Explanation Given data: The value of the source current is, i t = 5 cos 20 t A . The value of resistance ( R ) is 1.25 Ω . The value of inductance ( L ) is 200 mH . The value of Capacitance ( C ) is 50 mF . The energy stored in the circuit at t = 0 is zero. Formula used: Consider the general expression for current across the resistor in time domain. i R = v ( t ) R Consider the general expression for current across an inductor in time domain. i L = 1 L ∫ 0 t v ( t ) d t Consider the general expression for current across the capacitor in time domain. i C = C d v ( t ) d t Calculation: Find the Laplace Transform of input current I s as follows: I s = L { 5 cos 20 t } = 5 [ s s 2 + 20 2 ] = 5 s s 2 + 400 The given circuit is redrawn as shown in Figure 1. Write the equation for the given circuit using Kirchhoff’s Current Law as follows: i R + i L + i C = i t (1) Substitute v ( t ) R for i R , 1 L ∫ 0 t v ( t ) d t for i L and C d v d t for i C in equation (1) as follows: v ( t ) R + 1 L ∫ 0 t v ( t ) d t + C d v ( t ) d t = i t Refer to TABLE 12.1 in the textbook for an abbreviated list of Laplace transform pairs and TABLE 12.2 for an abbreviated list of operational transforms. Find the Laplace transform of the expression as follows: V ( s ) R + 1 L V ( s ) s + C [ s V ( s ) − v ( 0 − ) ] = I s { ∵ L [ K f ( t ) ] = K F ( s ) , L [ ∫ 0 1 f ( t ) d t ] = F ( s ) s a n d L [ d f ( t ) d t ] = s F ( s ) − f ( 0 − ) } Since the energy stored in the circuit at t = 0 is zero, substitute 0 for v ( 0 − ) as follows: V ( s ) R + 1 L V ( s ) s + C [ s V ( s ) − 0 ] = I s V ( s ) R + 1 L V ( s ) s + C s V ( s ) = I s V ( s ) ( 1 R + 1 s L + s C ) = I s (2) Rearrange the equation (2) to find V ( s ) as follows: V ( s ) = I s ( 1 R + 1 s L + s C ) = I s ( s L + R + s 2 R L C s R L ) = ( I s ) [ s R L R L C ( s R C + 1 L C + s 2 ) ] V ( s ) = s I s C ( s 2 + 1 R C s + 1 L C ) (3) Substitute 5 s s 2 + 400 for I s , 1.25 Ω for R , 200 mH for L and 50 mF for C in equation (3) as follows: V ( s ) = ( s 50 × 10 − 3 ) ( 5 s s 2 + 400 ) ( s 2 + 1 ( 1.25 × 50 × 10 − 3 ) s + 1 ( 200 × 10 − 3 × 50 × 10 − 3 ) ) = ( 5 s 2 s 2 + 400 ) ( 10 3 50 ) ( s 2 + 10 3 62.5 s + 10 6 10000 ) = ( 5 s 2 s 2 + 400 ) ( 20 ) ( s 2 + 16 s + 100 ) Simplify the expression as follows: V ( s ) = ( 5 s 2 s 2 + 400 ) ( 20 ) ( 1 s 2 + 16 s + 100 ) = 100 s 2 ( s 2 + 16 s + 100 ) ( s 2 + 400 ) = 100 s 2 ( s + 8 + j 6 ) ( s + 8 − j 6 ) ( s + j 20 ) ( s − j 20 ) Write the partial fraction expansion of V ( s ) as follows: V ( s ) = K 1 ( s + 8 − j 6 ) + K 1 * ( s + 8 + j 6 ) + K 2 ( s − j 20 ) + K 2 * ( s + j 20 ) (4) Substitute 100 s 2 ( s + 8 + j 6 ) ( s + 8 − j 6 ) ( s + j 20 ) ( s − j 20 ) for V ( s ) in equation (4). 100 s 2 ( s + 8 + j 6 ) ( s + 8 − j 6 ) ( s + j 20 ) ( s − j 20 ) = { K 1 ( s + 8 − j 6 ) + K 1 * ( s + 8 + j 6 ) + K 2 ( s − j 20 ) + K 2 * ( s + j 20 ) } Multiply both sides by ( s + 8 − j 6 ) and evaluate both sides at s = − 8 + j 6 to find the value of. K 1

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Electric Circuits (10th Edition)

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