Explanation It is given that the inductor of 0.5-H, the resistor of 6 Ω and the capacitor of 20-mF are connected in series with a generator. The voltage of the generator is E = 24 sin 10 t . The general equation for the above system is L D 2 q + R D q + q C = E . Substitute the values L = 0.5 , R = 6 , C = 0.02 and E = 24 sin 10 t in the above equation as follows. 0.5 D 2 q + 6 D q + 50 q = 24 sin 10 t D 2 q + 12 D q + 100 q = 48 sin 10 t Form an auxiliary equation and obtain the roots as follows: m 2 + 12 m + 100 = 0 m = − 12 ± ( − 12 ) 2 − 4 ( 100 ) 2 ( 1 ) ( ∵ for a x 2 + b x + c = 0 The value of x is − b ± b 2 − 4 a c 2 a ) m = − 12 ± 144 − 400 2 m = − 12 ± − 256 2 That is, m = − 12 ± 16 i 2 = − 6 ± 8 i . For complex roots m = a ± b i , the equation is of the from y = e a t ( c 1 cos b t + c 2 sin b t ) . As the roots are complex, the equation becomes q c = c 1 e − 6 t cos 8 t + c 2 e − 6 t sin 8 t . Consider the particular solution as q p = A cos 10 t + B sin 10 t . Obtain the first and the second derivatives as follows. D q p = − 10 A sin 10 t + 10 B cos 10 t D 2 q p = − 100 A cos 10 t − 100 B sin 10 t Substitute these terms in the equation, obtain the values of A and B as follows. − 100 A cos 10 t − 100 B sin 10 t + 12 ( − 10 A sin 10 t + 10 B cos 10 t ) + 100 A cos 10 t + 100 B sin 10 t = 48 sin 10 t − 100 A cos 10 t − 100 B sin 10 t − 120 A sin 10 t + 120 B cos 10 t + 100 A cos 10 t + 100 B sin 10 t = 48 sin 10 t − 120 A sin 10 t + 120 B cos 10 t = 48 sin 10 t − 120 A = 48 and 120 B = 0 ( On comparing ) A = − 48 120 = − 2 5 and B = 0 Thus, the particular solution is q p = − 0.4 cos 10 t . Hence, the general solution becomes q = q c + q p = c 1 e − 6 t cos 8 t + c 2 e − 6 t sin 8 t − 0.4 cos 10 t . Initially, at t = 0 , the value of q is 0. q ( 0 ) = c 1 e − 6 ( 0 ) cos 8 ( 0 ) + c 2 e − 6 ( 0 ) sin 8 ( 0 ) − 0

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 31, Problem 89RE

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To find: The value of q in terms of t where the initial charge and the initial current are zero.

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Chapter 31, Problem 88RE

Chapter 31, Problem 90RE

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Chapter 31 Solutions

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 31.1 - Show that is a solution of . Is it the general...Ch. 31.1 - Prob. 1E Ch. 31.1 - In Exercises 1 and 2, show that the indicated...Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 4E Ch. 31.1 - In Exercises 3–6, determine whether the given...Ch. 31.1 - Prob. 6E Ch. 31.1 - In Exercises 7–10, show that each function y =...Ch. 31.1 - Prob. 8E Ch. 31.1 - In Exercises 7–10, show that each function y =...

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