Explanation Given data: Refer to given circuit in the textbook. At time t = 0 the switch moves to position ‘b’. 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 Z M = s M (2) Here, L is the value of inductance, and M is the value of mutual inductance. Calculation: The given circuit is redrawn as shown in Figure 1. The T-equivalent circuit of the mutually connected series aiding inductors is shown in Figure 2. Implement the Figure 2 in Figure 1 as shown in Figure 3. Consider, L eq1 = L 1 + M L eq2 = L 2 + M M eq = − M Now, the Figure 3 is drawn as shown in Figure 4. For a DC circuit at steady state condition the switch is in positon ‘a’ for time t < 0 and the inductor acts like a short circuit. Now, the Figure 4 is reduced as shown in Figure 5. Refer to Figure 5, the short circuited inductor ( L eq2 ) is connected in parallel with the resistor R 3 . Since, the inductor is short circuited the entire current flows through it. Now, the Figure 5 is reduced as shown in Figure 6. Refer to Figure 6, the inductor L eq1 and L eq2 are connected in series. For a series connection, the current is same. i L 1 ( 0 − ) = i L 2 ( 0 − ) The current through inductors is calculated as follows: i L 1 ( 0 − ) = i L 2 ( 0 − ) = 90 V 5 Ω = 18 A The current through inductor is always continuous so that, i L 1 ( 0 ) = i L 1 ( 0 − ) = i L 1 ( 0 + ) = 18 A i L 2 ( 0 ) = i L 2 ( 0 − ) = i L 12 ( 0 + ) = 18 A For time t > 0 , the switch moves to position ‘b’ and the circuit in Figure 4 is reduced as shown in Figure 7. The Figure 7 can also be redrawn as shown in Figure 8. Use equation (1) to find Z R 2 . Z R 2 = 20 Ω Use equation (1) to find Z R 3 . Z R 3 = 10 Ω Use equation (2) to find Z L eq1 . Z L eq1 = s ( 4 H ) = 4 s Ω Use equation (2) to find Z L eq2 . Z L eq2 = s ( 3 H ) = 3 s Ω Use equation (2) to find Z M eq Z M eq = s ( − 1 H ) = − s Ω Convert the Figure 8 into s -domain as shown in Figure 9. The value of L eq1 i L 1 ( 0 ) is calculated as follows: L eq1 i L 1 ( 0 ) = ( 4 H ) ( 18 A ) = 72 V The value of L eq2 i L 2 ( 0 ) is calculated as follows: L eq2 i L 2 ( 0 ) = ( 3 H ) ( 18 A ) = 54 V Now, the Figure 9 is redrawn as shown in Figure 10 with the calculated values. Apply Kirchhoff’s current law at node V in Figure 10. V − 72 4 s + 20 + V − s + 10 + V + 54 3 s = 0 V 4 s + 20 − 72 4 s + 20 + V − s + 10 + V 3 s + 54 3 s = 0 V 4 s + 20 + V − s + 10 + V 3 s = 72 4 s + 20 − 54 3 s V ( 1 4 s + 20 + 1 − s + 10 + 1 3 s ) = 72 4 s + 20 − 54 3 s Simplify the above equation as follows: V ( 3 s ( − s + 10 ) + 3 s ( 4 s + 20 ) + ( 4 s + 20 ) ( − s + 10 ) 3 s ( 4 s + 20 ) ( − s + 10 ) ) = 72 ( 3 s ) − 54 ( 4 s + 20 ) 3 s ( 4 s + 20 ) V ( − 3 s 2 + 30 s + 12 s 2 + 60 s − 4 s 2 + 40 s − 20 s + 200 3 s ( 4 s + 20 ) ( − s + 10 ) ) = 216 s − 216 s − 1080 3 s ( 4 s + 20 ) V ( 5 s 2 + 110 s + 200 3 s ( 4 s + 20 ) ( − s + 10 ) ) = − 1080 3 s ( 4 s + 20 ) Simplify the above equation to find V

11th Edition

ISBN: 8220106795262

Author: Riedel

Publisher: YUZU

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Chapter 13, Problem 42P

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Find the current io for time t>0.

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EBK ELECTRIC CIRCUITS

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Find the current i o for time t &gt; 0 .

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Find the current io for time t>0.

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Find the current i o for time t > 0 .