Explanation Given data: Refer to given circuit in the textbook. The value of current ( i g ) is, i g = 5 u ( t ) mA (1) Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (2) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (3) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (4) Here, C is the value of capacitance. Calculation: The given circuit is redrawn as shown in Figure 1. Apply Laplace transform for equation (1) to find I g . I g = 5 mA s { ∵ L ( u ( t ) ) = 1 s } = 5 × 10 − 3 s Substitute 200 Ω for R in equation (2) to find Z R . Z R = 200 Substitute 40 mH for L in equation (3) to find Z L . Z L = s ( 40 mH ) = s ( 40 × 10 − 3 ) { ∵ 1 m = 10 − 3 } = 0.04 s Substitute 250 nF for C in equation (4) to find Z C . Z C = 1 s ( 250 nF ) = 1 s ( 250 × 10 − 9 F ) { ∵ 1 n = 10 − 9 } = 4 × 10 6 s Now, the Figure 1 is reduced as shown in Figure 2. Refer to Figure 2, the voltage across the impedance of capacitor is V ϕ . Apply voltage division rule in Figure 2 to find V ϕ . V ϕ = ( 4 × 10 6 s 200 + 4 × 10 6 s ) V o = ( 4 × 10 6 s 200 s + 4 × 10 6 s ) V o = ( 4 × 10 6 200 s + 4 × 10 6 ) V o Apply Kirchhoff’s current law in Figure 2. − 5 × 10 − 3 s + V o 200 + 4 × 10 6 s + 3.75 × 10 − 3 V ϕ + V o 0.04 s = 0 − 5 × 10 − 3 s + V o 200 s + 4 × 10 6 s + 3.75 × 10 − 3 V ϕ + V o 0.04 s = 0 s V o 200 s + 4 × 10 6 + 3.75 × 10 − 3 V ϕ + V o 0.04 s = 5 × 10 − 3 s Substitute ( 4 × 10 6 200 s + 4 × 10 6 ) V o for V ϕ in above equation. s V o 200 s + 4 × 10 6 + 3.75 × 10 − 3 ( ( 4 × 10 6 200 s + 4 × 10 6 ) V o ) + V o 0.04 s = 5 × 10 − 3 s s V o 200 s + 4 × 10 6 + 15000 V o 200 s + 4 × 10 6 + V o 0.04 s = 5 × 10 − 3 s ( s + 15000 ) V o 200 s + 4 × 10 6 + 25 V o s = 5 × 10 − 3 s V o [ ( s + 15000 ) 200 s + 4 × 10 6 + 25 s ] = 5 × 10 − 3 s Simplify the above equation as follows: V o [ s ( s + 15000 ) + 25 ( 200 s + 4 × 10 6 ) s ( 200 s + 4 × 10 6 ) ] = 5 × 10 − 3 s V o [ s 2 + 15000 s + 5000 s + 10 8 s ( 200 s + 4 × 10 6 ) ] = 5 × 10 − 3 s V o [ s 2 + 20000 s + 10 8 200 s ( s + 20000 ) ] = 5 × 10 − 3 s Simplify the above equation to find V o

11th Edition

ISBN: 9780134746968

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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