Two resistances of 100 Ω and 200 Ω are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100 Ω resistance, which gives reading as 1 V. The resistance of voltmeter must be _______ Ω. Numerical answer type. Work it out, then open the answer.

Answer: 200. The 200 Ω resistor gets 4 - 1 = 3 V, so I = 3/200 = 0.015 A. Combination 100 | RV = (1)/(0.015) = 200/3 Ω (100 RV)/(100 + RV) = 200/3 ⇒ 300 RV = 20000 + 200 RV RV = 200 Ω

Two resistances of 100 Ω and 200 Ω are connected in series with a… | JEE Main 2024 (30 Jan, Shift 2)

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Kirchhoff's Laws & Measuring Instruments

Two resistances of $100\ \Omega$ and $200\ \Omega$ are connected in series with a battery of $4\text{ V}$ and negligible internal resistance. A voltmeter is used to measure voltage across $100\ \Omega$ resistance, which gives reading as $1\text{ V}$ . The resistance of voltmeter must be _______ $\Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 200

The $200\ \Omega$ resistor gets $4 - 1 = 3\text{ V}$ , so $I = \dfrac{3}{200} = 0.015\text{ A}$ .
Combination $100 \| R_V = \dfrac{1}{0.015} = \dfrac{200}{3}\ \Omega$
$\dfrac{100 R_V}{100 + R_V} = \dfrac{200}{3} \Rightarrow 300 R_V = 20000 + 200 R_V$
$R_V = 200\ \Omega$

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