The battery that cannot deliver the bulb's rating
A student is given a '12 V, 36 W' car bulb and a battery pack. With a 0.50 A load the terminal voltage is 11.6 V; with a 2.00 A load it is 10.0 V. The student assumes the battery is ideal and connects the bulb directly, expecting 36 W. The bulb instead glows dimly, and after ten minutes of running at about 3 A the battery's terminal voltage at the same current has fallen by a further 0.3 V while the battery case is warm. The student's notes say: 'internal resistance = ΔV/ΔI = 1.6/1.5 = 1.07 Ω, so the battery can deliver any power I want; maximum power is at maximum current.' The lab has a variable resistor, an ammeter, a voltmeter, and a thermometer.
1. Find the emf and internal resistance from the two measurements, then predict the current and power in the bulb, taking its rated resistance as constant.
Read the worked answer
Terminal voltage V = ε − Ir for two points: 11.6 = ε − 0.50r 10.0 = ε − 2.00r Subtract: 1.6 = 1.50r, so r = 1.07 Ω and ε = 11.6 + 0.50(1.07) = 12.13 V (≈12.1 V).
The bulb's rated resistance is R = V²/P = 12²/36 = 4.0 Ω. Connected directly: I = ε/(R + r) = 12.13/(4.0 + 1.07) = 2.39 A, P_bulb = I²R = (2.39)² × 4.0 = 22.9 W, terminal voltage = 12.13 − 2.39 × 1.07 = 9.58 V.
The bulb runs well below its 36 W rating, which matches the dim glow. Caveat: a tungsten filament at 9.6 V is cooler and has a lower resistance than the rated 4.0 Ω, so the true current is slightly higher than 2.39 A and the true power somewhat above 22.9 W; the constant-R value is a lower-bound estimate. Also, r itself is not strictly constant (see below).
2. Evaluate the note's claim that maximum power occurs at maximum current, and find the load that actually extracts maximum power.
Read the worked answer
Power delivered to an external resistance R is P = I²R = ε²R/(R + r)². At short circuit (R = 0) the current is maximal (ε/r) but P = 0, because all the energy is dissipated inside the battery. So 'maximum power at maximum current' is false.
Differentiating, or using the AM–GM inequality, P is maximal when R = r. Then I = ε/(2r) = 12.13/2.14 = 5.67 A, P_max = ε²/(4r) = 147.1/4.27 = 34.5 W, and the efficiency is I²R/(εI) = R/(R + r) = 50%.
So the bulb's 4.0 Ω is not the worst possible load; the maximum transferable power is about 34.5 W, still below the bulb's 36 W rating, and half of it would be wasted as internal heating. Maximum power transfer is a load-matching criterion, not an efficiency or safety criterion; real batteries may also be damaged at 5–6 A.
3. Explain the further 0.3 V drop at fixed current, and outline a measurement that would demonstrate the cause.
Read the worked answer
At fixed current, V = ε − Ir falls by 0.3 V, so either r has increased (by about 0.3/3 ≈ 0.10 Ω, roughly 10%) or ε has fallen, or both. The battery is warm, which points to the internal resistance rising with temperature: electrolyte conductivity, electrode kinetics, and contact resistances are temperature-dependent, and sustained 3 A dissipation (I²r ≈ 9.6 W internally) heats the cell.
Test: set the variable resistor to hold I constant at 3.00 A, record terminal voltage and case temperature every 30 s for 10–15 min, then switch off, allow cooling, and repeat. A plot of V against temperature (or of r = (ε − V)/I against temperature) should show a monotonic trend tracking temperature and should retrace its starting value after cooling. Distinguish this from simple charge depletion by checking whether the drop is reversible on cooling and whether ε (measured at very small current, when Ir is negligible) changes. If ε also falls irreversibly, the cell is being discharged, not just heated.