By Lionel Warnes (auth.)
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9) Vl If v is in volts (V) and i in amperes (A), p is in watts (W). 10 shows that PR has a minimum value of zero: the power (which is lost as heat) produced in a resistance carrying a sinusoidal current is always positive. 5 Vmlm and an alternating component of twice the current's frequency. 2a shows the relationships of the waveforms. The average power is P av lJ = T Making the substitution cor T 0 v1,,iR dt lJ = T TVmI m sin 2 wt dt . 12) (1- cos2x)dx = T, x = mT = 2n. I~ 2 This form should be compared to the power, 12 R, developed by a direct current, /.
72) after rearranging terms. 73 (if in doubt, reverse the arrow and the sign on this source, which leaves its value unchanged). 1 (/2 - I 1) by Ohm's law (V4 is obviously just - V3 , but it is best to give it a separate identity in a different mesh). 76) Again the voltages are found by using Ohm's law: V7 20(/3 - I2 ). 78) And there are three unknown mesh currents. 7 A. 5 A). 5 kW. 78: the first comes from applying KVL to mesh 1, so the coefficient of I 1 is opposite in sign to those of the other two currents.
3 shows. 1 ---2 r--. 3 Graphs showing v(t), i(t) and p(t) for an inductance carrying AC. 22) Sinusoidally-excited circuits 44 Once again the average power over half a cycle is zero. 4. 23) FF = Vrm)Vav where vav• the time-averaged value, is given by vav = -1 JT lv(t)ldt = - W J2'TT'/W . 24) and Iv(t) I means the absolute value or magnitude - the rectified value. s. 1 What are Vav• Vrms• the FF and the CF for v(t) = Vmsinwt? 25) We integrate over only half the period because the rectified sinewave is used (integration 45 Chapter 2 over the whole period gives zero of course).