(a) Explanation Formula used: The expression for parallel combination of inductance is as follows. 1 L = ( 1 L 1 + 1 L 2 ) (1) Here, L is the total inductance and L 1 and L 2 are the inductances in the circuit. The expression for series combination of inductance is as follows. L = ( L 1 + L 2 ) (2) Here, L is the total inductance and L 1 and L 2 are the inductance s in the circuit. The expression for KCL at node A is as follows. v A R 1 − i 2 + v A R 2 + v A R 3 = 0 (3) Here, i 2 isthe current in the circuit, v A is the voltage at node A and R 1 , R 2 and R 3 are the resistances in the circuit. The expression for current is as follows. i = v R (4) Here, i is the current, v is the voltage and R is the resistance. The expression for series combination of resistance is as follows. R = ( R 1 + R 2 ) (5) Here, R is the total resistances and R 1 and R 2 are the resistances in the circuit. The expression for the circuit time constant is as follows. τ = L R (6) Here, τ is the circuit time constant, R is the equivalent resistance across the inductor and L is the value of inductance. The expression for the current through inductor is as follows. i ( t ) = i ( 0 ) e − t τ (7) Here, i ( t ) is the current flowing through inductor, i ( 0 ) is the current through inductor at t > 0 , t is the given time and τ is the circuit time constant. Calculation: The circuit diagram is redrawn as shown in Figure 1. Refer to the redrawn Figure 1. As L 2 and L 3 are in parallel therefore from equation (1). 1 L 4 = ( 1 L 2 + 1 L 3 ) 1 L 4 = 1 1 H + 1 2 H 1 L 4 = 1 2 3 H (8) Rearrange equation (8) for L 4 . L 4 = 2 3 H The circuit diagram is redrawn as shown in Figure 2. As L 1 and L 4 are in series therefore from equation (2), L = ( L 1 + L 4 ) = 3 H + 2 3 H = 11 3 H The circuit diagram is redrawn as shown in Figure 3 for t < 0 . Refer to the redrawn Figure 3. At t = 0 − the inductor in the R L circuit is connected for long time. So, inductor behaved as short circuit. Substitute 5 A for i 2 , 8 Ω for R 1 , 3 Ω for R 2 and 2 Ω for R 3 in the equation (3), v A 8 Ω − 5 A + v A 3 Ω + v A 2 Ω = 0 3 v A − 120 A+8 v A + 12 v A 24 Ω = 0 23 v A − 120 A 24 Ω = 0 (9) Rearrange equation (9) for v A

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ISBN: 9781260540666

Author: Hayt

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Chapter 8, Problem 29E

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Obtain expression for current i1(t) and current iL(t).

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Chapter 8, Problem 28E

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Obtain expression for current i 1 ( t ) and current i L ( t ) .

Solution Summary: The author explains how to obtain expressions for current and series combination of inductance.

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Obtain expression for current i1(t) and current iL(t).

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