Current Electricity Charging Discharging Of Capacitors Question 541
Question: An electric bulb is rated 60W, 220V. The resistance of its filament is
[MP PET 2003]
Options:
A) 708 W
B) 870 W
C) 807 W
D) 780 W
Show Answer
Answer:
Correct Answer: C
Solution:
$ R=\frac{V^{2}}{P}=\frac{{{(220)}^{2}}}{60}=807,\Omega $
BETA
