A battery of emf 1.4V and internal resistance $2 Ω$ is connected to a resistance of $100 Ω$ resistance through an ammeter. The resistance of ammeter is $4 / 3 Ω .$ A voltmeter has also been connected to find the potential difference across the resistor. The ammeter reads 0.02A. What is the resistance of voltmeter.

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$300 Ω$

$100 Ω$

$150 Ω$

$200 Ω$

answer is A.

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Detailed Solution

Here we consider that R be the resistance of voltmeter and the voltmeter is connected in parallel with $100 Ω$ resistance.

The effective resistance with voltmeter will be $R' = 100 R / (100 + R)$

Total resistance of the circuit is given as $R_{T} = 2 + \frac{4}{3} + \frac{100 R}{(100 + R)} \Rightarrow R_{T} = \frac{10}{3} + \frac{100 R}{(100 + R)} = \frac{1000 + 310 R}{3 (100 + R)}$

Current in the circuit is given as $i = \frac{1.4}{{\frac{1000 + 310 R}{3 (100 + R)}}} \Rightarrow 0.02 = \frac{(1.4) {3 (100 + R)}}{1000 + 310 R} \Rightarrow R = 200 Ω$