Explanation Given data: The given diagram is shown in Figure 1. The given diagram is shown in Figure 2. Calculation: The conversion of mH into H is given as, 1 mH = 1 × 10 − 3 H . The conversion of 5 mH into H is given as, 5 mH = 5 × 10 − 3 H Hence, the value of the inductor, L = 5 × 10 − 3 H From the Figure 2, it is observable that the given waveform is odd symmetric. The expression for the function f ( t ) is given by, i s ( t ) = { 2 0 ≤ t ≤ 0.1 − 2 0.1 ≤ t ≤ 0.2 The time period of the given function is, T 0 = 0.2 s Calculate the Fourier coefficient for the given function. The given function is odd symmetric. Hence, a 0 and a n are zero. The expression for b n is given by, b n = 4 T ∫ 0 T 2 f ( t ) sin n ω 0 t d t (1) Here, ω 0 is the angular frequency. The angular frequency ω 0 is given by, ω 0 = 2 π T 0 Substitute 0.2 s for T 0 in the above expression. ω 0 = 2 π 0.2 s = 10 π rad / s Substitute 0.2 s for T 0 , 10 π rad / s for ω 0 and 2 for f ( t ) in equation (1). b n = 4 0.2 ∫ 0 0.1 2 sin n 10 π t d t = 40 ∫ 0 0.1 sin n 10 π t d t = 40 [ − cos 10 n π t 10 n π ] 0 0.1 = 40 [ − cos 10 n π ( 0.1 ) 10 n π + 1 10 n π ] Further solve the above equation for b n . b n = 4 n π [ 1 − cos n π ] = 4 n π [ 1 − ( − 1 ) n ] For even n , b n = 8 n π For odd n , b n = 0 The expression for the Fourier series expansion is given by, i s ( t ) = a 0 + ∑ n = 1 ∞ a n cos n ω 0 t + b n sin n ω 0 t Substitute 0 for a 0 , 0 for a n , 10 π for ω 0 and 8 n π for b n within defined range of n in the above expression. i s ( t ) = 0 + ∑ n = 1 ∞ 0 cos n ω 0 t + 8 n π sin n 10 π t = ∑ n = odd ∞ 8 n π sin n 10 π t To find the forced response for n t h harmonic, convert the time domain into frequency domain as, I s n = 8 n π ∠ − 90 ο = − j 8 n π Here, I s n is the forced current for n t h harmonic in frequency domain. The expression for the inductive reactance is given by, X L = j n ω 0 L Substitute 5 × 10 − 3 H for L and 10 π for ω 0 in the above expression. X L = j n ( 10 π ) ( 5 × 10 − 3 H ) = j 0.05 π n The expression for the current at n t h harmonic frequency using current divider rule is given by, I f n = I s n ( 10 Ω + X L 10 Ω + 10 Ω + X L ) Substitute j 0.05 π n for X L and − j 8 n π for I s n in the above expression. I f n = − j 8 n π ( 10 Ω + j 0.05 π n 10 Ω + 10 Ω + j 0.05 π n ) = − j 8 n π ( 10 Ω + j 0.05 π n 20 Ω + j 0.05 π n ) The expression for the voltage at n t h harmonic frequency across 10 kΩ resistor is given by, V f n = 10 I f n Substitute − j 8 n π ( 10 Ω + j 0.05 π n 20 Ω + j 0.05 π n ) for I f n in the above expression. V f n = 10 ( − j 8 n π ( 10 Ω + j 0

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ISBN: 9780073545516

Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.

Publisher: Mcgraw-hill Education,

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Chapter 17, Problem 25E

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The value of steady state voltage.

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