Example 11 A 150 ft resistor in series with an inductor takes a
current of 2.5 A from a 500 V supply at a frequency of 50 Hz. Find
the inductance of the coil. The circuit is shown in Figure 11.12.
150 n ,
2.5 A
_500V .
50 Hz
Figure
11.12
True power
Reactive power
Apparent power
V 500
Z = - = — = 200 ft
/ 2.5
Z
2
= R
2
+ X\
X
2
L
= Z
2
- R
2
X\ = (200)
2
- (150)
2
X
L
= V[(200)
2
- (150)
2
] = 132.3 ft
132.3 = 2 x TT x 50 x L
132.3
L
= = 0.42 H = 420 mH
2 x TT x 50
V
R
= 2.5 x 150 = 375 V
VL = 2.5 x 132.3 = 330.75 V
= 937.5 W (V
R
f)
= 826.88 VA (V
L
l)
= 1250 VA (VI)
f —
K
_
_VR _
375
V
500
R 150
or
—
=
—
Z 200
VRI
937. ...
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