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

Answer

`2.99 V, 0.33%`

Explanation

Here epsilon `= 3V, r = 3 Omega`

Resistance of voltmeter `R = 990 Omega` , current in the circuit,
`I = (epsilon)/(R + r) = (3V)/((990 + 3) Omega) = (3)/(993) = 3.02 xx 10^(-3) A`
Therefore pot diff across the cell = pot diff across the voltmeter `V = IR`
` = (3.02 xx 10^(-3)) xx 990 = 2.99 V`
Voltmeter used to measure emf of a cell will record `2.99 V` instead of `3V` Hence the percentage error is
`= (epsilon - V)/(epsilon) xx 100 = (1 - (V)/(epsilon)) xx 100 = (1 - (2.99)/(3.0)) xx 100`
`= (1 - 0.9967) xx 100 = 0.33%`