A cell of emf 90 V is connected across a series combination of two resistors, each of 100 \( \Omega \) resistance. A voltmeter of resistance 400 \( \Omega \) is used to measure the potential difference across each resistor. The reading of the voltmeter will be:
The problem requires us to determine the reading of a voltmeter when measuring the potential difference across each resistor in a series circuit with a total emf of 90 V.
Let's analyze the circuit:
The power source is a cell with an emf of \(90 \, V\).
There are two resistors connected in series, each with a resistance of \(100 \, \Omega\).
The total resistance due to these two resistors in series is calculated as: \(R_{\text{total}} = 100 \, \Omega + 100 \, \Omega = 200 \, \Omega\).
A voltmeter with a resistance of \(400 \, \Omega\) measures the potential difference across one resistor. This voltmeter is connected in parallel with the resistor.
The correct answer is (B) : 40 V
Req =500400×100+100 =180Ω i=18090=21A Reading =21×500400×100 =40 volt
