In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 `kOmega` are used. The figure of merit of the galvanometer is 60 `muA//"division"`. In the absence of shunt resistance, the galvanometer produces a deflection of `theta =9` divisions when current flows in the circuit. The value of the shunt resistance that can cause the delflection of `theta//2`, is closet to :

Options

A. `550 Omega`
B. `220 Omega`
C. `55 Omega`
D. `110 Omega`

Answer

d

Explanation

Current flowing in the circuit in the absence of shunt.
` I = 9 xx 60 = 649 muA = 54 xx 10^(-5) a`
` I = e/(R + R_(G)) rArr 54 xx 10^(-5) = 6/(11000 + R_(G))`
`rArr " " R_(G) = 1000/9 Omega`

Deflection of `theta //2` means that after connecting shunt in parallel, current through galvanometer should become half.
`rArr " " I/2 = (epsi/(R + (R_(G)S)/(R_(G) + S))) xx (S/(S + R_(G)))`
`rArr " " I/2 = (epsiS)/(R(S + R_(G)) + R_(G) S)`
Substituting values we can solve for S.
` S = 110 Omega`
Hence option (d) is correct.