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

Correct answer: D. 10 ~W . Actual power of bulb (P_1)=40 ~W Actual voltage of bulb (V_1)=200 ~V and supply voltage (V_2)=100 ~V Power (P)=(V^2)/(R) V^2 . Therefore (P_1)/(P_2)=(V_1^2)/(V_2^2) or, (40)/(P_2)=((200)^2)/((100)^2)=4 or P_2=(40)/(4)=10 ~W (where P_2= power when voltage is 100 ~V ).

AIIMS 2002

Exam: AIIMS
Year: 2002
Subject: Physics
Chapter: Current Electricity

AIIMS 2002 · Physics · Current Electricity

An electric bulb marked 40 ~W and 200 ~V , is used in a circuit of...

An electric bulb marked

40 \mathrm{~W}

and

200 \mathrm{~V}

, is used in a circuit of supply voltage

100 \mathrm{~V}

. Now its power is

Options

A $100 \mathrm{~W}$
B $20 \mathrm{~W}$
C $40 \mathrm{~W}$
D $10 \mathrm{~W}$ .

Correct answer

D. $10 \mathrm{~W}$ .

Step-by-step solution

Actual power of bulb

\left(P_1\right)=40 \mathrm{~W}

Actual voltage of bulb \left(V_1\right)=200 \mathrm{~V}

and supply voltage \left(V_2\right)=100 \mathrm{~V}

Power

(P)=\frac{V^2}{R} \propto V^2

. Therefore

\frac{P_1}{P_2}=\frac{V_1^2}{V_2^2}

or,

\frac{40}{P_2}=\frac{(200)^2}{(100)^2}=4

P_2=\frac{40}{4}=10 \mathrm{~W}

(where

P_2=

power when voltage is

100 \mathrm{~V}

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