Explanation Given data: The natural response of an R L C circuit is described by, d 2 v d t 2 + 2 d v d t + v = 0 (1) The value of initial voltage v ( 0 ) is 350 V . The value of d v ( 0 ) d t is 0 . Calculation: Apply Laplace transform for equation (1) with initial conditions. [ s 2 V ( s ) − d v ( 0 − ) d t − s v ( 0 − ) + 2 ( s V ( s ) − v ( 0 − ) ) + V ( s ) ] = 0 { ∵ L ( d 2 v d t 2 ) = s 2 V ( s ) − d v ( 0 − ) d t − s v ( 0 − ) L ( d v d t ) = s V ( s ) − v ( 0 − ) } Simplify the above equation as follows, s 2 V ( s ) − d v ( 0 − ) d t − s v ( 0 − ) + 2 V ( s ) − 2 v ( 0 − ) + V ( s ) = 0 (2) Substitute 350 for v ( 0 − ) , and 0 for d v ( 0 − ) d t in equation (2). s 2 V ( s ) − 0 − s ( 350 ) + 2 s V ( s ) − 2 ( 350 ) + V ( s ) = 0 s 2 V ( s ) − 0 − 350 s + 2 s V ( s ) − 700 + V ( s ) = 0 s 2 V ( s ) + 2 s V ( s ) + V ( s ) = 350 s + 700 [ s 2 + 2 s + 1 ] V ( s ) = 350 s + 700 Simplify the above equation to find V ( s ) . V ( s ) = 350 s + 700 s 2 + 2 s + 1 (3) From equation (3), the characteristic equation is written as follows, s 2 + 2 s + 1 (4) Write an expression to calculate the roots of characteristic equation ( a s 2 + b s + c = 0 ) . s 1 , 2 = − b ± b 2 − 4 a c 2 a . (5) Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. Compare equation (4) with quadratic equation ( a s 2 + b s + c = 0 ) . a = 1 b = 2 c = 1 Substitute 1 for a , 2 for b , and 1 for c in equation (5) to find s 1 , 2

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Series and Parallel Circuit

Circuits can be classified as parallel and series depending on the manner in which components are connected to one another and the electrical source. The resultant output of the circuit depends on the electrical component.

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Chapter 16, Problem 3P

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Find the expression of voltage v(t) for t>0.

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Chapter 16, Problem 2P

Chapter 16, Problem 4P

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