Question

A voltmeter of resistance `R_1` and an ammeter of resistance `R_2` are connected in series across a battery oif negligible internal resistance. When as resistance `R` is connected in parallel to voltmeter reading of ammeter increases three times white that of voltmeter reduces to one third. Find `R_1` and `R_2` in terms of `R`.

Answer 1

Let `E` be the emf of the battery
In the second case main current increses three times while current through voltmeter will reduces to `i//3.` Hence, the remaining `3i-i//3=8i//3` passes though `R` as shown in figure.

`V_C-V_D=(i/3)R_=((8i)/3)R`
or `R_1=8R`

In the second case main current beomes three times, Therefore, total resistance becomes `1/3` times or
`R_2+(R R_1)/(R+R_1)=1/3(R_1+R_2)`
Substitutng `R_18R` we get `R_2=(8R)/3`