Explanation Calculation: The linear second order differential equation of LC series circuit is, L d 2 q d t 2 + 1 C q = E ( t ) Substitute the given value of E ( t ) from Question 61 in above equation, L d 2 q d t 2 + 1 C q = E 0 cos γ t d 2 q d t 2 + 1 L C q = E 0 L cos γ t Substitute k 2 = 1 L C in above equation, d 2 q d t 2 + k 2 q = E 0 L cos γ t So, the linear second order differential equation is, d 2 q d t 2 + k 2 q = E 0 L cos γ t (1) The auxiliary equation of the equation (1) is as, m 2 + k 2 = 0 Solution of auxiliary equation is, m 2 = − k 2 m = ± k i Roots of auxiliary equation are complex, m = ± k i The complementary function is expressed as, q c ( t ) = c 1 cos k t + c 2 sin k t (2) The particular integral is expressed as, q p ( t ) = A t cos k t + B t sin k t (3) Differentiate both sides of equation (3) with respect to t , d q p d t = − k A t sin k t + A cos k t + B sin k t + k B t cos k t Differentiate both sides of above equation with respect to t , d 2 q p d t 2 = − k A sin k t − k 2 A t cos k t + k B cos k t − k A sin k t + k B cos k t − k 2 B t sin k t = 2 k B cos k t − 2 k A sin k t − k 2 A t cos k t − k 2 B t sin k t Substitute the values of q p ( t ) , d q p d t and d 2 q p d t 2 in equation (1), ( 2 k B cos k t − 2 k A sin k t − k 2 A t cos k t − k 2 B t sin k t ) + k 2 ( − k A t sin k t + A cos k t + B sin k t + k B t cos k t ) = E 0 L cos k t 2 k B cos k t − 2 k A sin k t = E 0 L cos k t Compare both sides of above equation, − 2 k A = 0 and 2 k B = E 0 L Therefore, the value of A and B is as follows

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 5.1, Problem 62E

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To find: The current when the circuit is in resonance.

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Chapter 5.1, Problem 61E

Chapter 5.2, Problem 1E

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The current when the circuit is in resonance.

Solution Summary: The author calculates the linear second order differential equation of LC series circuit.

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To find: The current when the circuit is in resonance.

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Chapter 5 Solutions