Explanation Given: The equation is L 1 d i 2 d t + ( R 1 + R 2 ) i 2 + R 1 i 3 = E ( t ) and L 2 d i 3 d t + R 1 i 2 + R 1 i 3 = E ( t ) if R 1 = 6 Ω , R 2 = 5 Ω , L 1 = 1 h , L 2 = 1 h, E ( t ) = − 50 sin t V and i 2 ( 0 ) = 0 , i 3 ( 0 ) = 0 . Concept: 1. A general piecewise continuous function of the form f ( t ) = { g ( t ) , 0 ≤ t < a h ( t ) , t ≥ a is also expressed as f ( t ) = g ( t ) − g ( t ) U ( t − a ) + h ( t ) U ( t − a ) . 2. If F ( s ) = L { f ( t ) } and a > 0 , then L − 1 { e − a s F ( s ) } = f ( t − a ) U ( t − a ) and L − 1 { e − a s s } = U ( t − a ) . 3. If f ( t ) = L − 1 { F ( s ) } and a > 0 , then L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) and L { U ( t − a ) } = e − a s s . 4. The Laplace transformation of a linear differential equation with constant coefficient is expressed as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] Calculation: The equation is L 1 d i 2 d t + ( R 1 + R 2 ) i 2 + R 1 i 3 = E ( t ) and L 2 d i 3 d t + R 1 i 2 + R 1 i 3 = E ( t ) . Substitute the value of R 1 = 6 Ω , R 2 = 5 Ω , L 1 = 1 h , L 2 = 1 h, E ( t ) = − 50 sin t V and i 2 ( 0 ) = 0 , i 3 ( 0 ) = 0 in above equation. Therefore, the resulting equation is i ′ 2 + 11 i 2 + 6 i 3 = 50 sin t and i ′ 3 + 6 i 2 + 6 i 3 = 50 sin t respectively. Apply Laplace transformation on i ′ 2 + 11 i 2 + 6 i 3 = 50 sin t as follows: s L { i 2 } − i 2 ( 0 ) + 11 L { i 2 } + 6 L { i 3 } = 50 L { sin t } s L { i 2 } − 0 + 11 L { i 2 } + 6 L { i 3 } = 50 ( 1 s 2 + 1 ) ( s + 11 ) L { i 2 } + 6 L { i 3 } = 50 ( 1 s 2 + 1 ) ... ( 1 ) Apply Laplace transformation on i ′ 3 + 6 i 2 + 6 i 3 = 50 sin t as follows: s L { i 3 } − i 3 ( 0 ) + 6 L { i 2 } + 6 L { i 3 } = 50 L { sin t } s L { i 3 } − 0 + 6 L { i 2 } + 6 L { i 3 } = 50 ( 1 s 2 + 1 ) ( s + 6 ) L { i 3 } + 6 L { i 2 } = 50 ( 1 s 2 + 1 ) ... ( 2 ) From equation (1) and (2), the expression for L { i 2 } is given below. L { i 2 } = 50 s ( s + 2 ) ( s + 15 ) ( s 2 + 1 ) Apply partial fraction in above equation. 50 s ( s + 2 ) ( s + 15 ) ( s 2 + 1 ) = A s + 2 + B s + 15 + C s + D s 2 + 1 50 s = A ( s + 15 ) ( s 2 + 1 ) + B ( s + 2 ) ( s 2 + 1 ) + ( C s + D ) ( s + 2 ) ( s + 15 ) ... ( 3 ) Substitute − 15 for s in equation (3)

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.6, Problem 17E

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The system of equation L1di2dt+(R1+R2)i2+R1i3=E(t) and L2di3dt+R1i2+R1i3=E(t) if R1=6Ω,R2=5Ω, L1=1 h, L2=1 h, E(t)=−50sintV using Laplace transformation.

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Chapter 7.6, Problem 16E

Chapter 7.6, Problem 18E

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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The system of equation L 1 d i 2 d t + ( R 1 + R 2 ) i 2 + R 1 i 3 = E ( t ) and L 2 d i 3 d t + R 1 i 2 + R 1 i 3 = E ( t ) if R 1 = 6 Ω , R 2 = 5 Ω , L 1 = 1 h , L 2 = 1 h, E ( t ) = − 50 sin t V using Laplace transformation.

Solution Summary: The author explains how Laplace transformation solves the equations L_1di

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The system of equation L1di2dt+(R1+R2)i2+R1i3=E(t) and L2di3dt+R1i2+R1i3=E(t) if R1=6Ω,R2=5Ω, L1=1 h, L2=1 h, E(t)=−50sintV using Laplace transformation.

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