A Switched Lamp: Power Dissipation in a Bulb
1 · Predict
A 9 V battery lights a bulb through a closed switch. If you swap in a bulb with lower resistance, does it glow brighter or dimmer?
- A lower-resistance bulb draws more current and dissipates more power — it glows brighter.
- A lower-resistance bulb glows dimmer.
- Brightness doesn't depend on the bulb's resistance.
2 · Set Up
- Open the switched-lamp preset and press Reset. The switch starts closed, so the 9 V battery lights the lamp.
- Enable the power readout on the lamp.
- Set the lamp's resistance for each trial and record the power dissipated.
3 · Collect Data
| Bulb resistance R (Ω) | Current I (A) | Power dissipated P (W) |
|---|---|---|
| 50 | ||
| 100 | ||
| 150 |
Plot power P (y-axis) against 1/R (x-axis) for your three trials.
4 · Analyze
- For one trial, compute I = E/R using E = 9 V, then P = I²R (or equivalently E²/R). Compare to the table.
- Explain why a lower-resistance bulb, which draws more current at the same voltage, ends up dissipating more power — brightness in a real bulb scales with power.
5 · Extend
- A '100 W' light bulb at 120 V draws about 0.83 A and has a resistance of about 144 Ω when hot. Using P = E²/R, explain why a 'higher wattage' bulb rated for the same voltage must have LOWER resistance, not higher.
- Incandescent filaments often fail right when switched on, when they're cold and their resistance is lowest. Using your power relationship, explain why the current (and power) surge is highest at that moment.
The Physics Behind This Experiment
Electrical Power (from I and R)
Power dissipated as heat and light in a resistive load is P = I²R. At fixed voltage, a lower-resistance load draws more current, and power grows even faster since it depends on the square of current.