Explanation Given data: Refer to Figure 16.54 in the textbook. The voltage source is, v 1 ( t ) = 24 V (1) The current source is, i 1 ( t ) = 10 u ( t ) 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: Take Laplace transform of equation (1) and (2). V 1 ( s ) = 24 s { ∵ ℒ [ u ( t ) ] = 1 s } and I 1 ( s ) = 10 s { ∵ ℒ [ u ( t ) ] = 1 s } Using equation (3), (4), and (5), the given circuit is redrawn as shown in Figure 1 in s -domain. In Figure 1, the current source 10 s with parallel resistance 4 ohms is converted into voltage source with series resistance using source transformation method which is shown in Figure 2. Figure 2 shows the modified circuit diagram with initial conditions for t > 0 . Since, the initial voltage for the capacitor is equal to the source voltage 24 V. That is, v ( 0 ) = 24 V in Figure 2 which is shown in s -domain as 24 s . Apply Kirchhoff’s voltage law for the loop in Figure 2. − 24 s + 2 I + 5 s I + 4 I + 40 s + 5 s I + 24 s + 2 I = 0 8 I + 5 s I − 5 s I = − 40 s I ( 8 + 5 s − 5 s ) = − 40 s I = − 40 s ( 8 + 5 s − 5 s ) Reduce the equation as follows, I = − 40 8 s + 5 s 2 − 5 = − 40 5 ( s 2 + 1.6 s + 1 ) I = − 8 s 2 + 1.6 s + 1 (6) From equation (6), the roots of s 2 + 1.6 s + 1 is calculated as follows. s = − 1.6 ± 1.6 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) { ∵ x = − b ± b 2 − 4 a c 2 a } = − 1.6 ± − 1.44 2 = − 0.8 ± j 0.6 Therefore, the roots of s 2 + 1.6 s + 1 is ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) . Substitute ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) for s 2 + 1.6 s + 1 in equation (6). I = − 8 ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) (7) Expand I using partial fraction method. I = − 8 ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) = A s + 0.8 + j 0.6 + B s + 0.8 − j 0.6 (8) Here, A and B are the constants. Find the constants by using residue method. Constant A : A = ( s + 0.8 + j 0.6 ) I | s + 0.8 + j 0.6 = 0 (9) Substitute equation (8) in equation (9) to find the constant A . A = ( s + 0.8 + j 0.6 ) × ( − 8 ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) ) | s + 0.8 + j 0.6 = 0 = − 8 s + 0.8 − j 0.6 | s = − 0.8 − j 0.6 = − 8 − 0.8 − j 0.6 + 0.8 − j 0.6 = − 6.667 j Convert into polar form, A = 6.667 ∠ − 90 ° Constant B : B = ( s + 0.8 − j 0.6 ) I | s + 0.8 − j 0.6 = 0 (10) Substitute equation (8) in equation (10) to find the constant B . B = ( s + 0.8 − j 0.6 ) × ( − 8 ( s + 0.8 + j 0.6 ) ( s + 0.8 − j 0.6 ) ) | s + 0.8 − j 0.6 = 0 = − 8 s + 0.8 + j 0.6 | s = − 0.8 + j 0.6 = − 8 − 0.8 + j 0.6 + 0.8 + j 0.6 = 6.667 j Convert into polar form, B = 6.667 ∠ 90 ° Substitute 6.667 ∠ − 90 ° for A and 6.667 ∠ 90 ° for B in equation (8) to find I . I = 6.667 ∠ − 90 ° s + 0.8 + j 0.6 + 6.667 ∠ 90 ° s + 0.8 − j 0.6 (11) Take inverse Laplace transform for the equation (11). i ( t ) = [ ( 6.667 ∠ − 90 ° ) e − ( 0.8 + j 0.6 ) t u ( t ) + ( 6.667 ∠ 90 ° ) e − ( 0.8 − j 0.6 ) t u ( t ) ] { ∵ ℒ [ e − a t u ( t ) ] = 1 s + a } = [ ( 6.667 ∠ − 90 ° ) e − 0.8 t e − j 0.6 t + ( 6.667 ∠ 90 ° ) e − 0.8 t e j 0.6 t ] u ( t ) = [ ( 6.667 e − j 90 ° ) e − 0.8 t e − j 0.6 t + ( 6.667 e j 90 ° ) e − 0.8 t e j 0.6 t ] u ( t ) { ∵ ∠ θ = e j θ } = 6.667 e − 0.8 t [ e − j 90 ° e − j 0.6 t + e j 90 ° e j 0

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 31P

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Find the voltage v(t) and the current i(t) for t>0 in the circuit of Figure 16.54.

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Chapter 16, Problem 30P

Chapter 16, Problem 32P

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Find the voltage v ( t ) and the current i ( t ) for t > 0 in the circuit of Figure 16.54.

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

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Find the voltage v(t) and the current i(t) for t>0 in the circuit of Figure 16.54.

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Chapter 16 Solutions

Find the voltage v ( t ) and the current i ( t ) for t &gt; 0 in the circuit of Figure 16.54.

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

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Find the voltage v(t) and the current i(t) for t>0 in the circuit of Figure 16.54.

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Chapter 16 Solutions

Find the voltage v ( t ) and the current i ( t ) for t > 0 in the circuit of Figure 16.54.