A galvanometer coil has a resistance 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

Resistance of the galvanometer coil, `G = 15 Omega`
Current for which the galvanometer shows full scale deflection,
`I_(s) = 4mA = 4 xx 10^(-3)A`
Range of the ammeter is 0, which needs to be converted to 6A.
`:.` Current, `I = 6A`
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
`S = (I_(s)G)/(I-I_(s))`
`= (4 xx 10^(-3) xx 15)/(6-4 xx 10^(-3))`
`S = (6 xx 10^(-2))/(6-0.004) = (0.06)/(5.996)`
`= 0.01 Omega = 10 m Omega`
Hence, a `10 m Omega` shunt resistor is to be connected in parallel with the galvanometer.