UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:
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1 UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT ECE 1270 HOMEWORK #6 Solution Summer After being closed a long time, the switch opens at t = 0. Find i(t) 1 for t > 0. t = 0 10kΩ 60µ A 10kΩ i 1 3mH Step 1: (Redraw circuit at t=0 - and solve for i L. Inductor acts as a wire since it has sat for a long time) 60µ A 10kΩ i 1 10kΩ i L This circuit is a current divider: 60µ 10k il = = 30µ A 10k 10k 60µ A 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: 10kΩ i 1 30µ A i 1 = 30µ 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 i circuit: L i 1 =0A 60µ A 10kΩ i 1 60µ A 10kΩ i 1 10kΩ To find R eq the inductor is removed from the final circuit to find path from top of inductor to bottom of inductor: τ = L 3m 150nsec R = 10k 10k = eq Step 4: Plug values into general equation: i () t 0 30µ 0e t n = A= 30µ e t n A 1 [ ] /150 sec /150 sec :31:50 1/39 Unit3_Examples7(first).pdf (#1, 1/10)
2 2. After being open for a long time, the switch closes at t = 0. Find V(t) 1 for t > 0. t = 0 2kΩ 20V 4kΩ V 1 2µ F V C Step 1: (Redraw circuit at t=0 - and solve for V C. Capacitor acts as an open since it has been a long time) 2kΩ There is no source connected between V c so V 1 =V C =0 20V 4kΩ V 1 V C 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 V 1 0V V C V 1 = 0 V 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.) 2kΩ V1 is found by a voltage divider: 20 4k 80k 40 V1 = = = V 20V 4kΩ V 2k 4k 6k 3 1 V C To find R eq 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 = Req C = ( 4k 2k) 2µ = # 2µ sec 1 1 $ = % 4k 2k & 6 Step 4: Plug values into general equation: 40 ' 40( 6/16 t msec 40 6/16 t msec V1 () t = 0 e V ( 1 e ) V 3 ) = 3 *, :31:50 2/39 Unit3_Examples7(first).pdf (2/10)
3 3. After being open for a long time, the switch closes at t = 0. Find i(t) 1 for t > 0. 2ki 1 4mA 2µ H 8V 4kΩ 1kΩ 4kΩ i 1 Step 1: (Redraw circuit at t=0 - and solve for i L. Inductor acts as a wire since it has sat for a long time) When the circuit contains a dependent 2ki 1 1kΩ i source, an extra step is needed to determine L 8V the value of the dependent variable: 4mA 4kΩ 8 i kΩ Note: i1 4k 8 i1 4k = 0! i1 = = 1mA 1 8k Taking a current summation at the top node can be used to find i L gives: i = 2k i 4m! i = 2k 1m 4m= 2.004A L 1 L 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.) V 1 Solve the circuit for i 1. Mesh currents or node-voltage can be used. Node-voltage 2ki 1 8V method: 4mA 2.004A 4kΩ With dependent sources: solve for 1kΩ 4kΩ i 1 dependent variable in terms of either the mesh current or the node-voltage. " V1 # i1 = $ % & 4k ' " V1 8 # 2k i1 4m $ % i1 = 0 & 4k ' " V1 # " V1 8 # " V1 # 2k $ % 4m $ % $ % = 0 & 4k ' & 4k ' & 4k ' " 2k 1 1 # 8 " 4k # V1 $ % = 4m 2.004! V1 = $ % 4V & 4k 4k 4k ' 4k & 2.002k ' " V1 # 4 The desired variable is i 1 : i1 = $ % = = 1mA & 4k ' 4k :31:51 3/39 Unit3_Examples7(first).pdf (3/10)
4 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.) 2ki 1 4mA 8V 4kΩ This wire puts 0V across 4k ohm resistor so: i 1 =0 1kΩ 4kΩ i 1 V 1 2ki 1 4mA 8V V test I 4kΩ test = 1A 1kΩ - 4kΩ i 1 To find R eq the inductor is removed from the final circuit to find path from top of inductor to bottom of inductor(thevenin Resistance): Vtest Using a test source: Rth = Itest Setting I test =1A means that only V test needs to be found. V = V test 1! V1 " i1 = # $ % 4k &! V1 8 " 2k i1 4m 1 # $ i1 = 0 % 4k &! V1 "! V1 8 "! V1 " 2k # $ 4m 1 # $ # $ = 0 % 4k & % 4k & % 4k &! 2k 1 1 " 8! 4k " V1 # $ = 4m 1 ' V1 = # $ 2V % 4k 4k 4k & 4k % 2.002k & V = 2V R test th Vtest 2V = = = 2Ω I 1A test L 2µ τ = = = 1µ sec Req 2 Step 4: Plug values into general equation: () 0 /1 sec /1 sec 1 [ 1 0 t µ ] 1 t µ i t = m e A= me A :31:51 4/39 Unit3_Examples7(first).pdf (4/10)
5 4. After being open for a long time, the switch closes at t = 0. Find i(t) 1 for t > 0. 2mF 10V i b 3kΩ 99i b 5kΩ t = 0 10kΩ i 1 2kΩ Step 1: (Redraw circuit at t=0 - and solve for V C. Capacitor acts as an open since it has been a long time) 10V V C 10kΩ i b 3kΩ 99i b i 1 2kΩ 5kΩ Solving for the dependent variable: ib=0 which opens the dependent source => 99ib=0 Taking a V-loop to get V c value(be Careful- It is not 0 when there is a path with a V src.) 0 10 V 0= 0! V = 10V C C 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 i 1 : First find dependent variable in terms of nodevoltage variable, V 1. V C 10V 10kΩ i b -10V 3kΩ 99i b 2kΩ 5kΩ V 1 Using the first equation and solving for V 1 : V1 V1 V1 V2 = 10k 3k 2k 2k " # V2 V1 $ % = & 60k 60k 60k ' 2k V2 " 60k # 15V2 V1 = $ % = 2k & 4 ' 2 i 1 V 2 V1 ( 10) ( ( 10)) V1 ib = = 10k 10k Next, take current summation equation at V 1 node: V1 V ( V 1 1 V2) = 0 10k 3k 2k Current summation equation at V 2 node: ( V2 V1) V1 ( 0 V2) 99 = 0 2k 10k 5k Plugging into the second equation: V2 " 15V2 #" 1 # 1 " 15V2 # V2 $ % $ % 99 $ % = 0 2k & 4 '& 2k ' 10k & 4 ' 5k " (15) 1 # V2 $ % = 0! V2 = 0 & 2k 8k 40k 5k ' V = i = :31:51 5/39 Unit3_Examples7(first).pdf (5/10)
6 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.) 10V V C i b 3kΩ 99i b 5kΩ i b =0 this means that 99ib source becomes open. i 1 =0 Since i 1 is always the same => i 1 (t)=0. 10kΩ i 1 2kΩ 5. Using superposition, derive an expression for V that contains no circuit quantities other than i,r,r s 1 2,R,R 3 4, α,orvs. α i x i s V R 1 V s R 2 R3 i x R 4 i s Step 1: is=0(off), Vs=on V α i x Step 1: is=on, Vs=0(off) α i x V 1 i x R 1 V s i 1 i 2 R 2 R3 i x R 1 i 1 i 2 i 3 V R 2 R3 R 4 Using mesh currents: i = αi 6 i 1 2 = i x x Using a voltage loop and then substituting above eq: ( ) S ( α ) ( α ) i i R V i R = i i R V i R = 0 x x x 3 S x 4 i R R R = V i R V S x = V ( α R R R ) S 2 = αixr2 = ( α R R R ) S αvr Using mesh currents: i = i i i s = αi = i x x :31:51 6/39 Unit3_Examples7(first).pdf (6/10)
7 Using a voltage loop and then substituting above eq: i i R i R = 0 ( 2 3) ( αix ix) R3 ixr4 i ( α R R R ) = 0 x x 2 = 0 i = 0 i = ( ) V = i i R = i R s 2 The total V is the sum of both solutions: αvr S 2 V = isr2 α R R R ( ) 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 = mv 30kΩ v(t) 3kΩ 150 mh :31:52 7/39 Unit3_Examples7(first).pdf (7/10)
8 7. After being open for a long time, the switch closes at t=0. a) Write an expression for v c (t=0 ). b) Write an expression for v c (t>0) in terms of i,r,r s 1 2,R,and 3 C. R 1 C i s t = 0 R 2 v C R :31:52 8/39 Unit3_Examples7(first).pdf (8/10)
9 Use the circuit below for both problem 8 and Calculate the value of R L that would absorb maximum power. 3kΩ 3 ma 15V 24kΩ R L 2kΩ 22kΩ 9. Calculate that value of maximum power R L could absorb :31:53 9/39 Unit3_Examples7(first).pdf (9/10)
10 10. Using superposition, derive an expression for i that contains no circuit quantities other than i,r,r s 1 2,R,R 3 4, α,orvs. R 1 i s R 4 R 2 i α i x R 3 i x v s :31:53 10/39 Unit3_Examples7(first).pdf (10/10)
11 1270 Sp 06 HOMEWORK # µh t = 0 i(t) V After being closed for a long time, the switch closes at t = 0. Calculate the energy stored on the inductor as t. 2. For the circuit in problem 1, find i(t) for t > C t = 0 v C R 1 i s R 2 R 3 After being open for a long time, the switch closes at t = 0. Write an expression for v c (t 0) in terms of R 1, R 2, R 3, i s, and C :31:53 11/39 Unit3_Examples1.pdf (#2, 1/5)
12 4. 6 A R L 80 V a) Calculate the value of R L that would absorb maximum power. b) Calculate that value of maximum power R L could absorb. 5. i s i x R 1 i x v 1 v s R 2 Using superposition, derive an expression for v 1 that contains no circuit quantities other than i s, v S, R 1, R 2, and, where > :31:53 12/39 Unit3_Examples1.pdf (2/5)
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25 UNIVERSITY OF UTAH ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT ECE 1270 HOMEWORK #6 Solution Summer k Ω 45mA 200Ω 1 t=0 i(t) 200uH 2.4kΩ 2 After being open for a long time, the switch closes at t=0. a) Calculate the energy stored on the inductor as t. b) Write a numerical expression for i(t) for t> :31:59 25/39 Unit3_Examples4.pdf (#5, 1/11)
26 :31:59 26/39 Unit3_Examples4.pdf (2/11)
27 :31:59 27/39 Unit3_Examples4.pdf (3/11)
28 :31:59 28/39 Unit3_Examples4.pdf (4/11)
29 2. V C - C R1 Vs R3 t=0 R2 After being open for a long time, the switch becomes closed at t=0. a) Write an expression for V c (t=0 ). b) Write an expression for V c (t>0) in terms of R1, R2, R3, Vs, and C :31:59 29/39 Unit3_Examples4.pdf (5/11)
30 :31:59 30/39 Unit3_Examples4.pdf (6/11)
31 :32:00 31/39 Unit3_Examples4.pdf (7/11)
32 3. 2k Ω 1.2k Ω 20mA RL 10V 1.8k Ω a) Calculate the value of RL that would absorb maximum power. b) Calculate that value of maximum power RL could absorb :32:00 32/39 Unit3_Examples4.pdf (8/11)
33 :32:00 33/39 Unit3_Examples4.pdf (9/11)
34 4. R1 R2 Vx Is Vs Vx - i R3 Using superposition, derive an expression for i that contains no circuit quantities other than Is, Vs, R1, R2, R3, and, where > :32:00 34/39 Unit3_Examples4.pdf (10/11)
35 :32:00 35/39 Unit3_Examples4.pdf (11/11)
36 :32:00 36/39 Unit3_Examples5.pdf (#6, 1/4)
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UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:
UNIERSITY OF UTH ELECTRICL & COMPUTER ENGINEERING DEPRTMENT ECE 70 HOMEWORK #6 Soluton Summer 009. fter beng closed a long tme, the swtch opens at t = 0. Fnd (t) for t > 0. t = 0 0kΩ 0kΩ 3mH Step : (Redraw
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