Explanation Given: The values are as follows: R = 2.5 Ω and C = 0.08 f For the linear differential R C series circuit the expression for the potential is shown below in the graph. The initial value is q ( 0 ) = 0 . Concept: 1. A piecewise defined continuous function of the form f ( t ) = { g ( t ) , 0 ≤ t < a h ( t ) , t ≥ a can be expressed as f ( t ) = g ( t ) − g ( t ) U ( t − a ) + h ( t ) U ( t − a ) . If F ( s ) = L { f ( t ) } and a > 0 , then L { f ( t − a ) U ( t − a ) } = e − a s F ( s ) L { U ( t − a ) } = e − a s s Where, U ( t − a ) is called unit step function. 2. 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 ) ] 3. The formula for the Laplace transform is as follows: L { 1 } = 1 s 4. The formula for the inverse Laplace transform is as follows: L − 1 { 1 s − a } = e a t 5. If F ( s ) = L { f ( t ) } and a > 0 , then L − 1 { F ( s − a ) } = e a t f ( t ) 6. The linearity property of the Laplace transformation is written below. If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } Calculation: The expression for the linear differential R C series circuit is as follows: R d q d t + 1 C q = E ( t ) Substitute the value of R and C in above expression. ( 2.5 ) d q d t + 1 0.08 q = E ( t ) 5 2 d q d t + 25 2 q = E ( t ) The step function for the given graph is as follows, E ( t ) = { 0 , 0 ≤ t < 3 5 , t ≥ 3 Use the definition 1 to write the unit step function for the function E ( t ) . E ( t ) = 5 U ( t − 3 ) Use the definition 3 to write the Laplace transform of the above linear differential R C series circuit and substitute E ( t ) = 5 U ( t − 3 ) . 5 2 [ s L { q } − q ( 0 ) ] + 25 2 L { q } = L { 5 U ( t − 3 ) } ...... ( 1 ) Substitute the given initial value q ( 0 ) = 0 in equation (1). ( 5 2 s + 25 2 ) L { q } = 5 e − 3 s s ( s + 5 ) L { q } = 2 e − 3 s s L { q } = 2 e − 3 s s ( s + 5 ) Label the above equation as equation (2)

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 7.3, Problem 73E

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The expression for the charge $q(t)$ on the capacitor in an $RC$ series circuit using Laplace transform.

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Chapter 7.3, Problem 72E

Chapter 7.3, Problem 74E

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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 ) on the capacitor in an R C series circuit using Laplace transform.

Solution Summary: The author describes the expression for the charge q(t) on the capacitor in a RC series circuit using Laplace transform.

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The expression for the charge $q(t)$ on the capacitor in an $RC$ series circuit using Laplace transform.

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Chapter 7 Solutions

The expression for the charge q ( t ) on the capacitor in an R C series circuit using Laplace transform.

Solution Summary: The author describes the expression for the charge q(t) on the capacitor in a RC series circuit using Laplace transform.

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The expression for the charge q(t)<m:math><m:mrow><m:mi>q</m:mi><m:mrow><m:mo>(</m:mo><m:mi>t</m:mi><m:mo>)</m:mo></m:mrow></m:mrow></m:math> on the capacitor in an RC<m:math><m:mrow><m:mi>R</m:mi><m:mi>C</m:mi></m:mrow></m:math> series circuit using Laplace transform.

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Chapter 7 Solutions

The expression for the charge $q(t)$ on the capacitor in an $RC$ series circuit using Laplace transform.