Question

An ideal battery of emf 1V is connected in series with an ammeter and voltmeter of unknown internal resistance. If certain resistance is connected in parallel with voltmeter, the voltmeter and ammeter readings become half and twice of their initial readings respectively. Initial reading of the voltmeter was

Answer 1

$\epsilon =i{R}_A+V$ ...(1)
& $\epsilon =ni{R}_A+\frac{V}{n}$ ...(2)
Solving (1) & (2) we get
$V=\frac{n\epsilon }{n+1}=\frac{2\times 1}{2+1}=\frac{2}{3}=0.666$