Question

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220V$ and $100W$ is connected $(220\times 8)V$ sources, then the actual power would be

• A. Between $100\times (0.8{)}^{2}W$ and $100\times 0.8W$
• B. $100\times (0.8{)}^{2}W$
• C. $100\times 0.8W$
• D. Between $100\times 0.8W$ and $100W$

Answer 1

The correct option is A Between $100\times (0.8{)}^{2}W$ and $100\times 0.8W$

Let power developed across $220V$ is $P_1$ and across $(220\times 0.8V)$ be $P_2$, then
$P_1=\frac{(220{)}^2}{R_1}and{P}_2=\frac{(220\times 0.8{)}^2}{R_2}$
$\frac{P_2}{P_1}=\frac{(220\times 0.8{)}^2}{(220{)}^2}\times \frac{R_1}{R_2}$
$\frac{P_2}{P_1}=(0.8{)}^2\times \frac{R_1}{R_2}$
Hence, $R_2>R_1$
(Because voltage decreases from $220V\to 220\times 0.8V$ It means heat produced/power dissipated $\to$ decreases)
So $\frac{R_1}{R_2}>1$
$P_2>(0.8{)}^2{P}_1$
$P_2>(0.8{)}^2\times 100W$
Also
$\frac{P_2}{P_1}=\frac{(220\times 0.8)i_2}{220i_1}$,
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 $(100\times 0.8)W.$
Hence, option $\left(d\right)$ is correct.