The value of R in the above circuit to make the galvanometer measure 2A is

Answer Details

Given: Ig ${}_g$ = 50mA = 0.05A, I to be measured = 2A, r = 2Ω $\mathrm{\Omega }$, Is ${}_s$ = I - Ig ${}_g$ = 2 - 0.05 = 1.95A

Shunt(R) = IgIs $\frac{I_g}{I_s}$ x r

R =  0.051.95 $\frac{0.05}{1.95}$ x 10 = 0.2564Ω