Explanation Given data: Refer to Figure 16.52 in the textbook. Formula used: Write an expression to calculate the value of step input. u ( t ) = { 0 t < 0 1 t > 0 Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (1) 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 (2) 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 when time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. The value of voltage source is calculated as follows, Substitute − 1 for t in V s . V s = 35 [ 1 − u ( − 1 ) ] V = 35 ( 1 − 0 ) V { ∵ u ( t ) = 0 for t < 0 } = 35 ( 1 ) V = 35 V Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the voltage source ( V s ) , the resistor ( R ) and capacitor ( C ) is connected in parallel. For the parallel connection voltage is same. v C ( 0 − ) = 35 V Since the capacitor is open circuited, there is no current flow through inductor. i L ( 0 − ) = 0 A The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 0 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 35 V For time t > 0 , substitute 1 for t in V s . V s = 35 [ 1 − u ( 1 ) ] V = 35 ( 1 − 1 ) V { ∵ u ( t ) = 1 for t > 0 } = 35 ( 0 ) V = 0 V The value of source voltage is zero and it is short circuited. Now, the Figure 1 is reduced as shown in Figure 3. Refer to Figure 3, the short circuited voltage source and resistor are connected in parallel. Since the voltage source is short circuited full current flows through the resistor, therefore the resistor is short circuited. The reduced diagram of Figure 3 is shown in Figure 4. Substitute 1 16 F for C in equation (2) to find Z C . Z C = 1 s ( 1 16 ) = 16 s Substitute 1 4 H for L in equation (1) to find Z L . Z L = s ( 1 4 H ) = 1 4 s Using element transformation methods with initial conditions convert the Figure 4 into s -domain. Apply Kirchhoff’s voltage law for Figure 5. 16 s I ( s ) + v ( 0 ) s + s 4 I ( s ) − L i ( 0 ) = 0 I ( s ) [ 16 s + s 4 ] + v ( 0 ) s − L i ( 0 ) = 0 Substitute 35 for v ( 0 ) , and 0 for i ( 0 ) in above equation. I ( s ) [ 16 s + s 4 ] + 35 s − L ( 0 ) = 0 I ( s ) [ 16 ( 4 ) + s 2 4 s ] + 35 s − 0 = 0 I ( s ) [ s 2 + 64 4 s ] = − 35 s Simplify the above equation to find I ( s ) . I ( s ) = − 35 s ( 4 s s 2 + 64 ) = − 140 s 2 + 64 = − 140 s 2 + 8 2 = − 140 s 2 − ( j 8 ) 2 I ( s ) = − 140 ( s + j 8 ) ( s − j 8 ) { ∵ a 2 − b 2 = ( a + b ) ( a − b ) } Take partial fraction for above equation

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Question

Chapter 16, Problem 29P

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Find the expression of current i(t) for t>0 shown in Figure 16.52.

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Chapter 16, Problem 28P

Chapter 16, Problem 30P

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