JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Current Electricity

Two resistance of

100Ω

and

200Ω

are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across

100Ω

resistance, which gives reading as 1 V. The resistance of voltmeter must be

____

Ω

Answer: 200

Solution

The voltmeter of resistance

R_{v}

is connected across the

100 Ω

resistor, forming a parallel combination:

R_{p a r a l l e l} = \frac{R _{v} \times 100}{R _{v} + 100}

This combination is in series with the

200 Ω

resistor across the 4 V battery. The voltmeter reads 1 V across the parallel combination, so the remaining

3 V

appears across the

200 Ω

resistor. Since the same current flows through both parts of the series circuit, voltages divide in proportion to resistance:

\frac{1}{3} = \frac{R _{p a r a l l e l}}{200}

R_{p a r a l l e l} = \frac{200}{3} Ω

Substituting:

\frac{100 R _{v}}{R _{v} + 100} = \frac{200}{3}

300 R_{v} = 200 R_{v} + 20000

100 R_{v} = 20000

R_{v} = 200 Ω

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer