As shown in the figure below, the voltmeter reads 2 V across the 5 $$\Omega$$ resistor. The resistance of the voltmeter is ______ $$\Omega$$.

The voltmeter reads $$2\text{ V}$$ across the $$5\Omega$$ resistor, which means the potential difference across the entire parallel section is $$2\text{ V}$$

Voltage across the $$2\Omega$$ resistor ($$V_2$$), $$V_2 = V_{total} - V_p = 3\text{ V} - 2\text{ V} = 1\text{ V}$$

Since the $$2\Omega$$ resistor is in series with the rest of the circuit, the total current ($$I$$) flows through it. 

$$I = \frac{V_2}{R_2} = \frac{1\text{ V}}{2\Omega} = 0.5\text{ A}$$

The total current of $$0.5\text{ A}$$ splits between the $$5\Omega$$ resistor and the voltmeter.

$$I_5 = \frac{V_p}{5\Omega} = \frac{2\text{ V}}{5\Omega} = 0.4\text{ A}$$ (Current through the $$5\Omega$$ resistor)

$$I_v = I_{total} - I_5 = 0.5\text{ A} - 0.4\text{ A} = 0.1\text{ A}$$ (Current through the voltmeter)

$$R_v = \frac{V_p}{I_v} = \frac{2\text{ V}}{0.1\text{ A}} = 20\Omega$$