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Audel electrical course for apprentices and journeymen - Rosenberg P.

Rosenberg P. Audel electrical course for apprentices and journeymen - Wiley & sons , 2004. - 424 p.
ISBN: 0-764-54200-1
Download (direct link): audelelectricalcourseforapprentices2004.pdf
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Chapter 22
Resistance, Capacitance, and Inductance in Series and Parallel
This chapter will cover calculations of resistance, capacitance, and inductance in series and parallel.
RC Circuit
Figure 22-1 shows a 100-ohm noninductive resistor in series with a 4-^F capacitor, connected to a 130-Hz AC source of 2000 volts.
Figure 22-1 Resistance and capacitance in series.
Let us find,
1. Capacitive reactance in the circuit
2. Total impedance of the load
3. Current in the circuit
4. Power factor of the circuit
5. Angle of lead in degrees
11
Xr
306 ohms
2pfC 6.28 X 130 X 0.000004
Z = VR2 + X2C = 21002 + 3062 = 322 ohms I = E/Z — 200/322 = 6.21 amperes cosf = R/Z = 100/322 = 0.311 PF = cosf X 100 = 0.311 X 100 = 31.1% From the table of cosines at the back of this book,
0.311 = 72°
(1)
(2)
(3)
(4)
(5)
243
244 Chapter 22
This may be plotted vectorially. See Figure 22-2. Draw a line, AB, 100 units long for R. Then at 90° from AB, draw BC 306 units long for XC. Connect A and C by a line, measure the units of length in this line, and it will be found to be 322 units in length or 322 ohms impedance.
Figure 22-2 Vector value of R;XC and XL in series as shown in Figure 22-1.
With a protractor, measure angle f and it will be found to be
72°. Since this is a capacitive circuit, the power factor will be lead-
ing and PF = (R/Z) X 100 = (100/322) X 100 = 31.1%.
RLC Circuit
Next add a coil into the circuit as shown in Figure 22-3. This coil is
0.3 H.
1. Find the inductive reactance of the coil.
2. Find the total impedance of the circuit.
3. Find the current in the circuit.
4. Find the power factor.
5. Find the angle f in degrees.
XL = 6.28fL = 6.28 X 130 X 0.3 = 245 ohms (6)
Figure 22-3 Inductance added to the circuit of Figure 22-1.
Resistance, Capacitance, and Inductance in Series and Parallel 245
Z = Vr2 + (XC - Xl)2 = Vl002 + (306 - 245)2
= Vl002 + 612 = 117 ohms (7)
I = E/Z = 2000/117 = 17.09 amperes (8)
cosf = R/Z = 100/117 = 0.855
PF = cosf X 100 = 0.855 X 100 = 85.5% (9)
Cosine f is 0.855, and from the table is 31°. (10)
The capacitive reactance is greater than the inductive reactance, so the power factor is leading.
Now in Figure 22-3 measure CD up from C as 245 units of length (inductive reactance), which has to be subtracted from the capacitive reactance, BC. This leaves the effective capacitive reactance as BD, or 61 ohms XC. Angle can be measured by a protractor as 31° leading.
No circuit actually can have inductance without some resistance. So there are core losses and copper losses. With a capacitor there is a loss required to reverse the dielectric strain. The two problems just covered therefore did not include these losses. The values for a problem covering such losses are illustrated in Figure 22-4.
Figure 22-4 Resistance, inductive reactance, and capacitive reactance in series.
Find:
1. Total current
2. PF of the entire circuit
In conjunction with this problem, Figure 22-5 will be used. Line AB, 200 units long, represents the resistance of R. The inductive
246 Chapter 22
15.490
2000
(A) Resistor vector.
19.5Q
(B) Inductor vectors.
(C) Capacitor vectors.
Figure 22-5 Vector diagram for impedance of R, L, and C of Figure 22-4.
reactance of coil L is XL = 6.28 X 60 X 0.2 = 75.36 ohms. The impedance of coil L is
ZL = 2R2l + XL = 219.52 + 75.362 = 77.8 ohms
The impedance vector for inductor L is given by CE. The capacitive reactance of C is
v 1 1 t
Xc = 6.28fC = 6.28 X 60 X 0.000015 = 177 ohmS
There is no ohmic resistance in a capacitor, so the energy component of a capacitor can’t be expressed as such. It is, however, possible to express the energy component in equivalent ohms of the total impedance of the capacitor. To do this, the power factor of the capacitor must be obtained. This may be done with a voltmeter, ammeter, and a wattmeter as covered in Chapter 20: W/VA = cos f, and cos f X 100 = PF.
From these figures, vector diagram FGH may be constructed. Line GH was found to be 177 ohms. If the power factor is known, the energy component, FG, and the total impedance may be found. Now FH = GH/sin f 2.
Since neither angle f 2 nor its sine are known, this can’t be found directly. The power factor of the capacitor is cos fC, so if PF =
0.0872 = cos fC, the sine of fC may be found by using a table.
Resistance, Capacitance, and Inductance in Series and Parallel 247
Then, applying the above formula, FH = GH/sin = 177/0.9962 = 177.67 or ZC. ZC X cos ^C = 177.67 X 0.0872 = 15.49, which is line FG or the energy component of the capacitor.
To get the combined impedance of these three devices,
R = 200 + 15.49 + 19.5 = 234.99 ohms
Z = 2r2 + (XC - XL)2 = V234.992 + (177 - 75.36)2 = 256.22 ohms.
The current is I = E/Z = 2200/256.2 = 8.5 amperes.
PF = (R/Z) X 100 = 234.99/256.22 X 100 = 91.7%
Figure 22-6 is the combined vectors shown in Figure 22-5 for the impedance of R, L, and C in Figure 22-4. With vectors, the wattless
Figure 22-6 Vector diagram showing summation of impedances of R, L, and C in Figure 22-4.
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