UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:

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1 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 crcut at t=0 - and solve for L. Inductor acts as a wre snce t has sat for a long tme) 0kΩ 0kΩ L Ths crcut s a current dvder: 0k L = = 30µ 0k 0k Step : Intal alue (Redraw crcut at t=0 and solve for unknown varable. Inductor acts as a current source snce the current n the nductor has to reman the same.) 0kΩ Only one current n the branch: 0kΩ 30µ = 30µ Step 3: Fnal alue(redraw crcut at t=! and solve for unknown varable. Inductor acts as a wre snce t has sat for a long tme n ths poston.) 0kΩ There are no sources connected to the fnal crcut: L =0 0kΩ 0kΩ 0kΩ To fnd R eq the nductor s removed from the fnal crcut to fnd path from top of nductor to bottom of nductor: τ = L 3m 50nsec R = 0k 0k = eq Step 4: Plug values nto general equaton: () t 0 30µ 0e t n = = 30µ e t n [ ] /50 sec /50 sec :54:54 /0 Unt3_Eamples7(frst).pdf (#, /0)

2 . fter beng open for a long tme, the swtch closes at t = 0. Fnd (t) for t > 0. t = 0 0 µ F C Step : (Redraw crcut at t=0 - and solve for C. Capactor acts as an open snce t has been a long tme) There s no source connected between c so = C =0 0 C Step : Intal alue (Redraw crcut at t=0 and solve for unknown varable. Capactor acts as a voltage source snce the voltage across capactor has to reman the same.) 0 0 C = 0 Step 3: Fnal alue(redraw crcut at t=! and solve for unknown varable. Capactor acts as an open snce t has sat for a long tme n ths poston.) s found by a voltage dvder: 0 4k 80k 40 = = = 0 k 4k 6k 3 C To fnd R eq the capactor s removed from the fnal crcut(same crcut) to fnd path from top to bottom of capactor. Independent sources are removed and the equvalent resstance s found:! " 6m τ = Req C = ( 4k k) µ = # µ sec $ = % 4k k & 6 Step 4: Plug values nto general equaton: 40 ' 40( 6/6 t msec 40 6/6 t msec () t = 0 e ( e ) 3 ) = 3 *, :54:54 /0 Unt3_Eamples7(frst).pdf (/0)

3 3. fter beng open for a long tme, the swtch closes at t = 0. Fnd (t) for t > 0. k 4m µ H 8 kω Step : (Redraw crcut at t=0 - and solve for L. Inductor acts as a wre snce t has sat for a long tme) When the crcut contans a dependent k kω source, an etra step s needed to determne L 8 the value of the dependent varable: 4m Note: 4k 8 4k = 0! = = m 8k Takng a current summaton at the top node can be used to fnd L gves: = k 4m! = k m 4m=.004 L L Step : Intal alue (Redraw crcut at t=0 and solve for unknown varable. Inductor acts as a current source snce the current n the nductor has to reman the same.) Solve the crcut for. Mesh currents or node-voltage can be used. Node-voltage k 8 method: 4m.004 Wth dependent sources: solve for kω dependent varable n terms of ether the mesh current or the node-voltage. " # = $ % & 4k ' " 8 # k 4m.004 $ % = 0 & 4k ' " # " 8 # " # k $ % 4m.004 $ % $ % = 0 & 4k ' & 4k ' & 4k ' " k # 8 " 4k # $ % = 4m.004! =.998 $ % 4 & 4k 4k 4k ' 4k &.00k ' " # 4 The desred varable s : = $ % = = m & 4k ' 4k :54:55 3/0 Unt3_Eamples7(frst).pdf (3/0)

4 Step 3: Fnal alue(redraw crcut at t=! and solve for unknown varable. Inductor acts as a wre snce t has sat for a long tme n ths poston.) k 4m 8 Ths wre puts 0 across 4k ohm resstor so: =0 kω k 4m 8 test I test = kω - To fnd R eq the nductor s removed from the fnal crcut to fnd path from top of nductor to bottom of nductor(thevenn Resstance): test Usng a test source: Rth = Itest Settng I test = means that only test needs to be found. = test! " = # $ % 4k &! 8 " k 4m # $ = 0 % 4k &! "! 8 "! " k # $ 4m # $ # $ = 0 % 4k & % 4k & % 4k &! k " 8! 4k " # $ = 4m ' =.006 # $ % 4k 4k 4k & 4k %.00k & = R test th test = = = Ω I test L µ τ = = = µ sec Req Step 4: Plug values nto general equaton: () 0 / sec / sec [ 0 t µ ] t µ t = m e = me :54:55 4/0 Unt3_Eamples7(frst).pdf (4/0)

5 4. fter beng open for a long tme, the swtch closes at t = 0. Fnd (t) for t > 0. mf 0 b 99 b 5kΩ t = 0 0kΩ Step : (Redraw crcut at t=0 - and solve for C. Capactor acts as an open snce t has been a long tme) 0 C 0kΩ b 99 b 5kΩ Solvng for the dependent varable: b=0 whch opens the dependent source => 99b=0 Takng a -loop to get c value(be Careful- It s not 0 when there s a path wth a src.) 0 0 0= 0! = 0 C C Step : Intal alue (Redraw crcut at t=0 and solve for unknown varable. Capactor acts as a voltage source snce the voltage across capactor has to reman the same.) Usng node-voltage to solve ths crcut to fnd : Frst fnd dependent varable n terms of nodevoltage varable,. C 0 0kΩ b b 5kΩ Usng the frst equaton and solvng for : = 0k 3k k k " # $ % = & 60k 60k 60k ' k " 60k # 5 = $ % = k & 4 ' ( 0) ( ( 0)) b = = 0k 0k Net, take current summaton equaton at node: ( ) = 0 0k 3k k Current summaton equaton at node: ( ) ( 0 ) 99 = 0 k 0k 5k Pluggng nto the second equaton: " 5 #" # " 5 # $ % $ % 99 $ % = 0 k & 4 '& k ' 0k & 4 ' 5k " 5 99(5) # $ % = 0! = 0 & k 8k 40k 5k ' = 0 = :54:55 5/0 Unt3_Eamples7(frst).pdf (5/0)

6 Step 3: Fnal alue(redraw crcut at t=! and solve for unknown varable. Capactor acts as an open snce t has sat for a long tme n ths poston.) 0 C b 99 b 5kΩ b =0 ths means that 99b source becomes open. =0 Snce s always the same => (t)=0. 0kΩ 5. Usng superposton, derve an epresson for that contans no crcut quanttes other than,r,r s,r,r 3 4, α,ors. α s R s R R3 R 4 s Step : s=0(off), s=on α Step : s=on, s=0(off) α R s R R3 R 3 R R3 R 4 Usng mesh currents: = α 6 = Usng a voltage loop and then substtutng above eq: ( ) S ( α ) ( α ) R R = R R = 0 3 S 4 R R R = R S = ( α R R R ) S = αr = ( α R R R ) S αr Usng mesh currents: = 3 s = α = :54:55 6/0 Unt3_Eamples7(frst).pdf (6/0)

7 Usng a voltage loop and then substtutng above eq: R R = 0 ( 3) ( α ) R3 R4 ( α R R R ) = 0 = 0 = 0 = ( ) = R = R s The total s the sum of both solutons: αr S = sr α R R R ( ) fter beng closed for a long tme, the swtch opens at t=0. a) Calculate the energy stored on the nductor as t ->!. b) Wrte a numercal epresson for v(t) for t> 0. 0kΩ t = m 30kΩ v(t) 50 mh :54:56 7/0 Unt3_Eamples7(frst).pdf (7/0)

8 7. fter beng open for a long tme, the swtch closes at t=0. a) Wrte an epresson for v c (t=0 ). b) Wrte an epresson for v c (t>0) n terms of,r,r s,r,and 3 C. R C s t = 0 R v C R :54:56 8/0 Unt3_Eamples7(frst).pdf (8/0)

9 Use the crcut below for both problem 8 and Calculate the value of R L that would absorb mamum power. 3 m 5 R L 9. Calculate that value of mamum power R L could absorb :54:57 9/0 Unt3_Eamples7(frst).pdf (9/0)

10 0. Usng superposton, derve an epresson for that contans no crcut quanttes other than,r,r s,r,r 3 4, α,ors. R s R 4 R α R 3 v s :54:57 0/0 Unt3_Eamples7(frst).pdf (0/0)