ELEC 273 Part 12 Old Exam Questions 1. 2. 3. 4. Old Exam Questions with Solutions 2016 Final Exam 2017 Final Exam 2018 Final Exam Copyright © 2020 C.W. Trueman 1 Mesh Analysis Write mesh equations. Method: 1.Write a “constraint equation” for the current source. 2.Write a set of mesh equations, including the supermesh. Solution F E G For the current generators: B π =1 C H π −π = 3 I Mesh path ABCDA 4 π −π +3−2 π −π A −8 π −π Supermesh DCBEFGHIJD 8 π −π +2 π −π −2 π =0 + 4 − 6π − 10π − (−5) = 0 D J Node Analysis Method: 1.Write “constraint equations” for the voltage sources. 2.Write a KCL equation for the supernode. 3.Write the “controlling current” π in terms of the node voltages and then eliminate π from the node equations. Note that this circuit has a supernode that includes two voltage sources! Solution Constraint Equations π£ − 2π = π£ π£ − 12 = π£ Controlling current π£ π= 4 Eliminate π from π£ − 2π = π£ to get π£ π£ −2 =π£ 4 Which simplifies to π£ π£ − =π£ 2 KCL Equation KCL for the supernode: flows BOTH into and out of the supernode! − π£ π£ π£ −π£ π£ −π£ π£ − − + − =0 4 1 2 2 4 − π£ π£ π£ − − =0 4 1 4 Hence the three equations are: π£ π£ − =π£ 2 π£ − 12 = π£ π£ π£ π£ − − − =0 4 1 4 π£ −π£ 2 Thevenin Equivalent Circuit Is=2A Vs=8V π =8 πΌ =2 Find the Thevenin Equivalent Circuit at terminals a-b. Method: 1.Find the open-circuit voltage, π , and then π = π To find π : Either find the short-circuit current, πΌ and π = , or find the resistance of the “dead circuit”. This is a “judgement call” based on your experience. In general when the circuit has dependent sources, the short-circuit current method is better. Solution π£ π£ π£ −8 π£ 8 2 Is=2A Vs=8V + π£ − Open Circuit Voltage: KCL at π£ π£ π£ − π£ −8 π£ −π£ − − − =0 1 2 2 KCL at the supernode: π£ − π£ −8 π£ −π£ π£ +2+ − =0 2 2 2 Multiply 1st eqn by 2: −2π£ − π£ + π£ − 8 − π£ + π£ = 0 −4π£ + 2π£ = 8 8 − 2π£ π£ π£ = = −2 + −4 2 nd Multiply 2 eqn by 2: π£ − π£ −8 +4+π£ −π£ −π£ = 0 2π£ − 3π£ = −8 − 4 π£ 2(−2 + ) − 3π£ = −12 2 −4 + π£ − 3π£ = −12 −2π£ = −8 π£ =4 Solution 0 volts π£ −8 0 8 2 Is=2A Vs=8V π 0 0 volts Short circuit current: KCL at π£ is almost the same as for the open-circuit test, except that the voltage at terminal “a” is zero: π£ π£ − −8 π£ − − − =0 1 2 2 −2π£ − π£ + −8 − π£ = 0 −4π£ = 8 π£ = −2 KCL at the supernode: π£ − −8 π£ +2+ −π =0 2 2 Multiply by 2 π£ − −8 + 4 + π£ − 2π = 0 2π£ − 2π = −8 − 4 −2π = −12 − 2π£ π =6+π£ π = 6 + (−2) π =4 Thevenin Equivalent Circuit π£ = 4 volts π = 4 amps So π£ =4 π£ π = π Check: Dead circuit method π π π π = ( 2 β₯2 +1 β₯2 = 1+1 β₯2 =2β₯2 =1 = 4 =1Ω 4 Maximum Power Transfer Theorem Find R such that it dissipates the maximum amount of power. How much power does R dissipate? Method: 1.Find the Thevenin Equivalent. 2.Choose π = π 3.Power is π = Find the open-circuit voltage: π + π£ − π Current in the 80+20 branch: π = = 0.4 amps Current in the 10+90 branch: π = = 0.4 amps π£ =voltage across the 10 ohm resistor + voltage across the 80 ohm resistor π£ = +10π − 80π π£ = 10π₯0.4 − 80π₯0.4 = 4 − 32 = −28 volts Check: π£ =voltage across the 90 ohm resistor + voltage across the 20 ohm resistor π£ = −90π + 20π π£ = −90π₯0.4 + 20π₯0.4 = 36 − 8 = −28 volts Dead circuit method, easy for this circuit! 80 Ω 20 Ω π 10 Ω π = 80 β₯ 20 + 10 β₯ 90 = 80π₯20 10π₯90 + = 16 + 9 = 25 Ω 80 + 20 10 + 90 90 Ω Maximum Power: Find R such that it dissipates the maximum amount of power. How much power does R dissipate? π = −28 volts A π = 25 Ω B Find R such that it dissipates the maximum amount of power. How much power does R dissipate? Method: 1.Find the Thevenin Equivalent – done! 2.Choose π = π : Hence choose the load as π = 25 Ω 3.Power is π = : Hence the maximum power is π = ( ) = 7.84 watts. Op Amp Circuit The op-amps are ideal. Find the output voltage π£ . Method: The voltage across the input of the op-amp is zero. Use node analysis and write KCL equations. Solution π£ +0 − 0 π£ π£ The input resistance of the op-amp is high so π = 0 so the voltage across π is zero. The voltage across the input to the op-amp is zero. Hence the voltage at the “-” input equal to π£ . π π£ = (π + π )π£ KCL at the top of π : π +π π£ −π£ π£ π£ = π£ − =0 π π π π π£ −π π£ −π π£ =0 Solution 0 +0 − 0 0 π£ π +π π£ π KCL at the input of the 2nd op amp: π£ π£ + =0 π π π π£ +π π£ =0 π£ = π£ π£ π π£ π π π +π =− π£ π π =− First Order Transient For π‘ < 0, the switch has been in position 1 for a long time. At π‘ = 0 the switch changes to position 2 and stays there for π‘ > 0. 1.Find the initial condition π£ 0 . 2.Find the final condition π£ (∞). This is the “forced response”. 3.Find the time constant π. 4.Find the “natural response”. 5.Add up the forced response plus the natural response and evaluate the arbitrary constant using the initial condition. 1 2 For π‘ < 0, the switch has been in position 1 for a long time. At π‘ = 0 the switch changes to position 2 and stays there for π‘ > 0. 1.Find the initial condition π£ 0 . 1 2 For π‘ < 0 the switch has been in position #1 for a long time. Find π£ (0). After the switch has been in position 1 “for a long time”, the voltage across the capacitor becomes constant and the current π = πΆ is zero. The capacitor behaves as an open circuit. 6Ω 6Ω 1 1 0.5 6Ω 0.1 F + π£ (0) − 0.5 6Ω 0.1 πF + π£ 0 = −0.5x6 = −3 − For π‘ < 0, the switch has been in position 1 for a long time. At π‘ = 0 the switch changes to position 2 and stays there for π‘ > 0. 1.Find the initial condition π£ 0 . Just before the switch changes: 6Ω 1 0.5 + π£ 0 = −3 − 6Ω Just after the switch changes. So the initial condition is π£ (0) = −3 volts 1 2 2Ω 6Ω 0.5 The stored energy in the capacitor is π€ = πΆπ£ . The stored energy cannot change instantaneously when the switch changes position. So the voltage across the capacitor π£ = −3 volts just before the switch changes is the same as the voltage across the capacitor π£ = −3 volts just after the switch changes. 6Ω 0.1F + π£ 0 = −3 − 24 Ω 8Ω 1 64 2.Find the final condition π£ (∞). The final value as π‘ → ∞ is the “forced response”. As π‘ → ∞, the switch has been in position 2 for a long time, and the voltage across the capacitor becomes constant and the current π = πΆ is zero. The capacitor behaves as an open circuit. 2 24 Ω 2Ω 0.1 πF + π£ ∞ − 8Ω 64 Write a node equation: π£ π£ − 64 − −1− =0 8 24 −3π£ − 24 − π£ + 64 = 0 −4π£ = −40 π£ = 10 π£ 2 + π£ ∞ − 1 2Ω 8Ω 24 Ω 1 64 Since there is no current in the 2 Ω resistor, π£ (∞) = 10 3.Find the time constant π. 2 24 Ω 2Ω Use the ‘dead circuit’ method to find the resistance across the capacitor terminals. 8 parallel 24 = π 8Ω = =6 π =2+6 = 8Ω πΆ = 0.1 F The time constant is π£ π = π πΆ = 8π₯0.1 = 0.8 seconds 4.Find the “natural response”. 5.Add up the forced response plus the natural response and evaluate the arbitrary constant using the initial condition. Solution 4.The “natural response” for a first-order circuit is π£ π‘ = π΄π . 5.The total response is the sum of the forced response π£ (∞) = 10 plus the natural response π£ π‘ = π΄π . π£ π‘ = 10 + π΄π . Where the time constant is π = 0.8 seconds The initial condition is π£ (0) = −3 volts π£ 0 = 10 + π΄π = −3 10 + π΄ = −3 π΄ = −13 The complete response is π£ π‘ = 10 − 13π Note that this question did not ask you to find the differential equation! First Order Transient The capacitor is uncharged at π‘ = 0. Write a differential equation for π£ π‘ . Solve the differential equation to find π£ π‘ . Remark: This question ask you to write a differential equation and then solve it. So you cannot use the “magic formula” to answer this question! +π£ − 0.1 F 3A 2Ω 8Ω 3A +π£ − Write a differential equation for π£ π‘ . We can make the circuit much simpler with “source transformations”. Replace 3 A in parallel with 2 ohms with 3x2=6 volts in series with 2 ohms. 0.1 F 3A 2Ω 8Ω 3A Replace 3 A in parallel with 8 ohms with 3x8=24 volts in series with 8 ohms. KVL: 6 − 2π − π£ − 8π + 24 = 0 −10π − π£ = −30 10π + π£ = 30 ππ£ π = 0.1 ππ‘ ππ£ 10x0.1 + π£ = 30 ππ‘ ππ£ + π£ = 30 ππ‘ This is the differential equation! +π£ − 2Ω 0.1 F 8Ω 24 v 6v π Solve the differential equation to find π£ π‘ . ππ£ + π£ = 30 ππ‘ The D.E. has the form + π£ = constant So the time constant is π = 1. The forced response is always a constant with D.C. sources, so use π£ = π΅, where π΅ is a constant. ππ΅ + π΅ = 30 ππ‘ 0 + π΅ = 30 so π΅ = 30 Forced response: π£ = 30 Note that this is the “final value” π£(∞) The “natural response” is always an exponential and satisfies the homogeneous differential equation: π£ = π΄π ππ£ +π£=0 ππ‘ ππ΄π + π΄π = 0 ππ‘ ππ΄π + π΄π = 0 π+1=0 π = −1 π£ = π΄π +π£ − 2Ω 0.1 F 8Ω 24 v 6v π Complete response = forced + natural π£ = π΄π + 30 Initial condition: The capacitor is uncharged at π‘ = 0 so π£ 0 = 0 0 = π΄π + 30 π΄ = −30 π£ = 30(1 − π ) First Order Transient The capacitor is initially uncharged. 1. Find the response for 0 < π‘ < 1. 2. Find the response for π‘ > 1. You are expected to know that 3π’(π‘) switches on at π‘ = 0, and that 3π’ π‘ − 1 switches on at π‘ = 1. Hence for 0 < π‘ < 1 only one source is active, but for π‘ > 1, both sources are active. The capacitor is initially uncharged. 1.Find the response for 0 < π‘ < 1. For 0 < π‘ < 1 only one source is active. The initial condition is π£ = 0. +π£ − 0.1 F 2Ω 8Ω 3A 8Ω 3A The “final value” of π£ is π£ = −3π₯8 = 24 volts The resistance at the capacitor terminals is π = 2 + 8 = 10 Ω The time constant is π = π πΆ = 10π₯0.1 = 1 second. The “magic formula” for 1st order response is π£ π‘ =π£ + (π£ −π£ )π So π£ π‘ = 24 + (0 − 24)π π£ π‘ = 24(1 − π ) for 0 < π‘ < 1 +π£ − 2Ω − 24 + 2.Find the response for π‘ > 1. For 0 < π‘ < 1 the response is π£ π‘ = 24(1 − π ) At π‘ = 1, we have π£ 1 = 24(1 − π ) = 15.17 +π£ − 0.1 F 3A 2Ω 8Ω 3A The “final value” of π£ is π£ = 6 + 24 = 30 volts +π£ − The “magic formula” does not work because the initial condition is at π‘ = 1 second! π£ π‘ =π£ + π΄π − + where π΄ is an arbitrary constant. 24 8 Ω 6 3 A 2 Ω The “final value” of π£ is π£ = 30 volts + − π£ π‘ = 30 + π΄π At π‘ = 1, we have π£ 1 = 15.17 15.17 = 30 + π΄π 0.3679π΄ = 15.17 − 30 = −14.83 Check: π΄ = −40.31 π£ 1 = 30 − 40.31π = 15.17 π£ π‘ = 30 − 40.31π 3A Second Order Transient 1.Find the initial condition π£ 0 2.Find the initial conditions | 3.Find the differential equation for π£ π‘ . 4.Find the time constants or the attenuation constant and the damped natural frequency. 5.Find the final value π£ ∞ 6.Find the complete solution for π£ π‘ . 1.Find the initial condition For t<0: When the switch has been in position 1 for a long time: -the capacitor is an open circuit -the inductor is a short circuit. The inductor shorts out the capacitor, so the initial value is π£ 0 = 0 All the current from the source flows through the inductor. The initial value of the current in the inductor is 10 π 0 = =2 5 2.Find the initial conditions At t=0, just after the switch throws: π The initial value is π£ 0 = 0 The initial value of the current in the inductor is π 0 =2 Since π = πΆ ππ£ π = ππ‘ πΆ , we can find by evaluating The voltage across 1.1547 ohms is π£ 0 = 0 and so the current in the resistor is 0 amps. KCL at the top of the circuit gives −3 − π − π = 0 π = −3 − π = −3 − 2 = −5 amps = = . = −20 v/s 3.Find the differential equation for For t>0: π = 1.1547 Ω π£ π . KCL for the node at the top of the circuit: π£ −3 − − π − π = 0 π ππ£ π =πΆ ππ‘ π£ ππ£ −3 − − πΆ −π =0 π ππ‘ Differentiate: 1 ππ£ π π£ ππ +πΆ + =0 π ππ‘ ππ‘ ππ‘ π£ =πΏ and π£ =π£ 1 ππ£ π π£ π£ +πΆ + =0 π ππ‘ ππ‘ πΏ + + π£ =0 4.Find the time constants or the attenuation constant and the damped natural frequency. For t>0: π = 1.1547 Ω π . + π£ =0 The characteristic equation is π + π+ =0 π£ The attenuation constant is πΌ = 1.732 The damped natural frequency is π = 1 The time constant is π = = = 0.5774 + π + π+ =0 . . . π + 3.464π + 4 = 0 −3.464 ± 3.464 − 4π₯4 π= 2 −3.464 ± −4 π= 2 −3.464 ± π2 π= 2 π = −1.732 ± π1 4.Find the time constants or the attenuation constant and the damped natural frequency. π + 3.464π + 4 = 0 π = −1.732 ± π1 π = −πΌ ± ππ The attenuation constant is πΌ = 1.732 The damped natural frequency is π = 1 The time constant is π = = = 0.5774 . 5.Find the final value For The capacitor becomes an open circuit. The inductor becomes a short circuit. : The voltage across the capacitor is zero. π£ π£ ∞ =0 π =0 6.Find the complete solution for π£ π‘ = π΄ π cos π π‘ + π΄ π sin π π‘ where π = 0.5774 and π = 1. π£ 0 =0 ππ£ | = −20 v/s ππ‘ Set π£ 0 = 0: π£ 0 = π΄ π cos 0 + π΄ π π΄ π₯1 + π΄ π₯0 = 0 π΄ =0 π£ π‘ =π΄ π =− π΄ π sin 0 = 0 sin π π‘ sin π π‘+π π΄ π cos π π‘ | =− π΄ sin 0+π π΄ cos 0 = −20 π π΄ = −20 20 20 π΄ =− =− = −20 π 1 π£ π‘ = −20π . sin π‘ . Superposition Find the voltage across the capacitor. Remark: This circuit has a DC source and an AC source, So we must use Superposition to solve the circuit. 1.solve the circuit with the DC source 2.solve the circuit with the AC source 3.Add the DC response and the AC response to get the response to both sources “acting together”. 8 ohms 10 ohms 8 mF 3 cos(12π‘) volts 1H 30 ohms 2H 5 volts + π£ − 5 mF 1.Solve the circuit with the DC source 8 ohms At DC or zero frequency, the capacitors become open circuits and the inductors become short circuits. 10 ohms 30 ohms 8 ohms 10 ohms 8 mF 3 cos(12π‘) volts + π£ − 1H 5 volts 30 ohms 2H 5 volts + π£ − 5 mF The circuit becomes trivially simple! π= = 0.1042 amps π£ = 8π = 8π₯0.1042 = 0.8333 volts 2.Solve the circuit with the AC source 8 ohms The DC source is set to zero. Convert the C’s and L’s to impedance at π = 12 rad/sec. 8 mF 1 1 π = −π = −π ππΆ 12π₯8π₯10 5 mF 1 1 π = −π = −π ππΆ 12π₯5π₯10 1H π = πππΏ = ππ₯12π₯1 = π12 2H π = πππΏ = ππ₯12π₯2 = π24 10 ohms 8 mF = −π10.42 = −π16.67 3 cos(12π‘) volts 1H 30 ohms 2H 5 volts + π£ − 5 mF 3 volts π 8 ohms Node equations 3−π π π −π − − =0 −π10.42 30 + π24 10 + π12 10 π −π10.42 3−π π −π π + − =0 8 10 + π12 −π16.67 3 π12 30 π24 + π − −π16.67 Solve the node equations Eqn 1 3−π π π −π − − =0 −π10.42 30 + π24 10 + π12 π0.09597 3 − π − (0.02033 − π0.01626) π − 0.04098 − π0.04918 π − π = 0 −π0.9597π − 0.02033 − π0.01626 π − 0.04098 − π0.04918 π + 0.04098 − π0.04918 π = −π0.2878 (−0.06131 − π0.03053) π + 0.04098 − π0.04918 π = −π0.2878 Eqn 2 3−π π −π π + − =0 8 10 + π12 −π16.67 0.125(3 − π) + (0.04098 − π0.04918)(π − π) − π0.0999π = 0 −0.125π + 0.04098 − π0.04918 π − 0.04098 − π0.04918 π − π0.05999π = −0.375 −0.125π + 0.04098 − π0.04918 π − 0.04098 − π0.04918 π − π0.05999π = −0.375 0.04098 − π0.04918 π + −0.16598 − π0.01081 π = −0.375 1 π = (−0.375 − −0.16598 − π0.01081 π) 0.04098 − π0.04918 π = −3.750 − π4.500 + 1.530 + π2.100 π Eqn 1 (−0.06131 − π0.03053)( −3.750 − π4.500 + 1.530 + π2.100 π) + 0.04098 − π0.04918 π = −π0.2878 0.09253 + π0.3904 + −0.02970 − π0.1755 π + 0.04098 − π0.04918 π = −π0.2878 0.01128 − π0.2246 π = −0.09253 − π0.6782 π = 2.991 − π0.5621 = 3.044∠ − 10.6° hence π£ π‘ = 3.044 cos(12π‘ − 10.1°) Superposition: add the DC response and the AC response. 1.solve the circuit with the DC source π£ = 0.8333 volts 8 ohms 10 ohms 2.solve the circuit with the AC source π£ π‘ = 3.044 cos(12π‘ − 10.1°) By superposition, the complete response is π£ π‘ = 0.8333 + 3.044 cos(12π‘ − 10.1°) 8 mF 3 cos(12π‘) volts 1H 30 ohms 2H 5 volts + π£ − 5 mF Final Exam Tips: 1.Read the whole exam before you start. 2.Do the “easy” questions first. 3.Take your time. It is better to answer 3 questions correctly than 6 questions incorrectly. 4.Get the signs right. 5.If the first equation you write is wrong, the entire solution is wrong. Good Luck with the Final Exam!