Explanation Given: The equation is L d i 1 d t + R i 2 = E ( t ) and R C d i 2 d t + i 2 − i 1 = 0 if R = 50 Ω , C = 10 − 4 f , E = 60 V , L = 1 2 h , i 1 ( 0 ) = 0 , i 2 ( 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 L { f ( t ) } = F ( s ) and a > 0 , then L − 1 { F ( s − a ) } = e a t f ( t ) . (3) 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 d i 1 d t + R i 2 = E ( t ) and R C d i 2 d t + i 2 − i 1 = 0 . Substitute the value of R = 50 Ω , C = 10 − 4 f , i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 in above equation. Therefore, the resulting equation is 2 i ′ 1 + 50 i 2 = 60 and 0.005 i ′ 2 + i 2 − i 1 = 0 respectively. Apply Laplace transformation on 2 i ′ 1 + 50 i 2 = 60 . L { 2 i ′ 1 } + 50 L { i 2 } = L { 60 } L { i ′ 1 } + 25 L { i 2 } = 30 L { 1 } ( Divide by 2 on both sides ) s L { i 1 } − i 1 ( 0 ) + 25 L { i 2 } = 30 s ( i 1 ( 0 ) = 0 ) s L { i 1 } + 25 L { i 2 } = 30 s ...... ( 1 ) Apply Laplace transformation on 0.005 i ′ 2 + i 2 − i 1 = 0 . L { 0.005 i ′ 2 } + L { i 2 } − L { i 1 } = L { 0 } L { i ′ 2 } + 200 L { i 2 } − 200 L { i 1 } = 0 ( Divide by 0.005 on both sides ) s L { i 2 } − i 2 ( 0 ) + 200 L { i 2 } − 200 L { i 1 } = 0 ( i 2 ( 0 ) = 0 ) ( s + 200 ) L { i 2 } − 200 L { i 1 } = 0 ...... ( 2 ) Multiply equation (1) by 200 and equation (2) by s and add both equations. 200 s L { i 1 } + 5000 L { i 2 } = 6000 s − 200 s L { i 1 } + ( s + 200 ) s L { i 2 } = 0 _ ( 5000 + s 2 + 200 s ) L { i 2 } = 6000 s From above equation, the expression for L { i 2 } is given below. L { i 2 } = 6000 s ( 5000 + s 2 + 200 s ) Apply partial fraction in above equation. 6000 s ( 5000 + s 2 + 200 s ) = A s + B s + C 5000 + s 2 + 200 s 6000 = A ( 5000 + s 2 + 200 s ) + s ( B s + C ) 6000 = A s 2 + 200 A s + 5000 A + B s 2 + s C 6000 = ( A + B ) s 2 + ( 200 A + C ) s + 5000 A ...... ( 3 ) Compare the coefficient in equation (3). A + B = 0 0 = 200 A + C 6000 = 5000 A From the equation 6000 = 5000 A , the value of A is given below

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.6, Problem 19E

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The solution for the system of equation Ldi1dt+Ri2=E(t) and RCdi2dt+i2−i1=0 if R=50Ω,C=10−4f,E=60V,L=12h, i1(0)=0,i2(0)=0 using Laplace transformation.

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Chapter 7.6, Problem 18E

Chapter 7.6, Problem 20E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 10E

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The solution for the system of equation L d i 1 d t + R i 2 = E ( t ) and R C d i 2 d t + i 2 − i 1 = 0 if R = 50 Ω , C = 10 − 4 f , E = 60 V , L = 1 2 h , i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 using Laplace transformation.

Solution Summary: The author explains the Laplace transformation solution of the equation Ldi_1t+R 2=E(t)

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The solution for the system of equation Ldi1dt+Ri2=E(t) and RCdi2dt+i2−i1=0 if R=50Ω,C=10−4f,E=60V,L=12h, i1(0)=0,i2(0)=0 using Laplace transformation.

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Chapter 7 Solutions