An ammeter of zero resistance and a volt meter are connected in series to a battery of emf 6.0 volt and certain internal resistance. When a resistance R is connected in parallel with the voltmeter the readings of the voltmeter becomes half of its initial reading and the reading of the ammeter becomes 2 times to its initial reading (before R is connected). Find the voltmeter reading in volt after the connection of the resistance.

Explanation

Let `r` = internal resistance of the battery. We shall take the resistance fo the ammeter to be = 0 and that of volmeter to be `G`.
Initailly, `V = xi - Ir, I = (xi)/(r + G)`
So, `V = xi (G)/(r + G)` ...(1)
After the volmeter si shunterd
`(V)/(eta) = xi - (xi r)/(r + (R G)/(R + G))` (voltmeter) .....(2)
and `(xi)/(r + (R G)/(R + G)) = eta (xi)/(r + G)` (Ammeter) ...(3)
From (2) and (3) we have
`(V)/(eta) = xi (eta x xi)/(r + G)`
From (1) and (4)
`(G)/(eta) = r + G - eta r` or `G = eta r`
That (1) given the required reading
`(V)/(eta) = (xi)/(eta + 1)`