Basic_Electrical_Engineering_4th_edition
.pdf
A-C CIRCUITS |
65 |
or
or
or or
100 = 2nfi(2) - 2nf10, x6
|
|
|
|
|
= |
12.56{1 |
- |
7.96 x |
||
f12 - 7.96 f1 - 633 = 0 |
|
|
|
f1 |
||||||
5 |
|
|
|
|||||||
|
|
|
|
{ |
= |
29. |
Hz |
|
||
|
|
|
|
1 |
|
|
|
|
|
|
For the second case |
|
|
|
|
|
|
||||
|
|
|
Xc - XLr = 100 |
|
|
|
||||
|
|
|
- |
2nf (2) |
= |
100 |
|
|
|
|
2nf2 x |
20 |
|
2 |
= |
|
|
|
|
|
|
f2 |
+ 7.96 f |
- 633 |
0 |
|
|
|
|
|||
2 |
|
|
2 |
{2 |
= |
21.5 Hz |
|
|
||
|
|
|
|
|
|
|||||
20
103
|
One can check that by using these frequencies the difference of the two reactances in |
|||
both the cases comes out to be |
100 Q. |
|
|
|
|
Example 1.5. A 120 V, |
500 W lamp is used with a series choke on a 230 V, 60 Hz line, |
||
what is the inductance of the required choke if the choke has a Q of 2? |
|
|
||
is 230 |
Solution: Here we have a 500 W lamp rated at 120 volt whereas the supply voltage |
|||
V. Naturally we have to connect some circuit elementR, L or C so that the |
bulb |
gets 120 V. |
||
|
is given by |
|
||
The resistance ofthe lamp vR2 |
= 500 = 120R 2 |
|
||
or |
R |
= 1202 = 28.8 Q |
|
|
|
|
500 |
|
|
Here it is suggested to connect a choke which has both resistance as well inductance. It is also given that Q of the coil is 2
i.e. RroL |
= 2 where R is the resistance of the choke itself. |
||||
Now ro |
= |
2n x 60 |
|
376.8 rad/sec. |
|
|
376.8L = 2R |
|
|||
or |
|
= |
R = 188.4 L |
||
Now current in the circuit is |
|
||||
|
|
120 x I = 500 |
|
||
|
|
|
|
500 |
4.167A |
or |
|
|
|
I = 120 = |
|
Hence impedance of the circuit is
4.167230 = 55·2 Q
Also that impedance is
(28.8 + 188.4£)2 + (367.8L)2 = 55.22
The solution gives L = 87 mH.
