Find potential difference across the resistance 300 `Omega` in A and B.

Explanation

In (A) : Potential difference `=(100)/(200+300)=300`
`=60` volt
In (B) Potential difference
`=(100)/(200+(300xx600)/(300+600))xx[(300xx600)/(300+600)]=50`volt
We see that by connected voltmeter the voltage which was to be measured has charged . Such voltmeters are not good .If its resistance had been very large than `300Omega` then it would have not affected the voltage by much amount.