# qr code generator vb.net * See note 7 in Appendix. in .NET framework Print Code 128 Code Set C in .NET framework * See note 7 in Appendix.

* See note 7 in Appendix.
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CHAPTER 5 Sinusoidal Waves. rms Value
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Fig. 103. P VI cos h.
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Equation (117) gives power in terms of V and I, that is, in terms of the magnitudes " " of the rms values of voltage and current and the angle  between the V and I vectors. Now suppose that V and I are the rms values of voltage across, and current through, a pure resistance of R ohms; in that case  0 (see discussion for Fig. 98) and, since cos 0 1, eq. (117) becomes, if V is the voltage across and I the current through, a resistance of R ohms, P VI or, since V RI, P I 2R or, since I V=R, P V 2 =R 120 where, as before, P is the average power in watts. Before going on let us pause, just brie y, to comment on vector diagram notation. In a vector diagram the lengths of the vectors represent the magnitudes of the rms values of voltage and current, and the phase angles are represented graphically by actually drawing the vectors at the speci ed angles with respect to each other and to the reference vector. " " Thus, in the vector diagram of Fig. 103, V and I (not V and I ) denote the MAGNI" and the vector current I , while the phase " TUDES of the rms values of the vector voltage V shift is shown directly on the diagram by drawing the vector lengths at the required angle  with respect to a reference vector or reference line. (The notation is further illustrated in Fig. 101.) Problem 73 This is a continuation of problem 72, Fig. 102, using all the same values as given in that problem. (a) (b) The power produced in the 25-ohm resistive load is power produced by generator of voltage V1 is power produced by generator of voltage V2 is power produced by generator of voltage V3 is (c) (d) watts. watts, watts, watts. The power produced by each individual generator in Fig. 102 is as follows: 119 118
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The sum of the three answers found in part (b) must be equal to the answer found in part (a); check to see that your answers satisfy this requirement. Make a freehand sketch of the vector diagram showing the relationships in Fig. 102. Show and label the generator voltages, the output voltage, the current, and the various phase angles involved.
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The principle of superposition states that, in a network composed of linear bilateral elements and several generators, the total CURRENT at any point in the network is equal to the sum of the currents due to each generator considered separately, the other generators being replaced by their internal resistances.* The principle of superposition, however, cannot, in general, be applied in the same way to POWER calculations as it is in current calculations. This is because power is proportional to the SQUARE of current, P I 2 R. Thus the total power produced in a resistance R, due to the presence of several di erent components of current in R, is not equal to the sum of the powers due to each current component considered separately as if the other components were absent. (This assumes all generators have the same frequency.) For example, suppose a total current I is equal to the sum of two separate current components I1 and I2 ; that is, I I1 I2 . The total power PT produced in R is equal to PT I 2 R I1 I2 2 R thus
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2 2 PT I1 R I2 R 2I1 I2 R
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121
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