A 100, W , bulb, B_1 and two 60, W ,bulbs ,B_ | vedclass

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Question

$\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ so, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}} 	herefore \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{100}$

Vedclass · Accepted Answer

$P_1 < P_2 < P_3$

Vedclass · Answer

$\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ so, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}} 	herefore \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{100}$ and $ \mathrm{R}_{2}=\mathrm{R}_{3}=\frac{\mathrm{V}^{2}}{60}$

Now, $\mathrm{P}_{1}=\frac{(250)^{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}ight)^{2}} \cdot \mathrm{R}_{1}$ and

$\mathrm{P}_{2}=\frac{(250)^{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}ight)^{2}} \cdot \mathrm{R}_{2}$ and $\mathrm{P}_{3}=\frac{(250)^{2}}{\mathrm{R}_{3}}$

$\mathrm{P}_{1} \cdot \mathrm{P}_{2} \cdot \mathrm{P}_{3}=15: 25: 64 \Rightarrow \mathrm{P}_{1}<\mathrm{P}_{2}<\mathrm{P}_{3}$

A $100\, W \, bulb\, B_1$ and two $60\, W \,bulbs \,B_2$ and $B_3$, are connected to a $220\, V$ source, as shown in Figure. Now $P_1, P_2$ and $P_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$ respectively. Then

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