To determine Find the maximum amplitude of the current in 5 mH inductor. Explanation Given data: v = 150 cos 4000 t V . Formula used: Consider the general expression for inductive reactance. X L = ω L (1) Here, L is the inductor. ω is the angular frequency. Consider the general expression for capacitive reactance. X C = − 1 ω C (2) Here, C is the Capacitor. ω is the angular frequency. Consider the general expression for phasor voltage of capacitor. V = I Z (3) Here, I is the phasor current. Z is the impedance. Calculation: The Phasor voltage of the given voltage signal is, V = 150 ∠ 0 ° V . From given data, draw the circuit as shown in Figure 1. Find the equivalent impedance between x and y terminals in Figure 1 as follows. Z x y = ( 20 ∥ j ω L ) + ( 5 − j 1 ω C ) = 20 ( j ω L ) 20 + j ω L + ( 5 − j 1 ω C ) Simplify the equation as follows. Z x y = [ [ 20 ( j ω L ) ] ( 20 − j ω L ) ( 20 + j ω L ) ( 20 − j ω L ) ] + ( 5 − j 1 ω C ) = [ 400 j ω L − 20 j 2 ω 2 L 2 20 2 − j 2 ω 2 L 2 ] + ( 5 − j 1 ω C ) = [ 400 j ω L + 20 ω 2 L 2 400 + ω 2 L 2 ] + ( 5 − j 1 ω C ) { ∵ j 2 = − 1 } Z x y = 400 j ω L 400 + ω 2 L 2 + 20 ω 2 L 2 400 + ω 2 L 2 + ( 5 − j 1 ω C ) The impedance will be purely resistive, when the j terms of equation (4) as follows. 400 ω L 400 + ω 2 L 2 = 1 ω C Substitute 5 mH for L and 25 μ F for C as follows. 400 ω ( 5 mH ) 400 + ω 2 ( 5 mH ) 2 = 1 ( 25 μ F ) ω 400 ω ( 5 × 10 − 3 ) 400 + ω 2 ( 5 × 10 − 3 ) 2 = 1 ( 25 × 10 − 6 ) ω { ∵ 1 m = 1 × 10 − 3 1 μ = 1 × 10 − 6 } [ 400 ω ( 5 × 10 − 3 ) ] [ ( 25 × 10 − 6 ) ω ] = 400 + ω 2 ( 5 × 10 − 3 ) 2 ( 5 × 10 − 5 ) ω 2 = 400 + ω 2 ( 2

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ISBN: 9780133760033

Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 9.6, Problem 8AP

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Find the maximum amplitude of the current in 5 mH inductor.

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Electric Circuits (10th Edition)

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