Explanation Given: Given that L = 1 h, R =100 Ω , C = 0.0004 f and E ( t ) = 30 V and initial conditions q ( 0 ) = 0 C , i ( 0 ) = 2 A . Calculation: 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. L d 2 q d t 2 + R d q d t + 1 C q = E ( t ) ( 1 ) d 2 q d t 2 + ( 100 ) d q d t + ( 1 0.0004 ) q = 30 d 2 q d t 2 + 100 d q d t + 2 , 500 q = 30 Thus, the linear second order differential equation is, d 2 q d t 2 + 100 d q d t + 2 , 500 q = 30 (1) The auxiliary equation of the equation (1) is, m 2 + 100 m + 2 , 500 = 0 . Solution of auxiliary equation is, m = − b ± b 2 − 4 a c 2 a = − 100 ± 100 2 − 4 ( 1 ) ( 2 , 500 ) 2 ( 1 ) = − 100 ± 0 2 = − 100 2 The roots of auxiliary equation are m = − 50 , − 50 , are same. Thus, the complementary function is expressed as, q c ( t ) = c 1 e − 50 t + c 2 t e − 50 t (2) The particular integral is, q p ( t ) = A (3) Differentiate both sides of equation (3) with respect to t . d q p d t = 0 Differentiate again with respect to t . d 2 q p d t 2 = 0 Substitute the values of q p ( t ) , d q p d t and d 2 q p d t 2 in equation (1). d 2 q d t 2 + 100 d q d t + 2 , 500 q = 30 0 + 100 ( 0 ) + 2 , 500 A = 30 2 , 500 A = 30 A = 30 2 , 500 A = 0.012 Substitute the value A = 0.012 in equation (3). q p ( t ) = 0.012 (4) Thus, the general solution is, q ( t ) = q c ( t ) + q p ( t ) . Substitute the values of q c ( t ) and q p ( t ) from equation (2) and equation (4) in above equation. q ( t ) = c 1 e − 50 t + c 2 t e − 50 t + 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Question

Chapter 3.8, Problem 52E

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The charge on the capacitor, the current in LRC series circuit and the maximum charge on the capacitor.

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Chapter 3.8, Problem 51E

Chapter 3.8, Problem 53E

Chapter 3 Solutions

Advanced Engineering Mathematics

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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The charge on the capacitor, the current in LRC series circuit and the maximum charge on the capacitor.

Solution Summary: The author calculates the linear second order differential equation of LRC series circuit by subtracting the values of L,R,C and E(t).

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The charge on the capacitor, the current in LRC series circuit and the maximum charge on the capacitor.

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Chapter 3 Solutions