Current Electricity Charging Discharging Of Capacitors Question 537
Question: A bulb has specification of one kilowatt and 250 volts, the resistance of bulb is
[MP PMT 2002]
Options:
A) 125 W
B) 62.5 W
C) 0.25 W
D) 625 W
Show Answer
Answer:
Correct Answer: B
Solution:
$ R=\frac{V^{2}}{P}=\frac{{{(250)}^{2}}}{10^{3}}=62.5,\Omega $
BETA
