Electricity · Lesson 8 of 9
Electric Power
“Electric power measures how quickly appliances turn electrical energy into useful work and warm bills.”
• Interpret electric power as the rate of electrical energy transfer. • Select and apply the three forms of the electric power relationship. • Relate appliance ratings to current and resistance. • Calculate electrical energy from power and operating time. • Convert between joules and kilowatt-hours. • Calculate operating cost and compare appliance energy use.
A 1000 W heater and a 10 W lamp may run for the same time, but they do not transfer energy at the same rate. The heater transfers one hundred times as much energy each second. Power measures this rate, while energy records the total accumulated transfer over time.
The rate at which electrical energy is transferred or consumed in a circuit.
At a fixed supply voltage, a lower-resistance appliance draws a larger current and therefore greater power, as P = V²/R shows. At fixed current, a larger resistance has greater power, as P = I²R shows. The condition held constant determines the correct comparison.
Appliance Ratings
A rating such as 220 V, 100 W states the intended operating voltage and the power consumed at that voltage. Current can be found from I = P/V, and operating resistance from R = V²/P. If the same approximately constant resistance is used at a lower voltage, power falls with the square of voltage.
Electrical Energy
The electrical energy consumed by a power of one kilowatt operating for one hour.
To calculate cost, first express power in kilowatts, multiply by hours of operation to obtain kilowatt-hours, and then multiply by the tariff per kilowatt-hour. For repeated use, multiply by the number of days or cycles. Keep the distinction between 400 W and 0.400 kW.
Electrical companies charge for energy, not for electrons. Electrons circulate through conductors; the paid quantity is energy transferred to appliances. A kilowatt is power, while a kilowatt-hour is energy.
Problem-Solving Strategy
- Identify whether the required quantity is power, energy, current, resistance or cost.
- Choose P = VI, I²R or V²/R according to the known values.
- For joules use watts and seconds; for billing use kilowatts and hours.
- Multiply by repeated operating days only after finding daily energy.
- Check whether a higher-power or longer-running appliance should use more energy.
Problem
A bulb connected to 220 V carries 0.50 A. Find its power.
- 1.Given: V = 220 V and I = 0.50 A.
- 2.Use P = VI.
- 3.P = 220 × 0.50 = 110 W.
- 4.The bulb transfers 110 J of electrical energy each second.
Problem
A 400 W refrigerator operates 8 h per day for 30 days. Find energy in kWh and cost at ₹3 per kWh.
- 1.Convert power: 400 W = 0.400 kW.
- 2.Total time = 8 h/day × 30 days = 240 h.
- 3.Energy = Pt = 0.400 × 240 = 96 kWh.
- 4.Cost = 96 × ₹3 = ₹288.
- 5.The unit check kW × h = kWh confirms the billing calculation.
Problem
A 220 V, 100 W bulb is operated at 110 V. Assuming its resistance remains constant, find the new power.
- 1.First find rated resistance: R = V²/P = 220²/100 = 484 Ω.
- 2.At 110 V, use P = V²/R.
- 3.P = 110²/484 = 12,100/484 = 25 W.
- 4.Halving voltage reduces power to one quarter because P ∝ V² at constant R.
- 5.This calculation uses the stated constant-resistance assumption; a real filament resistance changes with temperature.
Energy Comparison Example
Problem
Which uses more energy: a 250 W television for 1 h or a 1200 W toaster for 10 min?
- 1.Television: E = 0.250 kW × 1 h = 0.250 kWh.
- 2.Toaster time = 10/60 h = 1/6 h. E = 1.200 × 1/6 = 0.200 kWh.
- 3.The television uses more energy despite its lower power because it operates longer.
- 4.Difference = 0.250 − 0.200 = 0.050 kWh.
Quiz
What does a power of 60 W mean?
Which expression does not represent electric power?
Which is an energy unit?
A 220 V motor takes 5 A. What is its power?
At fixed resistance, voltage is halved. What happens to power?
Practice Problems
- A device operates at 12 V and 3 A. Find power. Solution: P = VI = 12 × 3 = 36 W.
- A 1000 W heater works at 250 V. Find current. Solution: I = P/V = 1000/250 = 4 A.
- Find the resistance of a 100 W appliance rated at 200 V. Solution: R = V²/P = 200²/100 = 400 Ω.
- A 1.5 kW heater runs for 40 min. Find energy in kWh and joules. Solution: t = 40/60 = 2/3 h. E = 1.5 × 2/3 = 1.0 kWh = 3.6 × 10⁶ J.
- A 2 kW appliance runs 3 h daily for 20 days at ₹6 per kWh. Find cost. Solution: E = 2 × 3 × 20 = 120 kWh. Cost = 120 × ₹6 = ₹720.
Key Takeaways
• Electric power is the rate of electrical energy transfer. • P = VI, I²R and V²/R are equivalent forms under Ohm’s law. • Appliance ratings state intended voltage and power. • Energy equals power multiplied by time. • One kilowatt-hour equals 3.6 × 10⁶ joules. • Kilowatt is power; kilowatt-hour is energy. • Cost equals energy in kilowatt-hours multiplied by tariff. • Operating time can outweigh power when comparing total energy use.