A battery of emf 20.0 V and internal resistance of 3 Ω is connected across a load of resistance 7.0 Ω. The power delivered to the load is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 28 W
The internal resistance is in series with the load, so the current is set by the total resistance, and the power delivered to the load is \( I^{2}R \) using the load resistance alone.
\( \displaystyle I = \frac{\varepsilon}{R + r} = \frac{20.0\,\mathrm{V}}{7.0\,\Omega + 3\,\Omega} = 2.0\,\mathrm{A} \)
\( \displaystyle P = I^{2}R = (2.0\,\mathrm{A})^{2} \times 7.0\,\Omega \)
\( \displaystyle P = 28\,\mathrm{W} \)