In the circuit shown in the figure, the cell has negligible internal resistance and the ammeter has a resistance of \( 2\,\Omega \). When the voltmeter is connected across the \( 100\,\Omega \) resistor, the ammeter reading is 0.04 A. The voltage read by the voltmeter is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.2 V
From the figure, the circuit is a single loop. It has the cell of emf 3.4 V, the \( 3\,\Omega \) resistor, the ammeter (\( 2\,\Omega \)) and the \( 100\,\Omega \) resistor with the voltmeter across it. The voltmeter reads the potential difference across the \( 100\,\Omega \) resistor. By Kirchhoff’s loop rule, this reading is the emf minus the drop across the \( 3\,\Omega \) resistor and the ammeter. Both of them carry the full current 0.04 A.
The drop across the \( 3\,\Omega \) resistor and the ammeter is:
\( \begin{aligned} V_{\mathrm{s}} &= I\,(R+R_{\mathrm{A}}) \\ &= 0.04\,\mathrm{A}\times(3\,\Omega+2\,\Omega) \\ &= 0.2\,\mathrm{V} \end{aligned} \)
The internal resistance of the cell is negligible, so the voltmeter reading is:
\( \begin{aligned} V &= \varepsilon-V_{\mathrm{s}} \\ &= 3.4\,\mathrm{V}-0.2\,\mathrm{V} \\ &= 3.2\,\mathrm{V} \end{aligned} \)