1. The problem statement, all variables and given/known data A 12 V, 24 W lamp and a resistor of fixed value are connected in some way inside a box with two external terminals. In order to discover the circuit arrangement inside the box a student connects a variable DC power source and an ammeter in series with the box and obtains the folliwing results: Applied potential difference and current 1: 1.0 V and 1.17 A; Applied potential difference and current 2: 12 V and 4.00 A. (a) Draw a circuit diagram of the most likely arrangement inside the box, giving your reason. (b) Use the 12 V, 4.00 A reading to deduce the value of the fixed resistor. (c) What is the percentage increase in the resistance of the lamp as the applied potential difference changes from 1.0 V to 12 V? (d) When the applied potential difference is increased to 24 V the current is again found to be 4.00 A. Explain this observation. (You may neglect the internal resistance of the power supply and the resistance of the ammeter in all your calculations.) Answers: (b) 6.0 Ω, (c) 500 %. 2. The attempt at a solution (a) I would guess that they are connected in parallel in the box. But I am struggling to understand the logic of the whole circuit. If they are connected in parallel then they both have 12 V. How can they have 12 V (the lamp and the resistor) if the applied PD is 1 V? Also if we calculate the current of the lamp: P = V I so ILamp = 24 / 12 = 2 A. How can current of the lamp be 2 A when it is 1.17 A in total? But in general I would say that the circuit has a battery and the ammeter in series and the lamp and the resistor are in parallel in the box. (b) RTotal = 12 / 4 = 3 Ω. ILamp = P / V = 24 / 12 = 2 A. RLamp = V / ILamp = 12 / 2 = 6 Ω. 1 / RTotal = 1 / RLamp + 1 / RResistor → 1 / 3 = 1 / 6 + 1 / RResistor → RResistor = 6 Ω. (c) Shouldn't the resistance of the lamp stay the same? ILamp = P / V = 24 / 12 = 2 A. RLamp = V / ILamp = 12 / 2 = 6 Ω. (d) Didn't start this one, since I'm not sure on the first three.