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 , 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 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 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 1 2 i ′ 1 + 50 i 2 = 60 and 50 10000 i ′ 2 + i 2 − i 1 = 0 respectively. Apply Laplace transformation on 1 2 i ′ 1 + 50 i 2 = 60 as follows: L { 0.5 i ′ 1 } + 50 L { i 2 } = L { 60 } L { i ′ 1 } + 100 L { i 2 } = 120 L { 1 } ( Divide by 0.5 on both sides ) s L { i 1 } − i 1 ( 0 ) + 100 L { i 2 } = 120 s ( i 1 ( 0 ) = 0 ) s L { i 1 } + 100 L { i 2 } = 120 s ...... ( 1 ) Apply Laplace transformation on 50 10000 i ′ 2 + i 2 − i 1 = 0 as follows: 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

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ISBN: 9781284105902

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Publisher: Jones & Bartlett Learning

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Chapter 4.6, Problem 18E

To determine

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

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Chapter 4.6, Problem 17E

Chapter 4.6, Problem 19E

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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 , i 1 ( 0 ) = 0 , i 2 ( 0 ) = 0 using Laplace transformation.

Solution Summary: The author explains the Laplace transformation solution of the equation Ldi_1 dt+R

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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, i1(0)=0,i2(0)=0 using Laplace transformation.

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Chapter 4 Solutions