Explanation Given data: Refer to Figure 16.12 in the textbook. Refer to Example 16.4 in the textbook. The initial voltage v ( 0 ) = 5 V . The initial current i ( 0 ) = − 1 A . 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 of the resistor. 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 of the inductor. 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 of the capacitor. Calculation: Refer to Figure 16.12 (b) in the textbook. The current I through the inductor is calculated by using Superposition theorem. Letting the current I is I = I 1 + I 2 + I 3 . To find the current I 1 due to voltage source 10 s when acting alone. Figure 1 shows the modified circuit diagram. Apply Kirchhoff’s current law at node V 1 in Figure 1. V 1 − ( 10 s ) ( 10 3 ) + V 1 5 s + V 1 ( 10 s ) = 0 V 1 ( 10 3 ) − ( 10 s ) ( 10 3 ) + V 1 5 s + s V 1 10 = 0 3 V 1 10 + V 1 5 s + s V 1 10 = 3 s ( 3 10 + 1 5 s + s 10 ) V 1 = 3 s Rearrange the equation as follows, V 1 = ( 3 s ) ( 3 10 + 1 5 s + s 10 ) = ( 3 s ) ( 3 s + 2 + s 2 10 s ) = 30 s 2 + 3 s + 2 In Figure 1, the current I 1 is calculated as follows. I 1 = V 1 5 s Substitute 30 s 2 + 3 s + 2 for V 1 to find the current I 1 . I 1 = ( 30 s 2 + 3 s + 2 ) 5 s = 6 s ( s 2 + 2 s + 1 s + 2 ) = 6 s ( s 2 + 2 s + 1 s + 2 ) = 6 s ( s ( s + 2 ) + 1 ( s + 2 ) ) Simplify the equation as follows, I 1 = 6 s ( s + 1 ) ( s + 2 ) Expand I 1 using partial fraction. I 1 = 6 s ( s + 1 ) ( s + 2 ) = A s + B s + 1 + C s + 2 (4) Here, A , B , and C are the constants. Find the constants by using algebraic method. Consider the partial fraction, 6 s ( s + 1 ) ( s + 2 ) = A s + B s + 1 + C s + 2 6 s ( s + 1 ) ( s + 2 ) = A ( s + 1 ) ( s + 2 ) + B s ( s + 2 ) + C s ( s + 1 ) s ( s + 1 ) ( s + 2 ) 6 = A ( s + 1 ) ( s + 2 ) + B s ( s + 2 ) + C s ( s + 1 ) 6 = ( A s + A ) ( s + 2 ) + B s 2 + 2 B s + C s 2 + C s Reduce the equation as follows, 6 = A s 2 + A s + 2 A s + 2 A + B s 2 + 2 B s + C s 2 + C s 6 = ( A + B + C ) s 2 + ( A + 2 A + 2 B + C ) s + 2 A 6 = ( A + B + C ) s 2 + ( 3 A + 2 B + C ) s + 2 A (5) Equating the coefficients of s 2 in equation (5). A + B + C = 0 (6) Equating the coefficients of s in equation (5). 0 = 3 A + 2 B + C (7) Equating the coefficients of constant term in equation (5). 6 = 2 A A = 6 2 A = 3 (8) Substitute equation (8) in equation (7). 0 = 3 ( 3 ) + 2 B + C 2 B = − 9 − C B = − 9 − C 2 (9) Substitute equation (8) and (9) in equation (6). 3 + ( − 9 − C 2 ) + C = 0 3 − 9 2 − C 2 + C = 0 C 2 = − 3 + 9 2 C = 2 ( − 3 + 9 2 ) Reduce the equation as follows, C = 2 ( 3 2 ) C = 3 (10) Substitute equation (10) in equation (9). B = − 9 − 3 2 = − 6 Substitute 3 for A , − 6 for B , and 3 for C in equation (4) to find I 1 . I 1 = 3 s − 6 s + 1 + 3 s + 2 (11) To find the current I 2 due to current source i ( 0 ) s when acting alone. Figure 2 shows the modified circuit diagram. Apply Kirchhoff’s current law at node V 2 in Figure 2. V 2 ( 10 3 ) + V 2 5 s + V 2 ( 10 s ) = − i ( 0 ) s Substitute − 1 for i ( 0 ) to find V 2 . ( 3 10 + 1 5 s + s 10 ) V 2 = 1 s V 2 = ( 1 s ) ( 3 10 + 1 5 s + s 10 ) V 2 = ( 1 s ) ( 3 s + 2 + s 2 10 s ) V 2 = 10 s 2 + 3 s + 2 In Figure 2, the current I 2 is calculated as follows. I 2 = V 2 5 s + i ( 0 ) s Substitute 10 s 2 + 3 s + 2 for V 2 and − 1 for i ( 0 ) to find I 2 . I 2 = ( 10 s 2 + 3 s + 2 ) 5 s − 1 s = 2 s ( s 2 + 2 s + 1 s + 2 ) − 1 s = 2 s ( s ( s + 2 ) + 1 ( s + 2 ) ) − 1 s I 2 = 2 s ( s + 1 ) ( s + 2 ) − 1 s (12) Consider X = 2 s ( s + 1 ) ( s + 2 ) in equation (12). Expand X using partial fraction method

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16.3, Problem 5PP

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Calculate the current i(t) through the inductor for all time t>0 using superposition.

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Calculate the current i ( t ) through the inductor for all time t > 0 using superposition.

Solution Summary: The author calculates the current i(t) through the inductor for all time t>0 using superposition.

Calculate the current i(t) through the inductor for all time t>0 using superposition.

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Chapter 16 Solutions

Calculate the current i ( t ) through the inductor for all time t &gt; 0 using superposition.

Solution Summary: The author calculates the current i(t) through the inductor for all time t>0 using superposition.

Calculate the current i(t) through the inductor for all time t>0 using superposition.

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Chapter 16 Solutions

Calculate the current i ( t ) through the inductor for all time t > 0 using superposition.