Current Electricity Charging Discharging Of Capacitors Question 535
Question: An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 V. Now its power is
[AIIMS 2002]
Options:
A) 100W
B) 40W
C) 20W
D) 10W
Show Answer
Answer:
Correct Answer: D
Solution:
$ P=\frac{V^{2}}{R} $
$ \Rightarrow \frac{P _{2}}{P _{1}}=\frac{V _{2}^{2}}{V _{1}^{2}} $ ( $ \because R $ is constant)
$ \Rightarrow \frac{P _{2}}{P _{1}}={{( \frac{100}{200} )}^{2}}=\frac{1}{4} $
$ \Rightarrow P _{2}=\frac{P _{1}}{4}=\frac{40}{4}=10,W $
BETA
