The resistance of the filament of an electric | vedclass

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Question

(d) ${P_1} = \frac{{{{(220)}^2}}}{{{R_1}}}$ and ${P_2} = \frac{{{{(220 	imes 0.8)}^2}}}{{{R_2}}}$
$\frac{{{P_2}}}{{{P_1}}} = \frac{{{{(220 	imes 0.

Vedclass · Accepted Answer

Between $100 	imes {(0.8)^2}\, watt$ and $100 	imes 0.8\,watt$

Vedclass · Answer

(d) ${P_1} = \frac{{{{(220)}^2}}}{{{R_1}}}$ and ${P_2} = \frac{{{{(220 imes 0.8)}^2}}}{{{R_2}}}$ $\frac{{{P_2}}}{{{P_1}}} = \frac{{{{(220 imes 0.8)}^2}}}{{{{(220)}^2}}} imes \frac{{{R_1}}}{{{R_2}}}$ $⇒$ $\frac{{{P_2}}}{{{P_1}}} = {(0.8)^2} imes \frac{{{R_1}}}{{{R_2}}}$ Here $R_2 < R_1$ (because voltage decreases from $220 V ⇒ 220 × 0.8 \,V$ It means heat produced $⇒$ decreases) So $\frac{{{R_1}}}{{{R_2}}} > 1$ $⇒$ ${P_2} > {(0.8)^2}{P_1}$ $⇒$ ${P_2} > {(0.8)^2} imes 100\,W$ Also $\frac{{{P_2}}}{{{P_1}}} = \frac{{(220 imes 0.8){i_2}}}{{220\,{i_1}}},$ Since ${i_2} < {i_1}$ (we expect) So $\frac{{{P_2}}}{{{P_1}}} < 0.8$ $⇒$ ${P_2} < (100 imes 0.8)$ Hence the actual power would be between $100 imes {(0.8)^2}\,W$ and $(100 × 0.8)\, W$

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

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