EE 223 Chapter 8: Electrical Engineering Problem Solutions, Exams of Microelectronic Circuits

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Examples Chapter #8, EE 223, 2003: 28, 36, 50, 74

Chapter 8, Problem 28.

After being closed for a long time, the switch in the circuit of Fig. 8.59 is opened at t = 0. ( a ) Find vC ( t ) for t > 0. ( b ) Calculate values for i (^) A (-100 μ s) and i (^) A (100 μ s). ( c ) Verify your solution with PSpice.

(a) 106 /(10 50 200) (^20000)

vc (0) = 10V ∴ vc ( ) t = 10 e^ −^^ t^ +^ = 10 e − t V

(b)

iA (0+) = 10V [1 / (10Ω+40Ω)] 50Ω/250Ω = 40 mA (voltages sources add to zero, total current due to capacitor’s voltage is split due to current division rule) (c) PSpice Verification.

2

( 100 ) (0 ) 50mA

(100 ) 10 5.413mA

A A

A

i s i

i s e

μ

μ

−

−

− = = =

= ^  =

 + 

 

Chapter 8, Problem 36.

The value of i (^) s in the circuit of Fig. 8.67 is 1 mA for t < 0, and zero for t > 0. Find vx ( t ) for ( a ) t < 0; ( b ) t > 0.

(a) Capacitor: O.C., iC = 0; inductor: S.C., v (^) L = 0;

t < 0: is = 1mA ∴± vc (0) = 10V, ↓ iL (0) = −1mA ∴ v x (0) = 10V, t < 0

(b) t > 0: v C (t) = 10V e-t/10kΩ×20nF^ = 10V e-5000t

i L(t) = -1mA e -1kΩ×t/0.1H^ = 1mA e -10000t^ ⇒ vL(t) = 1kΩ i L = 1V e-10000t

v x (t) = vC - v L = 10V e-5000t^ - e -10000t^ for t > 0

Chapter 8, Problem 74.

In the circuit of Fig. 8.99, one switch is opened at t = 0, while the other switch is simultaneously closed. Sketch the power absorbed by the 1-kΩ resistor over the interval -1 ms ≤ t ≤ 7 ms. At t = 0, the 1-mA source is also turned off.

before switching: P1k(t≤ 0 −) = I 12 R 1 = 0.001 2 x 10 3 = 0.001 W = 1 mW

initial voltage: VC(0) = I 2 R 2 = 7mA 900Ω = 6.3 V power: P (^) 1k(t=0 +^ ) = V (^) C^2 |t=0 / R 1 = 6.3 2 x 10 -3^ = 39.7 mW

final voltage: VC(t→∞) = I 2 × R 1 ||R 2 = 7mA 900||1000Ω = 3.3 V power: P (^) 1k(t→∞) = V (^) C^2 |t→∞ / R 1 = 3.3 2 x 10 -3^ = 10.9 mW

time constant: τ = C × R1||R 2 = 3μF 473.68Ω = 1.4 ms

(^0) - 1 0 1 2 3 4 5 6 7

5

1 0

1 5

2 0

2 5

3 0

3 5

4 0

t i m e ( m s )

power (mW)

t = τ

Chapter 8, Problem 50.

With reference to the circuit of Fig. 8.79, ( a) compute the power absorbed by the 2-kΩ resistor at t = 1 ms; ( b ) determine the value of i ( t ) at 3 μ s; ( c ) determine the peak current through the 12-kΩ resistor. ( d ) Verify your answers with PSpice.

(a) 0 W

(b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is 12 || 11 = 5.739 kΩ. Thus, the circuit time constant is L/R = 1.307 μs. The final value of the 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- 10

(^6) t / 1. ] A. Solving at t = 3 ms, we find 2.810 A.

(c) PSpice verification