The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220, volt and 100, watt is connected (220

Q: The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220\, volt$ and $100\, watt$ is connected $(220 \times 0.8)$ $volt$ sources, then the actual power would be

d (d) ${P_1} = \frac{{{{(220)}^2}}}{{{R_1}}}$ and ${P_2} = \frac{{{{(220 \times 0.8)}^2}}}{{{R_2}}}$ $\frac{{{P_2}}}{{{P_1}}} = \frac{{{{(220 \times 0.8)}^2}}}{{{{(220)}^2}}} \times \frac{{{R_1}}}{{{R_2}}}$ $⇒$ $\frac{{{P_2}}}{{{P_1}}} = {(0.8)^2} \times \frac{{{R_1}}}{{{R_2}}}$ Here $R_2 1$ $⇒$ ${P_2} > {(0.8)^2}{P_1}$ $⇒$ ${P_2} > {(0.8)^2} \times 100\,W$ Also $\frac{{{P_2}}}{{{P_1}}} = \frac{{(220 \times 0.8){i_2}}}{{220\,{i_1}}},$ Since ${i_2} < {i_1}$ (we expect) So $\frac{{{P_2}}}{{{P_1}}} < 0.8$ $⇒$ ${P_2} < (100 \times 0.8)$ Hence the actual power would be between $100 \times {(0.8)^2}\,W$ and $(100 × 0.8)\, W$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220\, volt$ and $100\, watt$ is connected $(220 \times 0.8)$ $volt$ sources, then the actual power would be

A$100 \times 0.8\,watt$
B$100 \times {(0.8)^2}\,watt$
CBetween $100 \times 0.8\, watt$ and $100\, watt$
DBetween $100 \times {(0.8)^2}\, watt$ and $100 \times 0.8\,watt$

Diffcult

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

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