Why the Science-Fair Turbine Won't Light Its Lamp
A Transition Year group has built a small wind turbine for a science fair. The generator is a coil of 250 turns, each of area 4.0 × 10⁻³ m², rotating in a uniform magnetic field of 0.18 T. On the test stand the rotor turns at 8.0 revolutions per second. They connect the coil directly to a lamp marked "6.0 V, 3.0 W" and are disappointed: it glows a dim orange rather than white. A digital multimeter on the a.c. volts range, connected across the coil with the lamp disconnected, reads 6.4 V, and they argue from this that the lamp is being supplied at its rated voltage and must therefore be faulty. The coil is wound from 0.40 mm copper wire and has a measured resistance of 2.5 Ω. The lamp's resistance at normal working temperature is 12 Ω. The fair hall is warm and the turbine is running at a steady speed for the test.
1. Show that the peak emf is about 9 V and the rms emf about 6.4 V, and explain why the multimeter reading alone does not prove the lamp is being run at its rated voltage.
Read the worked answer
For a coil rotating in a uniform field, ε = NBAω sin(ωt), so the peak value is
ε₀ = NBAω = 250 × 0.18 × (4.0 × 10⁻³) × (2π × 8.0) = 45 × 4.0 × 10⁻³ × 50.3 = 9.05 V
ε(rms) = 9.05/√2 = 6.40 V - exactly the multimeter reading, so the numbers are consistent.
The multimeter was connected with the lamp disconnected. It therefore measured the emf, not the potential difference available to the lamp. Every real source has internal resistance, and once current flows, an amount Ir is dropped inside the coil, so the terminal pd falls below the emf. The rated 6.0 V on the lamp refers to the pd across it while it is working, so 6.4 V open-circuit is not evidence that the lamp is receiving 6.0 V. (A second, smaller point: an a.c. meter is calibrated to read rms for a sine wave; if the rotation is uneven or the field is not uniform the waveform is distorted and the reading is only approximate.)
2. Show that the lamp dissipates only about 2.3 W when connected directly, and explain where the missing power goes.
Read the worked answer
Lamp resistance at working temperature: R = V²/P = 6.0²/3.0 = 12 Ω (confirm: rated current 0.50 A).
The coil and lamp are in series: R(total) = 12 + 2.5 = 14.5 Ω I = ε(rms)/R(total) = 6.40/14.5 = 0.44 A
Power in the lamp: P = I²R = (0.44)² × 12 = 2.3 W Terminal pd across the lamp: V = IR = 0.44 × 12 = 5.3 V, well below the rated 6.0 V.
The "missing" power is dissipated as heat inside the coil itself: I²r = (0.44)² × 2.5 = 0.49 W. So of the 2.8 W generated, 2.3 W reaches the lamp and 0.5 W heats the copper winding. That is why the lamp glows orange - its filament is running at roughly 77% of rated power.
A refinement worth stating: a real filament's resistance falls as it cools, so at this lower temperature R is somewhat less than 12 Ω, the current is a little higher than 0.44 A, and the true power is slightly above 2.3 W. The constant-resistance estimate is a lower bound.
3. The students want full brightness. Evaluate two possible changes, stating the trade-off in each case, and explain why the lamp brightness is not steady.
Read the worked answer
Since ε₀ = NBAω, full brightness needs a higher emf and a smaller fraction lost internally.
Option 1 - spin faster or increase N, B or A. Emf scales linearly with ω, so raising the speed to about 8.0 × (6.0/5.3) ≈ 9 rev/s would bring the terminal pd to roughly 6 V. Trade-off: mechanical stress and noise rise, and the coil's reactance and the bearings limit how fast the group can run it; in a real turbine the wind speed is not under their control.
Option 2 - reduce the internal resistance by using thicker wire. But fewer turns or a shorter winding also reduces N (and hence the emf), so there is a genuine trade-off: you gain in r and lose in ε. Spacing the existing turns over a larger area is the better version of this idea.
A third, cleaner fix is simply to fit a 5 V lamp, matching the source to the load.
Brightness flicker: ε = NBAω sin(ωt) with ω = 2π × 8.0 = 50 rad/s, i.e. a 8 Hz alternating emf. Power ∝ ε², so it peaks twice per revolution at 16 Hz and falls to zero twice per cycle. 16 Hz is near the flicker-fusion threshold, so the lamp appears to shimmer rather than to blink - that is the visible symptom of the alternating output.