Two resistors of resistance 1000 Ω and 2000 Ω are joined in series with a 100 V supply. A voltmeter of internal resistance 4000 Ω is connected to measure the potential difference across 1000 Ω resistor. Calculate the reading shown by the voltmeter.

Solution
Given,

R₁ = 1000 Ω
R₂ = 2000 Ω
Internal resistance of voltmeter (R₃) = 4000 Ω
Emf (E) = 100 V

Since, R₁ and R₃ are in parallel combination, so their effective resistance,

[latex]R’ = \frac{R_1.R_3}{R_1+R_3}[/latex]

[latex]= \frac{1000\times 4000}{1000+4000}[/latex]

[latex]= \frac{4000000}{5000}[/latex]

= 800 Ω

Since, R’ & R₂ are in series combination, so, total resistance in the circuit,
R_T = R’ + R₂
= 800 + 2000
= 2800 Ω

Now, current flowing in the circuit,

[latex]I = \frac{E}{R_T}[/latex]

[latex]= \frac{100}{2800}[/latex]