Explanation Given data: Refer to Figure given in the textbook. The given values are, L = 25 mH C = 62.5 μ F R = 8 Ω The current source is 2 A. The initial current in the 25 mH inductor is 1 A and the initial voltage on the capacitor is 50 V. Formula used: Write the general expression for resonant radian frequency as follows, ω 0 = 1 L C (1) Write the general expression for neper frequency as follows, α = 1 2 R C (2) Write the general expression for damoing frequency as follows, ω d = ω 0 2 − α 2 (3) Calculation: In the given circuit, the initial voltage across the capacitor is v ( 0 ) = 50 V and the initial current through the 25 mH inductor is i ( 0 ) = 1 A Substitute 8 for R and 62.5 μ for C in equation (2) as follows, α = 1 2 ( 8 ) ( 62.5 × 10 − 6 ) { ∵ 1 μ = 10 − 6 } = 1000 rad s Substitute 62.5 μ for C and 20 m for L in equation (1) as follows, ω 0 = 1 ( 25 × 10 − 3 ) ( 62.5 × 10 − 6 ) { ∵ 1 μ = 10 − 6 , 1 m = 10 − 3 } = 800 rad s Substitute 800 for ω 0 and 640 for α in equation (3) as follows, ω d = ( 800 ) 2 − ( 640 ) 2 = 480 rad s As α 2 > ω 0 2 , the output response is overdamped. Hence, the output equation will be as follows, i L ( t ) = I f + A ′ 1 e s 1 t + A ′ 2 e s 2 t (4) In steady state condition, the inductor acts a short circuit. Hence, the current through the inductor is, I = I f = 2 A The general expression to find the roots is, s 1 , 2 = − α ± α 2 − ω 0 2 (5) Substitute 1000 for α and 800 for ω 0 in equation (5) as follows, s 1 , 2 = − 1000 ± 1000 2 − 800 2 s 1 , 2 = − 1000 ± 600 Therefore, the roots are, s 1 , 2 = − 400 , − 1600 rad s Substitute 2 for I f , − 400 for s 1 , and − 1600 for s 2 in equation (4) as follows, i L ( t ) = 2 + A ′ 1 e − 400 t + A ′ 2 e − 1600 t (6) Determine the values of the coefficient A ′ 1 and A ′ 2 by applying the initial condition to the above equation. Substitute 0 for t in equation (6) as follows, i L ( 0 ) = 2 + A ′ 1 e − 400 ( 0 ) + A ′ 2 e − 1600 ( 0 ) 1 = 2 + A ′ 1 + A ′ 2 { ∵ i L ( 0 ) = 1 A } A ′ 1 + A ′ 2 = − 1 (7) Differentiating equation (6) with respect to t as follows, d i L ( t ) d t = − 400 A ′ 1 e − 400 t − 1600 A ′ 2 e − 1600 t (8) The general expression for inductor voltage v L ( t ) is, v L ( t ) = L d i L ( t ) d t (9) Substitute equation (9) in equation (8) as follows, v L ( t ) L = − 400 A ′ 1 e − 400 t − 1600 A ′ 2 e − 1600 t (10) As the given circuit is a parallel RLC circuit, the voltage across the inductor and the capacitor is the same

10th Edition

ISBN: 9780133760033

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Question

Chapter 8, Problem 28P

To determine

Find the expression for iL(t) for t≥0.

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Chapter 8 Solutions

Electric Circuits (10th Edition)

Ch. 8.1 - The resistance and inductance of the circuit in...Ch. 8.2 - Use the integral relationship between iL and v to...Ch. 8.2 - Prob. 3AP Ch. 8.2 - Prob. 4AP Ch. 8.2 - Prob. 5AP Ch. 8.3 - Prob. 6AP Ch. 8.4 - The switch in the circuit shown has been in...Ch. 8.4 - Prob. 8AP Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P

Ch. 8 - The resistance in Problem 8.1 is decreased to80 Ω...Ch. 8 - Prob. 4P Ch. 8 - Prob. 5P Ch. 8 - The natural voltage response of the circuit in...Ch. 8 - Prob. 7P Ch. 8 - Prob. 8P Ch. 8 - The natural response for the circuit shown in Fig....Ch. 8 - Prob. 10P Ch. 8 - The two switches in the circuit seen in Fig.P8.11...Ch. 8 - The resistor in the circuit of Fig. P8.11 is...Ch. 8 - The resistor in the circuit of Fig.P8.11 is...Ch. 8 - The switch in the circuit of Fig. P8.17 has been...Ch. 8 - The inductor in the circuit of Fig. P8.17 is...Ch. 8 - The inductor in the circuit of Fig. P8.17 is...Ch. 8 - Design a parallel RLC circuit (see Fig. 8.1) using...Ch. 8 - Prob. 18P Ch. 8 - Prob. 19P Ch. 8 - Find υ(t) for t ≥ 0 in the circuit in Problem 8.19...Ch. 8 - Prob. 21P Ch. 8 - Prob. 22P Ch. 8 - The initial value of the voltage υ in the circuit...Ch. 8 - Prob. 24P Ch. 8 - Prob. 25P Ch. 8 - Prob. 26P Ch. 8 - Prob. 27P Ch. 8 - Prob. 28P Ch. 8 - Prob. 29P Ch. 8 - Prob. 30P Ch. 8 - The switch in the circuit in Fig. P8.31 has been...Ch. 8 - Prob. 32P Ch. 8 - There is no energy stored in the circuit in Fig....Ch. 8 - For the circuit in Fig. P8.30, find υo for t ≥...Ch. 8 - The switch in the circuit in Fig. P8.36 has been...Ch. 8 - Prob. 36P Ch. 8 - Prob. 37P Ch. 8 - Prob. 38P Ch. 8 - Prob. 39P Ch. 8 - Find the voltage across the 80 nF capacitor for...Ch. 8 - The initial energy stored in the 31.25 nF...Ch. 8 - In the circuit in Fig. P8.42, the resistor is...Ch. 8 - Design a series RLC circuit (see Fig. 8.3) using...Ch. 8 - Change the resistance for the circuit you designed...Ch. 8 - Prob. 45P Ch. 8 - Prob. 46P Ch. 8 - The switch in the circuit shown in Fig. P8.48 has...Ch. 8 - The switch in the circuit in Fig. P8.48 has been...Ch. 8 - The initial energy stored in the circuit in Fig....Ch. 8 - The resistor in the circuit shown in Fig. P8.50 is...Ch. 8 - The resistor in the circuit shown in Fig. P8.50 is...Ch. 8 - Prob. 52P Ch. 8 - The two switches in the circuit seen in Fig. P8.53...Ch. 8 - Prob. 55P Ch. 8 - Prob. 57P Ch. 8 - Prob. 58P Ch. 8 - Prob. 59P Ch. 8 - Prob. 60P Ch. 8 - Prob. 61P Ch. 8 - Derive the differential equation that relates the...Ch. 8 - The voltage signal of Fig. P8.63(a) is applied to...Ch. 8 - The circuit in Fig. P8.63 (b) is modified by...Ch. 8 - Prob. 65P Ch. 8 - Prob. 66P Ch. 8 - Prob. 67P Ch. 8 - Prob. 68P

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