chapter 08 - Chapter 8 The Complete Response of RL and RC...

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Chapter 8 – The Complete Response of RL and RC Circuit Exercises Ex. 8.3-1 Before the switch closes: After the switch closes: Therefore () 2 8 so 8 0.05 0.4 s 0.25 t R τ == = = . Finally, 2.5 ( ) ( (0) ) 2 V for 0 oc oc tt v v v e e t −− =+ = + > vt Ex. 8.3-2 Before the switch closes: After the switch closes: 8-1
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Therefore 26 8s o 0 . 7 5 0.25 8 t R τ == = = s . Finally, 1.33 11 ( ) ( (0) ) A for 0 41 2 sc sc tt i i i e e t −− =+ > it Ex. 8.3-3 At steady-state, immediately before t = 0: () 10 12 0 0 . 1 A 10 40 16 40||10 i   ++  After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be 22 20 1 so 0.3 A, 40 , s 40 2 Finally: ( ) (0.1 0.3) 0.3 0.3 0.2 A sc t t L IR R e e = = = =− +=− 8-2
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Ex. 8.3-4 At steady-state, immediately before t = 0 After t = 0, we have: 6 so 12 V, 200 , (200)(20 10 ) 4 ms oc t t VR R C τ == = =⋅ = 4 Finally: ( ) (12 12) 12 12 V t vt e =− += Ex. 8.3-5 Immediately before ti 0, (0) 0. After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the inductor current: 25 5 0.2 A, 45 , 45 9 sc t th L IR R = =Ω=== So it 1.8 () 0 .2 (1 ) A t e Now that we have the inductor current, we can calculate v ( t ): 1.8 1.8 1.8 ( ) 40 ( ) 25 ( ) 8(1 ) 5(1.8) 8 V f o r 0 tt t d dt ee et −− =+ + > 8-3
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Ex. 8.3-6 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 = 93.75 mA, 640 , .1 1 s 640 6400 sc t t IR L R τ =Ω === 6400 ( ) 406.25 93.75 mA t it e =+ Finally: 6400 6400 6400 ( ) 400 ( ) 0.1 ( ) 400 (.40625 .09375) 0.1( 6400) (0.40625 ) 37.5 97.5 V tt t d vt e e dt e −− = + + = Ex. 8.4-1 ( )( ) () 36 3 500 210 110 2 10 s ( ) 5 1.5 5 V t c e × 0.5 (0.001) 5 3.5 2.88 V c ve =− = So ( ) will be equal to at 1 ms if 2.88 V. cT v t v T == 8-4
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Ex. 8.4-2 () 500 500 (0) 1 mA, 10 mA 10 9 mA 500 , 500 ( ) = 300 = 3 2.7 e V Ls c t L L t t L RL iI it e L R vt τ == ⇒= =Ω = We require that 1.5 V at = 10 ms = 0.01 s R t = . That is 500 (0.01) 5 1.5 3 2.7 8.5 H 0.588 L eL =− = = Ex. 8.6-1 1 0 tt << / /(1)(.1) 10 ( ) () + w h e r e = ( 1 A ) ( 1 ) = 1 V 1 1 tRC v Ae v −− =∞ =+ 10 Now (0 ) (0 ) 0 1 1 1 V t vv A A e −+ = +⇒= ∴= 1 > .5 10( .5) (1)(.1) 11 10(0.5) 0.5 s, ( ) (0.5) Now (0.5) 1 0.993 V t t tv t v t e v e ve = −= 10( 0.5) ( ) 0.993 V t e = 8-5
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Ex. 8 6-2 0 no sources (0 ) (0 ) 0 tv −+ <∴ = v = 1 0 tt << 57 21 0( 1 0 ) / () ( ) t tR C vt v Ae v × =+ = + where for t = (steady-state) capacitor becomes an open v ( ) = 10 V 50 50 10 Now (0) 0 10 10 ( ) 10(1 ) V t t vA A e ==+⇒= =− 11 , .1 s tt t >= 50( .1) 50(.1) 50( .1) ( .1 ) where (.1) 10(1 ) 9.93 V ( ) 9.93 V t t v e ve e −− = = = Ex. 8.6-3 For t < 0 i = 0. For 0 < t < 0.2 s 10 10 KCL: 5 / 2 0 10 50 also: 0.2 5 (0) 0 5 5 so we have ( ) 5 (1 ) A t t vi di i di dt v dt it iA A e += = ∴= + For t > 0.2 s 10( .2) (0.2) = 4.32 A ( ) 4.32 A t ii t e 8-6
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Ex. 8.7-1 () t 10sin 20 V KVL: 10sin 20 10 .01 0 10 100sin 20 s vt dvt tv dt t dt =  −+ +   ⇒+= t = 10 12 1 Natural response: ( ) where ( ) Forced response: try ( ) cos20 sin 20 Plugging ( ) into the differential equation and equating like terms yields: 4 & tt nt f f A e RC A e B t B t B τ −− == =+ =− n = 2 10 10 2 Comp le te response : ( ) 4cos20 2 sin 20 Now (0 ) (0 ) 0 4 4 ( ) 4 4cos 20 2sin 20 V nf t t B v t Ae t t vv AA e t t = + = = ∴= + Ex. 8.7-2 ( ) () () 5 5 5 10 for 0 KCL at top node: 10 /10 0 Now 0.1 100 1000 t s t t it e t ei t di t di t e dt dt => −++ = =⇒ + = 100 5 55 5 1 Natural response: ( ) where ( ) Forced response: try ( )
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.