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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
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$10 \mathrm{~W}$.
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, $\frac{40}{P_2}=\frac{(200)^2}{(100)^2}=4$ or $P_2=\frac{40}{4}=10 \mathrm{~W}$
(where $P_2=$ power when voltage is $100 \mathrm{~V}$ ).
Power $(P)=\frac{V^2}{R} \propto V^2$. Therefore $\frac{P_1}{P_2}=\frac{V_1^2}{V_2^2}$ or, $\frac{40}{P_2}=\frac{(200)^2}{(100)^2}=4$ or $P_2=\frac{40}{4}=10 \mathrm{~W}$
(where $P_2=$ power when voltage is $100 \mathrm{~V}$ ).
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