Explanation Given data: Refer to Figure 16.59 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 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 at time t = 0 − the capacitor acts like open circuit and the inductor acts like short circuit. For time t < 0 , the current source i s ( t ) is calculated as follows: i s ( t ) = 7.5 ( 1 − 0 ) { ∵ u ( t ) = 0 for t < 0 } = 7.5 A Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the current flows through the inductor is same as the source current i s ( t ) . The capacitor is connected in parallel with resistor R 2 . For the parallel connection voltage is same. The current through inductor is calculated as follows: i L ( 0 − ) = 7.5 A Refer to Figure 2, the resistor R 1 is connected in parallel with the series combination of resistors R 2 and R 3 . Use current division rule to Figure 2 to find the he value of current through resistor R 2 . i R 2 = ( 7.5 A ) ( ( 10 Ω + 5 Ω ) ( 10 Ω ) 10 Ω + 5 Ω + 10 Ω ) ( 1 15 Ω ) = ( 7.5 A ) ( ( 15 Ω ) ( 10 Ω ) 25 Ω ) ( 1 15 Ω ) = 3 A The voltage across the resistor R 2 is calculated as follows: v R 2 = ( 5 Ω ) ( 3 A ) = 15 V Since the capacitor and the resistor R 2 is connected in parallel the voltage is same for both. The value of voltage across the capacitor is calculated as follows v C ( 0 − ) = v R 2 = 15 V The current through inductor and voltage across capacitor is always continuous so that, i ( 0 ) = i L ( 0 − ) = i L ( 0 + ) = 7.5 A v ( 0 ) = v C ( 0 − ) = v C ( 0 + ) = 15 V For time t > 0 the value of current source i s ( t ) is calculated as follows: i s ( t ) = 7.5 ( 1 − 1 ) { ∵ u ( t ) = 1 for t > 0 } = 0 A When the value of current source is zero, it is open circuited. Now, the Figure 1 is reduced as shown in Figure 3. Refer to Figure 3, the resistor R 2 is connected in parallel with the series combination of resistors R 1 and R 3 . The equivalent resistor is calculated as follows: R eq = ( 5 Ω ) ∥ ( 10 Ω + 10 Ω ) = ( 5 Ω ) ∥ ( 20 Ω ) = ( 5 Ω ) ( 20 Ω ) 5 Ω + 20 Ω = 4 Ω Now, the Figure 3 is reduced as shown in Figure 4. Substitute 4 Ω for R eq in equation (1) to find Z R eq . Z R eq = 4 Substitute 3 4 H for L in equation (2) to find Z L . Z L = s ( 3 4 H ) = 3 s 4 Substitute 1 3 F for C in equation (3) to find Z C . Z C = 1 s ( 1 3 F ) = 3 s Using element transformation methods with initial conditions convert the Figure 4 into s -domain. Apply Kirchhoff’s voltage law for the circuit shown in Figure 5. − v ( 0 ) s + 3 s I ( s ) + 3 s 4 I ( s ) − ( 3 s 4 ) i ( 0 ) s + 4 I ( s ) = 0 [ 3 s + 3 s 4 + 4 ] I ( s ) − 3 4 i ( 0 ) − v ( 0 ) s = 0 [ 12 + 3 s 2 + 16 s 4 s ] I ( s ) = 3 4 i ( 0 ) + v ( 0 ) s [ 3 s 2 + 16 s + 12 4 s ] I ( s ) = 3 4 i ( 0 ) + v ( 0 ) s Substitute 7

EE 98: Fundamentals of Electrical Circuits - With Connect Access

6th Edition

ISBN: 9781259981807

Author: Alexander

Publisher: MCG

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Chapter 16, Problem 36P

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Find the expression of current i(t) for t>0 in the circuit of Figure 16.59.

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I what happend if can IY = 125 lead, IB = 125 unity IR=89.91 unity IB = 125-> Lag 0.8 Iy=125 IB 36.86 lag 0.8 Ir< 36-86 IN=IR+IN+IB FR 89.910+125<-120-36.86+1252+120 -36.86

Chapter 16 Solutions

EE 98: Fundamentals of Electrical Circuits - With Connect Access

Ch. 16.2 - Determine vo(t) in the circuit of Fig. 16.6,...Ch. 16.2 - Prob. 2PP Ch. 16.2 - Prob. 3PP Ch. 16.3 - For the circuit shown in Fig. 16.12 with the same...Ch. 16.3 - Prob. 5PP Ch. 16.3 - The initial energy in the circuit of Fig. 16.17 is...Ch. 16.4 - Prob. 7PP Ch. 16.4 - Prob. 8PP Ch. 16.4 - Prob. 9PP Ch. 16.5 - Obtain the state variable model for the circuit...

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Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.59.

Solution Summary: The author explains the expression of current i(t) in the circuit of Figure 16.59.

Find the expression of current i(t) for t>0 in the circuit of Figure 16.59.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t &gt; 0 in the circuit of Figure 16.59.

Solution Summary: The author explains the expression of current i(t) in the circuit of Figure 16.59.

Find the expression of current i(t) for t>0 in the circuit of Figure 16.59.

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Chapter 16 Solutions

Find the expression of current i ( t ) for t > 0 in the circuit of Figure 16.59.