Explanation Given: The current i ( t ) in an RL circuit is governed by the differential equation d i d t + R L i = 1 L E ( t ) , where R and L are constants. There is no current flowing initially. Approach: The Laplace transform of n th derivative of function is given by, L { f n ( t ) } = s n L { f n ( t ) } − s n − 1 f ( 0 ) − ... s f ( n − 2 ) ( 0 ) − f ( n − 2 ) ( 0 ) Calculation: For, Y ( x ) = L { f } L { f ′ } = s Y ( s ) − Y ( 0 ) L { f ″ } = s 2 Y ( s ) − s Y ( s ) − Y ′ ( 0 ) Consider the given differential equation, d i d t + R L i = 1 L E ( t ) Apply Laplace transform to both the sides to get, L { d i d t + R L i } = L { 1 L E ( t ) } L { d i d t } + L { R L i } = 1 L L { E ( t ) } Use the initial conditions, To determine (b) To find: The current i ( t ) when E ( t ) = E 0 sin ω t , where ω is a constant.

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 10.4, Problem 30P

To determine

(a)

To find:

The current i(t) if E(t)=E0 is a constant.

To determine

(b)

To find:

The current i(t) when E(t)=E0sinωt, where ω is a constant.

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Chapter 10.4, Problem 29P

Chapter 10.4, Problem 31P

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