An electric bulb marked 40, W and 200, V, is used in a circuit of supply voltage 100, V. Now its power is

Q: An electric bulb marked $40\, W$ and $200\, V$, is used in a circuit of supply voltage $100\, V$. Now its power is

d (d) $P = \frac{{{V^2}}}{R}$ $ \Rightarrow \frac{{{P_2}}}{{{P_1}}} = \frac{{V_2^2}}{{V_1^2}}$ ( is constant) $ \Rightarrow \frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{100}}{{200}}} \right)^2} = \frac{1}{4}$$ \Rightarrow {P_2} = \frac{{{P_1}}}{4} = \frac{{40}}{4} = 10\,W$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

An electric bulb marked $40\, W$ and $200\, V$, is used in a circuit of supply voltage $100\, V$. Now its power is

A$100$
B$40$
C$20$
D$10$

AIIMS 2002, Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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