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Current Electricity question

2024 · 30 Jan · Shift 2 · Q86
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Current Electricity question

2024 · 30 Jan · Shift 2 · Q86

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
Two resistance of 100Ω100 \Omega100Ω and 200Ω200 \Omega200Ω are connected in series with a battery of 4 V4 \mathrm{~V}4 V and negligible internal resistance. A voltmeter is used to measure voltage across 100Ω100 \Omega100Ω resistance, which gives reading as 1 V1 \mathrm{~V}1 V. The resistance of voltmeter must be ‾Ω\underline{\hspace{2cm}}\Omega​Ω.
Numerical answer
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