Explanation Given data: Refer to Figure 16.67 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 at time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. For time t < 0 , the current source i s ( t ) is calculated as follows: i s ( t ) = 24 ( 0 ) { ∵ u ( t ) = 0 for t < 0 } = 0 When the value of current source is zero, it is open circuited. Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the short circuited inductor is connected in parallel with the resistor ( R 2 ) . Since the inductor is short circuited, the total current flows through it and the resistor ( R 2 ) is also short circuited. Now, the Figure 2 is reduced as shown in Figure 3. The current flow through the inductor is calculated as follows: i L ( 0 − ) = 120 V 10 Ω = 12 A Refer to Figure 3, since the inductor is short circuited there is no voltage across the capacitor. v C ( 0 − ) = 0 V The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 12 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 0 V For time t > 0 , the Figure 1 is reduced as shown in Figure 4. Use source transformation to convert the current source ( i 2 ) into voltage source ( v 2 ) . i 1 = 120 ( 1 ) V 10 Ω = 120 u ( t ) V 10 Ω { ∵ u ( t ) = 1 for t > 0 } = 120 u ( t ) A Now, the Figure 4 is reduced as shown in Figure 5. Refer to Figure 5, the current sources i 1 and i s ( t ) are parallel aided. The resistor R 1 and R 2 are connected in parallel. The total current is calculated as follows: i 2 = 12 u ( t ) A + 24 u ( t ) A i 2 = 36 u ( t ) A (4) The equivalent resistor is calculated as follows: R eq = ( 10 Ω ) ( 40 Ω ) 10 Ω + 40 Ω = 8 Ω Now, the Figure 5 is reduced as shown in Figure 6. Apply Laplace transform for equation (4) to find I 2 ( s ) . I 2 ( s ) = 36 s { ∵ L ( u ( t ) ) = 1 s } Substitute 8 Ω for R eq in equation (1) to find Z R eq . Z R eq = 8 Substitute 4 H for L in equation (2) to find Z L . Z L = s ( 4 H ) = 4 s Substitute 10 mF for C in equation (3) to find Z C . Z C = 1 s ( 10 mF ) = 1 s ( 10 × 10 − 3 F ) { ∵ 1 m = 10 − 3 } = 100 s Using element transformation methods with initial conditions convert the Figure 6 into s -domain. Apply nodal analysis at node V ( s ) for the circuit shown in figure 7. − 36 s + V ( s ) 8 + V ( s ) ( 100 s ) + V ( s ) 4 s + i ( 0 ) s = 0 V ( s ) 8 + s V ( s ) 100 + V ( s ) 4 s = 36 s − i ( 0 ) s V ( s ) [ 1 8 + s 100 + 1 4 s ] = 36 s − i ( 0 ) s V ( s ) [ 12.5 s + s 2 + 25 100 s ] = 36 s − i ( 0 ) s Substitute 12 for i ( 0 ) in above equation. V ( s ) [ 12.5 s + s 2 + 25 100 s ] = 36 s − 12 s V ( s ) [ s 2 + 12

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 44P

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Find the expression of current i(t) for t>0 in the circuit of Figure 16.67.

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Chapter 16, Problem 43P

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Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.67.

Solution Summary: The author explains the expression of current i(t) in the circuit of Figure 16.67.

Find the expression of current i(t) for t>0 in the circuit of Figure 16.67.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t &gt; 0 in the circuit of Figure 16.67.

Solution Summary: The author explains the expression of current i(t) in the circuit of Figure 16.67.

Find the expression of current i(t) for t>0 in the circuit of Figure 16.67.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.67.