The internal resistance of a battery of 100 V is `5 Omega`. When the emf of the battery is measured by a voltmeter `20%` error is found. What is the resistance of the voltmeter?

Explanation

Suppose,resistance of the voltmeter is R, Its reading is the potential of the external circuit.
According to the question, voltmeter reading =1R=80% of 100V=80V
`therefore` Lost volt=lr=100-80=20 V
`therefore(IR)/(Ir)=80/20or,R/r=4`
or,`R=4r=4times5=20 Omega`