A galvanometer with a coil of resistance 120 ohm shows full-scale deflection for a current of 2.5 mA. How will you convert the galvanometer into an ammeter of range 0 to 7.5A? Determine the net resistance of the ammeter. When the ammeter is put in a circuit, does it read slightly less or more than the actual current in the original circuit? Justify your answer

Explanation

`R_(6) = 120 omega , I_(g) =2.5 m A = 0.0025A, I = 7.5 A`
`R_(s)=(I_s)/(I-I_g)xx R_s = (0.0025)/(7.5 -0.0025) xx 120 =0.04 Omega`
By connecting a shunt of 0.04 `Omega` across the given galvanometer, we get an ammeter of range 0 to 7.5 A.
Net resitance of the ammeter `=(120 xx 0.04)/(120 +0.04)= 0.03998 Omega` When an ammeter is put in circuit, it reads slightly less than the actual current. An ammeter has a small resistance. When it is connected in the circuit, it decrease the current by a small amount.