Explanation Given data: Refer to Figure 16.78 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. 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 voltage source v s ( t ) is calculated as follows: Substitute − 1 for t in v s ( t ) . v s ( t ) = 20 u ( ( − 1 ) ) V = 20 ( 0 ) V { ∵ u ( t ) = 1 for t > 0 } = 0 V Since, the value of voltage source is zero, it is short 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 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 For the time t > 0 , the circuit remains same as shown in Figure 1. Apply Laplace transform for voltage source v s ( t ) to find V s ( s ) . V s ( s ) = 20 s { ∵ L ( u ( t ) ) = 1 s } Substitute 4 Ω for R 1 in equation (1) to find Z R 1 . Z R 1 = 4 Ω Substitute 6 Ω for R 2 in equation (1) to find Z R 2 . Z R 2 = 6 Ω Substitute 2 H for L 1 in equation (2) to find Z L 1 . Z L 1 = s ( 2 H ) = 2 s Substitute 2 H for L 2 in equation (2) to find Z L 2 . Z L 2 = s ( 2 H ) = 2 s Convert the Figure 1 into s -domain as shown in Figure 3. Apply nodal analysis at node V ( s ) in Figure 3. V ( s ) − 20 s 4 + V ( s ) 2 s + V ( s ) 6 + 2 s = 0 V ( s ) 4 − 20 4 s + V ( s ) 2 s + V ( s ) 6 + 2 s = 0 V ( s ) [ 1 4 + 1 2 s + 1 2 ( s + 3 ) ] = 20 4 s V ( s ) [ s ( s + 3 ) + 2 ( s + 3 ) + 2 s 4 s ( s + 3 ) ] = 20 4 s Simplify the above equation to find V ( s ) . V ( s ) = 20 4 s ( 4 s ( s + 3 ) s ( s + 3 ) + 2 ( s + 3 ) + 2 s ) = 20 ( ( s + 3 ) s 2 + 3 s + 2 s + 6 + 2 s ) = 20 ( s + 3 ) s 2 + 7 s + 6 Refer to Figure 2, the current I 1 ( s ) is calculated as follows: I 1 ( s ) = V ( s ) 2 s Substitute 20 ( s + 3 ) s 2 + 7 s + 6 for V ( s ) in above equation to find I 1 ( s ) . I 1 ( s ) = ( 20 ( s + 3 ) s 2 + 7 s + 6 ) 2 s = 10 ( s + 3 ) s ( s 2 + 6 s + s + 6 ) = 10 ( s + 3 ) s ( s ( s + 6 ) + 1 ( s + 6 ) ) = 10 ( s + 3 ) s ( s + 1 ) ( s + 6 ) Take partial fraction for above equation. I 1 ( s ) = 10 ( s + 3 ) s ( s + 1 ) ( s + 6 ) = A s + B s + 1 + C s + 6 (3) The equation (3) can also be written as follows: 10 ( s + 3 ) s ( s + 1 ) ( s + 6 ) = A ( s + 1 ) ( s + 6 ) + B s ( s + 6 ) + C s ( s + 1 ) s ( s + 1 ) ( s + 6 ) Simplify the above equation as follows: 10 ( s + 3 ) = A ( s + 1 ) ( s + 6 ) + B s ( s + 6 ) + C s ( s + 1 ) (4) Substitute 0 for s in equation (4) to find A

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 55P

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Find the expression of current i1(t) and i2(t) for t>0 in the circuit of Figure 16.78.

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Chapter 16, Problem 54P

Chapter 16, Problem 56P

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Chapter 16 Solutions

Fundamentals of Electric Circuits

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Find the expression of current i 1 ( t ) and i 2 ( t ) for t > 0 in the circuit of Figure 16.78.

Solution Summary: The author explains the formula used to calculate the value of step input in the circuit of Figure 16.78.

Find the expression of current i1(t) and i2(t) for t>0 in the circuit of Figure 16.78.

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Chapter 16 Solutions

Find the expression of current i 1 ( t ) and i 2 ( t ) for t &gt; 0 in the circuit of Figure 16.78.

Solution Summary: The author explains the formula used to calculate the value of step input in the circuit of Figure 16.78.

Find the expression of current i1(t) and i2(t) for t>0 in the circuit of Figure 16.78.

Want to see the full answer?

Chapter 16 Solutions

Find the expression of current i 1 ( t ) and i 2 ( t ) for t > 0 in the circuit of Figure 16.78.