Explanation Given: The given condition is: The equations of the RLC network: 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 160 50 I ′ 1 ( t ) + 800 q 3 ( t ) = 160 I 1 ( t ) = I 2 ( t ) + I 3 ( t ) and I 3 ( t ) = q ′ 3 ( t ) initially q 3 ( 0 ) = 0.5 , I 3 ( 0 ) = 0 . Procedure used: Differentiate the first two equation and eliminate I 1 and form the normal system. Convert into matrix form and then find the eigenvalues. If the real matrix A has complex conjugate eigenvalues α ± i β corresponding eigen vectors α ± i β , then two linearly independent real vector solutions to x ′ ( t ) = A x ( t ) are: e α t cos β t a − e α t sin β t b e α t sin β t a + e α t cos β t b Calculation: Consider the given equations of the RLC network: 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 160 50 I ′ 1 ( t ) + 800 q 3 ( t ) = 160 I 1 ( t ) = I 2 ( t ) + I 3 ( t ) By using 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 160 , we get: 5 I ′ 1 ( t ) = − 8 I 2 ( t ) + 16 I ′ 1 ( t ) = − 8 5 I 2 ( t ) + 16 5 By equation 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 160 and 50 I ′ 1 ( t ) + 800 q 3 ( t ) = 160 , we get: 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 50 I ′ 1 ( t ) + 800 q 3 ( t ) I 2 ( t ) − 10 q 3 ( t ) = 0 I ′ 2 ( t ) − 10 I 3 ( t ) = 0 I ′ 2 ( t ) = 10 I 1 ( t ) − 10 I 2 ( t ) Differentiate 50 I ′ 1 ( t ) + 80 I 2 ( t ) = 160 : 50 I ″ 1 ( t ) + 80 I ′ 2 ( t ) = 0 5 I ″ 1 ( t ) + 8 I ′ 2 ( t ) = 0 Differentiate 50 I ′ 1 ( t ) + 800 q 3 ( t ) = 160 : 50 I ″ 1 ( t ) + 800 I 3 ( t ) = 0 I ″ 1 ( t ) + 16 I 3 ( t ) = 0 Substitute I 3 ( t ) = I 1 ( t ) − I 2 ( t ) into 50 I ″ 1 ( t ) + 800 I 3 ( t ) = 0 : 50 I ″ 1 ( t ) + 800 ( I 1 ( t ) − I 2 ( t ) ) = 0 I ″ 1 ( t ) + 16 I 1 ( t ) − 16 I 2 ( t ) = 0 I ″ 1 ( t ) = − 16 I 1 ( t ) + 16 I 2 ( t ) Let, x 1 = I 1 x 2 = I ′ 1 x 3 = I 2 Then, x ′ 1 = − 8 5 x 3 − 16 5 x ′ 2 = − 16 x 1 + 16 x 3 x ′ 3 = 10 x 1 − 10 x 3 Write into matrix form, we get: [ x 1 x 2 x 3 ] ′ = [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] [ x 1 x 2 x 3 ] + [ − 16 5 0 0 ] x ′ ( t ) = A x ( t ) + f ( t ) The eigen value of matrix A is: | A − λ I | = 0 | [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | [ − λ 0 − 8 5 − 16 − λ 16 10 0 − 10 − λ ] | = 0 Further simplify as: − λ 3 − 10 λ 2 − 16 λ = 0 − λ ( λ 2 + 10 λ + 16 ) = 0 λ ( λ + 2 ) ( λ + 8 ) = 0 λ = 0 , − 2 , − 8 The eigenvector corresponding to eigenvalue λ = 0 : ( [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] − ( 0 ) [ 1 0 0 0 1 0 0 0 1 ] ) [ x 1 x 2 x 3 ] = 0 ( [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] ) [ x 1 x 2 x 3 ] = 0 It reduces by using row-echelon form: [ 0 0 − 8 5 − 16 0 16 10 0 − 10 ] → [ 1 0 0 0 0 1 0 0 0 ] It indicate as, x 1 = 0 , x 3 = 0 and x 2 is free

Fundamentals of Differential Equations [With CDROM] - 7th Edition

7th Edition

ISBN: 9780321410481

Author: Saff, Edward B., Snider, Arthur David, Nagle, R. Kent

Publisher: Addison Wesley

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Chapter 9.6, Problem 22E

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The currents I1,I2 and I3 for the given condition q3(0)=0.5,I3(0)=0.

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Chapter 9.6, Problem 21E

Chapter 9.6, Problem 23E

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Fundamentals of Differential Equations [With CDROM] - 7th Edition

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The currents I 1 , I 2 and I 3 for the given condition q 3 ( 0 ) = 0.5 , I 3 ( 0 ) = 0 .

Solution Summary: The author explains how the equations of the RLC network are determined.

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The currents I1,I2 and I3 for the given condition q3(0)=0.5,I3(0)=0.

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