ECE 201 Spring 2010Homework 25 SolutionsProblem 43(a)To find the initial conditions at t=0-, we can write the following KVL equa-tions,20 + 40iL(0−) − 60(0.1 − iL(0−)) = 0⇒ iL(0−) = iL(0+) = iC(0+) = −0.14 A⇒ vC(0+) = vC(0−) = 20 − 0.14 × 40= 14.4 VAfter t=0, the circuit is a series RLC circuit with the following characteristicequation,s2+RLs +1LC= 0s2+ 15s + 50 = 0⇒ s1= −5, s2= −10⇒ vC(t) = c1e−5t+ c2e−10t⇒ c1+ c2= 14.4(5 × 10−3)(5c1+ 10c2) = 0.14⇒ c1= 23.2, c2= −8.8⇒ vC(t) = (23.2e−5t− 8.8e−10t) VThe plot of vC(t) for 0 < t ≤ 1 is shown on the next page.10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1051015t(seconds)vC(t)(Volts)Plot of vC(t) (part (a)) for 0<t<=1s(b)In this case, the situation after t=0 can be visualized as a series RLC circuitwith a constant input voltage source (after source transformation) which willhave the value 0.5 × 60 = 30 V . The initial conditions and the characteristicequation do not change. However, the expression for vC(t) will now have aterm due to the constant input also. Thus,vC(t) = c1e−5t+ c2e−10t+ 30⇒ c1+ c2+ 30 = 14.45c1+ 10c2= 28⇒ vC(t) = (−36.8e−5t+ 21.2e−10t+ 30) VThe plot of vC(t) for 0 < t ≤ 1 follows.20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1141618202224262830t(seconds)vC(t)(Volts)Plot of vC(t) (part (b)) for 0<t<=1sProblem 49(a)At t= 0-, the capacitor is o pen circuit and the inductor is short circuit. Att=0+, the current source is off and t he voltage source is still operating. Thuswe can write the following equations,vC(0−) = vC(0+) = 10 ViL(0−) + 0.005 =101000⇒ iL(0+) = iL(0−) = 0.005 A10 − vL(0+) − vC(0+) = 0⇒ vL(0+) = 0iC(0+) +101000= 0.005⇒ iC(0+) = −0.005 A3(b)For t > 0 , we can write the following KVL and KCL equations,10 − LdiLdt− vC= 0iL= CdvCdt+vCR⇒d2vCdt2+1RCdvCdt+vCLC=10LCThus the characteristic equation iss2+ 2000s + 1.087 × 107= 0⇒ s1,2= −1000 ± j1000π⇒ vC(t) = e−1000t[A cos(1000πt) + B sin(1000πt)] + 10A = 0, B = −10π(Using initial conditions)⇒ vC(t) = 10 −10πe−1000tsin(1000πt) ViL(t) = CdvCdt+vCR= 0.0052 − e−1000t1πsin(1000πt) + cos(1000πt)A(c)vC(t) is computed above in part (b).(d)vL(t) = 10 − vC(t)=10πe−1000tsin(1000πt) V(e)iC(t) = CdvCdt=0.005πe−1000t[sin(1000πt) − π cos(1000πt)] AThe plots for the respective quantities follow.40 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5x 10−356789101112x 10−3t(seconds)iL(t)(A)Plot of iL(t) for 0<t<=5ms0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5x 10−37.588.599.51010.511t(seconds)vC(t)(V)Plot of vC(t) for 0<t<=5ms50 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5x 10−3−1−0.500.511.522.5t(seconds)vL(t)(V)Plot of vL(t) for 0<t<=5ms0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5x 10−3−5−4−3−2−10123x 10−3t(seconds)iC(t)(A)Plot of iC(t) for
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