Explanation Given: The value of resistance, inductance and capacitance are R = 10 Ω , L = 1 2 h and C = 0.01 f respectively. For the linear differential R L C series circuit the expression for the potential is as follows: E ( t ) = { 10 , 0 ≤ t < 5 0 , t ≥ 5 The initial values are q ( 0 ) = 0 and q ′ ( 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 expression for the linear differential R L C series circuit is as follows: L d 2 q d t 2 + R d q d t + 1 C q = E ( t ) Substitute 10 Ω for R , 1 2 h for L and 0.01 f for C in above expression. 1 2 d 2 q d t 2 + 10 d q d t + 1 0.01 q = E ( t ) 1 2 d 2 q d t 2 + 10 d q d t + 100 q = E ( t ) Use the above definition (1) to write the unit step function for the function E ( t ) . E ( t ) = 10 − 10 U ( t − 5 ) Use the above definition (4)v to write the Laplace transform of the above linear differential R L C series circuit and substitute 10 − 10 U ( t − 5 ) for E ( t ) . 1 2 [ s 2 L { q } − s q ( 0 ) − q ′ ( 0 ) ] + 10 [ s L { q } − q ( 0 ) ] + 100 L { q } = L { 10 − 10 U ( t − 5 ) } … … ( 1 ) Substitute initial values 0 for q ( 0 ) and 0 for q ′ ( 0 ) in equation ( 1 ) . ( s 2 + 20 s + 200 ) L { q } = 20 L { 1 } − 20 L { U ( t − 5 ) } ( s 2 + 20 s + 200 ) L { q } = 20 s − 20 s e − 5 s L { q } = 20 s ( s 2 + 20 s + 200 ) − 20 s ( s 2 + 20 s + 200 ) e − 5 s … … ( 2 ) The term 20 s ( s 2 + 20 s + 200 ) in equation ( 2 ) is simplified by partial fractions. 20 s ( s 2 + 20 s + 200 ) = a 0 s + a 2 s + a 1 s 2 + 20 s + 200 … … ( 3 ) Multiply s ( s 2 + 20 s + 200 ) to equation ( 3 ) . 20 s ( s 2 + 20 s + 200 ) s ( s 2 + 20 s + 200 ) = a 0 s ( s 2 + 20 s + 200 ) s + s ( a 2 s + a 1 ) ( s 2 + 20 s + 200 ) s 2 + 20 s + 200 20 = a 0 ( s 2 + 20 s + 200 ) + s ( a 2 s + a 1 ) … … ( 4 ) Substitute 0 for s in equation ( 4 ) . 20 = a 0 ( 0 2 + 20 ⋅ 0 + 200 ) + ( a 2 ⋅ 0 + a 1 ) ⋅ 0 20 = 200 a 0 a 0 = 1 10 ( Divide by 200 ) Substitute 1 10 for a 0 in equation ( 4 )

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, Problem 46RE

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The expression for the charge q(t) from the linear differential equation of RLC series circuit using Laplace transform.

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Chapter 7, Problem 45RE

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Chapter 7 Solutions

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 expression for the charge q ( t ) from the linear differential equation of R L C series circuit using Laplace transform.

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The expression for the charge q(t) from the linear differential equation of RLC series circuit using Laplace transform.

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Chapter 7 Solutions