Program Explanation Given information: The integral-differential equation for the R L C circuit is written as, L d i d t + R i + 1 C ∫ 0 t i ( τ ) d τ = e ( τ ) The initial values for the above system is given as, i ( 0 ) = 0 The function e ( t ) is defined as, e ( t ) = { 100 sin 10 t 0 ≤ t < π 0 t ≥ π The values for constant L , R and C are defined as, L = 1 R = 0 C = 10 − 4 Calculation: Substitute the function e ( t ) and values of the elements on the integral-differential equation of R L C circuit as shown below. ( 1 ) d i d t + ( 0 ) i + 1 10 − 4 ∫ 0 t i ( τ ) d τ = ( 1 − u ( t − π ) ) 100 sin 10 t d i d t + 10 4 ∫ 0 t i ( τ ) d τ = ( 1 − u ( t − π ) ) 100 sin 10 t d i d t + 10 4 ∫ 0 t i ( τ ) d τ = 100 sin 10 t − 100 u ( t − π ) sin 10 t Take the Laplace transform of R L C circuit integral-differential equation as, L { d i d t + 10 4 ∫ 0 t i ( τ ) d τ } = L { 100 sin 10 t − 100 u ( t − π ) sin 10 t } L { d i d t } + 10 4 L { ∫ 0 t i ( τ ) d τ } = L { 100 sin 10 t } − L { 100 u ( t − π ) sin 10 t } ( s L { i } − i ( 0 ) ) + 10 4 L { i } s = 1000 s 2 + 10 2 − 1000 e − π s s 2 + 10 2 s ( s L { i } − ( 0 ) ) + 10 4 L { i } = 1000 s s 2 + 10 2 − 1000 s

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ISBN: 9780137442782

Author: C. Edwards, David Penney

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Chapter 7.5, Problem 38P

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Program Description: Purpose of the problem is to solve the initial value problem Ldidt+Ri+1C∫0ti(τ)dτ=e(τ);i(0)=0 .

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Chapter 7.5, Problem 37P

Chapter 7.5, Problem 39P

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Chapter 7 Solutions

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Ch. 7.1 - Apply the definition in (1) to find directly tile...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.1 - Prob. 4P Ch. 7.1 - Prob. 5P Ch. 7.1 - Prob. 6P Ch. 7.1 - Prob. 7P Ch. 7.1 - Prob. 8P Ch. 7.1 - Prob. 9P Ch. 7.1 - Prob. 10P

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Program Description: Purpose of the problem is to solve the initial value problem L d i d t + R i + 1 C ∫ 0 t i ( τ ) d τ = e ( τ ) ; i ( 0 ) = 0 .

Solution Summary: The author explains the function e(t) and values of the elements on the integral-differential equation of RLC circuit.

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Program Description: Purpose of the problem is to solve the initial value problem Ldidt+Ri+1C∫0ti(τ)dτ=e(τ);i(0)=0 .

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Chapter 7 Solutions