Current Electricity Charging Discharging Of Capacitors Question 528
Question: An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V it draws a power P. Then
[KCET 2001]
Options:
A) $ P={{( \frac{V _{0}}{V} )}^{2}}P _{0} $
B) $ P={{( \frac{V}{V _{0}} )}^{2}}P _{0} $
C) $ P=( \frac{V}{V _{0}} ),P _{0} $
D) $ P=( \frac{V _{0}}{V} ),P _{0} $
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Answer:
Correct Answer: B
Solution:
$ P\propto V^{2} $
Therefore $ \frac{P}{P _{0}}={{( \frac{V}{V _{0}} )}^{2}}\Rightarrow P={{( \frac{V}{V _{0}} )}^{2}}P _{0} $
BETA
