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
Diffcult
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(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 $R_2 < R_1$
(because voltage decreases from $220 V ⇒ 220 × 0.8 \,V$
It means heat produced $⇒$ decreases)
So $\frac{{{R_1}}}{{{R_2}}} > 1$ $⇒$ ${P_2} > {(0.8)^2}{P_1}$ $⇒$ ${P_2} > {(0.8)^2} \times 100\,W$
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 $(100 × 0.8)\, W$
$\frac{{{P_2}}}{{{P_1}}} = \frac{{{{(220 \times 0.8)}^2}}}{{{{(220)}^2}}} \times \frac{{{R_1}}}{{{R_2}}}$ $⇒$ $\frac{{{P_2}}}{{{P_1}}} = {(0.8)^2} \times \frac{{{R_1}}}{{{R_2}}}$ Here $R_2 < R_1$
(because voltage decreases from $220 V ⇒ 220 × 0.8 \,V$
It means heat produced $⇒$ decreases)
So $\frac{{{R_1}}}{{{R_2}}} > 1$ $⇒$ ${P_2} > {(0.8)^2}{P_1}$ $⇒$ ${P_2} > {(0.8)^2} \times 100\,W$
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 $(100 × 0.8)\, W$
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