Explanation Given: The magnitude of force applied is 30 emf , inductance is 0.1 henry and the resistance is 50 ohms . Calculation: Consider the linear differential equation for the current i ( t ) . L d i d t + R i = E ( t ) (1) Substitute 0.1 henry for L , 50 ohms for R and 30 emf for E in Equation (1). 0.1 d i d t + 50 i = 30 d i d t + 500 i = 300 (2) Compare Equation (2) with d i d t + p ( t ) i = Q ( t ) . The value of p ( t ) is 500 and Q ( t ) is 300 . Calculate the integrating factor. I F = e ∫ p ( t ) d t = e ∫ 500 d t = e 500 t Multiply the integrating factor e 500 t in Equation (2) and integrate

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.7, Problem 29E

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Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

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Find the value of current i ( t ) and determine the current as t → ∞ .

Solution Summary: The author calculates the linear differential equation for the current i(t).

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Chapter 2 Solutions