A battery of 3.0 V is connected to a resistor dissipating of 0.5W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:
Option: 1 0.50W
Option: 2 0.072 W
Option: 3 0.10 W
Option: 4 0.125 W

Answers (1)

P_{R}=0.5 W
\Rightarrow P_R=\mathrm{i}^{2} \mathrm{R}=0.5 \mathrm{~W}
Also,

\\ \mathrm{V}=\mathrm{E}-\mathrm{ir} \\ \Rightarrow 2.5=3-\mathrm{ir} \\ \Rightarrow ir =0.5
Power dissipated across 'r' = P_{r}=i^{2} r
Now

iR =2.5 .....(1)
ir =0.5 .......(2)
From equation (1) and (2)

we get

\frac{R}{r}=5

Now,

\begin{array}{l} \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 \mathrm{~W} \end{array}

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