A voltmeter of resistance 600 Ω is used to measure the potential drop across 300 Ω resistor. The measured value of potential drop will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 50\,\mathrm{V} \)
A voltmeter is connected in parallel with the resistor it measures, and a voltmeter of finite resistance draws some current, so it lowers the resistance of that part of the circuit and the reading is less than the true drop. The cell is taken as ideal.
The voltmeter (\( 600\,\Omega \)) in parallel with the \( 300\,\Omega \) resistor:
\( \displaystyle R' = \frac{600 \times 300}{600 + 300} = 200\,\Omega \)
This is in series with the \( 200\,\Omega \) resistor:
\( \displaystyle R_{\mathrm{total}} = 200 + 200 = 400\,\Omega \qquad I = \frac{100}{400} = 0.25\,\mathrm{A} \)
The reading is the drop across the parallel combination:
\( \displaystyle V = I\,R' = 0.25 \times 200 = 50\,\mathrm{V} \)