Explanation Calculation: The given diagram is shown in Figure 1 The conversion from μ F into F is given by 1 μ F = 10 − 6 F The conversion from 100 μ F into F is given by, 100 μ F = 100 × 10 − 6 F The conversion from 1 mH into H is given by, 1 mH = 10 − 3 H The conversion from 5 mH into H is given by, 5 mH = 5 × 10 − 3 H For time t < 0 the circuit is in steady state and the inductor behave as a short circuit and capacitor behaves as open circuit, mark the values and redraw the circuit. The required diagram is shown in Figure 2 From above, the expression for the current through the inductor for time t = 0 − is calculated as, i L ( 0 − ) = − 1 V 10 Ω = − 0.1 A The expression for the voltage across the capacitor for time t = 0 − is given by, v C ( 0 − ) = − i L ( 0 − ) ( 10 Ω ) Substitute − 0.1 A for i L ( 0 − ) in the above equation. v C ( 0 − ) = − ( − 0.1 A ) ( 10 Ω ) = 1 V Mark the super node and redraw the circuit. The required diagram is shown in Figure 3 From the above circuit, the super node equation is given by, v C = v L − 1 V The expression for the voltage across the inductor is given by, v L = L d i L d t Substitute 5 × 10 − 3 H for L in the above equation. v L = ( 5 × 10 − 3 H ) d i L d t Apply KVL to the super node. v C 10 + 100 × 10 − 6 d v C d t + v L 10 + i L = 0 Substitute v L − 1 V for v C in the above equation. v L − 1 V 10 + 100 × 10 − 6 d ( v L − 1 V ) d t + ( v L − 1 V ) 10 + i L = 0 10 − 4 d v L d t + 1 5 v L + i L = 1 10 Substitute ( 5 × 10 − 3 H ) d i L d t for v L in the above equation. 10 − 4 d [ ( 5 × 10 − 3 H ) d i L d t ] d t + 1 5 [ ( 5 × 10 − 3 H ) d i L d t ] + i L = 1 10 d 2 i L d t 2 + 2000 d i L d t + 2 × 10 6 i L = 2 × 10 5 The characteristic equation for the above differential equation is given by, s 2 + 2000 s + 2 × 10 6 = 0 s 1 , 2 = − 1000 ± j 1000 The expression for the natural response of the system is given by, i L n = e − 1000 t [ α 1 sin ( 1000 t ) + α 2 cos ( 1000 t ) ] The forced response of the system is calculated as, i L f = 2 × 10 5 D 2 + 2000 D + 2 × 10 6 Substitute 0 for D in the above equation. i L f = 2 × 10 5 0 2 + 2000 ( 0 ) + 2 × 10 6 = 0.1 The expression for the total response of the system is given by, i L ( t ) = i L f + i L n Substitute 0.1 for i L f and e − 1000 t [ α 1 sin ( 1000 t ) + α 2 cos ( 1000 t ) ] for i L n in the above equation. i L ( t ) = 0.1 + e − 1000 t [ α 1 sin ( 1000 t ) + α 2 cos ( 1000 t ) ] ........(1) Substitute 0 for t in the above equation

Principles and Applications of Electrical Engineering

6th Edition

ISBN: 9780073529592

Author: Giorgio Rizzoni Professor of Mechanical Engineering, James A. Kearns Dr.

Publisher: McGraw-Hill Education

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Chapter 5, Problem 5.62HP

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The value of voltage vC across the capacitor and the current through the inductor iL.

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The value of voltage v C across the capacitor and the current through the inductor i L .

Solution Summary: The author calculates the value of voltage v_C across the capacitor and the current through the inductor.

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The value of voltage vC across the capacitor and the current through the inductor iL.

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