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UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT HOMEWORK #6 Solution ECE 1270 1. Summer 2009 After being closed a long time, the switch opens at t = 0. Find i1 (t) for t > 0. t=0 10kΩ i1 10kΩ 60µA 3mH Step 1: (Redraw circuit at t=0- and solve for iL. Inductor acts as a wire since it has sat for a long time) 10kΩ 60µA i1 10kΩ This circuit is a current divider: 60 µ ⋅10k iL = = 30 µ A 10k + 10k iL Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Inductor acts as a current source since the current in the inductor has to remain the same.) 10kΩ Only one current in the branch: 60µA 10kΩ 30µA i1 i1 = −30 µ Step 3: Final Value(Redraw circuit at t=∞ ∞ and solve for unknown variable. Inductor acts as a wire since it has sat for a long time in this position.) 10kΩ There are no sources connected to the final circuit: i i1 10kΩ i1=0A L 60µA 10kΩ 60µA 10kΩ i1 To find Req the inductor is removed from the final circuit to find path from top of inductor to bottom of inductor: L 3m τ= = = 150n sec R eq 10k + 10k Step 4: Plug values into general equation: i1 (t ) = 0 + [ −30µ − 0] e −t /150 n sec A = −30µ e−t /150 n sec A 1 2. After being open for a long time, the switch closes at t = 0. Find V1(t) for t > 0. t=0 2kΩ 20V + + V1 4kΩ 2µF + VC Step 1: (Redraw circuit at t=0- and solve for VC. Capacitor acts as an open since it has been a long time) 2kΩ There is no source connected between Vc so V1 =VC=0 + + + V1 20V VC 4kΩ Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Capacitor acts as a voltage source since the voltage across capacitor has to remain the same.) 2kΩ 20V + 4kΩ + V1 0V + + VC V1 = 0V ∞ and solve for unknown variable. Capacitor acts as an open Step 3: Final Value(Redraw circuit at t=∞ since it has sat for a long time in this position.) 2kΩ V1 is found by a voltage divider: 20 ⋅ 4k 80k 40 + + V1 = = = V 2 k + 4k 6 k 3 V1 20V + V 4kΩ C To find Req the capacitor is removed from the final circuit(same circuit) to find path from top to bottom of capacitor. Independent sources are removed and the equivalent resistance is found: 1 16m τ = R eq ⋅ C = ( 4k 2k ) ⋅ 2µ = 1 1 ⋅ 2µ = sec 6 4k + 2k Step 4: Plug values into general equation: 40 40 −6t /16 m sec 40 V1 (t ) = + 0 − e V= 1 − e−6t /16 m sec V 3 3 3 ( 2 ) 3. After being open for a long time, the switch closes at t = 0. Find i1 (t) for t > 0. 2ki1 + 8V 4mA 4kΩ 2µH 1kΩ i1 4kΩ Step 1: (Redraw circuit at t=0- and solve for iL. Inductor acts as a wire since it has sat for a long time) iL 2ki1 + 8V When the circuit contains a dependent source, an extra step is needed to determine the value of the dependent variable: + 4kΩ 4mA 1kΩ - 8 = 1mA 8k + Taking a current summation at the top node can be used to find iL gives: iL = 2k ⋅ i1 + 4m ⇒ iL = 2k ⋅1m + 4m = 2.004 A i1 - i1 4kΩ Note: −i1 ⋅ 4k + 8 − i1 ⋅ 4k = 0 ⇒ i1 = Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Inductor acts as a current source since the current in the inductor has to remain the same.) V1 Solve the circuit for i1. Mesh currents or node-voltage can be used. Node-voltage 2ki1 method: + 8V 4kΩ 2.004A 4mA • With dependent sources: solve for i1 1kΩ 4kΩ dependent variable in terms of either the mesh current or the node-voltage. V i1 = 1 4k V −8 −2k ⋅ i1 − 4m + 2.004 + 1 + i1 = 0 4k V V − 8 V1 −2k ⋅ 1 − 4m + 2.004 + 1 + = 0 4k 4 k 4k 1 8 −2 k 1 V1 + + = +4m − 2.004 + ⇒ 4k 4 k 4k 4 k The desired variable is i1: 4k V1 = −1.998 ⋅ ≈ 4V −2.002k V 4 i1 = 1 = = 1mA 4k 4k 3 Step 3: Final Value(Redraw circuit at t=∞ ∞ and solve for unknown variable. Inductor acts as a wire since it has sat for a long time in this position.) This wire puts 0V across 4k ohm resistor so: i1=0 2ki1 + 8V 4kΩ 4mA 1kΩ i1 4kΩ V1 2ki1 + Vtest 4mA - 1kΩ + 8V 4kΩ Itest= 1A i1 4kΩ To find Req the inductor is removed from the final circuit to find path from top of inductor to bottom of inductor(Thevenin Resistance): V Using a test source: Rth = test I test Setting Itest=1A means that only Vtest needs to be found. Vtest = V1 V i1 = 1 4k V −8 −2k ⋅ i1 − 4m − 1 + 1 + i1 = 0 4k V V − 8 V1 −2k ⋅ 1 − 4m − 1 + 1 + = 0 4k 4k 4k 1 8 −2k 1 V1 + + = +4m + 1 + 4k 4 k 4k 4 k Vtest = −2V Rth = 4k V1 = 1.006 ⋅ ≈ −2V −2.002k ⇒ Vtest −2V = = 2Ω I test 1A L 2µ = = 1µ sec R eq 2 Step 4: Plug values into general equation: i1 (t ) = 0 + [1m − 0] e−t /1µ sec A = 1me−t /1µ sec A τ= 4 4. After being open for a long time, the switch closes at t = 0. Find i1 (t) for t > 0. 2mF 5kΩ ib 99i b t=0 3kΩ 10V + i1 10kΩ 2kΩ Step 1: (Redraw circuit at t=0- and solve for VC. Capacitor acts as an open since it has been a long time) + VC 10V + Solving for the dependent variable: ib=0 which opens the dependent source => 99ib=0 Taking a V-loop to get Vc value(Be CarefulIt is not 0 when there is a path with a V src.) +0 + 10 + VC − 0 = 0 ⇒ VC = −10V 5kΩ ib 99i b 3kΩ 10kΩ i1 2kΩ Step 2: Initial Value (Redraw circuit at t=0+ and solve for unknown variable. Capacitor acts as a voltage source since the voltage across capacitor has to remain the same.) Using node-voltage to solve this circuit to find i1: First find dependent variable in terms of nodevoltage variable, V1. ib V1 − ( −10) − ( −( −10)) V1 + 5kΩ i = = + b -10V VC 10k 10k 99i b Next, take current summation equation at V1 node: 3kΩ V1 V1 (V1 − V2 ) 10V + + − =0 V2 10k 3k 2k Current summation equation at V2 node: i1 10kΩ 2kΩ (V2 − V1 ) − 99 ⋅ V1 − ( 0 − V2 ) = 0 2k 10k 5k V1 Using the first equation and solving for V1: Plugging into the second equation: V1 V1 V1 V2 V2 −15V2 1 1 −15V2 V2 + − = − − 99 ⋅ =0 + 10k 3k 2k 2k 2 k 4 2k 10k 4 5k 20 30 V2 6 1 15 99(15) 1 V1 + − V2 + + + = 0 ⇒ V2 = 0 = 60 k 60 k 60 k 2 k 40k 5k 2 k 8k V1 = V2 60k −15V2 = 2k −4 2 V1 = 0 i1 = 0 5 Step 3: Final Value(Redraw circuit at t=∞ ∞ and solve for unknown variable. Capacitor acts as an open since it has sat for a long time in this position.) ib=0 + VC 99i b Since i1 is always the same => i1(t)=0. i1 10kΩ 5. i1=0 3kΩ 10V + this means that 99ib source becomes open. 5kΩ ib 2kΩ Using superposition, derive an expression for V that contains no circuit quantities other than is , R1, R 2 , R 3 , R 4 , α, or Vs . αi x is ix + Vs R1 R4 + V R2 R3 Step 1: is=0(off), Vs=on αi x V1 i2 = ix + Vs R1 i1 + V Using mesh currents: i1 = −α ix + R2 ix i2 + R3 + R4 Using a voltage loop and then substituting above eq: ( +i1 − i2 ) R3 + VS − i2 R4 = 0 ( −α ix − ix ) R3 + VS − ix R4 = 0 ix ( +α R3 + R3 + R4 ) = VS ix = VS ( +α R3 + R3 + R4 ) V = α ix R2 = Step 1: is=on, Vs=0(off) αi x αVS R2 ( +α R3 + R3 + R4 ) Using mesh currents: i1 = is i2 = −α ix + is i1 + V R1 R2 ix i2 i3 + i3 = ix + R4 R3 + 6 Using a voltage loop and then substituting above eq: ( +i2 − i3 ) R3 − i3 R4 = 0 ( −α ix − ix ) R3 − ix R4 = 0 ix ( +α R3 + R3 + R4 ) = 0 ix = 0 i2 = 0 V = ( +i1 − i2 ) R2 = is R2 The total V is the sum of both solutions: αVS R2 V = is R2 + ( +α R3 + R3 + R4 ) 6. After being closed for a long time, the switch opens at t=0. a) Calculate the energy stored on the inductor as t -> ∞. b) Write a numerical expression for v(t) for t> 0. 20kΩ t=0 300 mV + 30kΩ + v(t) 150 mH 7 3kΩ 7. After being open for a long time, the switch closes at t=0. a) Write an expression for vc(t=0+). b) Write an expression for vc(t>0) in terms of is , R1, R 2 , R 3 , and C. t=0 R1 is C vC + R3 8 R2 3kΩ Use the circuit below for both problem 8 and 9. 8. Calculate the value of RL that would absorb maximum power. + 3 mA 24kΩ 2kΩ 22kΩ 9. Calculate that value of maximum power RL could absorb. 9 15V RL 10. Using superposition, derive an expression for i that contains no circuit quantities other than is , R1, R 2 , R 3 , R 4 , α, or Vs . is R1 R2 i αi x + R4 R3 10 ix vs +