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A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated ( in W) within the internal resistance is:

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detailed solution

Correct option is @

$P_{R} = 0.5 W$

$\Rightarrow i^{2} R = 0.5 W$

$Also, V = E - ir$ $\Rightarrow ir = 0.5$

$\begin{array}{l} P = (E - V) i, 0 .5 = \frac{V^{2}}{R} \Rightarrow R = 12 .5 Ω \\ i = \frac{V}{R} = \frac{1}{5} A \end{array}$

$Now iR = 2.5 and ir = 0.5$

$On dividing : \frac{R}{r} = 5$

$\frac{P_{R}}{P_{r}} = \frac{i^{2} R}{i^{2} r} \Rightarrow \frac{P_{R}}{P_{r}} = \frac{R}{r} \Rightarrow \frac{P_{R}}{P_{r}} = 5$

$\Rightarrow P_{r} = \frac{P_{R}}{5}$

$\Rightarrow P_{r} = \frac{0.50}{5} \Rightarrow P_{r} = 0.10 W$

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detailed solution

Correct answer is 0.1

$P_{R} = 0.5 W$

$\Rightarrow i^{2} R = 0.5 W$

$Also, V = E - ir$ $\Rightarrow ir = 0.5$

$\begin{array}{l} P = (E - V) i, 0 .5 = \frac{V^{2}}{R} \Rightarrow R = 12 .5 Ω \\ i = \frac{V}{R} = \frac{1}{5} A \end{array}$

$Now iR = 2.5 and ir = 0.5$

$On dividing : \frac{R}{r} = 5$

$\frac{P_{R}}{P_{r}} = \frac{i^{2} R}{i^{2} r} \Rightarrow \frac{P_{R}}{P_{r}} = \frac{R}{r} \Rightarrow \frac{P_{R}}{P_{r}} = 5$

$\Rightarrow P_{r} = \frac{P_{R}}{5}$

$\Rightarrow P_{r} = \frac{0.50}{5} \Rightarrow P_{r} = 0.10 W$

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