Question

The current flowing in the galvanometer G when key k2 is open is I .on closing key K2 current become I/n .Obtain expression for Rg in terms of R,S andn

Answer 1

all resistors R, Rg and S are connected in circuit as shown in figure.

when K₂ is open , current I flows through only bigger loop.
So, I = E/(R + Rg) [Let emf of battery is E ]

When K₂ is closed , current flows through both loops.
first of all we should find equivalent resistance
Req = R + R' [ because S and Rg are in parallel ∴ R' = SRg/(S + Rg) ]
Now, current flows through circuit is I'
I' = E/(R + R') -----(1)

this current I' is divided in two parts for S and Rg ,
both S and Rg are in parallel ,
So, current divides in inverse ratio of Rg and S.
e.g., Ig = S/(S + Rg) × I' and Is = Rg/(S + Rg) × I'
so, current in Rg { galvanometer } will be Ig = S/(S + Rg) × I'
Ig = S/(S + Rg) × E/(R + R') [ from equation (1) ]
According to question,
Ig = I/n
so, I/n = E/(R + Rg)n = S/(S + Rg) × E/(R + R')
⇒E/n(R + Rg) = ES/(S + Rg) × 1/(R + SRg/(S + Rg)
⇒1/n(R + Rg) = S/(S + Rg) × (S + Rg)/(RS + RRg + SRg)
⇒RS + RRg + SRg =n(RS + SRg)
⇒RRg = (n - 1)RS + (n -1)SRg
⇒Rg[ R -(n -1)S ] = (n -1)RS
⇒Rg = (n -1)RS/[R + (1 - n)S ]

Hence,Rg = (n-1)RS/[R + (1 - n)S ]