A galvanometer has a coil of resistance $$200\Omega$$ with a full scale deflection at $$20\mu A$$. The value of resistance to be added to use it as an ammeter of range $$(0 - 20)$$ mA is :

To convert a galvanometer to an ammeter, we connect a small resistance (shunt) in parallel.

The galvanometer resistance is $$G = 200\Omega$$, the full scale deflection current is $$I_g = 20\mu A = 20 \times 10^{-6}$$ A, and the desired range of the ammeter is $$I = 20$$ mA = $$20 \times 10^{-3}$$ A.

The shunt resistance formula is:

$$S = \frac{I_g \cdot G}{I - I_g}$$

$$S = \frac{20 \times 10^{-6} \times 200}{20 \times 10^{-3} - 20 \times 10^{-6}}$$

$$= \frac{4000 \times 10^{-6}}{19.98 \times 10^{-3}} = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.2002 \approx 0.20\Omega$$

The correct answer is Option 2: $$0.20\Omega$$.