DEPARTMENT OF EECTRICA AND COMPUTER ENGINEERING RUTGERS UNIVERSITY 330:222 Principles of Electrical Engineering II Spring 2002 Exam 1 February 19, 2002 SOUTION NAME OF STUDENT: Student ID Number (last 4 digits ONY): Section (please circle): 1: M-W6, Marshall 2: T-Th4 Yates DO A 8 PROBEMS. This is a closed-book exam. Except calculators, no other material is permitted. Do all your work on these sheets. If more space is required, work on the back of these sheets and indicate accordingly. Please make sure that there are 10 pages in this booklet, including the cover page. Problem 1 2 3 4 5 6 7 8 Maximum Score 25 25 25 25 25 25 25 25 Your Score 1
Problem 1: The inductor current i (t) has an initial value I 0 at time t = 0. Please answer the following questions. + i R v - 1(a) State which Kirchoff law you are using, and use it to obtain a differential equation for i (t). Because the sum of vltage drops around a loop is zero, Dividing through by, di + i R = 0 di + i =R = 0 1(b) Assuming that the inductor current takes the form i (t) =Ae st, obtain the characteristic equation for the circuit. Show all steps. Substituting the solution form i (t) =Ae st, we obtain sae st + Aest τ = 0 where τ = =R. Dividing through by Ae st, we obtain s + 1=τ = 0 1(c) Assuming I 0 = 2A, R = 3Ω, and = 1H, what is i (t)? Since s = 1=τ and since there are no sources, the solution is i (t) =I 0 e t =τ = 2e 3t A 2
Problem 2: Consider the network shown with = 0:5 H and R = 5 Ω. et the initial value of i at time t = 0be 20 A. Write an expression for i as a function of t for t 0. Sketch i(t) as a function of t. Carefully label the t-axis of your plot. 100 V R i(t) To solve this problem, we use the solution form t =τ i(t) =i f +(i 0 i f )e In particular, the initial current i 0 = 20 A is given. At t =, we again have steady state, the inductor is shorted, and the final current is The time constant is The current is 25 20 15 10 i f = 100=R = 20 A τ = =R = 0:5=5 = 0:1 i(t) =20 +( 20 20)e t =τ = 20 40e 10t You should all be able to plot the graph from the given function. i(t) 5-5 -10-15 -20-25 t 3
Problem 3: The following circuit is in steady state at time t = 0 with the switch closed. At time t = 0, the switch opens. The circuit parameters are = 10mH, R 0 = 10 Ω, R 1 = 40Ω, C = 5 µf. Please answer the following questions. 10 V i(t) t=0 R 0 R 1 C + - v(t) 3(a) What is the capacitor voltage v(t) at time t = 0? At t = 0, the circuit is in steady state, the inductor is a short and the capacitor is open. With the switch closed, the capacitor is directly across the source. Thus v(0 )=10 V 3(b) For t > 0, find v(t). For t > 0, the switch is open and C and R 1 form a first order RC circuit without sources. The time constant is τ = R 1 C = 40 5 10 6 = 1 5000 The capacitor simply discharges through the resistor; the solution is v(t) =v(0)e t =τ = 10e 5000t 4
3(c) What is the inductor current i(t) at time t = 0? At t = 0, the inductor is a short, and the capacitor is open. The inductor current goes through R 0 and R 1 in parallel. We can replace the parallel resistors with an equivalent resistor R 0 = R 0 R 1 r 0 + R 1 = 8 Ω The inductor current is i(0 )= 10 R 0 = 1:25 A 3(d) For t > 0, find i(t). When the switch opens, and R 0 form a simple R circuit. As usual, we write t =tau i(t) =i f +(i 0 i f )e The time constant is The final voltage is The inductor current is τ = =R 0 = 1 1000 i f = 10 R 0 = 1 A i(t) =1 +(1:25 1)e t =τ = 1 + 1 4 e 1000t 5
Problem 4: For the parallel RC circut shown to the right, please answer the following questions. + R C v - 4(a) The characteristic equation of the parallel RC circuit is of the form s 2 +2αs +ω0 2 = 0: Find its solutions in terms of α if α = ω 0. This confused people because it was so simple. Replacing ω 0 by α in the characteristic equation yields s 2 + 2αs + α 2 = 0 which can also be written as (s + α) 2 = 0 The only solution is s 1 = α, as you would expect for the critically damped circuit. 4(b) For the above circuit, the voltage, v(t), has the form v(t) =(D 1 t + D 2 )e αt. If α = 1=2, v(0) =10, and dv(0)= = 4, find v(t) by solving for the unknowns D 1 and D 2. To find D 1 and D 2, we apply the initial conditions. First, v(0) =D 2 = 10 Second, At t = 0, or Thus, dv = α(d 1 t + D 2 )e αt + D 1 e αt dv(0) = αd 2 + D 1 = 4 D 1 = 4 + αd 2 = 4 + 10 2 = 9 t =2 v(t) =(9t + 10)e 6
Problem 5: The energies in the inductor and capacitor of the series RC circuit are 0 prior to the closing of the switch at t = 0. The characteristic equation is of the form s 2 + 2αs + ω0 2 = 0. For the underdamped case considered here, the roots are s 1;2 = α ± jω d, in which ω d qω = 0 2 α2 is the damped frequency. The step response has the form, v(t) =V f + e αt (B 1 cosω d t + B 2 sinω d t). The initial conditions (resulting from the zero initial energy assumption) determine B 1 and B 2. The circuit parameters are R = 2Ω, = 1H, C = 0:2F, and V 0 = 10V. 5(a) What are α and ω d? V 0 t=0 C R + v - i For the series RC circuit, This implies α R = 2 = 1 1 1 ω2 0 = C = 0:2 = 5 q ω d = ω0 2 α2 = 2 5(b) Find v(t) by solving for B 1, B 2, and V f. Show your work. For the underdamped series RC circuit with constant source, the general solution form is v(t) =v f + e αt (B 1 cosω d t + B 2 sinω d t) To find the final voltage v f, we examine the steady state at t =. The capacitor becomes an open circuit, implying no current throught the resistor. Thus v f = V 0 = 10 V and v(t) =10 + e t (B 1 cos2t + B 2 sin2t) In addition, we were told that there was no initial energy stored in the capacitor, impplying v(0) =0. Similarly, no initial stored energy in the inductor says that i(0) =0. However, i is also the capacitor current so that i = Cdv=. This dv(0)= = i(0)=c = 0. Applying these initial conditions, we have v(0) =10 + B 1 = 0 ) B 1 = 10 Also, dv(0) = B 1 + 2B 2 = 0 ) B 2 = B 1 =2 = 5 Putting the pieces together, we have v(t) =10 e t [10cos2t + 5sin2t] 7
Problem 6: At time t = 0, the inductor current is zero, i.e., i (0)=0. At time t = 0, the switch closes. As explained below, the switch opens at a future time t 1 and then stays open. The circuit parameters are V 0 = 50mV, = 20mH, and R = 0:2Ω. V 0 t=0 i R 6(a) For 0» t» t 1, give a differential equation for i (t). Show that the inductor current i (t) is increasing after the switch closes at t = 0. Hint: Examine the circuit with the switch closed and solve for di =. With the switch closed, by summing the voltages around the loop of the source and the inductor, we have V 0 + di = 0 or di V 0 = = 2:5 The inductor current is increasing because the derivative of i is positive. What s curious is that the the differential equation doesn t depend on the resistor R. 6(b) For 0» t» t 1, solve the differential equation to obtain an expression for i (t). The solution is the ramp function i (t) =2:5t 6(c) The switch opens (and stays open forever) at time t 1 when i (t) reaches 10A. Find the time, t 1, when this occurs. The switch opens at time t 1 when i(t 1 )=2:5t 1 = 10 ) t 1 = 4 s 6(d) Write and expression for i(t) for t t 1. If you are unable to find t 1 in the previous part, you may leave t 1 in your answer. When the switch opens at time t 1, we have an ordinary first order R circuit with the initial condition i (t 1 )=10. The time constant is τ = =R = 0:1 s. Since the source is now disconnected, the solution for t t 1 is i (t) =i (t 1 )e (t t 1)=τ = 10e 10(t 4) 8
Problem 7: The following circuit is in steady state at time t = 0 with the switch open. At time t = 0, the switch closes. The circuit parameters are = 1mH, R 0 = 12:5 Ω, R 1 = 50 Ω, C = 2:5 µf. Please answer the following questions. 10 V i(t) t=0 R 0 R 1 C + - v(t) 7(a) What are the inductor current i(t) and capacitor voltage v(t) at time t = 0? At t = 0, the switch is open. In the RC circuit on the right, the capacitor has discharged and v(0 )=0. In the R circuit on the left, the inductor is a short and i(0 )=10=R 0 = 0:8 A. 7(b) For t > 0, find i(t). For t > 0, the switch is closed and we have a parallel RC circuit. The two resistors in parallel can be combined into an effective resistor The parallel RC has In this case, α 2 = ω 2 0 R 0 = R 0 R 1 R 0 + R 1 = 10 Ω α = 1 2R 0 C = 2 104 = 20;000 ω 2 0 = 1 C = 4 108 and the circuit is critically damped with i(t) =i f +(D 1 t + D 2 )e 20;000t To find i f, we examine the steady state at t =. Here, is a short, C is open, and i f = 10=R 0 = 1. Thus, i(t) =1 +(D 1 t + D 2 )e 20;000t We also know from the previous part that i(0) =0:8 A, yielding i(0) =1 + D 2 = 0:8 ) D 2 = 0:2 Further, at t = 0 +, the voltage loop around the inductor and capacitor yields 10 + di(0+ ) + v(0 + )=0 Since v(0 + )=0, we have di(0 + )= = 1= = 10;000. Since di(0 + ) = 20;000D 2 + D 1 = 10;000 ) D 1 = 6;000 The complete solution is i(t) =1 +(6;000t 0:2)e 20;000t 9
t=0 Problem 8: The switch in this circuit has been closed for a long time before it is opened at t = 0. The circuit parameters are = 50 µh, R = 50 mω, C = 125mF and V 0 = 0:1V. V 0 C + v - R i 8(a) What are the inductor current i(t) and capacitor voltage v(t) at time t = 0? At t = 0, the capacitor is shorted, so v(0 )=0. Also, the indcutor is a short, so i(0 )= V 0 R = 2 A 8(b) Find i(t) for t > 0. When the switch opens, we have aseries RC circuit with α R = 2 = 500 1 ω2 0 = = 16 104 C The circuit is overdamped and the characteristic equation has roots s 1;2 = α ±p 25;000 16;000 = 500 ± 300 The roots are s 1 = 200 and s 2 = 800 and the general solution form is i(t) =i f + A 1 e 200t + A 2 e 800t At t +, the capacitor is open and the steady state final current is i f = 0. Next, we use the fact that the initial current is i(0) =2, yielding i(0) =A 1 + A 2 = 2 Also, at t = 0 +, V 0 + v(0)+i(0)r + di(0) = 0 Since v(0) =0 and since i(0)r = V 0, we have that di(0)= = 0. That is, di(0) = 0 = 200A 1 800A 2 ) A 1 = 4A 2 Finally, since A 1 + A 2 = 2, we have A 2 = 2=3 and A 1 = 8=3. The complete solution is i(t) = 8 3 e 200t 2 3 e 800t 10