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CHAPTER 5 Sinusoidal Waves. rms Value
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" " " " " in which I1 and I2 are unknown rms vector currents and V1 , V2 , V3 are known (given) peak values of the given generator voltages. At this point let us pause to note that the original Fig. 104, which is expressed in terms of the fundamental independent variable time, can now be redrawn in terms of the vector quantities of eqs. (129) and (130), as in Fig. 105.
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Fig. 105
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" " In the gure, I 2 is the vector rms current through the 12-ohm resistor; hence, Vo is the vector rms output voltage, and therefore, by Ohm s law, " " magnitude of rms output voltage jV o j 12jI2 j 131
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Thus, to nd the required ANSWER to our problem ( nd the voltmeter reading across " the 12-ohm resistance), all we now need to do is to solve eqs. (129) and (130) for I2 . The " by determinants are as follows, using the details of the solution of these equations for I2 standard procedure of section 3.5.   " "  25:456 V1 V2     11:314 V V  11:314V 14:142V 25:456V "1 "2 "3 "2 "3 "  I2  132  25:456 11:314  591:991    11:314 28:284  The next step is to nd the SUM OF THE VECTOR QUANTITIES in the numerator of the fraction to the right above. This can be done by recalling that the horizontal component Vh and the vertical component Vv of the resultant sum of a number of vectors is equal, respectively, to the sum of the horizontal components and the sum of the vertical components of the individual vectors (see discussion with Figs. 96 and 97). Making use of this fact, and eqs. (109) and (110) (and (128)), we have, for the case of eq. (132) (note angle in 3rd gradient), " " " 11:314V1 14:142V2 25:456V3 1388:719=236:388 hence " j I2 j thus " jVo j 12 2:346 28:150 volts; answer: 1388:719 2:346 amperes 591:991
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CHAPTER 5 Sinusoidal Waves. rms Value
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Some comments regarding the foregoing problem are as follows. Note that we began with Fig. 104, to which eqs. (122) and (123) apply, these being expressed in terms of the independent variable time, t. Then, by requiring that only rms values need be calculated, we were able to eliminate the variable time, thus producing Fig. 105 and the corresponding vector eqs. (129) and (130). (It should be mentioned that an experienced engineer would not generally bother with Fig. 104 at all, but would start directly with Fig. 105 and eqs. (129) and (130).) We might also comment once again on the sense of direction given to the voltage and current arrows in a network. In Fig. 104, for example, 43 sin !t was chosen to be the REFERENCE VOLTAGE WAVEFORM, with a voltage arrow drawn beside the generator. The arrow is drawn with the meaning of Fig. 89, denoting the polarity to during the times that 43 sin !t is positive in value. Next in Fig. 104 consider, for example, the voltage waveform having the peak value of 64 volts. It is GIVEN in the problem that this voltage waveform leads the reference voltage wave by =3 radians in the sense to indicated by the voltage arrow given alongside the generator. Next, the CURRENT ARROWS de ne the sense in which positive currents ow in a network. In our work we ll usually draw all the current arrows around the loops in the same sense (let us say all clockwise), because this will tend to produce voltage equations that are somewhat more symmetrical in form. When a negative current, I sin !t a , occurs in a solution, it means that that particular current wave actually ows in a counterclockwise sense during the times that I sin !t a is positive in value. Problem 74 " In Fig. 105, nd the magnitude of Vo if it had been given that the voltage arrow for "3 should point from left to right instead of from right to left. V " (Answer: jVo j 37:992 volts)
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