1. The problem statement, all variables and given/known data In the following circuit, cell A has an EMF of 10 V and an internal resistance of 2 Ω; cell B has an EMF of 3 V and an internal resistance of 3 Ω. (a) Show that the currents through A and B are 65 / 71 amps and 14 / 71 amps respectively. What is the magnitude of the current through GF? (b) Determine the power dissipated as heat in the resistor FE. If the circuit is switched on for 30 minutes, calculate the energy dissipated in FE in kilowatt-hours. (c) What is the potential difference (i) across the terminals of cell A, (ii) across the terminals of cell B? (d) Calculate the rate at which energy is being supplied (or absorbed) by cells A and B. (e) If the contact F can be moved along the resistor GE, find the value of the resistance GF when no current is flowing through cell B. (f) At what setting of F would cell B (i) be discharging at the maximum possible rate and (ii) be charging at the maximum possible rate? Answers: (a) 0.718 A, (b) 4.19 W, 2.095 * 10-3 kWh, (c) 8.17 V, 3.59 V, (d) 9.15 W, 0.592 W, (e) 3.60 Ω. 2. The attempt at a solution (a) 10 - 5 (I1 - I3) - 5 I1 - 2 I1 = 0 → I3 = 2.4 I1 - 2. Plug into: -3 -3 I3 + 5 I1 - 5 I3 = 0 → I1 = 0.9155 A (65 / 71), I3 = 0.197 A and I2 = 0.718 A. (b) P = V I and V = R I = 5 * (65 / 71) = 4.6 V. Then P = 4.6 * (65 / 71) = 4.19 W. W = V I t = 4.6 * (65 / 71) * (30 * 60) = 7543 J. 1 kWh = 3.6 * 106 J so 7543 / 3.6 * 106 = 2.095 * 10-3 kWh. (c) This is where difficulties come. As I understand, I should calculate the voltage of the circuits. (i) V = I R = (65 / 71) * (5 + 2) = 6.4 V of the 5 Ω and 2 Ω resistors around A. (ii) We calculte the voltage of the 3 Ω resistor: V = (14 / 71) * 3 = 0.59 V plus 3 V is 3.59 V for the 5 Ω resistor since it is parallel. I didn't continue since I am not sure on (c). What's wrong with it?