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25,000 X 0.8 = 20,000 kW true power from the system. The system equipment from alternators through transmission lines, distribution lines, etc., has to be large enough to handle the current associated with 25,000 kVA, while 20,000 kW is all that is working.
Alternators will motor, and when motoring, if the field is overexcited, the alternator will act as a capacitor, causing the current to lead the impressed voltage. Such units are used to correct power factors of highly inductive loads. These are called synchronous or rotary capacitors.
Capacitors may be installed at motors or elsewhere on a system, in parallel with the load, and thus compensate for the inductive reactance.
Utilities have power factor clauses in their energy rates to large users, that is, when the power factor drops, say, to 90%, the rate goes up. As was stated earlier, wattmeters register true power and not apparent power, so billings for electrical energy ignore the power factor involved and the utilities are entitled to some compensation for this. Power factor meters are installed to indicate or record power factor for rate determination.
Consider the parallel circuit in Figure 22-10 with values as given in the illustration. Find
1. Current in branch A
2. Current in branch B
3. Current in branch C
4. Total current
5. Impedance of branch B
6. Impedance of branch C
252 Chapter 22
Figure 22-10 Resistance, inductance, and capacitance in parallel.
7. Combined impedances
8. Power factor
Branch A has 200 ohms resistance. Branch B has XL = 6.28fL =
6.28 X 60 X 0.2 = 75.36 ohms inductive reactance.
ZB = VR2 + X2l = V19.52 + 75.362 = 77.8 ohms (11)
^ 1,000,000 1,000,000 XC
6.28fC 6.28 X 60 X 15
= 177 ohms capacitive reactance
Cos f for 8.72% PF = 0.0872 and from the table, sin f = 0.996.
177/0.996 = 177.71 ohms
ZR = plain R, as it is a noninductive circuit
IA = E/Zr = 2200/200 = 11 amperes (12a)
IB = E/Zb = 2200/77.8 = 28.3 amperes (12b)
IC = E/Zc = 2200/177.71 = 12.4 amperes (12c)
The power factor of branch B is
cos f = R/Xl = 19.5/77.8 = 0.2506 or 24.6%
The energy component of IB is
I = IB cos f = 28.3 X 0.2509 = 7.1 amperes The wattless component of IB (from the table, sin fB = 0.966) is I = IB sin fB = 28.3 X 0.966 = 27.3 amperes
Resistance, Capacitance, and Inductance in Series and Parallel 253
The power factor of branch C is 8.72%, so cos = 0.0872, and from the table, sin $C = 0.996.
The energy component of IC is
IC cos fC = 12.3 X 0.0876 = 1.07 amperes.
The wattless component of IC is
IC sin fC = 12.3 X 0.0996 = 12.25 amperes.
The total current in the circuit will be the sum of the currents in A, B, and C, taking into account the phase angles involved.
The energy components of all three branches are in phase, so they may be added. For A, this is 11 amperes; for B, this is 7.1 amperes; and for C, this is 1.07 amperes: 11 + 7.1 + 1.07 = 19.17 amperes of energy component or Ie.
The wattless current for A is zero; for B, 27.3 amperes; and for C, 12.25 amperes. The wattless current for C will be subtracted from the wattless current for B or 27.3 - 12.25 = 15.05 amperes.
z = Vr2 + (Xl - Xc)2
Xl = 6.28/L
Xc = 6.28/C
IC = E/Zc
Il = E/Xl
1. A coil has 5 ohms resistance and 7.5 ohms inductive reactance. A capacitor with capacitive reactance of 20 ohms and a power factor of 0.06 is connected in parallel with the coil on a 220-volt circuit.
(a) What is their combined impedance?
(b) What is the current in each device?
(c) What is the total current?
(d) What is the overall power factor?
254 Chapter 22
2. What causes resonance?
3. With capacitive reactance and inductive reactance in parallel, what happens to the line current as XC approaches XL?
4. With XL and XC in series, what happens to the line current and voltage as resonance is approached?
5. Does the frequency affect resonance?
6. Is current resonance desirable?
7. What effect does current resonance have on power factor?
8. In a 60-Hz circuit of 120 volts, the current is 12 amperes, and the current lags the voltage by 60°. Find (a) the power factor,
(b) the power in volt-amperes, and (c) the power in watts.
9. Give two methods of power factor correction.
10. The cosine of f is 0.866. What is the sine of f ?
Thus far, only single-phase (1f) AC circuits have been considered. Single-phase has a very important role in the use of electricity; by the same token, polyphase circuits are also very important.
This discussion will primarily cover three-phase (written 3f) circuits. Mention must be made that at one time it was felt that two-phase would be the most practical polyphase system. It was found, however, that three-phase was more practical and economical. The only two-phase equipment currently being made is for replacement purpose. There are also six-phase systems. Six-phase systems are sometimes used in special rectifier circuits, but the six-phase is derived from rotary converters with a three-phase input and a six-phase output or from special transformers with three-phase input and six-phase output.