An ammeter and a voltmeter are connected in series to a battery of emf `E=6.0V`. When a certain resistance is connected in parallel with the voltmeter, the readding of the voltmeter decreases two times, whereas the reading of the ammeter increases the same number of times. Find the voltmeter reading after the connection of the resistance.

Explanation

Before connection of the resistance,
I (ammeter reading) `=(E)/(r+R)`
and V (voltmeter reading) `=(E)/(r+R)xxR`
After connection of the resistance (S) in parallel to R, `i` (ammeter reading)
`=i^(')xx(RS)/(R+S)=(E(R+S))/(r(R+S)+RS)xx(RS)/(R+S)`
`=(ERS)/(r(R+S)+RS)`
It is given that `i^(')=ni` and `nV^(')=V`
`therefore(E(R+S))/(r(R+S)+RS)=n(E)/(r+R)` or
`(R+S)(r+R)=n{r(R+S)}RS}` ...(i)
and `nxx(ERS)/(r(R+S)+RS)=(ER)/(R+r)` ltbr or `nS(R+r)=r(R+S)+RS` ..(ii)
from (i) and (ii), `S=(R)/(n^(2)-1)`
and putting this value of S in (i) we have `R=nr`
`thereforeR,R=n:1`
`V^(')=(ERS)/(r(R+S)+RS)`
`=(ExxRxx(R)/(n^(2)-1))/((R)/(n)(R+(R)/(n^(2)-1))+Rxx(R)/(n^(2)-1))`
`=(ER^(2))/(R^(2)(n+1))` or `V^(')(E)/(n+1)=(6)/(2+1)=2` volts