In the circuit shown, value of current in the circuit if the ammeter shown is made by a galvanometer of resistance 60.00 Ω and a shunt resistance of 0.02 Ω is:

0.99 A

The correct answer is Option (2) → 0.99 A

Let $R_g$ be the resistance of galvanometer -

Current (I) = $\frac{ε}{R+R_g}=\frac{3}{3+60}=0.048A$

Now, the ammeter is shunted with resistance s.

$∴R_{eff}=\frac{gs}{g+s}=\frac{60×0.02}{60+0.02}≃0.02Ω$

Current in the circuit $(I)' = \frac{ε}{R+R_{eff}}=\frac{3}{3+0.02}$

$≃0.99A$