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where we have selected the positive value of arccos (pf1 ) because pf1 is lagging, and the negative value of arccos (pf2 ) because pf2 is leading Now we compute the reactive power at each load: |S1 | = |S2 | = 105 P P = = 1429 105 VA = pf 1 cos( 1 ) 07 P P 5 104 = = = 1634 104 VA pf 2 cos( 2 ) 095 1 = arccos(pf 1 ) = 0795 2 = arccos(pf 2 ) = 0318
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and from these values we can calculate Q as shown below: Q1 = |S1 | sin( 1 )
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Q1 = 1429 10 sin(0795) = 102 kVAR Q2 = |S2 | sin( 2 )
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Q2 = 5263 10 sin( 0318) = 1643 kVAR where, once again, 1 is positive because pf1 is lagging, 2 is negative because pf2 is leading (see Table 72) The total reactive power is therefore Q = Q1 + Q2 = 856 kVAR To compute the reactance needed for the power factor correction, we observe that we need to contribute a negative reactive power equal to 856 kVAR This requires a negative reactance, and therefore a capacitor with QC = 856 kVAR The reactance of such a capacitor is given by the expression: XC = and, since C= 1 XC |VS |2 (480)2 = = 2692 QC 856 105
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1 1 = 9853 F = XC 377 ( 2692)
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~ IC ~ I2 ~ IS ~ I1 ~ IL
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2 To draw the phasor diagram, we need only to compute the capacitor current, since we have already computed the other two: ZC = j XC = j 2692 VS = 1783 (1571) A IC = ZC The total current is IS = I1 + I2 + IC = 3125 0 A The phasor diagram corresponding to these three currents is shown in Figure 736
Focus on Computer-Aided Tools: A le containing the computer-generated solution to
this problem may be found in the CD-ROM that accompanies this book
Check Your Understanding
76 Compute the power factor for the load of Example 76 with and without the inductance in the circuit 77 Show that one can also express the instantaneous power for an arbitrary complex load Z = |Z| as p(t) = I 2 |Z| cos + I 2 |Z| cos(2 t + )
~ VS
+ ~
78 Determine the power factor for the load in the circuit of Figure 737, and state whether it is leading or lagging for the following conditions: a vS (t) = 540 cos( t + 15 ) V i(t) = 2 cos( t + 47 ) A b vS (t) = 155 cos( t 15 ) V i(t) = 2 cos( t 22 ) A 79 Determine whether the load is capacitive or inductive for the circuit of Figure 737
if a b c d pf = 087 (leading) pf = 042 (leading) vS (t) = 42 cos( t) i(t) = 42 sin( t) vS (t) = 104 cos( t 12 ) i(t) = 04 cos( t 12 )
710 Prove that the power factor is indeed 1 after the addition of the parallel capacitor in Example 78 711 Compute the magnitude of the current drawn from the source after the power factor correction in the circuit of Example 78
TRANSFORMERS
AC circuits are very commonly connected to each other by means of transformers A transformer is a device that couples two AC circuits magnetically rather than through any direct conductive connection and permits a transformation of the voltage and current between one circuit and the other (for example, by matching a high-voltage, low-current AC output to a circuit requiring a low-voltage, highcurrent source) Transformers play a major role in electric power engineering and
Part I
Circuits
are a necessary part of the electric power distribution network The objective of this section is to introduce the ideal transformer and the concepts of impedance re ection and impedance matching The physical operations of practical transformers, and more advanced models, will be discussed in 16 The Ideal Transformer The ideal transformer consists of two coils that are coupled to each other by some magnetic medium There is no electrical connection between the coils The coil on the input side is termed the primary, and that on the output side the secondary The primary coil is wound so that it has n1 turns, while the secondary has n2 turns We de ne the turns ratio N as n2 N= (732) n1 Figure 738 illustrates the convention by which voltages and currents are usually assigned at a transformer The dots in Figure 738 are related to the polarity of the coil voltage: coil terminals marked with a dot have the same polarity Since an ideal inductor acts as a short circuit in the presence of DC currents, transformers do not perform any useful function when the primary voltage is DC However, when a time-varying current ows in the primary winding, a corresponding time-varying voltage is generated in the secondary because of the magnetic coupling between the two coils This behavior is due to Faraday s law, as will be explained in 16 The relationship between primary and secondary current in an ideal transformer is very simply stated as follows: V2 = N V1 I1 I2 = N (733)
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