Explanation Given data: Refer to Figure given in the textbook. The value of inductor is L = 25 mH . The value of capacitor is C = 25 nF . The value of resistor is R = 500 Ω . The current source is 24 mA. 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 value of voltage across the capacitor V 0 and inductor current I 0 are both zero as the initial stored energy is zero. That is, v ( 0 + ) = 0 V i L ( 0 + ) = i L ( 0 − ) = 0 A As the given circuit is a parallel RLC circuit, the voltage across the inductor and the capacitor is the same. That is, v L ( t ) = v C ( t ) And, v L ( 0 ) = v C ( 0 ) = 0 V As t → ∞ , the capacitor acts as an open circuit and inductor behaves like a short circuit that shunts the resistor. Therefore, all of the current from the source is in the inductor. Therefore, I f = 24 mA Substitute 400 for R and 25 n for C in equation (2) as follows, α = 1 2 × ( 500 ) × ( 25 × 10 − 9 ) { ∵ 1 n = 10 − 9 } α = 40000 rad s Squaring the above expression as follows, α 2 = 40000 2 rad s Substitute 25 n for C and 25 m for L in equation (1) as follows, ω 0 = 1 ( 25 × 10 − 9 ) ( 25 × 10 − 3 ) { ∵ 1 n = 10 − 9 , 1 m = 10 − 3 } ω 0 = 40000 rad s Squaring the above expression as follows, ω 0 2 = 40000 2 rad s As α 2 = ω 0 2 , the output response is critically damped. Hence, the output equation will be as follows, i L ( t ) = I f + D 1 ′ t e − α t + D ′ 2 e − α t , t ≥ 0 (4) Substitute 24 m for I f and 40000 for α in equation (4) as follows, i L ( t ) = ( 24 × 10 − 3 ) + D 1 ′ t e − 40000 t + D 2 ′ e − 40000 t (5) The general expression for inductor voltage v L ( t ) is, v L ( t ) = L d i L ( t ) d t (6) Substitute 0 for t in equation (6) as follows, v ( 0 + ) = L d i L ( 0 + ) d t v ( 0 + ) L = d i L ( 0 + ) d t Substitute 25m for L in the above expression as follows, d i L ( 0 + ) d t = v ( 0 + ) 25 × 10 − 3 = 0 25 × 10 − 3 { ∵ v ( 0 + ) = 0 } = 0 Determine the values of the coefficient D 1 ′ and D 2 ′ by applying the initial condition to the above equation

10th Edition

ISBN: 8220100801792

Author: Riedel

Publisher: YUZU

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Chapter 8, Problem 26P

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Find the expression for v(t) for t≥0.

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Chapter 8 Solutions

EBK ELECTRIC CIRCUITS

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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