1. The problem statement, all variables and given/known data A battery of internal resistance 0.50 Ω is connected (as shown below) through a switch S to a resistor X, which is initially at 0 °C. When S is closed, the voltmeter reading falls immediated from 12.0 V to 10.0 V. The reading then rises gradually to a steady value of 10.5 V. (a) Explain these observations. (b) Calculate: (i) the initial current when S is closed; (ii) the initial resistance of X; (iii) the resistance of X when he steady state has been reached. Answers: (b) (i) 4.0 A, (ii) 2.5 Ω, (iii) 3.5 Ω. 2. The attempt at a solution (a) Maybe it is the temperature that gives an increase over time to 10.5 V? The initial temperature increases from zero to some unknown value. (b) (i) I tried to find the current via I = V / R = 12 / 0.5 = 24 A, but the answer is wrong. (b) (ii) E = I (r + R) → 12 = 4 (from the answer) * (0.5 + R) → R = 2.5 Ω. (b) (iii) I guess I'll need a new current to correctly calculate this part. 10.5 = 4 (0.5 + R) → R = 2.125 Ω, which is wrong. So it's mostly about (a) and (b) (i). Is it due to the temperature in (a) and how to find current in (b) (i)?