Explanation Given data: Refer to Figure 16.69 in the textbook. Formula used: Write an expression to calculate the value of step input. u ( t ) = { 0 t < 0 1 t > 0 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 time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. The value of current source i s ( t ) calculated as follows: Substitute − 1 for t in v s ( t ) . i s ( t ) = 5 e − 2 ( − 1 ) u ( ( − 1 ) ) A = 5 e 2 ( 0 ) A { ∵ u ( t ) = 1 for t > 0 } = 0 A Since the value of current source is zero, it is open circuited and the Figure 1 becomes as shown in Figure 2. Refer to Figure 2, there is no source connected in a circuit. Therefore, there is no current flow through inductor and the voltage across the capacitor is zero. Thus, the initial conditions are zero. The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 0 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 0 V For the time t > 0 , the circuit remains same as shown in Figure 1. Apply Laplace transform for current source i s ( t ) to find I s ( s ) . I s ( s ) = 5 ( 1 s + 2 ) { ∵ L ( e − a t u ( t ) ) = 1 s + a } = 5 s + 2 Substitute 2 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 2 Ω Substitute 1 Ω for R 2 in equation (1) to find Z R 2 . Z R 2 = 1 Ω Substitute 2 H for L in equation (2) to find Z L . Z L = s ( 2 H ) = 2 s Substitute 1 F for C in equation (3) to find Z C . Z C = 1 s ( 1 F ) = 1 s Convert the Figure 1 into s -domain as shown in Figure 3. Apply nodal analysis at node V o ( s ) in Figure 3. V o ( s ) 2 + 1 s − 5 s + 2 + V o ( s ) 2 s + 1 = 0 V o ( s ) 2 + 1 s + V o ( s ) 2 s + 1 = 5 s + 2 V o ( s ) [ s 2 s + 1 + 1 2 s + 1 ] = 5 s + 2 V o ( s ) [ s + 1 2 s + 1 ] = 5 s + 2 Simplify the above equation to find V o ( s )

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 46P

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Find the expression of current io(t) in the circuit of Figure 16.69.

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Chapter 16, Problem 45P

Chapter 16, Problem 47P

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Find the expression of current i o ( t ) in the circuit of Figure 16.69.

Solution Summary: The author explains the expression of current i_o(t) in the circuit of Figure 16.69.

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Find the expression of current io(t) in the circuit of Figure 16.69.

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Chapter 16 Solutions