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

Explanation

Let S be the shunt connected in parallel with the galvanometer to convert it into an ammeter of range O to 6A.

If G is the resistance of the galvanometer and `I_(g)` is the current flowing the salvanometer out of total current I, then
` (I - I_(g))S = I_(g) G`
` rArr" " S = (I_(g) G)/(I - I_(g))`
Here , `G = 15 Omega , I_(g) = 4 mA = 0.004 A, I = 6 A`
` S = (0.004 xx 15)/(6 - 0.004) = (0.06)/(5 . 996) `
` = 10 m Omega`
So, shunt resistance ` = 10 m Omega`