To determine Find the time domain expression of i o for time t > 0 . Explanation Given data: Refer to given circuit in the textbook. The switch is in position ‘a’ for long time and moved to position ‘b’ at time t = 0 . 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 (2) Here, L is the value of 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. For a DC circuit, at steady state condition for time t < 0 the switch is in position ‘a’ and the capacitor acts like open circuit and the inductor acts like short circuit. Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the capacitor is connected in parallel with the resistor R 3 . For the parallel connection, the voltage is same. Apply Kirchhoff’s current law in Figure 2. v o ( 0 − ) − 500 V 5 Ω + v o ( 0 − ) 25 Ω + v o ( 0 − ) 100 Ω = 0 v o ( 0 − ) 5 Ω − 500 V 5 Ω + v o ( 0 − ) 25 Ω + v o ( 0 − ) 100 Ω = 0 v o ( 0 − ) [ 1 5 Ω + 1 25 Ω + 1 100 Ω ] = 500 V 5 Ω v o ( 0 − ) [ 20 + 4 + 1 100 Ω ] = 500 V 5 Ω v o ( 0 − ) [ 25 100 Ω ] = 500 V 5 Ω Simplify the above equation to find v o ( 0 − ) . v o ( 0 − ) = ( 500 V 5 Ω ) ( 100 Ω 25 ) = 400 V Refer to Figure 2, the series connected inductor and resistor is connected in parallel with the resistor R 3 . For the parallel connection, the voltage is same and for the series connection current is same. The current through inductor is calculated as follows: i L ( 0 − ) = v o ( 0 − ) 25 Ω Substitute 400 V for v o ( 0 − ) in above equation to find i L ( 0 − ) . i L ( 0 − ) = 400 V 25 Ω = 16 A The capacitor voltage and inductor current is always continuous so that, v ( 0 ) = v o ( 0 − ) = v o ( 0 + ) = 400 V i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 16 A For time t > 0 , the switch is moved to position ‘b’ and the Figure 1 is reduced as shown in Figure 3. Use equation (1) to find Z R 1 . Z R 1 = 5 Use equation (1) to find Z R 2 . Z R 2 = 25 Use equation (1) to find Z R 3 . Z R 3 = 100 Substitute 25 H for L in equation (2) to find Z L . Z L = s ( 25 H ) = 25 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 3 into s -domain with initial conditions as shown in Figure 4. Redraw the Figure 4 as shown in Figure 5 with the appropriate values. Apply Kirchhoff’s current law in Figure 5. V o + 400 25 + 25 s + V o 100 + V o − 400 s ( 100 s ) = 0 V o 25 + 25 s + 400 25 + 25 s + V o 100 + V o ( 100 s ) − ( 400 s ) ( 100 s ) = 0 V o 25 + 25 s + 400 25 + 25 s + V o 100 + s V o 100 − 400 100 = 0 V o 25 + 25 s + V o 100 + s V o 100 = 400 100 − 400 25 + 25 s Simplify the above equation as follows: V o [ 1 25 ( s + 1 ) + 1 100 + s 100 ] = 4 − 400 25 + 25 s V o [ 4 + s + 1 + s ( s + 1 ) 100 ( s + 1 ) ] = 4 ( 25 + 25 s ) − 400 25 + 25 s V o [ 4 + s + 1 + s 2 + s 100 ( s + 1 ) ] = 100 + 100 s − 400 25 ( s + 1 ) V o [ s 2 + 2 s + 5 100 ( s + 1 ) ] = 100 s − 300 25 ( s + 1 ) Simplify the above equation to find V o . V o = ( 100 s − 300 25 ( s + 1 ) ) ( 100 ( s + 1 ) s 2 + 2 s + 5 ) = ( 100 ( s − 3 ) 25 ( s + 1 ) ) ( 100 ( s + 1 ) s 2 + 2 s + 5 ) = 400 ( s − 3 ) s 2 + 2 s + 5 Refer to Figure 5, the current I o is calculated as follows: I o = V o − 400 s ( 100 s ) Substitute 400 ( s − 3 ) s 2 + 2 s + 5 for V o in above equation to find I o

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

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Find the time domain expression of i o for time t > 0 .