Two resistance 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 .

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Aicharya · Accepted Answer

Voltage across 200 Ω = 4 - 1 = 3 V . Therefore, current through the battery, 3 200 Now, equivalent resistance of $R_V \& 100 \, \Omega = \frac{R_v 100}{R_v + 100}$ For voltmeter, $\frac{R_v 100}{R_v + 100} 	imes \frac{3}{200} = 1$ $\Rightarrow 3 R_v = 2 R_v + 200$ $\Rightarrow R_v = 200 \, \Omega$

Two resistance of 100 Ω and 200 Ω are connected in series with a battery of 4 V and negligible internal…

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