Two resistances {R_1} and {R_2} when connecte | vedclass

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Question

(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})}}

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(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 $ $\frac{{{R_1}}}{{{R_2}}} = \frac{1}{1}$

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

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