Physics · Electricity

Electric Power

Electric power is the product of voltage and current; it states how much energy a device converts per second.

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Formula

P = U \cdot I

LaTeX: P = U \cdot I

P in watts [W] = [V·A] · U in volts [V] · I in amperes [A]

Variables & units – Electric Power

Symbol	Meaning	Unit
P	Electric power	W (Watt)
U	Electric voltage	V
I	Electric current	A

Derivation & background – Electric Power

Power is energy conversion per time: P = E/t. With Ohm's law the variants P = U²/R and P = I²·R follow; the latter describes the heat loss in cables (Joule heating). The electricity bill counts energy in kilowatt hours: 1 kWh = 3.6×10⁶ J.

Exam blueprint

Validity range

P = U·I holds generally for direct current; for alternating current it is the instantaneous power, and on average the power factor cos(φ) enters. The variants P = U²/R and P = I²R assume an ohmic load.

Derivation steps

Power is energy per time; voltage is energy per charge and current is charge per time.

1Energy per charge times charge per time: P = (E/Q)·(Q/t) = U·I.
2With U = R·I follow P = I²·R and P = U²/R.

Rearrangements

Current from power and voltage

I = \frac{P}{U}

This is how you check whether a fuse (e.g. 16 A) is sufficient.

Power from resistance and voltage

P = \frac{U ^{2}}{R}

Useful when no current is given; only for ohmic loads.

Energy from power and time

E = P \cdot t

The basis of electricity cost calculations in kilowatt hours.

Task variant

A fan heater has P = 2,300 W on the 230 V mains. Find I.

I = P/U = 2,300 W / 230 V = 10 A. A 16 A fuse is sufficient.

A 100 W lamp runs for 5 hours. How much energy does it use in kWh?

E = P·t = 100 W × 5 h = 500 Wh = 0.5 kWh.

Common mistakes

Confusing power (watts) with energy (watt hours/joules).

Power is the rate; energy = power × time.

Applying P = U²/R to non-ohmic loads.

The R variants only hold when R is constant.

Mixing kW and W when substituting.

Convert everything to watts first: 2 kW = 2,000 W.

Exam context

Classic in household problems (fuses, electricity costs) and combined with Ohm law in circuit analysis.

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Formula cluster

Energy in circuits

Connects circuit quantities with energy and cost considerations.

U = R \cdot I

Ohm's Law

P = \frac{W}{t}

Mechanical Power

E = \frac{1}{2} C U^{2}

Energy Stored in a Capacitor

Worked example

A kettle on the 230 V mains draws a current of I = 8.7 A: P = 230 V × 8.7 A ≈ 2,000 W = 2 kW.

Applications

Sizing household appliances, fuse rating, electricity cost calculation (kWh), sizing of cable cross-sections

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Curated exam set for "Electric Power":

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Which formula describes Electric Power?

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Question (front)

How do you rearrange P = U·I for Current from power and voltage?

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Question (front)

Which common mistake happens with Electric Power?

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Scientific sources

[1]Halliday, D., Resnick, R., & Walker, J. (2017). Halliday Physik. Wiley-VCH.

Common notations & search queries

P=U*IP=UIP=U^2/RP=I^2*RWatt Formelelektrische Leistung berechnenelectric power formulaLeistung Spannung Strom

Related formulas

U = R \cdot I

Ohm's Law

P = \frac{W}{t}

Mechanical Power

E = \frac{1}{2} C U^{2}

Energy Stored in a Capacitor

More Physics formulas

F = m \cdot a

Newton's Second Law

F_{G} = G \cdot \frac{m _{1} \cdot m _{2}}{r ^{2}}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^{2}

Kinetic Energy

E_{p o t} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Frequently asked questions about Electric Power

How do you calculate the electric power of a device?+

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.

What is the difference between a watt and a kilowatt hour?+

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.

When do you use P = U²/R and when P = I²·R?+

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.

How do you check whether a fuse is sufficient for a device?+

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.

How do you calculate the electricity cost of a device?+

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.

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How do you calculate with Electric Power?

Here is how to work through a typical Electric Power (P = U·I) task step by step:

1
Task
A fan heater has P = 2,300 W on the 230 V mains. Find I.
Solution path
I = P/U = 2,300 W / 230 V = 10 A. A 16 A fuse is sufficient.
2
Task
A 100 W lamp runs for 5 hours. How much energy does it use in kWh?
Solution path
E = P·t = 100 W × 5 h = 500 Wh = 0.5 kWh.