Resistors R1, R2 and R3 has a resistance of 1,2 and 3 ohms respectivley. The internal resistance is 1 ohm and the emf E= 7V. Calculate the current i = 7V/7 ohms gives 1A. The voltmeter should read p.d = iR2 = 1x2 = 2V. Try getting the p.d using the left loop across the battery and R1 and R2. Consider one point to have 0V, the lower point (right of R3). Since current is leftwards, the drop from 0V gives - 3V(1A x 3ohms). Upwards across the battery is 7V higher including a drop of 1V across internal resistance which gives 3V( -3 +7-1) at positive terminal of battery. Across the resistor of R1 is a drop of 1V as current flows to the right. The potential at the top end of voltmeter is now only 2V(3V - 1V). Comparing the top end(2 V) and the lower end (0 V) gives a p.d of 2 V as expected voltmeter reading in the first instance.
