Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $$\Omega$$ is used to measure the potential difference across 500 $$\Omega$$ resister, the reading of the voltmeter will be ______ V.

When the voltmeter of resistance $$R_v = 2000\ \Omega$$ is connected across the $$500\ \Omega$$ resistor, they form a parallel combination.

Equivalent resistance ($$R_p$$) of this combination:    $$R_p = \frac{500 \times 2000}{500 + 2000} = \frac{1,000,000}{2500} = 400\ \Omega$$

This combination is now in series with the remaining $$600\ \Omega$$ resistor. Total equivalent resistance ($$R_{\text{total}}$$) of the circuit:  $$R_{\text{total}} = R_p + 600 = 400 + 600 = 1000\ \Omega$$

Total circuit current ($$I$$) drawn from the $$20\ \text{V}$$ battery:   $$I = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{20}{1000} = 0.02\ \text{A}$$

The voltmeter reading is the voltage drop across the parallel network ($$R_p$$):   $$V = I \times R_p = 0.02 \times 400 = 8\ \text{V}$$