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17.11 Find the average power in a resistance R 10
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, if the current in Fourier series form is i 10 sin !t 5 sin 3!t 2 sin 5!t (A).
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The current has an e ective value Ieff 2 power is P Ieff R 64:5 10 645 W.
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q p 2 2 2 1 1 1 64:5 8:03 A. Then the average 2 10 2 5 2 2
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Another Method The total power is the sum of the harmonic powers, which are given by 1 Vmax Imax cos . 2 voltage across the resistor and the current are in phase for all harmonics, and n 0. Then, vR Ri 100 sin !t 50 sin 3!t 20 sin 5!t and P 1 100 10 1 50 5 1 20 2 645 W. 2 2 2
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17.12 Find the average power supplied to a network if the applied voltage and resulting current are v 50 50 sin 5 103 t 30 sin 104 t 20 sin 2 104 t
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i 11:2 sin 5 10 t 63:48 10:6 sin 10 t 458 8:97 sin 2 104 t 26:68 A
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FOURIER METHOD OF WAVEFORM ANALYSIS
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[CHAP. 17
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The total average power is the sum of the harmonic powers: P 50 0 1 50 11:2 cos 63:48 1 30 10:6 cos 458 1 20 8:97 cos 26:68 317:7 W 2 2 2
17.13 Obtain the constants of the two-element series circuit with the applied voltage and resultant current given in Problem 17.12.
The voltage series contains a constant term 50, but there is no corresponding term in the current series, thus indicating that one of the elements is a capacitor. Since power is delivered to the circuit, the other element must be a resistor. q Ieff 1 11:2 2 1 10:6 2 1 8:97 2 12:6 A 2 2 2
2 2 The average power is P Ieff R, from which R P=Ieff 317:7=159:2 2
. 4 At ! 10 rad/s, the current leads the voltage by 458. Hence,
1 tan 458
1 !CR
1 50 mF 104 2
Therefore, the two-element series circuit consists of a resistor of 2
and a capacitor of 50 mF.
17.14 The voltage wave shown in Fig. 17-35 is applied to a series circuit of R 2 k
and L 10 H. Use the trigonometric Fourier series to obtain the voltage across the resistor. Plot the line spectra of the applied voltage and vR to show the e ect of the inductance on the harmonics. ! 377 rad/s.
Fig. 17-35
The applied voltage has average value Vmax =, as in Problem 17.5. The wave function is even and hence the series contains only cosine terms, with coe cients obtained by the following evaluation integral: an 1  =2 300 cos !t cos n!t d !t
=2
600 cos n=2  1 n2
Here, cos n=2 has the value 1 for n 2; 6; 10; . . . ; and 1 for n 4; 8; 12; . . . . For n odd, cos n=2 0. However, for n 1, the expression is indeterminate and must be evaluated separately.   300 !t sin 2!t =2 300 V  2 4 2 =2 =2   300  2 2 2 1 cos !t cos 2!t cos 4!t cos 6!t V v  2 3 15 35 a1 1  =2 300 cos2 !t d !t
Thus,
In Table 17-3, the total impedance of the series circuit is computed for each harmonic in the voltage expression. The Fourier coe cients of the current series are the voltage series coe cients divided by the Zn ; the current terms lag the voltage terms by the phase angles n .
CHAP. 17]
FOURIER METHOD OF WAVEFORM ANALYSIS
Table 17-3 n 0 1 2 4 6 n!, rad/s 0 377 754 1508 2262 I0 R, k
2 2 2 2 2 300= mA 2 300=2 cos !t 628 4:26 mA n!L; k
0 3.77 7.54 15.08 22.62 Zn ; k
2 4.26 7.78 15.2 22.6 n 08 628 75.18 82.458 84.928
i1
i2 Then the current series is i
600=3 cos 2!t 75:18 mA 7:78 ....................................
300 300 600 cos !t 628 cos 2!t 75:18 2 2 4:26 3 7:78 600 600 cos 4!t 82:458 cos 6!t 84:928 15 15:2 35 22:6
mA
and the voltage across the resistor is vR Ri 95:5 70:4 cos !t 628 16:4 cos 2!t 75:18 1:67 cos 4!t 82:458 0:483 cos 6!t 84:928
Figure 17-36 shows clearly how the harmonic amplitudes of the applied voltage have been reduced by the 10-H series inductance.
Fig. 17-36
17.15 The current in a 10-mH inductance has the waveform shown in Fig. 17-37. Obtain the trigonometric series for the voltage across the inductance, given that ! 500 rad/s.
Fig. 17-37
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