Current Electricity Charging Discharging Of Capacitors Question 516
Question: A bulb rated at (100W ? 200V) is used on a 100V line. The current in the bulb is
[JIPMER 1999]
Options:
A) $ \frac{1}{4} $ amp
B) 4 amp
C) $ \frac{1}{2} $ amp
D) 2 amp
Show Answer
Answer:
Correct Answer: A
Solution:
$ P=\frac{V^{2}}{R}\Rightarrow 100=\frac{{{(200)}^{2}}}{R} $
Therefore $ R=\frac{4\times 10^{4}}{10^{2}}=400,\Omega $ Now, $ i=\frac{V}{R}=\frac{100}{400}=\frac{1}{4}amp $
BETA
