Question

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $$220 \ V$$ and $$100 \ W$$ is connected to $$(220\times 0.8) \ V$$ source, then the power would be :

Answer 1

Correct option is D. Between $$100\times (0.8)^{2}watt$$ and $$100\times 0.8watt$$ watt

Since Power $$ P = \dfrac{V^2}{R}$$

For the first case, when $$V=220 \ V$$

$$P_{1}=\dfrac{(220)^2}{R_1} $$

For the second case, when $$V=220 \times 0.8 \ V$$

$$ P_{2}=\dfrac{(220\times 0.8)^{2}}{R_{2}}$$

Now, $$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$$

$$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$$

Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the resistance, from Ohm's law.

Hence, $$R_{2}< R_{1}$$

$$\Rightarrow \dfrac{R_1}{R_2} > 1$$

$$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$$

$$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$$

Also, Since Power $$P = Vi$$

Hence$$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$$

Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the current, from Ohm's law.

Hence, $$i_{2}< i_{1}$$

$$\Rightarrow \dfrac{i_2}{i_1} <1$$

So $$\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$$

Hence the actual power would be between $$100 \times (0.8)^{2} \ W$$ and $$100\times (0.8) \ W$$