Electric Power

Multiply the applied voltage in volts by the current in amperes: P = U·I, and the result comes out in watts. A kettle on the 230 V mains with I = 8.7 A delivers P = 230 × 8.7 ≈ 2,000 W. If you know the resistance instead of the current, use the variants P = U²/R or P = I²·R, which follow from Ohm law. Nameplates usually state the power; from it you find the current backwards with I = P/U, which is important to check whether a 16 A fuse is sufficient. Remember to convert kilowatts to watts before calculating.

The watt measures power, the rate of energy conversion, meaning how much energy a device converts per second. The kilowatt hour, by contrast, measures the energy itself: it is the amount of energy a 1 kW device converts in one hour, so 1 kWh = 1,000 W × 3,600 s = 3.6×10⁶ J. That is why the electricity bill counts kilowatt hours, not watts. Worked example: a 100 W lamp burning for 5 hours uses E = 0.1 kW × 5 h = 0.5 kWh. The classic mistake is to say "watts per hour", which is physically meaningless. Power is already a rate; energy is power times time.

Both follow from P = U·I with Ohm law and hold for ohmic loads. Choose the form whose quantities are given: if a known voltage is applied (socket, battery), take P = U²/R. If the current is known or the same for all components (series circuit), take P = I²·R. The second form explains heat losses in cables: for a fixed transmitted power the current drops when the voltage rises, which is why transmission lines run at 380 kV, since half the current means a quarter of the line losses. In a parallel circuit the same voltage lies across all components, so P = U²/R is the fastest way to compare them.

Find the current the device draws with I = P/U and compare it with the rated current of the fuse. Example: a 2,300 W fan heater on the 230 V mains draws I = 2,300/230 = 10 A, so a common 16 A fuse is sufficient. It gets critical when several devices share the same circuit: the currents add up. A kettle (2,000 W ≈ 8.7 A) plus a fan heater (10 A) already gives 18.7 A and trips the fuse. The fuse protects the cable from overheating, not the device. In such problems always calculate with the mains voltage of 230 V unless stated otherwise.

First compute the energy: E = P·t; with the power in kilowatts and the time in hours this directly gives kilowatt hours. Then multiply by the electricity price per kWh. Example: a 2 kW kettle running 10 minutes daily uses E = 2 kW × (1/6) h ≈ 0.33 kWh per day, about 122 kWh per year. At 0.35 €/kWh that is roughly 43 € annually. The same calculation pays off for standby devices: 5 W of continuous operation gives 5 W × 8,760 h = 43.8 kWh per year. The most common mistake is inserting watts instead of kilowatts and being off by a factor of 1,000.