A moving coil galvanometer of resistance $$100\Omega$$ shows a full scale deflection for a current of 1 mA. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of 5 mA, is ____ $$\Omega$$

The galvanometer behaves like a resistor $$R_g = 100\Omega$$ that can carry a maximum (full-scale) current of $$I_g = 1\text{ mA} = 0.001\text{ A}$$. To measure larger currents we connect a small resistance $$R_s$$ in parallel with the galvanometer. This combination is called a shunt-type ammeter.

Let the required full-scale current of the ammeter be $$I = 5\text{ mA} = 0.005\text{ A}$$. When this current enters the parallel combination, part $$I_g$$ flows through the galvanometer and the remainder $$I_s$$ flows through the shunt:

$$I = I_g + I_s$$

Because the galvanometer and the shunt are in parallel, the voltage across each branch is the same:

$$I_g\,R_g = I_s\,R_s$$ $$-(1)$$

From the current relation, the shunt current is

$$I_s = I - I_g = 0.005 - 0.001 = 0.004\text{ A}$$

Substitute $$I_g$$, $$I_s$$ and $$R_g$$ into equation $$-(1)$$ to obtain $$R_s$$:

$$R_s = \frac{I_g\,R_g}{I_s} = \frac{0.001 \times 100}{0.004} = \frac{0.1}{0.004} = 25\Omega$$

Therefore, the resistance that must be connected in parallel with the galvanometer to convert it into an ammeter of full-scale 5 mA is $$25\Omega$$.

Option C is correct.