Heating effect and electric power
When current I flows through a resistor with potential difference V for time t, the electrical energy converted (mostly into heat) is W = VIt. This is Joule heating: H = I²Rt.
Electric power is the rate of energy conversion: P = VI = I²R = V²/R. Unit: watt (W) = J/s.
- When resistors are in series, the current is the same, so use P = I²R: the larger resistance gets more power.
- When resistors are in parallel, the voltage is the same, so use P = V²/R: the smaller resistance gets more power.
Rating of appliances: a bulb marked “100 W, 220 V” takes 100 W when connected to 220 V. Its resistance is R = V²/P = 220²/100 = 484 Ω and its current I = P/V ≈ 0.45 A. If it is run at 110 V (same R), power = 110²/484 = 25 W, one-fourth of the rated value.
Commercial unit of energy: 1 kilowatt-hour (kWh, the “unit” on an electricity bill) = 1000 W × 3600 s = 3.6 × 10⁶ J.
Power transmission: power lost in transmission lines is I²R. For a given power P = VI, transmitting at high voltage means low current and therefore much smaller loss. This is why power is sent over long distances at very high voltages and stepped down near homes.
Key points
- P = VI = I²R = V²/R; H = I²Rt.
- 1 kWh = 3.6 × 10⁶ J.
- Series: higher R dissipates more power. Parallel: lower R dissipates more power.
- High-voltage transmission reduces I²R losses.
Exam tips
- From a rating, first find R = V²/P; R stays the same at other voltages.
- Electricity-bill questions: units = kW × hours.
Common mistakes
- Assuming rated power is drawn at any voltage.
- Using P = I²R in parallel combinations where the voltage is known, then mixing up which resistor gets more power.
- Forgetting to convert W to kW when calculating units.