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 $B_3$ $(60\, W)$:

Vedclass · Accepted Answer

$P_1 < P_2 < P_3$

Vedclass · Answer

(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 $B_3$ $(60\, W)$: $R_2 = R_3 = \frac{V^2}{60}$.From the circuit diagram,bulbs $B_1$ and $B_2$ are in series,and this combination is in parallel with bulb $B_3$ across the $220\, V$ source.Bulb $B_3$ is connected directly across the $220\, V$ source,so its power $P_3 = 60\, W$.For the series combination of $B_1$ and $B_2$,the current $I = \frac{V}{R_1 + R_2} = \frac{V}{\frac{V^2}{100} + \frac{V^2}{60}} = \frac{V}{V^2(\frac{3+5}{300})} = \frac{300}{8V} = \frac{37.5}{V}$.The power consumed by $B_1$ is $P_1 = I^2 R_1 = (\frac{37.5}{V})^2 \cdot \frac{V^2}{100} = \frac{1406.25}{100} = 14.06\, W$.The power consumed by $B_2$ is $P_2 = I^2 R_2 = (\frac{37.5}{V})^2 \cdot \frac{V^2}{60} = \frac{1406.25}{60} = 23.44\, W$.Comparing the values: $P_1 = 14.06\, W$,$P_2 = 23.44\, W$,and $P_3 = 60\, W$.Therefore,$P_1 < P_2 < P_3$.

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

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