Two resistances {R_1} and {R_2} when connected in series and parallel with 120, V line, power consumed will be 25, W and 100, W respectively.

Q: Two resistances ${R_1}$ and ${R_2}$ when connected in series and parallel with $120\, V$ line, power consumed will be $25\, W$ and $100\, W$ respectively. Then the ratio of power consumed by ${R_1}$ to that consumed by ${R_2}$ will be

a (a) $P = \frac{{{V^2}}}{R}$$ \Rightarrow $ $\frac{{{P_P}}}{{{P_S}}} = \frac{{{R_S}}}{{{R_P}}} = \frac{{({R_1} + {R_2})}}{{{R_1}{R_2}/({R_1} + {R_2})}} = \frac{{{{({R_1} + {R_2})}^2}}}{{{R_1}\,{R_2}}}$ $ \Rightarrow $ $\frac{{100}}{{25}} = \frac{{{{({R_1} + {R_2})}^2}}}{{{R_1}{R_2}}}$$ \Rightarrow $ $\frac{{{R_1}}}{{{R_2}}} = \frac{1}{1}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Two resistances ${R_1}$ and ${R_2}$ when connected in series and parallel with $120\, V$ line, power consumed will be $25\, W$ and $100\, W$ respectively. Then the ratio of power consumed by ${R_1}$ to that consumed by ${R_2}$ will be

A$1:1$
B$1:2$
C$2:1$
D$1:4$

Medium

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

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