A galvanometer coil has a resistance of `15Omega` and the meter shows FSD for a current of 4mA. How will you convert the meter into an ammeter of range 0 to 6A.

Explanation

Given: `G=15Omega, I_(g)=4 times 10^(-3)A, I=6A`
Using the formula, `" "S=(I_(g)G)/(I-I_(g))`
we get, `" "S=(4 times 10^(-3) times 15)/(6-0.004)=(60 times 10^(-3))/6`
`" "S=10 times 10^(-3)Omega=0.010Omega`
A shunt resistance `0.010Omega` is connected in parallel with MCG, to convert it into an ammeter of range 0 to 6A.