Explanation Given data: Refer to Figure 16.66 in the textbook. The voltage source is, v 1 ( t ) = 36 V (1) The current source is, i 1 ( t ) = 9 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: For t < 0 : In the given circuit, the inductor reaches steady state and acts like an short circuit whereas the capacitor will behaves as an open circuit which is shown in Figure 1. Figure 1 shows the modified circuit diagram for t < 0 . In Figure 1, the initial current through the inductor is calculated as follows. i ( 0 ) = 36 V 4 Ω + 9 A = 9 A + 9 A { ∵ 1 A = 1 V 1 Ω } = 18 A Take Laplace transform of i ( 0 ) (initial inductor current). I ( 0 ) = 18 s { ∵ ℒ [ u ( t ) ] = 1 s } Take Laplace transform of equation (1) and (2). V 1 ( s ) = 36 s { ∵ ℒ [ u ( t ) ] = 1 s } and I 1 ( s ) = 9 s { ∵ ℒ [ u ( t ) ] = 1 s } For t > 0 : Using equation (3), (4), and (5), the given circuit is redrawn as shown in Figure 2 in s -domain for t > 0 . Apply Kirchhoff’s current law at node V in Figure 2. V 5 s + V ( 20 s ) + V 5 = 9 s − 18 s ( 1 5 s + s 20 + 1 5 ) V = − 9 s V = − ( 9 s ) ( 1 5 s + s 20 + 1 5 ) V = − ( 9 s ) ( 4 + s 2 + 4 s 20 s ) Reduce the equation as follows, V = − 180 s 2 + 4 s + 4 = − 180 s 2 + 2 s + 2 s + 4 = − 180 s ( s + 2 ) + 2 ( s + 2 ) = − 180 ( s + 2 ) ( s + 2 ) Simplify the equation as follows, V = − 180 ( s + 2 ) 2 (6) In Figure 2, the current through the inductor is calculated as follows. I = V 5 s + 18 s (7) Substitute equation (6) in equation (7). I = − 180 ( s + 2 ) 2 5 s + 18 s = − 180 5 s ( s + 2 ) 2 + 18 s I = − 36 s ( s + 2 ) 2 + 18 s (8) Expand − 36 s ( s + 2 ) 2 using partial fraction method. − 36 s ( s + 2 ) 2 = A s + B s + C ( s + 2 ) 2 (9) Here, A , B , and C are the constants. Find the constants by using algebraic method. Reduce the equation (9) as follows, − 36 s ( s + 2 ) 2 = A ( s + 2 ) 2 + ( B s + C ) s s ( s + 2 ) 2 − 36 = A ( s + 2 ) 2 + ( B s + C ) s − 36 = A ( s 2 + 4 s + 4 ) + B s 2 + C s { ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 } − 36 = A s 2 + 4 A s + 4 A + B s 2 + C s Reduce the equation as follows, − 36 = ( A + B ) s 2 + ( 4 A + C ) s + 4 A (10) Equating the coefficients of s 2 in equation (10)

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 43P

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Find the current i(t) for t>0 in the circuit of Figure 16.66.

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Chapter 16, Problem 42P

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