Current Electricity Charging Discharging Of Capacitors Question 471
Question: An electric bulb of 100 watt is connected to a supply of electricity of 220 V. Resistance of the filament is
[EAMCET 1981, 82; MP PMT 1993, 97]
Options:
A) $ 484,\Omega $
B) $ 100,\Omega $
C) $ 22000,\Omega $
D) $ 242,\Omega $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{V^{2}}{R}=P, $
Therefore $ R=\frac{V^{2}}{P}=\frac{220\times 220}{100}=484\Omega $
BETA
