A voltmeter of resistance `995 Omega` is connected across a cell of emf 3V and internal resistance `5 Omega`. Find the potential difference across the voltmeter, that across the terminals of the cell and percentage error in the reading of voltmeter.

Explanation

Here, `R= 995 Omega, epsilon = 3V, r= 5 Omega`
Current in the circuit, `I=epsilon/R+r = 3/995+5 = 3 xx 10^(-3) A`
Potential differences across the voltmeter `V= IR=(3 xx 10^(-3)) xx 995=2.985 V`
The potential differences across the terminals of the cell, `V'=epsilon - Ir=3-(3xx 10^(-3))xx5=2.985 V`
As the voltmeter used is to measure the emf of the cell but will read 2.985 V. Hence percentage error is
`= (epsilon-V)/(epsilon) xx 100 = (3 -2.985)/3 xx 100=0.5%`