Explanation Given data: Initial voltage across the capacitor is given as 12 V and Initial current through the inductor is given as 2 A . Calculation: Redraw the circuit as shown in Figure 1. The expression for the voltage across the capacitor is as follows. v C ( t ) = 1 C ∫ 0 t i C d t ' + v C ( 0 ) (1) Here, v C ( t ) is the voltage across the capacitor. C is the capacitance of the capacitor. i C is the current flowing in the capacitor v C ( 0 ) is the capacitor voltage at time 0 s . The expression for the voltage across the inductor. v L = L d i L d t (2) Here, v L is the voltage across the inductor. L is the inductance of the inductor. i L is the current flowing in the inductor. Apply mesh analysis in loop 1. − v s + i 20 R 1 + v C ( t ) = 0 (3) Here, R 1 is the resistance. v s is the source voltage. i 20 is the current in the loop 1. Apply mesh analysis in loop 2. − v C + i L R 2 + v L = 0 (4) Here, R 2 is the resistance. The current through the capacitor is as follows. i C = i 20 − i L Refer to the redrawn Figure 1. For loop 1, Substitute 1 C ∫ 0 t i C d t ' + v C ( 0 ) for v C ( t ) in equation (3). − v s + i 20 R 1 + 1 C ∫ 0 t i C d t ' + v C ( 0 ) = 0 (5) Substitute i 20 − i L for i C in equation (5). − v s + i 20 R 1 + 1 C ∫ 0 t ( i 20 − i L ) d t ' + v C ( 0 ) = 0 (6) Substitute 20 Ω for R 1 , 5 × 10 − 6 F for C and 12 V for v C ( 0 ) in equation (6). − v s + i 20 ( 20 Ω ) + 1 5 × 10 − 6 ∫ 0 t ( i 20 − i L ) d t ' + 12 V = 0 For loop 2. Substitute L d i L d t for v L and 1 C ∫ 0 t i C d t ' + v C ( 0 ) for v C ( t ) in equation (4)

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ISBN: 9780073545516

Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.

Publisher: Mcgraw-hill Education,

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Chapter 7, Problem 46E

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Write the mesh equations for the circuit.

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Chapter 7, Problem 45E

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Engineering Circuit Analysis

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