Two resistors of \( 300\,\Omega \) and \( 600\,\Omega \) are connected in series with a 6 V battery of negligible internal resistance. The potential difference measured by a voltmeter of resistance \( 600\,\Omega \) across the \( 300\,\Omega \) resistor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.5 V
A voltmeter is joined in parallel with the resistor it measures. So its own resistance combines with that resistor and changes the current in the circuit. The reading is the potential difference across this parallel pair.
The \( 600\,\Omega \) voltmeter in parallel with the \( 300\,\Omega \) resistor gives:
\( \begin{aligned} R_{\mathrm{p}} &= \frac{300\times600}{300+600}\,\Omega \\ &= 200\,\Omega \end{aligned} \)
This is in series with the \( 600\,\Omega \) resistor, so the current from the battery is:
\( \begin{aligned} I &= \frac{V}{R_{\mathrm{p}}+600\,\Omega} \\ &= \frac{6\,\mathrm{V}}{800\,\Omega} \\ &= 7.5\times10^{-3}\,\mathrm{A} \end{aligned} \)
The voltmeter reads:
\( \begin{aligned} V_{\mathrm{p}} &= IR_{\mathrm{p}} \\ &= 7.5\times10^{-3}\,\mathrm{A}\times200\,\Omega \\ &= 1.5\,\mathrm{V} \end{aligned} \)