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(D) Given:Electromotive force $E = 3.0 \ V$Terminal voltage $V = 2.5 \ V$Power dissipated by external resistor $P_R = 0.5 \ W$We know that the termina

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$0.10$

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(D) Given:Electromotive force $E = 3.0 \ V$Terminal voltage $V = 2.5 \ V$Power dissipated by external resistor $P_R = 0.5 \ W$We know that the terminal voltage $V$ is given by $V = E - ir$,where $i$ is the current and $r$ is the internal resistance.$2.5 = 3.0 - ir$$ir = 0.5 \ V$The power dissipated in the internal resistance is $P_r = i^2r = (ir) \cdot i$.Since $V = iR = 2.5 \ V$,we have $i = \frac{2.5}{R}$.Also,$P_R = i^2R = 0.5 \ W$.Substituting $i = \frac{2.5}{R}$ into $P_R = i^2R$:$0.5 = (\frac{2.5}{R})^2 \cdot R = \frac{6.25}{R}$$R = \frac{6.25}{0.5} = 12.5 \ \Omega$.Now,find the current $i$:$i = \frac{V}{R} = \frac{2.5}{12.5} = 0.2 \ A$.Finally,calculate the power dissipated in the internal resistance $P_r$:$P_r = i^2r = i(ir) = 0.2 \ A 	imes 0.5 \ V = 0.10 \ W$.

$A$ 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 within the internal resistance is $....... \ W$.

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