Explanation Formula used: Consider the general form of sinusoid. v ( t ) = V m sin ( ω t + θ ) Here, V m is the maximum amplitude. θ is the phase angle. Consider the trigonometric identity for A cos α + B sin α as follows: A cos α + B sin α = A 2 + B 2 cos ( α + tan − 1 ( − B A ) ) Consider the voltage across the capacitor. i C = C d v C d t Calculation: Refer to Figure in the respective question. Redraw Figure as shown in Figure 1. Apply KVL to loop of Figure 1. 3 cos 4 t = v C + 5 i C Substitute C d v C d t for i C as follows. 3 cos 4 t = v C + 5 ( C d v C d t ) Substitute 2 F for C as follows. 3 cos 4 t = v C + 5 [ ( 2 ) d v C d t ] 3 cos 4 t = v C + ( 10 ) d v C d t (1) Consider, v C = V 1 cos 4 t + V 2 sin 4 t (2) Differentiate with respect to t on both sides of equation. d v C d t = d d t ( V 1 cos 4 t + V 2 sin 4 t ) = − 4 V 1 sin 4 t + 4 V 2 cos 4 t Substitute V 1 cos 4 t + V 2 sin 4 t for v C and − 4 V 1 sin 4 t + 4 V 2 cos 4 t for d v C d t in equation (1) as follows. 3 cos 4 t = V 1 cos 4 t + V 2 sin 4 t + ( 10 ) ( − 4 V 1 sin 4 t + 4 V 2 cos 4 t ) 3 cos 4 t = V 1 cos 4 t + V 2 sin 4 t − 40 V 1 sin 4 t + 40 V 2 cos 4 t Separate the cos and sin functions in terms of two unknown equations as follows. V 1 + 40 V 2 = 3 (3) − 40 V 1 + V 2 = 0 (4) Solve equation (3) and equation (4) as follows. V 1 = 0.00187 V V 2 = 0.0749 V Substitute 0.00187 V for V 1 and 0.0749 V for V 2 in equation (2) as follows. v C = 0.00187 cos 4 t + 0.0749 sin 4 t = ( 0.00187 ) 2 + ( 0.0749 ) 2 cos ( 4 t + tan − 1 − 0.0749 0.00187 ) = 5.6135 × 10 − 3 cos ( 4 t + tan − 1 ( − 40

9th Edition

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 10, Problem 14E

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Find the energy stored in capacitor at t=785 ms and t=1.57 s

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Find the energy stored in capacitor at t = 785 ms and t = 1 .57 s

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Find the energy stored in capacitor at t=785 ms and t=1.57 s

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