Explanation Given data: Refer to given circuit in the textbook. 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. 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. 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. If the inductor value is zero it acts as a short circuit. Therefore, the inductor L 1 − M acts as a short circuit. Consider, L eq2 = L 2 − M Now, the circuit in Figure 3 is reduced as shown in Figure 4. For a DC circuit, at steady condition for time t < 0 the switch is closed and the capacitor acts like open circuit and the inductor acts like short circuit. Now, the Figure 4 is reduced as shown in Figure 5. Refer to Figure 5, the short circuited switch is connected in parallel with the open circuited inductor. The entire current flows through the short circuited path and there is no current flow and voltage across the capacitor. v C ( 0 − ) = 0 V Refer to Figure 5, the resistor R 2 is connected in parallel with the short circuited inductor M . Since, the inductor is short circuited the entire current flows through it. Now, the Figure 5 is reduced as shown in Figure 6. The current through inductor ( M ) is calculated as follows: i 2 ( 0 − ) = 48 V 4 Ω = 12 A The capacitor voltage and inductor current is always continuous so that, v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 0 V i 2 ( 0 ) = i 2 ( 0 − ) = i 2 ( 0 + ) = 12 A For time t > 0 , the switch is opened and the Figure 4 is drawn as shown in Figure 7. Apply Laplace transform for voltage source v . V ( s ) = 48 s { ∵ ℒ ( u ( t ) ) = 1 s } Use equation (1) to find Z R 1 . Z R 1 = 4 Use equation (1) to find Z R 2 . Z R 2 = 20 Use equation (2) to find Z L eq2 . Z L eq2 = s ( 0.8 H ) = 0.8 s Use equation (2) to find Z M Z M = s ( 0.8 H ) = 0.8 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 Convert the Figure 7 into s -domain with initial conditions as shown in Figure 8. Apply Kirchhoff’s current law at node V 1 in Figure 8. V 1 − 48 s 100 s + 4 + V 1 + M i 2 ( 0 ) 0.8 s + V 1 0.8 s + 20 = 0 s V 1 − 48 s 100 + 4 s s + V 1 + M i 2 ( 0 ) 0.8 s + V 1 0.8 s + 20 = 0 s V 1 − 48 100 + 4 s + V 1 + M i 2 ( 0 ) 0.8 s + V 1 0.8 s + 20 = 0 s V 1 100 + 4 s − 48 100 + 4 s + V 1 0.8 s + M i 2 ( 0 ) 0.8 s + V 1 0.8 s + 20 = 0 Substitute 0.8 H for M , and 12 A for i 2 ( 0 ) in above equation. s V 1 100 + 4 s − 48 100 + 4 s + V 1 0.8 s + ( 0.8 ) ( 12 ) 0.8 s + V 1 0.8 s + 20 = 0 s V 1 100 + 4 s − 48 100 + 4 s + V 1 0.8 s + 9.6 0.8 s + V 1 0.8 s + 20 = 0 s V 1 4 ( s + 25 ) + V 1 0.8 s + V 1 0.8 ( s + 25 ) = 48 100 + 4 s − 9.6 0.8 s V 1 [ s 4 ( s + 25 ) + 1 0.8 s + 1 0.8 ( s + 25 ) ] = 48 4 ( s + 25 ) − 9.6 0.8 s Simplify the above equation as follows: V 1 [ s 2 + 5 ( s + 25 ) + 5 s 4 s ( s + 25 ) ] = 48 s − 9.6 ( 5 ( s + 25 ) ) 4 s ( s + 25 ) V 1 [ s 2 + 5 s + 125 + 5 s 4 s ( s + 25 ) ] = 48 s − 48 s − 1200 4 s ( s + 25 ) V 1 [ s 2 + 10 s + 125 4 s ( s + 25 ) ] = − 1200 4 s ( s + 25 ) Simplify the above equation to find V 1 . V 1 = ( − 1200 4 s ( s + 25 ) ) ( 4 s ( s + 25 ) s 2 + 10 s + 125 ) = − 1200 s 2 + 10 s + 125 Apply voltage division rule in Figure 8 to find V o . V o = ( 20 0.8 s + 20 ) V 1 Substitute − 1200 s 2 + 10 s + 125 for V 1 in above equation. V o = ( 20 0.8 s + 20 ) ( − 1200 s 2 + 10 s + 125 ) = ( 20 0.8 ( s + 25 ) ) ( − 1200 s 2 + 10 s + 125 ) V o = − 30000 ( s + 25 ) ( s 2 + 10 s + 125 ) (4) Write an expression to solve quadratic equation. s 1 , 2 = − b ± b 2 − 4 a c 2 a (5) Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. From equation (4), the characteristic equation of denominator is written as follows: s 2 + 10 s + 125 Compare the above equation with the quadratic equation ( a s 2 + b s + c = 0 ) . a = 1 b = 10 c = 125 Substitute 1 for a , 10 for b , and 125 for c in equation (5) to find s 1 , 2 . s 1 , 2 = − 10 ± ( 10 ) 2 − 4 ( 1 ) ( 125 ) 2 ( 1 ) = − 10 ± 100 − 500 2 = − 10 ± − 400 2 = − 10 ± j 20 2 Simplify the equation. s 1 , 2 = − 10 + j 20 2 , − 10 − j 20 2 = − 5 + j 10 , − 5 − j 10 Therefore, the roots of characteristic equations are real and imaginary. s 1 = − 5 + j 10 s 2 = − 5 − j 10 Now, the equation (4) becomes, V o = − 30000 ( s + 25 ) ( s − ( − 5 + j 10 ) ) ( s − ( − 5 − j 10 ) ) = − 30000 ( s + 25 ) ( s + 5 − j 10 ) ( s + 5 + j 10 ) Take partial fraction for above equation

Electric Circuits, Student Value Edition Format: Unbound (saleable)

11th Edition

ISBN: 9780134747170

Author: NILSSON, James W.^riedel, Susan

Publisher: Prentice Hall

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Chapter 13, Problem 43P

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Electric Circuits, Student Value Edition Format: Unbound (saleable)

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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Chapter 13 Solutions