A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A is:

- 2 Ω
- 0.1 Ω
- 3 Ω
- 0.01 Ω

**Solution**

Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.

If resistance of galvanometer is R_{g} and it gives full-scale deflection when current I_{g} is passed through it. Then,

V = I_{g}R_{g}

V = 1 mA × 100 Ω

Let a shunt of resistance (R_{s}) is connected in parallel to galvanometer. If total current through the circuit is I.

I = 10 A = I_{s} + I_{g}

V = I_{s}R_{s} = (I - I_{g})R_{s}

(I - I_{g})R_{s} = I_{g}R_{g}

(10 - 10^{-3})R_{s} = 100 × 10^{-3}

R_{s} ≈ 0.01 Ω

**The correct option is D.**