Explanation Given: The given integral equation of R C series circuit is R i + 1 C ∫ 0 t i ( τ ) d τ = E ( t ) . The given values are R = 10 Ω , C = 0.5 f . For the given R C series circuit the expression for the potential is E ( t ) = 2 ( t 2 + t ) . Concept used: The integral property of the Laplace transform : If L { f ( t ) } = F ( s ) then L { ∫ 0 t f ( t ) d t } = F ( s ) s . The formula for the Laplace transform is L { t n } = n s n + 1 . The formula for the inverse Laplace transform is L − 1 { 1 s − a } = e a t . Linearity property of the Laplace transformation 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: From the given integral equation of R C series circuit the equation for potential is as shown below. R i + 1 C ∫ 0 t i ( τ ) d τ = E ( t ) Substitute the value E ( t ) = 2 ( t 2 + t ) in above expression as follows. R i + 1 C ∫ 0 t i ( τ ) d τ = 2 ( t 2 + t ) Use the definition of Laplace transformation to write the Laplace transform of the above integral R C series circuit as shown below. R L { i } + 1 C L { i } s = 2 L { t 2 } + 2 L { t } R L { i } + 1 C L { i } s = 2 ⋅ 2 s 3 + 2 ⋅ 1 s 2 Substitute the value of R and C in above expression. 10 L { i } + 2 L { i } s = 2 ⋅ 2 s 3 + 2 ⋅ 1 s 2 ( 10 s + 2 ) L { i } = 4 s 2 + 2 s ( 10 s + 2 ) L { i } = 2 ( s + 2 ) s 2 L { i } = ( s + 2 ) s 2 ( 5 s + 1 ) That is, L { i } = ( s + 2 ) s 2 ( 5 s + 1 ) (1) Simplify the equation (1) using partial fraction as follows. s + 2 s 2 ( 5 s + 1 ) = a 0 s + a 1 s 2 + a 2 5 s + 1 (2) Multiply with s 2 ( 5 s + 1 ) on both side of above expression

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4, Problem 43CR

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The expression for the current i(t) from the given integral equation of RC series circuit using Laplace transform.

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The expression for the current i ( t ) from the given integral equation of R C series circuit using Laplace transform.

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Chapter 4 Solutions