A voltmeter of resistance \( 998\,\Omega \) is connected across a cell of emf 2 V and internal resistance \( 2\,\Omega \). The potential difference across the voltmeter is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.996 V
The voltmeter acts as an external resistance \( R \) across the cell. The current is the emf divided by the total resistance \( R+r \), and the voltmeter reads the potential difference \( IR \) across itself.
\( \begin{aligned} I &= \frac{\varepsilon}{R+r} \\ &= \frac{2\,\mathrm{V}}{(998+2)\,\Omega} \\ &= 2\times10^{-3}\,\mathrm{A} \end{aligned} \)
\( \begin{aligned} V &= IR \\ &= 2\times10^{-3}\,\mathrm{A}\times998\,\Omega \\ &= 1.996\,\mathrm{V} \end{aligned} \)