(a) Explanation Given: The values are R = 50 Ω , C = 0.01 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 2. The Laplace transformation of a linear differential equation with constant coefficient is express as, 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 as 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: 50 d q d t + 1 0.01 q = E ( t ) 50 d q d t + 100 q = E ( t ) The step function for the given graph is as follows: E ( t ) = { 0 , 0 ≤ t < 1 E 0 , 1 ≤ t < 3 0 , t ≥ 3 Use the definition 1 to write the unit step function for the function E ( t ) . E ( t ) = E 0 [ U ( t − 1 ) − U ( t − 3 ) ] Use the definition 2 to write the Laplace transform of the above linear differential R C series circuit and substitute E ( t ) = E 0 [ U ( t − 1 ) − U ( t − 3 ) ] ,then equation becomes 50 [ s L { q } − q ( 0 ) ] + 100 L { q } = E 0 [ L { U ( t − 1 ) } − L { U ( t − 3 ) } ] … … . ( 1 ) Substitute the given initial value q ( 0 ) = q 0 in equation (1) and solve. ( 50 s + 100 ) L { q } = E 0 [ L { U ( t − 1 ) } − L { U ( t − 3 ) } ] ( 50 s + 100 ) L { q } = E 0 [ e − s s − e − 3 s s ] L { q } = E 0 50 [ e − s s ( s + 2 ) − e − 3 s s ( s + 2 ) ] From the above expression, the result obtained is as follows: L { q } = E 0 50 [ e − s s ( s + 2 ) − e − 3 s s ( s + 2 ) ] … … ( 2 ) The term 1 s ( s + 2 ) in equation (2) is simplified using partial fraction (b) To determine To draw: The graph for the expression of charge q ( t ) for 0 ≤ t ≤ 6 , and then find the maximum value of charge from the graph.

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 76E

(a)

To determine

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

(b)

To determine

To draw: The graph for the expression of charge q(t) for 0≤t≤6, and then find the maximum value of charge from the graph.

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Chapter 4.3, Problem 75E

Chapter 4.3, Problem 77E

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

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

To draw: The graph for the expression of charge q(t) for 0≤t≤6, and then find the maximum value of charge from the graph.

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Chapter 4 Solutions