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

Q: 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

d $P = \frac{{{V^2}}}{R}$ 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 ${W_1} < {W_2} < {W_3}$

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 $250\, V$ 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

A${W_1} > {W_2} = {W_3}$
B${W_1} > {W_2} > {W_3}$
C${W_1} < {W_2} = {W_3}$
D${W_1} < {W_2} < {W_3}$

IIT 2002, Medium

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

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