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 t 0 ≤ t < 1 0 t ≥ 1 The values for constant L , R and C are defined as, L = 1 R = 150 C = 2 × 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 + ( 150 ) i + 1 2 × 10 − 4 ∫ 0 t i ( τ ) d τ = ( 1 − u ( t − 1 ) ) 100 t d i d t + 150 i + 5000 ∫ 0 t i ( τ ) d τ = ( 1 − u ( t − 1 ) ) 100 t d i d t + 150 i + 5000 ∫ 0 t i ( τ ) d τ = 100 t − u ( t − 1 ) 100 t Take the Laplace transform of R L C circuit integral-differential equation as, L { d i d t + 150 i + 5000 ∫ 0 t i ( τ ) d τ } = L { 100 t − u ( t − 1 ) 100 t } L { d i d t } + 150 L { i } + 5000 L { ∫ 0 t i ( τ ) d τ } = L { 100 t } − L { u ( t − 1 ) 100 t } [ ( s L { i } − i ( 0 ) ) + 150 L { i } + 5000 L { i } s ] = [ 100 s 2 − 100 L { ( t − 1 ) u ( t − 1 ) } − 100 L { u ( t − 1 ) } ] ( s L { i } − ( 0 ) ) + 150 L { i } + 5000 L { i } s = 100 s 2 − 100 e − s s 2 − 100 e − s s Further, solve the above equation as, ( s + 150 + 5000 s ) L { i } = 100 s 2 − 100 e − s s 2 − 100 e − s s ( s 3 + 150 s 2 + 5000 s ) L { i } = 100 − 100 e − s − 100 s e − s L { i } = 100 s ( s + 100 ) ( s + 50 ) − 100 e − s s ( s + 100 ) ( s + 50 ) − 100 s e − s s ( s + 100 ) ( s +<

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Author: Calvis

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 7.5, Problem 39P

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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 38P

Chapter 7.5, Problem 40P

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Chapter 7 Solutions

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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 the 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