Solutions to Assignment34

Assignments 3 and 4 SOLUTIONS2.(a)(b)6− 10 t /(10+ 50 200) −20000tv (0) = 10V ∴ v ( t) = 10e = 10eVcc− 10iA( − 100 µ s) = iA(0 ) = = 50mA200⎛ 1 ⎞ 50i µ s = e ⎜ ⎟ =⎝10 + 40 ⎠250−2A(100 ) 10 5.413mA3.(a)vCLeft(0) = 20V, v (0) = 80VCRIGHT6 6−10 t/8 −10 t/ 0.8CR∴ v = 20 e , v = 80eCL∴ v = v − v = e − e t>out CR CL−1,250,000t−125,000t80 20 V, 0(b)v v µ s e e+ −1.25 −0.125out(0 ) = 60V;out(1 ) = 80 − 20 = 5.270Vv µ s e e−6.25 −0.625out(5 ) = 80 − 20 =−10.551V4. (a) 0 W(b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is12 || 11 = 5.739 kΩ. Thus, the circuit time constant is L/R = 1.307 µs. The final value ofthe total current flowing into the parallel inductor combination is 50/12 mA = 4.167 mA.This will be divided between the two inductors, so that i(∞) = (4.167)(30)/ (30 + 10) =3.125 mA.We may therefore write i(t) = 3.125[1 – e- 106 t/ 1.307 ] A. Solving at t = 3 µs,we find 2.810 A.5.(a)− 30 3−ix(0 ) = × = 3A, iL(0 ) = 4A7.5 4+ +(b) i (0 ) = i (0 ) = 4AxL(c)i ( ∞ ) = i ( ∞ ) = 3AxL∴ i t = + e = + e ∴i−10 t/ 0.5 −20tx() 3 1 3 Ax(0.04)= + =−0.83 e 3.449ACIRCUIT THEORY 1 (ELEC 243-09)Dr. Kalyana Veluvolu

Assignments 3 and 4 SOLUTIONS8.R 80vc(0) = 50+ 80× 2= 210V, iL(0) = 0, α = = = 202L 42 1002ωo= = 500 : ωd= 500 − 20 = 102∴ v t = e t+ t ∴ =−20tc( ) (A1cos10 A2sin10 ) A1210V− 20t+ 1 +∴ vc( t) = e (210cos10t+ A2sin10 t); vc′(0 ) = ic(0 ) = 0C∴ = − = ∴ v t = e t+420sin10 t)−20t0 10A2 20(210), A2420c( ) (210cos10∴ v = + =−0.8c(40ms) e (210cos 0.4 420sin 0.4) 160.40V−20tAlso, iL= e (B1cos10t+B2sin10 t),+ 1 + 1 + 1iL(0 ) = vL(0 ) = [0 − vc(0 )] = × 210L 2 2+∴ i ′ (0 ) =− 105 = 10B ∴ B = 10.5L2 2∴ i t =− e t t>−20tL( ) 10.5 sin10 A, 0∴ v t = i = e−20tR( ) 80L840 sin10tV∴ v =− =−−0.8R(40ms) 840esin 0.4 146.98Vv () t =−v () t −v () t −v () t ∴vL c c R L(40ms) =− 160.40 + 146.98 =−13.420 V= − − +−20t[check: vLe ( 210cos 420sin 840sin)−20t= e ( − 210cos10t+ 420sin10 t) V, t>0∴ v = − − +−0.8L(40ms) e ( 210cos 420sin 840( − 210cos10t+ 420sin10 t)V, t>0∴ V (40ms) = eL−0.8sin) = e(420sin 0.4 − 210cos 0.4) =−13.420 V Checks]−20tCIRCUIT THEORY 1 (ELEC 243-09)Dr. Kalyana Veluvolu

Assignments 3 and 4 SOLUTIONS9. (a)1 8× 10 8× 10 × 13α= = = ω = = ×2RC 2× 4×10 4∴ω = − × = =Ld6 62 61000, 26 103o326 1 10 5000, vc(0) 8Vi (0) = 8mA, v = 0c,f∴ = (A cos1000 + A sin 5000 )−1000tvce1t2t1 8∴ = ′ = = × − − =C 4000∴5000A − 1000× 8 = 0, A = 1.6+ +6A18; vc(0 ) ic(0 ) 8 10 (0.01 0.008) 02 2So v c (t) = e -1000t (8 cos 1000t + 1.6 sin 1000t) V, t > 0(b)CIRCUIT THEORY 1 (ELEC 243-09)Dr. Kalyana Veluvolu

Assignments 3 and 4 SOLUTIONS10. (a)R 100Series, driven: α = = = 500,2L 0.262 1 10×10ωo= = = 250,000LC 40∴ Crit. damp i ( f) = 3(1 − 2) =−3,i (0) = 3, v (0) = 300VLL1cL∴ i =− + e t+ ∴ =− + =−500t3 (A1 A2) 3 3 A2, A26A+ 1+iL(0 ) = A1− 300 = [ vc(0) − vR(0 )] = 0L∴ A = 3000 e ∴ i ( t) =− 3+e(3000t+ 6), t>0−5000t−500tL∴ i t = u − t + − + e t+u t−500tL( ) 3 ( ) [ 3 (3000 6)] ( )A(b)− 500 te o(3000t + 6) = 3; by SOLVE, t = 3.357msoo6.evvxL ,xc ,−1000t=−2000t200eV= −−1000t100(1 e )Vv = v − v = 0x x, L x,c∴ 200e= 100 −100e−2000t−1000t∴ + − =−1000t−1000t2100e200( e ) 100 0,− 100 ± 10,000 + 80,000= =− 0.25 ± 0.75400e 0.5, t 0.6931ms−1000t∴ = =CIRCUIT THEORY 1 (ELEC 243-09)Dr. Kalyana Veluvolu