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Two equal resistances $R$ are joined with voltage source $V$ in (i) series (ii) parallel, the ratio of electrical power consumed in two cases wll be
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Verified Answer
The correct answer is:
$1: 4$
(i) In series: total resistance $=R+R=2 R$
Power consumed $P_1=\frac{V^2}{2 R} \quad \ldots$ (i)
(ii) In parallel : potential difference across each resistance will be $V$
So, power consumed in each resistance is
$P=\frac{V^2}{R}$
So total power consumed in two resistance is
$P_2=\frac{V^2}{R}+\frac{V^2}{R}=\frac{2 V^2}{R}$ \quad \ldots$ (ii)
$\begin{array}{r}\text { Hence, required ratio }=\frac{P_1}{P_2} \\ =\frac{V^2 \times R}{2 R \times 2 V^2}=\frac{1}{4}\end{array}$
Power consumed $P_1=\frac{V^2}{2 R} \quad \ldots$ (i)
(ii) In parallel : potential difference across each resistance will be $V$
So, power consumed in each resistance is
$P=\frac{V^2}{R}$
So total power consumed in two resistance is
$P_2=\frac{V^2}{R}+\frac{V^2}{R}=\frac{2 V^2}{R}$ \quad \ldots$ (ii)
$\begin{array}{r}\text { Hence, required ratio }=\frac{P_1}{P_2} \\ =\frac{V^2 \times R}{2 R \times 2 V^2}=\frac{1}{4}\end{array}$
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