Explanation Given data: Refer to given circuit in the textbook. Now energy stored in the circuit at the instant the switch closes. Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of capacitance. Calculation: The given circuit is redrawn as shown in Figure 1. Since no energy stored in the circuit at instant the switch closes, the current through inductor and voltage across the capacitor is equal to zero. i ( 0 ) = 0 A v ( 0 ) = 0 V For time t > 0 , the switch closes and the circuit in Figure 1 is reduced as shown in Figure 2. Use equation (1) to find Z R 1 . Z R 1 = 1 k Ω = 1 × 10 3 { ∵ 1 k = 10 3 } = 1000 Use equation (1) to find Z R 2 . Z R 2 = 1 k Ω = 1 × 10 3 { ∵ 1 k = 10 3 } = 1000 Substitute 10 H for L in equation (2) to find Z L . Z L = s ( 10 H ) = 10 s Substitute 2 μ F for C in equation (3) to find Z C . Z C = 1 s ( 2 μ F ) = 1 s ( 2 × 10 − 6 F ) { ∵ 1 μ = 10 − 6 } = 5 × 10 5 s Convert the Figure 2 into s -domain as shown in Figure 3. To obtain the Thevenin impedance short circuit the voltage source and remove the load capacitor in Figure 3. Now, the Figure 3 is reduced as shown in Figure 4. Refer to Figure 4, the impedance of resistor R 2 is connected in series with the parallel combination of impedance of resistor R 1 and inductor L . The Thevenin impedance Z Th is calculated as follows: Z Th = 1000 + ( 1000 ∥ 10 s ) = 1000 + 10000 s 1000 + 10 s = 1000 ( 1000 + 10 s ) + 10000 s 1000 + 10 s = 1000 ( 1000 + 10 s + 10 s ) 1000 + 10 s Simplify the above equation as follows: Z Th = 1000 ( 1000 + 20 s ) 1000 + 10 s = 20000 ( s + 50 ) 10 ( s + 100 ) = 2000 ( s + 50 ) s + 100 To find the Thevenin voltage open circuit the load capacitor in Figure 3. Now, the Figure 3 is reduced as shown in Figure 5. Apply voltage division rule in Figure 5 to find V Th . V Th = ( 40 s ) ( 10 s 10 s + 1000 ) = ( 40 s ) ( 10 s 10 ( s + 100 ) ) = 40 s + 100 Now, the Thevenin equivalent circuit is drawn as shown in Figure 6. Refer to Figure 6, the Thevenin impedance and the impedance of capacitor is connected in series. For the series connection, the voltage is same. The current ( I ) is calculated as follows: I = V Th Z Th + 5 × 10 5 s Substitute 40 s + 100 for V Th , and 2000 ( s + 50 ) s + 100 for Z Th in above equation to find I . I = ( 40 s + 100 ) 2000 ( s + 50 ) s + 100 + 5 × 10 5 s = ( 40 s + 100 ) 2000 s ( s + 50 ) + 5 × 10 5 ( s + 100 ) s ( s + 100 ) = 40 s 2000 s ( s + 50 ) + 5 × 10 5 ( s + 100 ) = 40 s 2000 s 2 + 1 × 10 5 s + 5 × 10 5 s + 5 × 10 7 Simplify the above equation as follows: I = 40 s 2000 s 2 + 6 × 10 5 s + 5 × 10 7 = 40 s 2000 ( s 2 + 300 s + 25000 ) I = 0.02 s s 2 + 300 s + 25000 (4) Write an expression to solve quadratic equation. s 1 , 2 = − b ± b 2 − 4 a c 2 (5) Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. From equation (4), the characteristic equation of denominator is written as follows: s 2 + 300 s + 25000 = 0 Compare the above equation with the quadratic equation ( a s 2 + b s + c = 0 ) . a = 1 b = 300 c = 25000 Substitute 1 for a , 300 for b , and 25000 for c in equation (5) to find s 1 , 2 . s 1 , 2 = − 300 ± ( 300 ) 2 − 4 ( 1 ) ( 25000 ) 2 ( 1 ) = − 300 ± 90000 − 100000 2 = − 300 ± − 10000 2 = − 300 ± j 100 2 Simplify the equation. s 1 , 2 = − 300 + j 100 2 − 300 − j 100 2 = − 150 + j 50 , − 150 − j 50 Therefore, the roots of characteristic equations are real and imaginary. s 1 = − 150 + j 50 s 2 = − 150 − j 50 Now, the equation (4) is written as follows: I = 0.02 s ( s − ( − 150 + j 50 ) ) ( s − ( − 150 + j 50 ) ) = 0

10th Edition

ISBN: 9781292060545

Author: James W. Nilsson, Susan Riedel

Publisher: Pearson Education Limited

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Chapter 13, Problem 35P

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Find the time domain expression of i(t) for t>0 using Thevenin equivalent circuit.

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Chapter 13, Problem 33P

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Chapter 13 Solutions

Electric Circuits, Global Edition

Ch. 13.2 - Prob. 1AP Ch. 13.2 - The parallel circuit in Example 13.1 is placed in...Ch. 13.3 - Prob. 3AP Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.3 - The dc current and dc voltage sources are applied...Ch. 13.3 - Prob. 6AP Ch. 13.3 - Using the results from Example 13.7 for the...Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.4 - Derive the numerical expression for the transfer...Ch. 13.5 - Find (a) the unit step and (b) the unit impulse...

Ch. 13.5 - The unit impulse response of a circuit is υo(t) =...Ch. 13.7 - The current source in the circuit shown is...Ch. 13.7 - For the circuit shown, find the steady-state...Ch. 13 - Prob. 1P Ch. 13 - Prob. 2P Ch. 13 - Prob. 3P Ch. 13 - Prob. 4P Ch. 13 - An 8 kΩ resistor, a 25 mH inductor, and a 62.5 pF...Ch. 13 - Prob. 6P Ch. 13 - Find the poles and zeros of the impedance seen...Ch. 13 - Find the poles and zeros of the impedance seen...Ch. 13 - Prob. 9P Ch. 13 - Prob. 10P Ch. 13 - Prob. 13P Ch. 13 - Prob. 15P Ch. 13 - There is no energy stored in the circuit in Fig....Ch. 13 - There is no energy stored in the circuit in Fig....Ch. 13 - Prob. 25P Ch. 13 - Prob. 28P Ch. 13 - The switch in the circuit seen in Fig. P13.32 has...Ch. 13 - Prob. 31P Ch. 13 - Prob. 33P Ch. 13 - Prob. 35P Ch. 13 - Prob. 46P Ch. 13 - Prob. 47P Ch. 13 - Find the transfer function H(s) − Vo/Vi for the...Ch. 13 - Prob. 49P Ch. 13 - Prob. 50P Ch. 13 - Prob. 51P Ch. 13 - Prob. 53P Ch. 13 - Prob. 54P Ch. 13 - The operational amplifier in the circuit in Fig....Ch. 13 - Find the transfer function Io/Ig as a function of...Ch. 13 - Prob. 58P Ch. 13 - Prob. 59P Ch. 13 - Prob. 60P Ch. 13 - Prob. 61P Ch. 13 - Assume the voltage impulse response of a circuit...Ch. 13 - Prob. 68P Ch. 13 - The input voltage in the circuit seen in Fig....Ch. 13 - Find the impulse response of the circuit shown in...Ch. 13 - Prob. 73P Ch. 13 - Prob. 74P Ch. 13 - Prob. 75P Ch. 13 - The op amp in the circuit seen in Fig. P13.81 is...Ch. 13 - Prob. 78P Ch. 13 - The transfer function for a linear time-invariant...Ch. 13 - Prob. 80P Ch. 13 - Prob. 81P Ch. 13 - Prob. 82P Ch. 13 - Prob. 84P Ch. 13 - Prob. 85P Ch. 13 - The parallel combination of R2 and C2 in the...Ch. 13 - Show that if R1C1 = R2C2 in the circuit shown in...Ch. 13 - The switch in the circuit in Fig P13.91 has been...Ch. 13 - Prob. 90P Ch. 13 - Prob. 91P

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Find the time domain expression of i ( t ) for t &gt; 0 using Thevenin equivalent circuit.

Find the time domain expression of i(t) for t>0 using Thevenin equivalent circuit.

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Chapter 13 Solutions

Find the time domain expression of i ( t ) for t > 0 using Thevenin equivalent circuit.