Explanation Given data: Refer to Figure 16.65 in the textbook. 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. For a DC circuit, at steady state condition when the switch is opened at time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the resistors R 1 and R 2 are connected in series with the voltage source v s . The current flow through the inductor is calculated as follows: i L ( 0 − ) = − 180 2 Ω + 1 Ω = − 60 A Refer to Figure 3, the capacitor is connected in parallel with the resistor R 1 . For the parallel connection the voltage is same. v C ( 0 − ) = ( 60 A ) ( 1 Ω ) = 60 V For time t > 0 , the switch is closed and the Figure 1 is reduced as shown in Figure 3. Refer to Figure 3, the short circuit path is connected in parallel with the resistor R 2 and the voltage source v s . The total current flows through the short circuit path so the resistor R 2 and the voltage source v s is also short circuited. Now, the Figure 3 is reduced as shown in Figure 4. Substitute 1 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 1 Substitute 1 H for L in equation (2) to find Z L . Z L = s ( 1 H ) = s Substitute 1 4 F for C in equation (3) to find Z C . Z C = 1 s ( 1 4 F ) = 4 s Using element transformation methods with initial conditions convert the Figure 4 into s -domain. Apply nodal analysis at node V ( s ) for the circuit shown in Figure 5. V ( s ) 1 + V ( s ) ( 4 s ) − ( 60 s ) ( 4 s ) + V ( s ) s + i ( 0 ) s = 0 V ( s ) 1 + s V ( s ) 4 − 15 + V ( s ) s + i ( 0 ) s = 0 V ( s ) [ 1 + s 4 + 1 s ] = 15 − i ( 0 ) s V ( s ) [ 4 s + s 2 + 4 4 s ] = 15 s − i ( 0 ) s Substitute − 60 for i ( 0 ) in above equation. V ( s ) [ 4 s + s 2 + 4 4 s ] = 15 s − ( − 60 ) s V ( s ) V ( s ) [ 4 s + s 2 + 4 4 s ] = 15 s + 60 s V ( s ) [ s 2 + 4 s + 4 4 s ] = 15 ( s + 4 ) s Simplify the above equation to find V ( s ) . V ( s ) = 15 ( s + 4 ) s ( 4 s s 2 + 4 s + 4 ) = 60 ( s + 4 ) s 2 + 2 s + 2 s + 4 = 60 ( s + 4 ) s 2 + 2 s + 2 s + 4 = 60 ( s + 4 ) s ( s + 2 ) + 2 ( s + 2 ) Simplify the above equation to find V ( s ) . V ( s ) = 60 ( s + 4 ) ( s + 2 ) ( s + 2 ) = 60 ( s + 4 ) ( s + 2 ) 2 Take partial fraction for above equation. V ( s ) = 60 ( s + 4 ) ( s + 2 ) 2 = A s + 2 + B ( s + 2 ) 2 (4) The equation (4) can also be written as follows: 60 ( s + 4 ) ( s + 2 ) 2 = A ( s + 2 ) + B ( s + 2 ) 2 Simplify the above equation as follows: 60 ( s + 4 ) = A ( s + 2 ) + B 60 s + 240 = s A + 2 A + B (5) Equate the co-efficient of s term in equation (5) to find A . A = 60 Equate the co-efficient of constant term in equation (5) to find B . 2 A + B = 240 Rearrange the above equation to find B . B = 240 − 2 A Substitute 60 for A in above equation to find B . B = 240 − 2 ( 60 ) = 240 − 120 = 120 Substitute 60 for A , and 120 for B in equation (4) to find V ( s ) . V ( s ) = 60 s + 2 + 120 ( s + 2 ) 2 Take inverse Laplace transform for above equation to find v ( t ) . v ( t ) = 60 e − 2 t u ( t ) + 120 t e − 2 t u ( t ) { ∵ L − 1 ( 1 s + a ) = e − a t u ( t ) , L − 1 ( 1 ( s + a ) n ) = t n − 1 ( n − 1 ) ! e − a t u ( t ) } = [ 60 e − 2 t + 120 t e − 2 t ] u ( t ) V Refer to Figure 5, the current I ( s ) is calculated as follows: I ( s ) = V ( s ) s + i ( 0 ) s Substitute 60 ( s + 4 ) ( s + 2 ) 2 for V ( s ) , and − 60 for i ( 0 ) in above equation to find I ( s )

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 42P

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Find the expression of current i(t) and voltage v(t) for t>0 in the circuit of Figure 16.65.

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Chapter 16, Problem 41P

Chapter 16, Problem 43P

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Find the expression of current i ( t ) and voltage v ( t ) for t > 0 in the circuit of Figure 16.65.

Solution Summary: The author explains the formula used to calculate the impedance of a resistor in s -domain.

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Find the expression of current i(t) and voltage v(t) for t>0 in the circuit of Figure 16.65.

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Chapter 16 Solutions

Find the expression of current i ( t ) and voltage v ( t ) for t &gt; 0 in the circuit of Figure 16.65.

Solution Summary: The author explains the formula used to calculate the impedance of a resistor in s -domain.

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Find the expression of current i(t) and voltage v(t) for t>0 in the circuit of Figure 16.65.

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Chapter 16 Solutions

Find the expression of current i ( t ) and voltage v ( t ) for t > 0 in the circuit of Figure 16.65.