V rms Vc = 220∠90° V rms One phase current is IA = Va 220∠−30° = = 14.67∠ − 83.1° A rms Z 9+ j12 The other phase currents are I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms 4 Section 12-6: Balanced Three-Phase Circuits P 12.6-1 The English Channel Tunnel rail link is supplied at 25 kV rms from the United Kingdom and French grid systems. When there is a grid supply failure, each end is capable of supplying the whole tunnel but in a reduced operational mode. The tunnel traction system is a conventional catenary (overhead wire) system similar to the surface main line electric railway system of the United Kingdom and France. What makes the tunnel traction system different and unique is the high density of traction load and the end-fed supply arrangement. The tunnel traction load is considerable. For each half tunnel, the load is 180 MVA (Barnes and Wong, 1991). Assume that each line-to-line voltage of the Y-connected source is 25 kV rms and the three-phase system is connected to the traction motor of an electric locomotive. The motor is a Y-connected load with Z = 150 ∠25° Ω. Find the line currents and the power delivered to the traction motor. Solution: Va = IA 25 ×103 ∠0° Vrms 3 25 ×103 ∠0° Va 3 = = = 96∠ − 25° A rms 150 ∠25° Z ⎛ 25 ⎞ P = 3 Va I A cos (θ v -θ I ) = 3 ⎜ ×103 ⎟ 96 cos(0 − 25°) = 3.77 MW ⎝ 3 ⎠ P 12.6-2 A three-phase source with a line voltage of 45 kV rms is connected to two balanced loads. The Y-connected load has Z = 10 + j20 Ω, and the Δ load has a branch impedance of 50 Ω. The connecting lines have an impedance of 2 Ω. Determine the power delivered to the loads and the power lost in the wires. What percentage of power is lost in the wires? Solution: Convert the delta load to an equivalent Y connected load: ˆ = 50 Ω ZΔ ⇒ Z Y 3 To get the per-phase equivalent circuit shown to the right: The phase voltage of the source is Z Δ = 50 Ω Va = 45×103 ∠0° = 26∠0° kV rms 3 The equivalent impedance of the load together with the line is 50 3 + 2 = 12 + j 5 = 13∠22.6° Ω Z eq = 50 10 + j 20 + 3 (10 + j 20 ) The line current is Ι aA Va 26 × 103 ∠0° = = = 2000∠ − 22.6° A rms Z eq 13∠22.6° The power delivered to the parallel loads (per phase) is 50 ⎫ ⎧ ⎪ (10 + j 20 ) 3 ⎪ 2 6 PLoads = I aA × Re ⎨ ⎬ = 4 ×10 × 10 = 40 MW 50 ⎪10 + j 20 + ⎪ 3 ⎭ ⎩ The power lost in the line (per phase) is PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW 2 The percentage of the total power lost in the line is PLine 8 × 100% = × 100% = 16.7% PLoad + PLine 40 +8 P 12.6-3 A balanced three-phase source has a Y-connected source with va = 5 cos (2t + 30°) connected to a three-phase Y load. Each phase of the Y-connected load consists of a 4-Ω resistor and a 4-H inductor. Each connecting line has a resistance of 2 Ω. Determine the total average power delivered to the load. Solution: Ia = Va 5∠30° = = 0.5∠ − 23° A ∴ I a = 0.5 A Z T 6 + j8 2 PLoad I = 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W 2 also (but not required) : PSource = 3 (5) (0.5) cos(−30 − 23) = 2.25 W 2 2 Pline I = 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W 2 Section 12.7 Load P 12.7-1 Instantaneous and Average Power in a Balanced Three-Phase Find the power absorbed by a balanced three-phase Y-connected load when VCB = 208 ∠15° V rms and IB = 3 ∠110° A rms The source uses the abc phase sequence. Answer: P = 620 W Solution: Assuming the abc phase sequence: VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms Then VA = VAB 208∠315° 208 = = ∠285° V rms 3∠30° 3∠30° 3 also I B = 3∠110° A rms ⇒ I A = 3∠230° A rms Finally P = 3 VA I A cos (θ V − θ I ) = 3( 208 ) (3) cos(285° − 230°) = 620 W 3 P 12.7-2 A three-phase motor delivers 20 hp operating from a 480-V rms line voltage. The motor operates at 85 percent efficiency with a power factor equal to 0.8 lagging. Find the magnitude and angle of the line current for phase A. Hint: 1 hp = 745.7 W Solution: Assuming a lagging power factor: cos θ = pf = 0.8 ⇒ θ = 36.9° The power supplied by the three-phase source is given by Pin = Pout η = Pin = 3 I A VA pf 20 ( 745.7 ) = 17.55 kW where 1 hp = 745.7 W 0.85 ⇒ IA = 17.55 ×103 Pin = = 26.4 A rms 3 VA pf ⎛ 480 ⎞ 3⎜ ⎟ ( 0.8 ) ⎝ 3⎠ 480 ° I A = 26.4∠ − 36.9° A rms when VA = ∠0 V rms 3 1 P 12.7-3 A three-phase balanced load is fed by a balanced Y-connected source with a line-toline voltage of 220 V rms. It absorbs 1500 W at 0.8 power factor lagging. Calculate the phase impedance if it is (a) Δ-connected and (b) Y-connected. Solution: (a) For a Δ-connected load, Eqn 12.7-5 gives PT 1500 = = 4.92 A rms 3 VP pf 3( 220 )(.8) 3 The phase current in the Δ-connected load is given by PT = 3 VP I L pf ⇒ IL = I IL 4.92 ⇒ IP = L = = 2.84 A rms 3 3 3 The phase impedance is determined as: IP = Z= V VL VL 220 = ∠ (θ V − θ I ) = L ∠ cos −1 pf = ∠ cos −1 0.8 = 77.44∠36.9° Ω IP IP IP 2.84 (b) For a Y-connected load, Eqn 12.7-4 gives PT = 3 VP I L pf ⇒ I L = PT 1500 = = 4.92 A rms 220 3 VP I L pf 3( )(.8) 3 The phase impedance is determined as: 220 V V V Z = P = P ∠ (θ V − θ I ) = P ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω 4.92 IP IP IP 2 P 12.7-4 A 600-V rms three-phase Y-connected source has two balanced Δ loads connected to the lines. The load impedances are 40 ∠30° Ω and 50 ∠–60° Ω, respectively. Determine the line current and the total average power. Solution: Parallel Δ loads ZΔ = Z1Z 2 (40∠30° ) (50∠−60° ) = = 31.2 ∠−8.7° Ω ° ° Ζ1 + Ζ 2 40∠30 + 50∠− 60 VL = VP , Ι P = VP 600 = = 19.2 A rms, ZΔ 31.2 IL = 3 Ι P = 33.3 A rms So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW P 12.7-5 A three-phase Y-connected source simultaneously supplies power to two separate balanced three-phase loads. The first total load is Δ connected and requires 39 kVA at 0.7 lagging. The second total load is Y connected and requires 15 kW at 0.21 leading. Each line has an impedance 0.038 + j0.072 Ω/phase. Calculate the line-to-line source voltage magnitude required so that the loads are supplied with 208-V rms line-to-line. Solution: We will use In our case: S = S ∠θ = S cos θ + j S sin θ = S pf + j S sin ( cos −1 pf ) S 1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA 15 sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA S 2 = 15 + 0.21 S S 3φ = S 1 + S 2 = 42.3 − j 42.0 kVA ⇒ S φ = 3φ = 14.1− j 14.0 kVA 3 The line current is * ⎛ S ⎞ (14100+ j 14000) S = Vp I L ⇒ I L = ⎜ ⎟ = = 117.5 + j 116.7 A rms = 167 ∠45° A rms ⎜V ⎟ 208 ⎝ p⎠ 3 208 ∠0° = 120∠0° V rms. The source must The phase voltage at the load is required to be 3 provide this voltage plus the voltage dropped across the line, therefore * = 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms V Sφ Finally V = 116.6 V rms Sφ 3 P 12.7-6 A building is supplied by a public utility at 4.16 kV rms. The building contains three balanced loads connected to the three-phase lines: (a) (b) (c) Δ-connected, 500 kVA at 0.85 lagging Y-connected, 75 kVA at 0.0 leading Y connected; each phase with a 150-Ω resistor parallel to a 225-Ω inductive reactance The utility feeder is five miles long with an impedance per phase of 1.69 + j0.78 Ω/mile. At what voltage must the utility supply its feeder so that the building is operating at 4.16 kV rms? Hint: 41.6 kV is the line-to-line voltage of the balanced Y-connected source. Solution: The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the Δ-connected load. The apparent power per phase 500 kVA required by the Δ-connected load is S1 = = 167 kVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA and * 3 ⎛ S1 ⎞ ⎛ (167 ×10 ) ∠31.8° ⎞ ⎜ ⎟ = 69.6∠ − 31.8° = 59 − j36.56 A rms ⇒ I1 = ⎜ ⎟ = 3 ⎜ ⎟ V × ∠ ° 2.402 10 0 ( ) ⎝ P⎠ ⎝ ⎠ * S1 = VP I1 * Let I2 be the line current required by the first Y-connected load. The apparent power per phase 75 kVA required by this load is S 2 = = 25 kVA . Then, noticing the leading power factor, 3 S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA and * 3 ⎛ S 2 ⎞ ⎛ ( 25 ×10 ) ∠ − 90° ⎞ ⎟ = 10.4∠90° = j10.4 A rms ⇒ I2 = ⎜ ⎟ = ⎜ 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ * S 2 = VP I 2 * Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3 to be 2402∠0° 2402∠0° I3 = + = 16 − j 10.7 A rms 150 j 225 The line current is I L = I1 + I 2 + I 3 = 75− j 36.8 A rms 4 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms Finally VSL = 3 (3179) = 5506 Vrms 5 P 12.7-7 The diagram shown in Figure P 12.7-7 has two three-phase loads that form part of a manufacturing plant. They are connected in parallel and require 4.16 kV rms. Load 1 is 1.5 MVA, 0.75 lag pf, Δ-connected. Load 2 is 2 MW, 0.8 lagging pf, Y-connected. The feeder from the power utility’s substation transformer has an impedance of 0.4 + j0.8 Ω/phase. Determine the following: (a) (b) (c) The required magnitude of the line voltage at the supply. The real power drawn from the supply. The percentage of the real power drawn from the supply that is consumed by the loads. Three-phase supply from utility 0.4 Ω j0.8 Ω 0.4 Ω j0.8 Ω 0.4 Ω j0.8 Ω Load 2 Load 1 Figure P 12.7-7 Solution: The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the Δ-connected load. The apparent power per phase 1.5 MVA required by the Δ-connected load is S1 = = 0.5 MVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA and * 6 ⎛ S1 ⎞ ⎛ ( 0.5 ×10 ) ∠41.4° ⎞ ⎟ = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms ⇒ I1 = ⎜ ⎟ = ⎜ 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ * S1 = VP I1 * Let I2 be the line current required by the first Y-connected load. The complex power, per phase, is 0.67 S 2 = 0.67 + sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA 0.8 6 * 6 ⎛ S 2 ⎞ ⎛ ( 0.67 + j 0.5 ) ×106 ⎞ ⎛ ( 0.833 ×10 ) ∠ − 36.9° ⎞ ⎟ =⎜ ⎟ I2 = ⎜ ⎟ = ⎜ 3 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ = 346.9∠ − 36.9° = 277.4 − j 208.3 A rms The line current is I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms * * 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms Finally VSL = 3 (2859.6) = 4953 Vrms The power supplied by the source is PS = 3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW The power lost in the line is PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW The percentage of the power consumed by the loads is 3.49 − 0.369 × 100% = 89.4% 3.49 7 P 12.7-8 The balanced three-phase load of a large commercial building requires 480 kW at a lagging power factor of 0.8. The load is supplied by a connecting line with an impedance of 5 + j25 mΩ for each phase. Each phase of the load has a line-to-line voltage of 600 V rms. Determine the line current and the line voltage at the source. Also, determine the power factor at the source. Use the line-to-neutral voltage as the reference with an angle of 0°. Solution: The required phase voltage at the load is VP = 600 ∠0° = 346.4∠0° Vrms . 3 Let I be the line current required by the load. The complex power, per phase, is S = 160 + j 160 sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA 0.8 The line current is * * ⎛ S ⎞ ⎛ (160 + j 120 ) ×103 ⎞ I=⎜ ⎟ =⎜ ⎟ = 461.9 − j 346.4 A rms 346.4∠0° ⎝ VP ⎠ ⎝ ⎠ 600 ∠0° = 346.4∠0° Vrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms Finally VSL = 3 (357.5) = 619.2 Vrms The power factor of the source is pf = cos (θ V − θ I ) = cos (1.6° − ( − 37°)) = 0.78 8 Section 12-8: Two-Wattmeter Power Measurement P 12.8-1 The two-wattmeter method is used to determine the power drawn by a three-phase 440-V rms motor that is a Y-connected balanced load. The motor operates at 20 hp at 74.6 percent efficiency. The magnitude of the line current is 52.5 A rms. The wattmeters are connected in the A and C lines. Find the reading of each wattmeter. The motor has a lagging power factor. Hint: 1 hp = 745.7 W Solution: W = 14920 W hp P 14920 Pin = out = = 20 kW 0.746 η Pout = 20 hp × 746 Pin = 3 VL I L cos θ ⇒ cos θ = Pin 20 × 103 = = 0.50 3 VL I L 3 (440) (52.5) ⇒ θ cos -1 ( 0.5 ) = 60° The powers read by the two wattmeters are P1 = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0 and P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW 1 P 12.8-2 A three-phase system has a line-to-line voltage of 4000 V rms and a balanced Δ-connected load with Z = 40 + j30 Ω. The phase sequence is abc. Use the two wattmeters connected to lines A and C, with line B as the common line for the voltage measurement. Determine the total power measurement recorded by the wattmeters. Answer: P = 768 kW Solution: VP = VL = 4000 V rms IP = VP 4000 = = 80 A rms ZΔ 50 Z Δ = 40 + j 30 = 50 ∠36.9° Ι L = 3 I P = 138.6 A rms pf = cos θ = cos (36.9° ) = 0.80 P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW PT = P1 + P2 = 767.9 kW Check : PT = 3 Ι L VL cos θ = 3 (4000) (138.6) cos 36.9° = 768 kW which checks 2 P 12.8-3 A three-phase system with a sequence abc and a line-to-line voltage of 200 V rms feeds a Y-connected load with Z = 70.7 ∠45° Ω. Find the line currents. Find the total power by using two wattmeters connected to lines B and C. Answer: P = 400 W Solution: Vp = Vp = 200 = 115.47 Vrms 3 VA =115.47∠0° V rms, VB = 115.47∠−120° V rms and VC = 115.47∠120° V rms IA = VA 115.47∠0° = = 1.633∠ − 45° A rms Z 70.7∠45° I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms PT = 3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W 3 P 12.8-4 A three-phase system with a line-to-line voltage of 208 V rms and phase sequence abc is connected to a Y-balanced load with impedance 10 ∠–30° Ω and a balanced Δ load with impedance 15 ∠30° Ω. Find the line currents and the total power using two wattmeters. Solution: ZY = 10∠ − 30° Ω and Z Δ = 15∠30° Ω Convert Z Δ to Z Yˆ → Z Yˆ = then Zeq = ZΔ = 5∠30° Ω 3 (10∠−30° ) ( 5∠30° ) = 10∠−30°+5∠30° 208 Vp = Vp = = 120 V rms 3 VA = 120∠0° V rms ⇒ I A = 50∠0° = 3.78∠10.9° Ω 13.228 ∠−10.9° 120∠0° = 31.75 ∠−10.9° 3.78 ∠10.9° I B = 31.75∠−130.9° I C = 31.75∠109.1° PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW W1 = VL I L cos (θ −30°) = 6.24 kW W2 = VL I L cos (θ + 30°) = 4.99 kW P 12.8-5 The two-wattmeter method is used. The wattmeter in line A reads 920 W, and the wattmeter in line C reads 460 W. Find the impedance of the balanced Δ-connected load. The circuit is a three-phase 120-V rms system with an abc sequence. Answer: ZΔ = 27.1 ∠–30° Ω P12.8-5 PT = PA + PC = 920 + 460 = 1380 W tan θ = 3 ( −460 ) = −0.577 ⇒ θ = −30° PC − PA = 3 PC + PA 1380 PT = 3 VL I L cos θ so I L = PT 1380 = =7.67 A rms 3 VL cos θ 2 ×120×cos( −30 ) IL = 4.43 A rms 3 120 = 27.1 Ω ο r Z Δ = 27.1 ∠−30° 4.43 IP = ∴ ZΔ = 4 P 12.8-6 Using the two-wattmeter method, determine the power reading of each wattmeter and the total power for Problem P 12.5-1 when Z = 0.868 + j4.924 Ω. Place the current coils in the Ato-a and C-to-c lines. Solution: Z = 0.868 + j 4.924 = 5∠80° VL = 380 V rms, VP = I L = I P and I P = ⇒ θ = 80° 380 = 219.4 V rms 3 VP = 43.9 A rms Z P1 = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W PT = P1 + P2 = 5017 W 5 Section 12.9 How Can We Check …? P 12.9-1 A Y-connected source is connected to a Y-connected load (Figure 12.3-1) with Z = 10 + j4 Ω. The line voltage is VL = 416 V rms. A student report states that the line current IA = 38.63 A rms and that the power delivered to the load is 16.1 kW. Verify these results. Solution: 416 = 240 V = VA 3 Z = 10 + j4 = 10.77 ∠21.8° Ω VA = VA 240 = = 22.28 A rms ≠ 38.63 A rms Z 10.77 38.63 = 22.3 . It appears that the line-to-line voltage was The report is not correct. (Notice that 3 mistakenly used in place of the phase voltage.) IA = P 12.9-2 A Δ load with Z = 40 + j30 Ω has a three-phase source with VL = 240 V rms (Figure 12.3-2). A computer analysis program states that one phase current is 4.8 ∠–36.9° A. Verify this result. Solution: VL = VP = 240∠0° Vrms Z = 40 + j 30 = 50 ∠36.9° Ω IP = VP 240∠0° = = 4.8 ∠−36.9° A rms ° Z 50∠36.9 The result is correct. 1 PSpice Problems SP 12-1 Use PSpice to determine the power delivered to the load in the circuit shown in Figure SP 12-1. a 5Ω 5 mH A 20 Ω 60 mH b 5Ω 5 mH B 20 Ω 60 mH C 20 Ω + – 110 cos (377t) N + n – 110 cos (377t +120°) c 5Ω 5 mH 60 mH + – 110 cos (377t + 240°) Source Line Load Figure SP 12-1 Solution: FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01 FREQ IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 1 6.000E+01 3.142E+00 -4.443E+01 FREQ 6.000E+01 VM(N01496) 2.045E-14 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 3.142E+00 7.557E+01 7.829E-01 3.043E+00 VP(N01496) 2.211E+01 2.244E+00 VR(N01496) 1.895E-14 -2.200E+00 VI(N01496) 7.698E-15 3.1422 20 = 98.7 W 2 3.1422 20 = 98.7 W I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB = 2 3.1422 20 = 98.7 W I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC = 2 I A = 3.142∠ − 43.43° A and RA = 20 Ω ⇒ PA = P = 3 ( 98.7 ) = 696.1 W 2 SP 12-2 2. Use PSpice to determine the power delivered to the load in the circuit shown in SP 12a 10 Ω 5 mH A 20 Ω 60 mH b 10 Ω 5 mH B 30 Ω 25 mH C 60 Ω + – 110 cos (377t) – N + n 110 cos (377t +120°) c 10 Ω 5 mH 20 mH + – 110 cos (377t + 240°) Source Line Load Figure SP 12-2 Solution: FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00 FREQ 6.000E+01 IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 2.537E+00 -3.748E+01 2.013E+00 -1.544E+00 3 FREQ 6.000E+01 VM(N01496) VP(N01496) 1.215E+01 -1.439E+01 VR(N01496) VI(N01496) 1.177E+01 -3.018E+00 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 2.858E+00 1.084E+02 -9.023E-01 2.712E+00 2.537 2 20 = 64.4 W 2 2.8582 30 = 122.5 W I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB = 2 1.6122 60 = 78 W I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC = 2 I A = 2.537∠ − 37.48° A and RA = 20 Ω ⇒ PA = P = 64.4 + 122.5 + 78 = 264.7 V 4 Design Problems DP 12-1 A balanced three-phase Y source has a line voltage of 208 V rms. The total power delivered to the balanced Δ load is 1200 W with a power factor of 0.94 lagging. Determine the required load impedance for each phase of the Δ load. Calculate the resulting line current. The source is a 208-V rms ABC sequence. Solution: P = 400 W per phase, 0.94 = pf = cos θ θ = cos -1 ( 0.94 ) =20° ⇒ 208 I L 0.94 ⇒ I L = 3.5 A rms 3 I I Δ = L = 2.04 A rms 3 V 208 Z = L = = 101.8 Ω IΔ 2.04 400 = Z = 101.8 ∠20° Ω DP 12-2 A three-phase 240-V rms circuit has a balanced Y-load impedance Z. Two wattmeters are connected with current coils in lines A and C. The wattmeter in line A reads 1440 W, and the wattmeter in line C reads zero. Determine the value of the impedance. Solution: VL = 240 V rms PA = VL I L cos (30° + θ ) = 1440 W PC = VL I L cos (30° − θ ) = 0 W ⇒ 30−θ = 90° or θ = −60° then 1440 = 240 I L cos (−30° ) I L = 6.93 A rms IL = IP = VP Z ⇒ ⇒ 240 V Z = P = 3 = 20 Ω IP 6.93 Finally, Z = 20 ∠ − 60° Ω 1 DP 12-3 A three-phase motor delivers 100 hp and operates at 80 percent efficiency with a 0.75 lagging power factor. Determine the required Δ-connected balanced set of three capacitors that will improve the power factor to 0.90 lagging. The motor operates from 480-V rms lines. Solution: Pin = Pout η 100 hp × (746 = W ) hp 0.8 = 93.2 kW, P = φ Pin = 31.07 kW 3 VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor. 0.75 = pf = cos θ ⇒ θ = cos-1 ( 0.75) = 41.4° 480 I p ( 0.75 ) ⇒ I p = 149.5 Arms 3 480 VP 3 = 1.85 Ω Z = = IP 149.5 31070 = Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω The capacitance required to correct the power factor is given by ⎡⎣ tan (cos −1 0.75) − tan (cos −1 0.9) ⎤⎦ 1.365 C= = 434 μ F × 1.3652 +1.2042 377 (Checked using LNAPAC 6/12/03) 2 DP 12-4 A three-phase system has balanced conditions so that the per-phase circuit representation can be utilized as shown in Figure DP 12-4. Select the turns ratio of the step-up and step-down transformers so that the system operates with an efficiency greater than 99 percent. The load voltage is specified as 4 kV rms, and the load impedance is 4/3 Ω. 2.5 Ω 1 : n1 + 20 kV – j40 Ω IL n2 : 1 + + + V1 V2 VL – – – Load Figure DP 12-4 Solution: VL = 4∠0° kV rms n2 25 VL = 4000∠0° = 100∠0° kVrms n1 1 Try n2 = 25 then V2 = VL 4×103∠0° = = 3∠0° kA rms 4 ZL 3 3000∠0° = 120∠0° A rms The line current in 2.5 Ω is I = 25 Thus V1 = ( R + j X ) I + V2 IL = = (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV Step need : n1 = Ploss = I 2 100.4 kV = 5.02 ≅ 5 20 kV R = 120 2 (2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW 12 − .036 × 100% = 99.7 % of the power supplied by the source 12 is delivered to the load. ∴η = 3 Chapter 13: Frequency Response Exercises Exercise 13.2-1 The input to the circuit shown in Figure E 13.2- R 1 is the source voltage, vs, and the response is the capacitor voltage, vo. Suppose R = 10 kΩ and C = 1 μF. What are the values + vs + – vo C – of the gain and phase shift when the input frequency is ω = 100 rad/s? Figure E 13.2-1 Answer: 0.707 and –45° Solution: H (ω ) = gain = 1 Vo (ω ) = 1 + jω C R Vs (ω ) 1 1 + (ω C R) 2 phase shift = − tan −1 (ω C R ) When R = 104 Ω, ω = 100 rad/s, and C = 10−6 F, then gain = 1 = 0.707 and phase shift = − 45o 2 13Ex-1 Exercise 13.2-2 The input to the circuit shown in Figure E 13.2-2 L is the source voltage, vs, and the response is the resistor voltage, vo. + R = 30 Ω and L = 2 H. Suppose the input frequency is adjusted until vs + vo R – i the gain is equal to 0.6. What is the value of the frequency? – Figure E 13.2-2 Answer: 20 rad/s Solution: H (ω ) = gain = Vo (ω ) R = Vs (ω ) R + jω L R R 2 + (ω L) 2 2 0.6 = 30 30 + (2ω ) 2 2 ⎛ 30 ⎞ 2 ⎜ ⎟ − 30 ⎝ .6 ⎠ ⇒ ω= = 20 rad s 2 Exercise 13.2-3 The input to the circuit shown in Figure E 13.2-2 L is the source voltage, vs, and the response is the mesh current, i. R = 30 Ω and L = 2 H. What are the values of the gain and phase shift when the input frequency is ω = 20 rad/s? + vs + – R i vo – Figure E 13.2-2 Answer: 0.02 A/V and –53.1° Solution: H (ω ) = gain = I (ω ) 1 = R + jω L Vs (ω ) 1 R + (ω L) 2 2 phase shift = − tan −1 ωL R When R = 30 Ω, L = 2 H, and ω = 20 rad/s, then gain = 1 30 + 40 2 2 = 0.02 A ⎛ 40 ⎞ and phase shift = − tan −1 ⎜ ⎟ = − 53.1° V ⎝ 30 ⎠ 13Ex-2 Exercise 13.2-4 The input to the circuit shown in Figure E 13.2-1 R is the source voltage, vs, and the response is the capacitor voltage, + vo. Suppose C = 1 μF. What value of R is required to cause a phase + – vs vo C – shift equal to –45° when the input frequency is ω = 20 rad/s? Answer: R = 50 kΩ Figure E 13.2-1 Solution: H (ω ) = gain = Vo (ω ) 1 = Vs (ω ) 1 + jω C R 1 1 + (ω C R) 2 phase shift = − tan −1 ω C R −1 −6 −45° = − tan (20 ⋅10 ⋅ R ) tan (45° ) ⇒ R= = 50 ⋅103 = 50 kΩ −6 20⋅10 Exercise 13.2-5 The input to the circuit shown in Figure E 13.2-1 R is the source voltage, vs, and the response is the capacitor voltage, vo. Suppose C = 1 μF. What value of R is required to cause a gain + + – vs vo C – equal to 1.5 when the input frequency is ω = 20 rad/s? Figure E 13.2-1 Answer: No such value of R exists. The gain of this circuit will never be greater than 1. Solution: H (ω ) = gain = Vo (ω ) 1 = Vs (ω ) 1 + jω C R 1 1 + (ω C R) 2 ω , C , and R are all positive, or at least nonnegative, so gain ≤ 1. These specifications cannot be met. 13Ex-3 Exercise 13.3-1 (a) Convert the gain|Vo/Vs| = 2 to decibels. (b) Suppose |Vo/Vs| = –6.02 dB. What is the value of this gain “not in dB”? Answer: (a) + 6.02 dB (b) 0.5 Solution: (a) dB = 20 log (2) = 6.02 dB −6.02 20 (b) 10 = 0.5 Exercise 13.3-2 In a certain frequency range, the magnitude of the network function can be approximated as H = 1/ω2. What is the slope of the Bode plot in this range, expressed in decibels per decade? Answer:–40 dB/decade Solution: ⎛ 1 ⎞ 20 log H = 20 log ⎜ 2 ⎟ = 20 log (ω )-2 = −40 log ω ⎝ω ⎠ ⎛ω ⎞ = − 40 log ω2 + 40 log ω1 = − 40 log⎜ 2 ⎟ ⎝ ω1 ⎠ let ω2 = 10 ω1 to consider 1 decade, then slope = − 40 log10 = − 40 dB decade slope = 20 log H (ω2 ) − 20 log H (ω1 ) 13Ex-4 Exercise 13.3-3 Consider the network function H (ω ) = jω A B + jωC Find (a) the corner frequency, (b) the slope of the asymptotic magnitude Bode plot for ω above the corner frequency in decibels per decade, (c) the slope of the magnitude Bode plot below the corner frequency, and (d) the gain for ω above the corner frequency in decibels. Answer: (a) ω0 = B/C (b) zero (c) 20 dB/decade (d) 20 log10 = A C Solution: When ω C >> B, H (ω ) − (d ) (b) jω A A = jω C C ⎛ A⎞ H (ω ) in dB = 20 log10 H (ω ) = 20 log10 ⎜ ⎟ ⎝C ⎠ H (ω ) does not depend on ω so slope = 0 When ω C << B, H (ω ) − jω A ⎛ A⎞ = jω ⎜ ⎟ B ⎝B⎠ ⎛ A⎞ H (ω ) in dB = 20 log10 H (ω ) = 20 log10ω +20 log10 ⎜ ⎟ ⎝B⎠ (c) The slope is the coefficient of log10 ω , that is, slope = 20 dB decade (a) The break frequency is the frequency at which ω0 C = B, that is, ω0 = B C 13Ex-5 Exercise 13.4-1 For the RLC parallel resonant circuit when R = 8 kΩ, L = 40 mH, and C = 0.25 μF, find (a) Q and (b) bandwidth. Answer: (a) Q = 20 (b) BW = 500 rad/s Solution: a) Q = ω 0 RC = R b) BW = ωo = Q C 2.5 × 10−7 = 8000 = 20 L 40 × 10−3 1 1 = = 500 rad s Q LC 20 (40×10−3 ) (2.5×10−7 ) Exercise 13.4-2 A high-frequency RLC parallel resonant circuit is required to operate at ω0 = 10 Mrad/s with a bandwidth of 200 krad/s. Determine the required Q and L when C = 10 pF. Answer: Q = 50 and L = 1 mH Solution: ω0 107 = = 50 Q= BW 2 × 105 1 1 1 ω0 = ⇒ L = = = 1 mH 2 7 2 LC ω0 C (10 ) (10 × 10−12 ) Exercise 13.4-3 A series resonant circuit has L = 1 mH and C = 10 μF. Find the required Q and R when it is desired that the bandwidth be 15.9 Hz. Answer: Q = 100 and R = 0.1 Ω Solution: 1 1 = = 104 rad s 1 LC ⎡(10−3 )(10−5 ) ⎤ 2 ⎣ ⎦ ω0 = Q = R = ω0 BW ω 0L Q = = 104 = 100 2π (15.9) (104 )(10−3 ) = 0.1 Ω 100 13Ex-6 Exercise 13.4-4 A series resonant circuit has an inductor L = 10 mH. (a) Select C and R so that ω0 = 106 rad/s and the bandwidth is BW = 103 rad/s. (b) Find the admittance Y of this circuit for a signal at ω = 1.05×106 rad/s. Answer: (a) (b) Solution: a) C = 100pF, R = 10 Ω Y= 10 1 + j 97.6 ω0 = Q= b) Q = Y= 1 1 ⇒ C = = 100 pF 6 2 LC (10 ) (0.01) ω0 BW ω0 BW = 1 103 BW ⇒ R= = = 10 Ω ω 0 RC ω 02 C (106 )2 (10−10 ) = 106 = 1000 103 1 ⎛ ω ω0 ⎞ − 1+ j Q⎜ ⎟ ⎜ω0 ω ⎟ ⎝ ⎠ = 1 ⎡1.05×106 106 ⎤ − 1+ j 1000 ⎢ ⎥ 6 1.05×106 ⎦⎥ ⎣⎢ 10 = 1 1+ j 97.6 13Ex-7 Section 13-2: Gain, Phase Shift, and the Network Function P 13.2-1 The input to the circuit shown in Figure P 13.2-1 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across the parallel connection of the capacitor and 10-Ω resistor. Determine the network function, H(ω) = Vo(ω)/Vi(ω), of this circuit. Answer: H (ω ) = 0.