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.

Q: 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

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

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$P_1 > P_2 = P_3$
B$P_1 > P_2 > P_3$
C$P_1 < P_2 = P_3$
D$P_1 < P_2 < P_3$

Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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