A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be: (a) 40 V (b) 90 V (c) 45 V (d) 80 V

Answer: (a). 100 Ω | 400 Ω = 80 Ω Total R = 100 + 80 = 180 Ω, I = 90/180 = 0.5 A. Reading = 0.5 × 80 = 40 V So the answer is (a) 40 V.

A cell of emf 90 V is connected across series combination of two… | JEE Main 2023 (24 Jan, Shift 2)

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Kirchhoff's Laws & Measuring Instruments

A cell of emf 90 V is connected across 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:

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Answer: (a)

$100\ \Omega \| 400\ \Omega = 80\ \Omega$
Total $R = 100 + 80 = 180\ \Omega$ , $I = \dfrac{90}{180} = 0.5\text{ A}$ .
Reading $= 0.5 \times 80 = 40\text{ V}$
So the answer is (a) 40 V.

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