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 of 100 μ F in to F is given by, 100 μ F = 100 × 10 − 6 F The switch is closed at t > 0 . The required diagram is shown in Figure 2. Apply KVL in the above circuit. − 20 sin ( 100 t ) + ( 1 ) d i ( t ) d t + 400 i ( t ) + 1 100 × 10 − 6 ∫ i ( t ) d t = 0 d 2 i ( t ) d t 2 + 400 i ( t ) + 10 4 ∫ i ( t ) d t = 20 sin ( 100 t ) Differentiate the above equation with respect to t . d 2 i ( t ) d t 2 + 400 d i d t + 10000 i ( t ) = 20 ( 100 ) cos ( 100 t ) ........ (1) The general expression for the differential equation is given by, d 2 i p ( t ) d t 2 + 2 α d i p ( t ) d t + ω 0 2 i p ( t ) = f ( t ) ......... (2) The equation for the particular integral is given by, i p ( t ) = A cos ( 100 t ) + B sin ( 100 t ) ......... (3) Substitute A cos ( 100 t ) + B sin ( 100 t ) for i p ( t ) , 400 for 2 α , 2 , 000 cos ( 100 t ) for f ( t ) and 10 , 000 for ω 0 2 i p ( t ) in equation (2). [ − A ( 10 4 ) cos ( 100 t ) + − B ( 10 4 ) sin ( 100 t ) + 400 [ − A ( 100 ) sin ( 100 t ) + B ( 100 ) cos ( 100 t ) ] + ( 10 4 ) [ A cos ( 100 t ) + B sin ( 100 t ) ] ] = 2 , 000 cos ( 100 t ) The coefficients of sine terms is calculated as, − B ( 10 4 ) + A ( 100 ) ( 400 ) + 10 4 B = 0 − A ( 100 ) ( 400 ) = 0 A = 0 The coefficients of cosine terms is calculated as, − A ( 10 4 ) + B ( 100 ) ( 00 ) + 10 4 A = 2 , 000 − ( 0 ) ( 10 4 ) + B ( 100 ) ( 400 ) + ( 10 4 ) ( 0 ) = 2000 B = 2 , 000 ( 100 ) ( 400 ) B = 0.05 Substitute 0.05 for B and 0 for A in equation i p ( t ) = ( 0 ) cos ( 100 t ) + 0.05 sin ( 100 t ) = 0.05 sin ( 100 t ) The attenuation constant α is calculated as, α = R 2 L Substitute 400 Ω for R and 1 H for L in the above equation. α = 400 2 ( 1 ) = 200 The natural frequency is calculated as, ω 0 = 1 L C Substitute 100 × 10 − 6 F for C and 1 H for L in the above equation. ω 0 = 1 ( 1 ) ( 100 × 10 − 6 ) = 100 The damping ratio is calculated as, ξ = α ω 0 Substitute 100 for ω 0 and 200 for α in the above equation. ξ = 200 100 = 2 The expression for the complementary solution is given by, i c ( t ) = K 1 e − s 1 t + K 2 e − s 2 t

Electrical Engineering: Principles & Applications (7th Edition)

7th Edition

ISBN: 9780134484143

Author: Allan R. Hambley

Publisher: PEARSON

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Chapter 4, Problem 4.69P

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The expression for i(t) for t>0.

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Electrical Engineering: Principles & Applications (7th Edition)

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