Explanation Given that, L = 1 h, R =2 Ω , C = 0.25 f and E ( t ) = 50 cos t V . The linear second order differential equation of LRC series circuit is, L d 2 q d t 2 + R d q d t + 1 C q = E ( t ) Substitute the given values of L , R , C and E ( t ) in above equation. ( 1 ) d 2 q d t 2 + ( 2 ) d q d t + ( 1 0.25 ) q = 50 cos t d 2 q d t 2 + 2 d q d t + 4 q = 50 cos t So, the linear second order differential equation is, d 2 q d t 2 + 2 d q d t + 4 q = 50 cos t (1) The auxiliary equation of the equation (1) is, m 2 + 2 m + 4 = 0 . Solution of auxiliary equation is, m = − b ± b 2 − 4 a c 2 a = − 2 ± 2 2 − 4 ( 1 ) ( 4 ) 2 ( 1 ) = − 2 ± 4 − 16 2 = − 2 ± 2 3 i 2 Roots of auxiliary equation are m = − 1 ± i 3 , is complex. The complementary function is expressed as, q c ( t ) = e − t ( c 1 cos 3 t + c 2 sin 3 t ) . The particular integral is, q p ( t ) = A cos t + B sin t (2) Differentiate both sides of equation (2) with respect to t . d q p d t = − A sin t + B cos t Differentiate again with respect to t . d 2 q p d t 2 = − A cos t − B sin t Substitute the values of q p ( t ) , d q p d t and d 2 q p d t 2 in equation (1)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.8, Problem 53E

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The steady state charge and the steady state current in LRC series circuit.

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Chapter 3.8, Problem 52E

Chapter 3.8, Problem 54E

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Chapter 3 Solutions

Advanced Engineering Mathematics

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The steady state charge and the steady state current in LRC series circuit.

Solution Summary: The author explains the steady state charge and steady-state current in LRC series circuit. The linear second order differential equation is L,R,C and E(t).

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The steady state charge and the steady state current in LRC series circuit.

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Chapter 3 Solutions