A dry cell delivering a current of 2 A has a terminal voltage of 1.41 V. What is the internal resistance of the cell if its open-circuit voltage is 1.59 V?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.09\,\Omega \)
The open-circuit voltage of a cell is its emf \( \varepsilon \). When the cell delivers a current \( I \), part of the emf is used up across its internal resistance \( r \). So the terminal voltage drops to \( V=\varepsilon-Ir \).
\( \varepsilon=1.59\,\mathrm{V} \)
\( V=1.41\,\mathrm{V} \)
\( I=2\,\mathrm{A} \)
\( \begin{aligned} r &= \frac{\varepsilon-V}{I} \\ &= \frac{1.59\,\mathrm{V}-1.41\,\mathrm{V}}{2\,\mathrm{A}} \\ &= \frac{0.18\,\mathrm{V}}{2\,\mathrm{A}} \\ &= 0.09\,\Omega \end{aligned} \)