Explanation Given: From the given information, Voltage source, E ( t ) = 10 cos 20 t V Resistor, R = 120 Ω Inductor, L = 4 H Capacitor, C = 1 2200 F Approach: The RLC series circuit is defined by the differential equation. L d I d t + R I + q C = E ... ( 1 ) The auxiliary equation of a differential equation a n y ( n ) + a n − 1 y ( n − 1 ) + ⋅ ⋅ ⋅ + a 1 y ( 1 ) + a 0 is a n r n + a n − 1 r n − 1 + ⋅ ⋅ ⋅ + a 1 r 1 + a 0 ... ( 2 ) . Here, y ( n ) denotes the n th derivative of y . If x = a is a zero of the polynomial a n x n + a n − 1 x n − 1 + ⋅ ⋅ ⋅ + a 1 x 1 + a 0 , then ( x − a ) is a factor of the polynomial ... ( 3 ) . If the auxiliary equation has roots r 1 , r 2 then the solution of the differential equation is y = A e r 1 t + B e r 2 t ... ( 4 ) If the auxiliary equation has same root r 1 = r 2 = r , then the solution of the differential equation is y = A e r t + t B e r t The formula for resonance frequency is f = ω 4 π Calculation: Differentiate with respect to t on both sides. L d 2 I d t 2 + R d I d t + I C = d E d t Substitute 4 for L , 120 for R , 1 2200 for C , 10 cos 20 t 20 for E , in the above equation. 4 d 2 I d t 2 + 120 d I d t + 2200 I = d d t ( 10 cos 20 t ) d 2 I d t 2 + 30 d I d t + 550 I = − 50 sin 20 t The auxiliary equation is r 2 + 30 r + 550 = 0 ( 6 ) Let p ( r ) = r 2 + 30 r + 550 r = − 30 ± 30 2 − 4 × 1 × 550 2 × 1 = − 30 ± 1300 i 2 = − 15 ± 325 i This implies I h ( t ) = A e − 15 t cos 325 t + B e − 15 t sin 325 t ( 7 ) As t → ∞ , I h → 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 5.7, Problem 3E

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The steady state current and the resonance frequency.

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Fundamentals of Differential Equations and Boundary Value Problems

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The steady state current and the resonance frequency.

Solution Summary: The author explains the steady state current and resonance frequency. The auxiliary equation of a differential equation is y(n).

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To find: The steady state current and the resonance frequency.

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Chapter 5 Solutions

To find:

The steady state current and the resonance frequency.