(a) Explanation Given data: Refer to Figure given in the textbook. The value of the inductor is L = 400 mH . The value of the capacitor is C = 10 μ F The value of the voltage source is 100 V. The value of the current source is 4 A. 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 = R 2 L (2) Write the general expression for damping frequency as follows, ω d = ω 0 2 − α 2 (3) Calculation: Consider the circuit for t < 0 . In Figure 1, for t < 0 , the capacitor acts as an open circuit and inductor as a short circuit to find the initial conditions. The initial value of current is calculated by using ohms law as follows, i o ( 0 − ) = 100 50 = 2 A The initial value of voltage is calculated by using ohms law as follows, v o ( 0 − ) = 100 × ( − 4 ) = − 400 V Consider the circuit for t > 0 . Substitute 500 for R and 400 m for L in equation (2) as follows, α = 500 2 × 400 × 10 − 3 { ∵ 1 m = 10 − 3 } = 625 rad s Squaring the above expression as follows, α 2 = 625 2 rad s Substitute 10 μ for C and 400 m for L in equation (1) as follows, ω 0 = 1 ( 10 × 10 − 6 ) ( 400 × 10 − 3 ) { ∵ 1 m = 10 − 3 } ω 0 = 500 rad s Squaring the above expression as follows, ω 0 2 = 500 2 rad s As α 2 < ω 0 2 , the output response is under damped. Hence, the output equation will be as follows, i L ( t ) = A 1 e − s 1 t + A 2 e − s 2 t (4) Consider the general expression for the roots of the characteristic equation, s 1 , 2 = − α ± α 2 − ω 0 2 Substitute 625 for α and 500 for ω 0 in the above equation as follows, s 1 , 2 = − 625 ± 625 2 − 500 2 = − 625 ± 375 Therefore, the roots are, s 1 = − 250 s 2 = − 1000 Substitute − 250 for s 1 and − 1000 for s 2 in equation (4) as follows, i L ( t ) = A 1 e − 250 t + A 2 e − 1000 t (5) Substitute 0 for t in equation (5) as follows, i L ( 0 ) = A 1 e − 250 ( 0 ) + A 2 e − 1000 ( 0 ) i L ( 0 ) = A 1 + A 2 { ∵ e 0 = 1 } 2 = A 1 + A 2 { i o ( 0 ) = i L ( 0 ) = 2 A } (6) Differentiate equation (6) with respect to t (b) To determine Find the expression for the voltage v o ( t ) for t ≥ 0 .

10th Edition

ISBN: 8220100801792

Author: Riedel

Publisher: YUZU

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Chapter 8, Problem 45P

(a)

To determine

Find the expression for the current io(t) for t≥0.

(b)

To determine

Find the expression for the voltage vo(t) for t≥0.

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Chapter 8, Problem 44P

Chapter 8, Problem 46P

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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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