A $100\, W$ bulb $B_1$, and two $60\,W$ bulbs $B_2$ and $B_3$, are connected to a $250\, V$ source, as shown in the figure. Now $ W_1, W_2$ and $W_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$, respectively. Then
IIT 2002, Medium
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$P = \frac{{{V^2}}}{R}$ so $R = \frac{{{V^2}}}{P}$
${R_1} = \frac{{{V^2}}}{{100}}$ and ${R_2} = {R_3} = \frac{{{V^2}}}{{60}}$
Now ${W_1} = \frac{{{{(250)}^2}}}{{{{({R_1} + {R_2})}^2}}}.{R_1}$, ${W_2} = \frac{{{{(250)}^2}}}{{{{({R_1} + {R_2})}^2}}}.{R_2}$
and ${W_3} = \frac{{{{(250)}^2}}}{{{R_3}}}$
${W_1}:{W_2}:{W_3} = 15:25:64$ or ${W_1} < {W_2} < {W_3}$
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