Explanation Given data: Refer to given circuit in the textbook. The polarity dot is placed at the top of the 20 H inductor. No initial energy stored in the circuit when the switch is closed. 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 and mutual inductance 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: From the given data, the circuit is drawn as shown in Figure 1. The T-equivalent circuit of the mutually connected aided inductors is shown in Figure 2. Implement the Figure 2 in Figure 1 as shown in Figure 3. Since no energy stored in the circuit at the time when the switch is closed, the initial current through the inductor is equal to zero. i ( 0 ) = 0 A Consider, L eq1 = L 1 − M L eq2 = L 2 − M For time t > 0 , the switch is closed and the Figure 3 is drawn as shown in Figure 4. Apply Laplace transform for voltage source v . V ( s ) = 180 s { ∵ ℒ ( u ( t ) ) = 1 s } Use equation (1) to find Z R 1 . Z R 1 = 100 Use equation (1) to find Z R 2 . Z R 2 = 40 Use equation (1) to find Z R 3 . Z R 3 = 160 Use equation (2) to find Z L eq1 . Z L eq1 = s ( 5 H ) = 5 s Use equation (2) to find Z L eq2 . Z L eq2 = s ( 10 H ) = 10 s Use equation (2) to find Z M Z M = s ( 10 H ) = 10 s Convert the Figure 4 into s -domain as shown in Figure 5. Apply mesh analysis for Loop 1 in Figure 5. − 180 s + 100 I 1 + 5 s I 1 + 10 s ( I 1 − I 2 ) = 0 100 I 1 + 5 s I 1 + 10 s I 1 − 10 s I 2 = 180 s 100 I 1 + 15 s I 1 − 10 s I 2 = 180 s ( 100 + 15 s ) I 1 − 10 s I 2 = 180 s (3) Apply mesh analysis for Loop 2 in Figure 5. 10 s I 2 + 40 I 2 + 160 I 2 + 10 s ( I 2 − I 1 ) = 0 10 s I 2 + 200 I 2 + 10 s I 2 − 10 s I 1 = 0 20 s I 2 + 200 I 2 − 10 s I 1 = 0 ( 20 s + 200 ) I 2 − 10 s I 1 = 0 10 s I 1 = ( 20 s + 200 ) I 2 Simplify the above equation to find I 1 . I 1 = 10 ( 2 s + 20 ) I 2 10 s I 1 = ( 2 s + 20 ) I 2 s (4) Substitute equation (4) in equation (3). ( 100 + 15 s ) ( ( 2 s + 20 ) I 2 s ) − 10 s I 2 = 180 s ( 1 s ) ( 100 + 15 s ) ( 2 s + 20 ) I 2 − 10 s I 2 = 180 s ( 1 s ) ( 200 s + 2000 + 30 s 2 + 300 s ) I 2 − 10 s I 2 = 180 s ( 1 s ) ( 30 s 2 + 500 s + 2000 ) I 2 − 10 s I 2 = 180 s Simplify the above equation as follows: ( 30 s 2 + 500 s + 2000 − 10 s 2 s ) I 2 = 180 s ( 20 s 2 + 500 s + 2000 s ) I 2 = 180 s ( 20 ( s 2 + 25 s + 100 ) s ) I 2 = 180 s Simplify the above equation to find I 2

ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

11th Edition

ISBN: 9780134894300

Author: NILSSON

Publisher: PEARSON

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ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

Ch. 13.2 - The parallel circuit in Example 13.1 is placed in...Ch. 13.3 - Prob. 2AP Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.3 - The dc current and dc voltage sources are applied...Ch. 13.3 - Prob. 6AP Ch. 13.3 - Using the results from Example 13.7 for the...Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.4 - Derive the numerical expression for the transfer...Ch. 13.5 - Find (a) the unit step and (b) the unit impulse...Ch. 13.5 - The unit impulse response of a circuit is υo(t) =...

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