V2;avg V1;avg V1;avg 2V1;avg 2Vm = And so, or p V2;eff Vm = 2

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Use the results of Problems 6.5 and 6.13 to nd V2;eff .

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2 2 2 2 2 V2;eff V1;eff V1;eff 2V1;eff Vm =2

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WAVEFORMS AND SIGNALS

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The rms value of v2 t can also be derived directly. Because of the squaring p operation, a full-recti ed cosine function has the same rms value as the cosine function itself, which is Vm = 2.

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A 100-mH inductor in series with 20- resistor [Fig. 6-18(a)] carries a current i as shown in Fig. 6-18(b). Find and plot the voltages across R, L, and RL.

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Fig. 6-18 8 > 10 < i 10 1 103 t A > : 0 8 > 200 V < vR Ri 200 1 103 t V > : 0 8 for t < 0 >0 di < 4 for 0 < t < 10 3 s 10 A=s dt > : 0 for t > 10 3 s 8 for t < 0 >0 di < vL L 1000 V for 0 < t < 10 3 s dt > : 0 for t > 10 3 s

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Since the passive elements are in series, vRL vR vL and so 8 < 200 V 2 105 t 800 : 0 for t < 0 for 0 < t < 10 3 s for t > 10 3 s

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The graphs of vL and vRL are given in Fig. 6-18(c) and (d), respectively. The plot of the resistor voltage vR has the same shape as that of the current [see Fig. 6-18(b)], except for scaling by a factor of 20 .

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CHAP. 6]

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A radar signal s t , with amplitude Vm 100 V, consists of repeated tone bursts. Each tone burst lasts Tb 50 ms. The bursts are repeated every Ts 10 ms. Find Seff and the average power in s t .

p Let Veff Vm 2 be the e ective value of the sinusoid within a burst. The energy contained in a single 2 2 burst is Wb Tb Veff . The energy contained in one period of s t is Ws Ts Seff . Since Wb Ws W, we obtain p 2 2 2 2 Seff Tb =Ts Veff Seff Tb =Ts Veff 40 Tb Veff Ts Seff Substituting the values of Tb , Ts , and Veff into (40), we obtain q p Seff 50 10 6 = 10 10 3 100= 2 5 V Then W 10 2 25 0:25 J. The average power in s t is

2 2 P W=Ts Ts Seff =Ts Seff 25 W 2 2 The average power is represented by Seff and its peak power by Veff . The ratio of peak power to p of s t average power is Ts =Tb . In this example the average power and the peak power are 25 W and 5000 W, respectively.

An appliance uses Veff 120 V at 60 Hz and draws Ieff 10 A with a phase lag of 608. Express v, i, and p vi as functions of time and show that power is periodic with a dc value. Find the frequency, and the average, maximum, and minimum values of p.

p v 120 2 cos !t p i 10 2 cos !t 608

p vi 2400 cos !t cos !t 608 1200 cos 608 1200 cos 2!t 608 600 1200 cos 2!t 608 The power function is periodic. The frequency f 2 60 120 Hz and Pavg 600 W, pmax 600 1200 1800 W, pmin 600 1200 600 W.

A narrow pulse is of 1-A amplitude and 1-ms duration enters a 1-mF capacitor at t 0, as shown in Fig. 6-19. The capacitor is initially uncharged. Find the voltage across the capacitor.

Fig. 6-19 The voltage across the capacitor is 1 VC C t

8 <0 i dt 106 t : 1V

for t < 0 for 0 < t < 1 ms (charging period) for t > 1 ms

If the same amount of charge were deposited on the capacitor in zero time, then we would have v u t (V) and i t 10 6 t (A).

The narrow pulse is of Problem 6.18 enters a parallel combination of a 1-mF capacitor and a 1-M resistor (Fig. 6-20). Assume the pulse ends at t 0 and that the capacitor is initially uncharged. Find the voltage across the parallel RC combination.