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

Explanation

`"In (A) :Potential difference ="100/(200+300)xx300=60" volt"`
`"In (B) :Potential difference ="100/(300+600)xx(300xx600)/(300xx600)=50" volt"`
300+600
We see that by connecting voltmeter the which was to be measured has changed. Such voltmeters are not good. If its resistance had been very large than `300 Omega` then it would not have affected the voltage by much amount.