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

Download EE 223 Chapter 8: Electrical Engineering Problem Solutions and more Exams Microelectronic Circuits in PDF only on Docsity! 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 iA(-100 µs) and iA(100 µs). (c) Verify your solution with PSpice. (a) 610 /(10 50 200) 20000(0) 10V ( ) 10 10 Vt tc cv v t e e − + −= ∴ = = (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 10( 100 ) (0 ) 50mA 200 1 50(100 ) 10 5.413mA 10 40 250 A A A i s i i s e µ µ − − − = = =  = = +  Chapter 8, Problem 36. The value of is 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., vL = 0; 0: 1mA (0) 10V, (0) 1mA (0) 10V, 0s c L xt i v i v t< = ∴± = ↓ = − ∴ = < (b) t > 0: vC(t) = 10V e-t/10kΩ×20nF = 10V e-5000t iL(t) = -1mA e-1kΩ×t/0.1H = 1mA e-10000t ⇒ vL(t) = 1kΩ iL = 1V e-10000t vx(t) = vC - vL = 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−) = I12 R1 = 0.0012 x 103 = 0.001 W = 1 mW initial voltage: VC(0) = I2 R2 = 7mA 900Ω = 6.3 V power: P1k(t=0+) = VC2|t=0 / R1 = 6.32 x 10-3 = 39.7 mW final voltage: VC(t→∞) = I2 × R1||R2 = 7mA 900||1000Ω = 3.3 V power: P1k(t→∞) = VC2|t→∞ / R1 = 3.32 x 10-3 = 10.9 mW time constant: τ = C × R1||R2 = 3µF 473.68Ω = 1.4 ms - 1 0 1 2 3 4 5 6 7 0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 t i m e ( m s ) po w er (m W ) t = τ