PHY044 · Electricity & Circuits · Classes 10–12
Electrical power
Objective: To measure the electrical power of a bulb or resistor and see how it depends on voltage.
🧰 Apparatus
- Low-voltage bulb (12 V)
- Supply (0–24 V DC)
- Ammeter
- Voltmeter
- Wires
📐 Formulas and SI units
P = VI = I²R = V²/R (W)
Energy = Pt (J); 1 kWh = 3.6 MJ
💡 Theory
Electrical power is the rate at which energy is transferred: P = VI. Using Ohm's law it is also I²R and V²/R, so doubling the voltage on the same resistance gives four times the power.
🪜 Procedure
- Connect the bulb with the ammeter in series and the voltmeter across it.
- Increase the voltage in steps.
- Record V and I and calculate P.
- Note the brightness each time.
- Plot P against V².
📋 Observations Simulated values
✅ Result and conclusion
12 V across 6 Ω gives 2 A and 24 W; doubling V quadruples P.
🔬 In a real lab
A bulb's filament resistance rises as it heats, so its P–V² graph curves; a fixed resistor gives a straight line.
⚠️ Safety
Bulbs get very hot; let them cool before touching. Stay within the bulb's rated voltage.
❓ Quick check
1. A 60 W bulb runs for 10 hours. Energy?
2. Doubling the current through a resistor makes the power…
3. Unit of electrical power?
🏋️ Practice
Easy230 V, 5 A. P?
Answer: 1150 W
Medium2 kW heater for 3 h. Energy?
Answer: 6 kWh
ChallengeP = 100 W at 230 V. I?
Answer: ≈ 0.43 A