Chap_8_Solut_Dorf

Chap_8_Solut_Dorf - Chapter 8 The Complete Response of RL...

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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 == Ω = = . Finally, 2.5 ( ) ( (0) ) 2 V for 0 oc oc tt vt v v v e e t −− =+ = + > Ex. 8.3-2 Before the switch closes:
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After the switch closes: Therefore 26 8s o 0 . 7 5 0.25 8 t R τ == Ω = = s . Finally, 1.33 11 ( ) ( (0) ) A for 0 41 2 sc sc tt it i i i e e t −− =+ > Ex. 8.4-1 ( )( ) () 36 3 500 210 110 2 10 s ( ) 5 1.5 5 V t c vt 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 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 e L R ⇒= = We require that 1.5 V at = 10 ms = 0.01 s R t = . That is
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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 vt v Ae v v t −− =∞ Ω =+ 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 = Ex. 8 6-2 0 no sources (0 ) (0 ) 0 <∴ = v = 1 57 21 0( 1 0 ) / t tR C v 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 ==+⇒= , .1 s t >=
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50( .1) 50(.1) 50( .1) () ( .1 ) where (.1) 10(1 ) 9.93 V ( ) 9.93 V t t vt v e ve e −− = =− = = Ex. 8.7-1 t1 0 s i n 2 0 V s = Apply KVL: ( ) 10sin 20 10 .01 0 10 100sin 20 dvt tv t dt t dt ⎛⎞ −+ + = ⎜⎟ ⎝⎠ ⇒+= Natural Response: 10 ( ) where ( ) tt nt vt A e RC e τ == n = Forced Response: 12 try ( ) cos20 sin 20 f B t B t = + Plugging ( ) into the differential equation and equating like terms yields: f 1 2 20 20 cos20 10 10 sin 20 100 sin 20 B tB t B t B t ++ = t 21 1 2 20 10 0 and 20 10 100 BB B B += = 4 and 2 = −= Complete Response: 10 ( ) ( ) ( ) 4cos20 2 sin 20 t nf v t Ae t t =+ + Now (0 ) (0 ) 0 4 4 vv A A = = 10 ( ) 4 4cos 20 2sin 20 V t e t t +
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Problems Section 8.3: The Response of a First Order Circuit to a Constant Input P8.3-1 Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the initial capacitor voltage, v (0). By voltage division () 6 01 2 666 v == 4 V + + Next, consider the circuit after the switch closes. The closed switch is modeled as a short circuit. We need to find the Thevenin equivalent of the part of the circuit connected to the capacitor. Here’s the circuit used to calculate the open circuit voltage, V oc . oc 6 12 6 V 66 V + Here is the circuit that is used to determine R t . A short circuit has replaced the closed switch. Independent sources are set to zero when calculating R t , so the voltage source has been replaced by a short circuit. ( )( ) t 3 R = + Then ( ) t 3 0.25 0.75 s RC τ = Finally, () ( ) oc oc 1.33 / 0 6 2 V for > 0 t t vt V v V e e t =+ =
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P8.3-2 Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit.
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Chap_8_Solut_Dorf - Chapter 8 The Complete Response of RL...

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