1. The problem statement, all variables and given/known data A power supply used in a laboratory has an EMF of 5000 V. When, however, a voltmeter of resistance 20 kΩ is connected to the terminal of the power supply a reading of only 40 V is obtained. (a) Explain this observation. (b) Calculate (i) the current flowing in the meter and (ii) the internal resistance of the power supply. Answers: (b) (i) 2 mA, (ii) 2.48 MΩ 2. The attempt at a solution (b) (i) I = V / R = 40 / 20 000 = 2 * 10-3 A. (b) (ii) E = I (r + R) → 5000 = 2 * 10-3 (r + 20 000) → r = 2 480 000 Ω. (a) I am not sure why EMF and the PD are different. Maybe it has something to do with the differences between the general differences between the EMF and the PD. As far as I know, their difference is in the fact that PD is work done to move a charge between two points in a circuit and EMF is the total work done to move charge throught a complete circuit. And the formula is also different: V = I R, (R = total resistance), E = I (R + r), (R + r) = total external and internal resistance. But not sure whether this is applied to this situation.