Explanation Given: R 1 = R 2 = 1 Ω , C = 1 F, L = 1H . The electrical circuit is shown below. Calculation: The three equations for the unknown currents are: I 2 R 2 − q C = 0 , I 1 = d q d t , − I 2 R 2 + I 3 R 1 + L d I 3 d t = 0 Substitute given values in above equations. I 2 = q ( 1 ) I 1 = d q d t = d I 2 d t ( 2 ) − I 2 + I 3 + d I 3 d t = 0 ( 3 ) By Kirchoff Current Law, I 1 + I 2 + I 3 = 0 . I 3 = − I 1 − I 2 ( 4 ) Substitute Equation ( 1 ) and Equation ( 2 ) in Equation ( 4 ) . I 3 = − d I 2 d t − q ( 5 ) Differentiate Equation ( 5 ) . d I 3 d t = − d 2 I 2 d t 2 − d q d t = q − I 3 = q + d I 2 d t + q = 2 q + d I 2 d t d I 3 d t = − d 2 I 2 d t 2 − d q d t I 2 − I 3 = − d 2 I 2 d t 2 − d q d t q + d I 2 d t + q = − d 2 I 2 d t 2 − d q d t d 2 I 2 d t 2 + d I 2 d t + d q d t + 2 q = 0 d 2 q d t 2 + 2 d q d t + 2 q = 0 ( 6 ) The auxiliary equation of Equation ( 6 ) is, r 2 + 2 r + 2 = 0 ( 7 ) Roots of Equation ( 7 ) are r = − 1 ± i . So, the general solution of Equation ( 6 ) is, q ( t ) = A 1 e − t cos ( t ) + B 1 e − t sin ( t ) Here, A 1 and B 1 are arbitrary constants

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Chapter 5.RP, Problem 17RP

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The general solution I 1 ( t ) , I 2 ( t ) and I 3 ( t ) for the systems of differential equations.

Solution Summary: The author explains the three equations for the unknown currents: lI_2R

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