Problem 5.37 Pror to t =, capactor C 1 n the crcut of Fg. P5.37 was uncharged. For I = 5 ma, R 1 = 2 kω, = 5 kω, C 1 = 3 µf, and C 2 = 6 µf, determne: (a) The equvalent crcut nvolvng the capactors for t. Specfy υ 1 () and υ 2 (). (b) (t) for t. (c) υ 1 (t) and υ 2 (t) for t. Soluton: 1 I R 1 C 1 υ 1 t = 2 C 2 υ 2 (a) At t = I R 1 C 2 υ 2 ( ) = I R 1 (b) At t > 2 C 1 υ 1 C 2 υ 2 C eq υ eq Fgure P5.37 (a) At t =, υ 2 ( ) = I R 1 = 5 1 3 2 1 3 = 1 V. υ 1 ( ) = (Gven). At t =, crcut s as shown n Fg. P5.37(b), wth: C eq = C 1C 2 = 3 6 µf = 2 µf, C 1 +C 2 3+6
and υ eq () = υ 2 () υ 1 () = 1 = 1 V, τ = C eq = 5 1 4 2 1 6 =.1 s. (b) For t, (c) υ eq (t) = υ eq ( )+[υ eq () υ eq ( )]e t/τ = +(1 )e 1t = 1e 1t. (t) = C eq d dt υ eq = 2 1 6 d dt (1e 1t ) =.2e 1t (ma). υ 1 (t) = υ 1 ()+ 1 t (t) dt C 1 = 1 3 1 6 t υ 2 (t) = υ 2 ()+ 1 t (t) dt C 2 = 1+ 1 6 1 6.2e 1t 1 3 dt = 6.7(1 e 1t ) (V), for t, t.2e 1t 1 3 dt = [6.7+3.3e 1t ] (V), for t.
Problem 5.38 The swtch n the crcut of Fg. P5.38 had been closed for a long tme before t was opened at t =. Gven that V s = 1 V, R 1 = 2 kω, = 1 kω, C 1 = 6 µf, and C 2 = 12 µf, determne (t) for t. Soluton: (a) V s + _ R 1 t = C 1 C 2 (b) At t = V s R 1 + _ C 1 υ 1 C 2 υ 2 (c) At t > C 1 v 1 C 2 υ 2 (d) C eq υ eq Fgure P5.38 At t =, the crcut s as depcted n Fg. P5.38(b). ( ) = υ 1 ( ) = υ 2 ( ) = V s = 1 V. At t >, the crcut becomes as shown n Fg. P5.38(c). Hence, ( ) ( ) C1 C 2 6 12 τ = = 1 5 1 6 =.4 s, C 1 +C 2 6+12 but C eq = C 1C 2 C 1 +C 2 = 4 µf, υ eq ( ) = υ 2 ( ) υ 1 ( ) = 1 1 =. Consequently, because the two capactors have equal ntal voltages at t =, the net voltage s zero, resultng n no current flow through. That s, and υ 1 (t) = υ 2 (t) = 1 V for t, (t) =.
Problem 5.44 Gven that n Fg. P5.44, I 1 = 4 ma, I 2 = 6 ma, R 1 = 3 kω, = 6 kω, and C =.2 mf, determne υ(t). Assume the swtch was connected to termnal 1 for a long tme before t was moved to termnal 2. Soluton: 1 2 (a) I 1 R 1 C t = υ I 2 1 (b) At t = I 1 R 1 C υ( ) 2 (c) At t > C υ I 2 Fgure P5.44 At t = (Fg. P5.44(b)), υ( ) = I 1 R 1 = 4 1 3 3 1 3 = 12 V. At t, τ = C = 6 1 3.2 1 3 = 1.2 s, υ() = υ( ) = 12 V, υ( ) = I 2 = 6 1 3 6 1 3 = 36 V. Hence, for t : υ(t) = υ( )+[υ() υ( )]e t/τ = [36+[12 36]e t/1.2 ] = [36 24e t/1.2 ] (V).
Problem 5.48 Determne (t) for t gven that the crcut n Fg. P5.48 had been n steady state for a long tme pror to t =. Also, I = 5 A, R 1 = 2 Ω, = 1 Ω, R 3 = 3 Ω, R 4 = 7 Ω, and L =.15 H. Soluton: 1 L (a) t = 2 I R 1 R 4 R 3 (b) At t = 1 L R 3 I R 1 (c) At t > 2 L R 3 R 4 Fgure P5.48 At t =, current dvson n the crcut of Fg. P5.48(b) gves ( ) = I R 1 R 1 + R 3 = 5 2 2+3 = 2 A. Hence, At t =, At t, ( ) = τ = L R eq () = ( ) = 2 A. (no actve sources). R eq = (R 3 + R 4 ) 1 (3+7) = + R 3 + R 4 1+3+7 = 5.
τ =.15 =.3 s. 5 L (t) = [( )+[() ( )]e t/τ ] = 2e 1t/3 (A).
Problem 5.49 For the crcut n Fg. P5.49, determne L (t) and plot t as a functon of t for t. The element values are I = 4 A, R 1 = 6 Ω, = 12 Ω, and L = 2 H. Assume that L = before t =. Soluton: (a) Crcut t = t =.5 s I R 1 L L1 (b) At < t <.5s I R 1 L L (c) At t >.5 s I R 1 L Fgure P5.49 1. Tme segment t.5 s Before closng the swtches, L was not part of any closed crcut. Hence Contnuty requres that L ( ) =. L1 () = L ( ) =. After closng the frst swtch at t =, the crcut looks as n Fg. P5.48(b). The response of the crcut can be calculated as f no change wll happen at t =.5 s. Hence, L1 ( ) = I. 2. Tme segment t >.5 s At t =.5 s, τ 1 = L R 1 = 2 6 = 1 3 s. L1 (t) = [ L1 ( )+[ L1 () L1 ( )]e t/τ ] = I (1 e R 1t/L ) = 4(1 e 3t ) (A). L1 (.5) = 4(1 e 3.5 ) = 3.11 A. L2 (.5) = L1 (.5) = 3.11 A.
τ 2 = L 2 = ( ) = 2 R eq R1 4 =.5 s R 1 + L2 ( ) = I = 4 A, L2 (t) = { L2 ( )+[ L2 (.5) L2 ( )]e (t.5)/τ 2 } u(t.5) = [4+(3.11 4)e 2(t.5) ] u(t.5) = [4.89e 2(t.5) ] u(t.5) (A).
Problem 5.53 In the crcut of Fg. P5.53(a), R 1 = = 2 Ω, R 3 = 1 Ω, and L = 2.5 H. Determne (t) for t gven that υ s (t) s the step functon descrbed n Fg. P5.53(b). Soluton: (a) Crcut υ s (t) + _ R 1 R 3 L υ s (t) (b) υ s (t) 12 V t At t = 8 υ s + _ R 1 R 3 1 2 L Fgure P5.53 Because υ s (t) was zero before t =, and contnuty requres that ( ) =, () = ( ) =. The crcut n Fg. P5.53(c) pertans to the condtons at t =. Soluton leads to: Also, Hence, for t : υ s + R 1 1 + ( 1 2 ) = ( 2 1 )+R 3 2 = ( ) = 2 ( ) =.3 A. R eq = R 3 + R 1 2 2 = 1+ = 2 Ω. R 1 + 2+2 τ = L = 2.5 R eq 2 =.125 s (t) = [( )+[() ( )]e t/τ ] =.3(1 e 8t ) (A).
+ _ Problem 5.59 The nput-voltage waveform shown n Fg. P5.59(a) s appled to the crcut n Fg. P5.59(b). Determne and plot the correspondng υ out (t). 12 V υ (a) Waveform of υ (t) 2 4 6 8 1 12 t (s) 12 V 2 μf 5 kω (b) Op-amp crcut v υ out V cc = 6 V υ out (t).6 V.6 V 2 4 6 8 1 12 t (s) Fgure P5.59 Soluton: The crcut n Fg. P5.59(a) s a dfferentator crcut wth RC = 5 1 3 2 1 6 =.1. For the tme segment between t = and t = 2 s, the slope of the nput sgnal s (12/2) = 6 V/s. The output voltage s gven by υ out = RC dυ dt =.1 6 =.6 V. Hence, υ out s a square wave wth an ampltude of.6 V, as shown n the fgure.
Problem 5.6 Relate υ out to υ n the crcut of Fg. P5.6. Assume υ () =. υ υ p υ n n = + _ υ out C R Fgure P5.6 Soluton: υ n = υ p = υ υ n R +C d dt (υ n υ out ) = Hence, d dt υ out = 1 RC υ + dυ. dt Integratng all terms from t = to t, t ( ) dυout dt = 1 t υ dt + dt RC t ( ) dυ dt, dt whch smplfes to or υ out (t) t = 1 t υ dt + υ (t) t RC υ out (t) = υ out ()+υ (t)+ 1 t υ dt. RC
Problem 5.62 Relate υ out to υ n the crcut of Fg. P5.62. Assume υ C = at t =. υ C + _ 2 C R 1 1 n = υ υ n υ p _ + υ out Fgure P5.62: Crcut for Problem 5.62. Soluton: But υ n = υ p =, and whch leads to 1 = υ n υ R 1 υ n υ out = 2 + 1 C t 2 = 1 = υ R 1, 2 dt ( R2 υ out = υ + 1 t ) υ dt. R 1 R 1 C
Problem 5.68 The two-stage op-amp crcut n Fg. P5.68 s drven by an nput step voltage gven by υ (t) = 1u(t) mv. If V cc = 1 V for both op amps and the two capactors had no charge pror to t =, determne and plot: (a) υ out1 (t) for t (b) υ out2 (t) for t Soluton: (a) Hence, υ out1 (t) = 1 t υ dt R 1 C 1 R 1 C 1 = 5 1 3 4 1 6 =.2. t υ out1 (t) = 5 1 1 3 dt =.5t (V), for t. (b) For the second stage: υ out2 (t) = 1 t υ out1 (t) dt C 2 C 2 = 1 1 6 5 1 6 = 5 υ out2 (t) = 1 5 t (.5t) dt =.1 t2 2 =.5t2 (V), for t. Plots of υ (t), υ out1 (t), and υ out2 (t) are shown below. We note that υ out1 (t) reaches saturaton at V cc = 1 V after 2 s, and υ out2 (t) reaches saturaton at V cc = +1 V at t = 14.14 s.
(a) 4 μf 5 μf υ 5 kω _ υ out1 1 MΩ _ υ out2 + V + cc = 1 V V cc = 1 V υ (t) 1 mv υ out1 (t) 2 4 6 8 1 12 14 16 18 2 22 t (s) (b) -2 V -4 V -6 V -8 V -1 V -12 V 2 4 6 8 1 12 14 16 18 2 22 t (s) saturaton @ -V cc υ out2 (t) 1 V 8 V 6 V 4 V 2 V 2 4 6 8 1 12 14 16 14.14 s saturaton @ +V cc 18 2 22 t (s) Fgure P5.68