A 100, W bulb B_1,and two 60, W bulbs B_2 and | vedclass

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(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For bulb $B_1$ $(100\, W)$,$R_1 = \frac{V^2}{100}$.For bulbs $B_2$ and $B_3$ $(60\, W)$,$

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$W_1 < W_2 < W_3$

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(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For bulb $B_1$ $(100\, W)$,$R_1 = \frac{V^2}{100}$.For bulbs $B_2$ and $B_3$ $(60\, W)$,$R_2 = R_3 = \frac{V^2}{60}$.Bulbs $B_1$ and $B_2$ are in series,and this combination is in parallel with bulb $B_3$ across the $250\, V$ source.The power dissipated in $B_3$ is $W_3 = \frac{(250)^2}{R_3} = \frac{(250)^2}{V^2/60} = 60 	imes \frac{(250)^2}{V^2}$.The current through the series branch ($B_1$ and $B_2$) is $I = \frac{250}{R_1 + R_2} = \frac{250}{\frac{V^2}{100} + \frac{V^2}{60}} = \frac{250}{V^2(\frac{3+5}{300})} = \frac{250 	imes 300}{8V^2} = \frac{75000}{8V^2} = \frac{9375}{V^2}$.The power in $B_1$ is $W_1 = I^2 R_1 = (\frac{9375}{V^2})^2 	imes \frac{V^2}{100} = \frac{9375^2}{100 V^2}$.The power in $B_2$ is $W_2 = I^2 R_2 = (\frac{9375}{V^2})^2 	imes \frac{V^2}{60} = \frac{9375^2}{60 V^2}$.Comparing $W_1$ and $W_2$,since $R_1 < R_2$,$W_1 < W_2$. Since $B_3$ is connected directly across the source,it operates at its rated voltage,while $B_1$ and $B_2$ operate at lower voltages,thus $W_2 < W_3$. Therefore,$W_1 < W_2 < W_3$.

$A$ $100\, W$ 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

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