Explanation Given data: Refer to Figure 16.58 in the textbook. The voltage source is, v ( t ) = 25 e − t u ( t ) V (1) The current source is, i ( t ) = 4.5 u ( t ) A (2) Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (3) Here, R is the value of resistance of the resistor. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (4) Here, L is the value of inductance of the inductor. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (5) Here, C is the value of capacitance of the capacitor. Calculation: Take Laplace transform of equation (1) and (2). V o ( s ) = 25 s + 1 { ∵ ℒ [ e − a t u ( t ) ] = 1 s + a } and I ( s ) = 4.5 s { ∵ ℒ [ u ( t ) ] = 1 s } Using equation (3), (4) and (5), the given circuit is redrawn as shown in Figure 1 in s -domain. Apply Kirchhoff’s current law at node V o in Figure 1. V o − ( 25 s + 1 ) s + V o ( 1 2 s ) + V o 4 = 4.5 s V o s − 25 s ( s + 1 ) + 2 s V o + 0.25 V o = 4.5 s ( 1 s + 2 s + 0.25 ) V o = 4.5 s + 25 s ( s + 1 ) ( 1 + 2 s 2 + 0.25 s s ) V o = 4.5 ( s + 1 ) + 25 s ( s + 1 ) Reduce the equation as follows, ( 2 s 2 + 0.25 s + 1 ) V o = 4.5 ( s + 1 ) + 25 ( s + 1 ) V o = 4.5 ( s + 1 ) + 25 2 ( s 2 + 0.125 s + 0.5 ) ( s + 1 ) V o = 2.25 ( s + 1 ) + 12.5 ( s 2 + 0.125 s + 0.5 ) ( s + 1 ) V o = 2.25 s + 2.25 + 12.5 ( s 2 + 0.125 s + 0.5 ) ( s + 1 ) Simplify the equation as follows, V o = 2.25 s + 14.75 ( s + 1 ) ( s 2 + 0.125 s + 0.5 ) (6) From equation (6), the roots of s 2 + 0.125 s + 0.5 is calculated as follows: s = − 0.125 ± 0.125 2 − 4 ( 1 ) ( 0.5 ) 2 ( 1 ) { ∵ x = − b ± b 2 − 4 a c 2 a } = − 0.125 ± − 1.984375 2 = − 0.0625 ± j 0.7043 Therefore, the roots of s 2 + 0.125 s + 0.5 is ( s + 0.0625 + j 0.7043 ) ( s + 0.0625 − j 0.7043 ) . Substitute ( s + 0.0625 + j 0.7043 ) ( s + 0.0625 − j 0.7043 ) for s 2 + 0.125 s + 0.5 in equation (6). V o = 2.25 s + 14.75 ( s + 1 ) ( s + 0.0625 + j 0.7043 ) ( s + 0.0625 − j 0.7043 ) (7) Expand V o using partial fraction method. V o = A s + 1 + B s + 0.0625 + j 0.7043 + C s + 0.0625 − j 0.7043 (8) Here, A , B , and C are the constants. Now, to find the constants by using residue method. Constant A : A = ( s + 1 ) V o | s + 1 = 0 (9) Substitute equation (7) in equation (9) to find the constant A . A = ( s + 1 ) × 2.25 s + 14.75 ( s + 1 ) ( s + 0.0625 + j 0.7043 ) ( s + 0.0625 − j 0.7043 ) | s + 1 = 0 = 2.25 s + 14.75 ( s + 0.0625 + j 0.7043 ) ( s + 0.0625 − j 0.7043 ) | s = − 1 = 2.25 ( − 1 ) + 14.75 ( − 1 + 0.0625 + j 0.7043 ) ( − 1 + 0.0625 − j 0.7043 ) = 9

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ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 35P

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Find the voltage vo(t) in the circuit of Figure 16.58.

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Chapter 16, Problem 34P

Chapter 16, Problem 36P

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Find the voltage v o ( t ) in the circuit of Figure 16.58.

Solution Summary: The author explains how to calculate the voltage v_o(t) in the circuit of Figure 16.58.

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Find the voltage vo(t) in the circuit of Figure 16.58.

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Chapter 16 Solutions