A 100V battery has an internal resistance `3 Omega`. What is the reading of a voltmeter having resistance `200 Omega`, when placed across the terminals of the battery? What should be the minimum value of the voltmeter resistance so that the error in finding the emf of the battery may not be more than 1%?

Explanation

Main current `I=E/(R+r)=100/(200+3)=100/203A`
So lost volt `=lr=100/203times3=300/203=1.48V`
Therefore, reading of the voltemeter,
`V=E-Ir=100-1.48=98.52V`
In the second case, error =1%=`1/100`
Reading of the voltmeter
`V=E-E/100=100-1=99V`
SO, lost volt =100-99=1 V
i.e., `Ir=1or,I=1/r=1/3A`
This current passes throught the voltmeter. SO resistance of the voltmeter `=V/I=99/(1/3)=297Omega`