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Fig. 96
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Algebraically, one way of nding the magnitude and phase angle of V in Fig. 96, given the magnitudes of V1 and V2 and the phase angle a, is to rst resolve V1 and V2 into their individual HORIZONTAL AND VERTICAL COMPONENTS; then the horizontal component of V sum of horizontal components of V1 and V2 vertical component of V sum of vertical components of V1 and V2
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* If you wish to study a review of vectors, see note 4 in Appendix.
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CHAPTER 5 Sinusoidal Waves. rms Value
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The above rule is correct because the horizontal components all lie along the same straight line, and the vertical components also all lie along a straight line (perpendicular to the line containing the horizontal components). Hence, upon using the above procedure, we now have the horizontal and vertical components of the resultant vector V, which let us denote by Vh and Vv respectively. Thus, using this notation, Fig. 96 becomes Fig. 97.
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Fig. 97
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Since Fig. 97 is a right triangle, the complete description of the resultant vector V, in both magnitude and phase angle, is given by the equations jVj q 2 2 Vh Vv 109 110 *
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h arctan Vv =Vh
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In the same way, the resultant vector, V, of THREE or more vectors is equal to the vector sum of the horizontal components and the vertical components of the individual vectors. Example 1
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As previously mentioned, ac voltmeters and ammeters read MAGNITUDE OF RMS voltage and current. Going back to Fig. 91, suppose the two series-connected voltmeter readings are V1 58 volts, V2 112 volts, and suppose it is also known that V1 and V2 are 65 degrees out of phase. (It is understood, as always, that we re dealing with sinusoidal waves of the same frequency.) (a) (b) Taking V1 to be the reference vector, with V2 leading by 658, nd the magnitude and phase angle of the vector sum, V , of V1 and V2 . What would be the reading of voltmeter V in the gure
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Solution (a) First, the values of the horizontal and vertical components are as follows. For V1 , Vh 58 volts and Vv 0 volts. Next, for V2 , Vh 112 cos 658 47:333 volts, and Vv 112 sin 658 101:507 volts.
* In Fig. 97, by de nition, tan h Vv =Vh ; eq. (110) simply says that h is the ANGLE whose tangent is equal to Vv divided by Vh . For example, since tan 608 1:7321 we have the inverse statement that 608 arctan 1:7321
CHAPTER 5 Sinusoidal Waves. rms Value
Hence the horizontal component of V is 58 47:333 105:333 volts, and the vertical component of V is 101.507 volts. Thus, p by eq. (109), jVj 21 398:71 146:283 volts, approx., and by eq. (110), h arctan 101:507=105:333 arctan 0:963 68 43:948. The above answers can be combined together in the convenient polar form, thus " V 146:283=43:948 " in which V denotes that V is a vector quantity, which can be read as vector V. Thus the above answer can be read as V is a vector quantity of magnitude 146.283 at angle of 43.94 degrees. (b) 146.283 volts, answer, because ac meters read the magnitude of rms values. Let us next consider certain details about the CURRENT and POWER that are produced by ac generators working into a purely RESISTIVE load. We begin our discussion with Fig. 98, in which a single ac generator, of peak voltage Vp , produces an ac current of peak value Ip amperes in a load resistance of R ohms, as shown.
Fig. 98
Let us now SUPPOSE that a phase shift of  radians exists between the current wave and the voltage wave, as shown in the gure. As we can see from the gure, the generator voltage v is at all times equal to the voltage drop across R, which is equal to Ri by Ohm s law. Thus at all times v Ri Vp sin !t RIp sin !t  The above equation is true for all values of time, including t 0, and upon setting t 0 we have that (since sin 0 0) 0 RIp sin  which can be true only if  0, which shows that there is ZERO PHASE SHIFT between the voltage and current waves in a purely resistive circuit. This fact is shown in phasor diagrams, for the case of Fig. 98, in Fig. 99, in which Fig. 99A and B are combined in the single diagram of Fig. 99C.
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