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?

2 · Set Up

  1. Open the switched-lamp preset and press Reset. The switch starts closed, so the 9 V battery lights the lamp.
  2. Enable the power readout on the lamp.
  3. Set the lamp's resistance for each trial and record the power dissipated.

3 · Collect Data

Bulb resistance R (Ω)Current I (mA)Power dissipated P (W)
50.00
100.00
150.00

Plot power P (y-axis) against 1/R (x-axis) for your three trials.

4 · Analyze

  1. For one trial, compute I = E/R using E = 9 V, then P = I²R (or equivalently E²/R). Compare to the table.
  2. 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

  1. 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.
  2. 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, which at fixed voltage is the same as P = V²/R. A lower-resistance load draws more current and dissipates more power, but power and current scale by the same factor as resistance changes — halving R doubles both, not quadruples power, because R itself is shrinking even as I grows.

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