Explanation Formula used: The differential equation of an electric circuit containing C, R, L and E is as L d 2 q d t 2 + R d q d t + q c = E The quadratic equation a x 2 + b x + c = 0 is calculated as x = − b ± b 2 − 4 a c 2 a . Theorem used: “If r 1 = α + i β and r 2 = α − i β are the two complex roots of the auxiliary equation a r 2 + b r + c = 0 , then y = e α x ( c 1 cos β x + c 2 sin β x ) is the general solution of a y ″ + b y ′ + c y = 0 ”. Calculation: An electric circuit containing capacitor C = 0.500 μF , inductor L = 8.00 mH , resistor R = 0 Ω and electric field E = 20.0 e − 200 t mV . So, the characteristic equation is as follows: L d 2 q d t 2 + R d q d t + q c = E Substitute the value of C = 0.500 μF L = 8.00 mH R = 0 Ω and E = 20.0 e − 200 t mV in L d 2 q d t 2 + R d q d t + q c = E . 8 × 10 − 3 d 2 q d t 2 + 0 d q d t + q 5 × 10 − 7 = 2 × 10 − 2 e − 200 t d 2 q d t 2 + 2.50 × 10 8 q = 2.50 e − 200 t The auxiliary equation is as follows: m 2 + 2.5 × 10 8 = 0 Compare the equation m 2 + 2.5 × 10 8 = 0 with a x 2 + b x + c = 0 here a = 1 , b = 0 , c = 2.5 × 10 8 to calculate the value of m = − b ± b 2 − 4 a c 2 a m = − 0 ± 0 2 − 4 × 1 × 2.5 × 10 8 2 × 1 m = − 0 ± 0 − 10 × 10 8 2 m = ± 1.58 × 10 4 i The auxiliary equation is as follows: q c = e α t ( c 1 sin β t + c 2 cos β t ) Here, α = 0 , β = 1.58 × 10 4 in m = ± 1.58 × 10 4 i Substitute the value of α = 0 , β = 1.58 × 10 4 in the auxiliary equation q c = e α t ( c 1 sin β t + c 2 cos β t ) q c = e 0 t ( c 1 sin 1.58 × 10 4 t + c 2 cos 1.58 × 10 4 t ) = c 1 sin 1.58 × 10 4 t + c 2 cos 1.58 × 10 4 t To solve the non-homogeneous solution of the form, let q p = A e − 200 t . Differentiate the equation q p = A e − 200 t is as follows: D q p = − 200 A e − 200 t Differentiate the equation D q p = − 200 A e − 200 t is as follows: D 2 q p = 400 A e − 200 t = 4 × 10 4 A e − 200 t Substitute the value of D q p = − 200 A e − 200 t , q p = A e − 200 t , D 2 q p = 4 × 10 4 A e − 200 t in d 2 q d t 2 + 2.50 × 10 8 q = 2

11th Edition

ISBN: 9780134508290

Author: Evans

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 31.10, Problem 23E

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The relation between the current and time.

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Chapter 31.10, Problem 22E

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Chapter 31 Solutions

EBK BASIC TECHNICAL MATHEMATICS

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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