Chapter 6_Capacitance and Inductance

# Chapter 6_Capacitance and Inductance - The Analysis and...

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The Analysis and Design of Linear Circuits Seventh Edition 6 Capacitance and Inductance 6.1 Exercise Solutions Exercise 6–1. A 1- μ F capacitor has no voltage across it at t = 0. A current flowing through the capacitor is given as i C = 2 u ( t ) 3 u ( t 2) + u ( t 4) μ A. Find the voltage across the capacitor at t = 4 s. Apply the capacitor i - v relationship to find the voltage. v C ( t ) = v C (0) + 1 C integraldisplay t 0 i C ( x ) dx v C (4) = 0 + 1 10 - 6 integraldisplay 4 0 i C ( x ) dx v C (4) = 10 6 bracketleftbiggintegraldisplay 2 0 (2 × 10 - 6 ) dx integraldisplay 4 2 (1 × 10 - 6 ) dx bracketrightbigg v C (4) = [(2)(2 0) (1)(4 2)] v C (4) = 2 V Exercise 6–2. (a). The voltage across a 10- μ F capacitor is 25[sin(2000 t )] u ( t ) V. Derive an expression for the current through the capacitor. Apply the capacitor i - v relationship to find the voltage. i C ( t ) = C dv C ( t ) dt i C ( t ) = 10 × 10 - 6 d dt [25 sin(2000 t )] = (10 × 10 - 6 )(2000)(25) cos(2000 t ) i C ( t ) = 500 cos(2000 t ) mA , t > 0 (b). At t = 0 the voltage across a 100-pF capacitor is 5 V. The current through the capacitor is 10[ u ( t ) u ( t 10 - 4 )] μ A. What is the voltage across the capacitor for t > 0? v C ( t ) = v C (0) + 1 C integraldisplay t 0 i C ( x ) dx v C ( t ) = 5 + 1 100 × 10 - 12 integraldisplay t 0 (10 × 10 - 6 )[ u ( t ) u ( t 10 - 4 )] dx v C ( t ) = 5 + 10 5 t V , for 0 < t < 10 - 4 v C ( t ) = 5 + (10 5 )(10 - 4 ) = 5 + 10 = 5 V , for t 10 - 4 Exercise 6–3. For t 0 the voltage across a 200-pF capacitor is 5 e - 4000 t V. (a). What is the charge on the capacitor at t = 0 and t = + ? Solution Manual Chapter 6 Page 6-1

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The Analysis and Design of Linear Circuits Seventh Edition Find the voltage at the given times and then compute the charge. q ( t ) = Cv C ( t ) q (0) = Cv C (0) = (200 × 10 - 12 )(5)( e 0 ) = 1 nC q ( ) = Cv C (0) = (200 × 10 - 12 )(5)( e -∞ ) = 0 C (b). Derive an expression for the current through the capacitor for t 0. The current through a capacitor is the capacitance times the derivative of the voltage. i C ( t ) = C dv C ( t ) dt = (200 × 10 - 12 ) d dt ( 5 e - 4000 t ) = (200 × 10 - 12 )(5)( 4000) e - 4000 t = 4 e - 4000 t μ A (c). For t > 0 is the device absorbing or delivering power? Compute the power and examine the sign. p C ( t ) = i C ( t ) v C ( t ) = ( 4 e - 4000 t μ A ) ( 5 e - 4000 t V ) = 20 e - 8000 t μ W The power is always negative, so the capacitor is always delivering power. Exercise 6–4. Find the power and energy for the capacitors in Exercise 6–2. (a). Compute the power. p C ( t ) = i C ( t ) v C ( t ) = [0 . 5 cos(2000 t )] [25 sin(2000 t ) u ( t )] = 12 . 5 cos(2000 t ) sin(2000 t ) u ( t ) W = 6 . 25 sin(4000 t ) u ( t ) W Compute the energy. w C ( t ) = 1 2 Cv 2 C ( t ) = 1 2 (10 - 5 ) [25 sin(2000 t ) u ( t )] 2 = 3 . 125 sin 2 (2000 t ) u ( t ) mJ (b). Compute the power for the two time intervals. p C ( t ) = i C ( t ) v C ( t ) = (10 - 5 )( 5 + 10 5 t ) = 50 × 10 - 6 + t W for 0 < t < 0 . 1 ms p C ( t ) = (0)(5) = 0 W for t 0 . 1 ms Compute the energy for the two time intervals. w C ( t ) = 1 2 Cv 2 C ( t ) = 1 2 (10 - 10 )(25 10 6 t + 10 10 t 2 ) = 1 . 25 (5 × 10 4 ) t + (5 × 10 8 ) t 2 nJ for 0 < t < 0 . 1 ms w C ( t ) = 1 2 (10 - 10 )(5) 2 = 1 . 25 nJ for t 0 . 1 ms Exercise 6–5. Find the power and energy for the capacitor in Exercise 6–3.
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