Physics · Electricity

Electrical power

Q: How do you calculate the electric power of a device?

For constant DC voltage and current, $P=UI$. Both quantities must describe the same load. With current directed into the positive voltage terminal, positive power means absorption and negative power delivery. An ohmic resistor also permits current squared times resistance or voltage squared divided by resistance. AC instead requires the average of the simultaneous product. For sinusoidal quantities, rms values and power factor give real power. The retained kettle case 230 V and 8.7 A yields about 2 kW if its input is approximately resistive. Electrical input and useful mechanical or thermal output must be distinguished for real devices.

Q: What is the difference between a watt and a kilowatt hour?

A watt measures energy transfer rate: one watt is one joule per second. A kilowatt-hour instead measures energy, specifically 3.6 million joules. A constant 75 W load running for 40 minutes absorbs 0.050 kWh or 180 kJ. Without duration, 75 W does not determine consumption. Conversely, one energy quantity does not define a unique power profile: short high-power and long low-power operation can transfer the same energy. For varying loads, add energy from operating phases or integrate instantaneous power. Electricity usage is ordinarily billed as energy; a fixed fee is a separate tariff component.

Q: When do you use $P =\frac{ U^{2}}{R}$ and when $P = I^{2}\cdot{}R$?

Both variants follow for an ohmic resistor from $P=UI$ and its component relation. If current is known, current squared times resistance is often direct, for example when calculating line heat loss. If voltage across that exact resistor is known, use voltage squared divided by resistance. At fixed current dissipation increases with R; at fixed voltage it decreases. $\frac{A 6 V}{24} \Omega{}$ model gives 0.25 A and 1.5 W by both methods. These variants do not describe a motor simply through winding resistance: back emf and useful work additionally belong in its complete energy balance.

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

Electrical operating current can be derived from absorbed real power in the appropriate DC model. Sinusoidal AC additionally requires the power factor; multiplying rms values initially gives apparent power. This calculation identifies electrical load but does not establish suitability of a particular fuse. Inrush, trip characteristics, conductor conditions and coordinated protection are additional requirements. The kettle example gives about 8.7 A in a purely resistive 230 V model. That is not a general fuse approval. Several parallel loads may jointly draw substantially more current even if each individual device has only moderate power.

Q: How do you calculate the electricity cost of a device?

First determine actual absorbed electrical energy. A constant load uses power times duration; kilowatt-hours require kilowatts and hours. Multiply by the explicitly stated price per kilowatt-hour. The original $\frac{75 W}{40} \min{}$ case gives 0.050 kWh; at the chosen model tariff €$\frac{0.28}{kWh}$ this costs €0.014 in usage charges. That tariff is a calculation input, not a current market quotation. Fixed charges, time-variable tariffs and other contract terms need separate treatment. Nameplate power times total on-time may overestimate a cycling device; use its actual operating phases or an appropriate energy measurement.

$P=UI$ applies to constant DC quantities; $p=ui$ is the more general instantaneous relation. Resistor forms require ohmic behaviour. Sinusoidal AC rms values require the power factor.

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Formula

Calculate

P = U \cdot I

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

Variables & units

Variables & units
Symbol	Meaning	Unit
$P$	Electric power	W
$U$	Electric voltage	V
$I$	Electric current	A

LaTeX

P = U \cdot I

QUANTA STUDYquanta-study.de

Closed circuit with 18 and 36 ohm parallel branches; signed power bars alongside show minus 12, plus 8 and plus 4 watts. — Two parallel loads across 12 V absorb 8 W and 4 W. The source delivers 12 W; its absorbed power is negative.

Electrical power calculator

Find power, voltage or current for direct current, purely resistive parts or alternating current with a power factor.

Calculation model

Find

Voltage U

DC voltage across the device, e.g. 230 V or 12 V in a car.

Current I

Current through the device, e.g. 2 A.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for P, substitute, evaluate

1. Rearrange for P
$P=UI$
2. Substitute values (in SI units)
$P=230\cdot2=460\,\mathrm W$
3. Result
P = 460 W

Positive passive power means energy absorption; zero means no net power here.

Result: Power P

P = 460 W

With U = 230 V and I = 2 A, P = 460 W.

Quick tips

Remember P = U · I. At 230 V each ampere is about 230 W, so a 2 kW kettle draws just under 9 A.
If you only know the resistance, replace U or I using Ohm’s law: P = I² · R or P = U² / R.

Assumptions and limits

Constant DC values with passive port orientation: positive power is absorption, negative power delivery. This does not replace averaging arbitrary varying signals.

Derivation & background

Direct current into the positive voltage terminal of the component. Positive p then denotes absorbed power. Voltage describes energy per transferred charge: $\mathrm dE=u\,\mathrm dq$ . Divide by the corresponding time interval and use $i=\frac{\mathrm dq}{\mathrm dt}$ . Thus $p(t)=u(t)i(t)$ instantaneously. Constant DC quantities give $P=UI$ . For an ohmic resistor, inserting $U=RI$ gives $P=I^2R=\frac{U^2}{R}$ . For sinusoidal AC, average the simultaneous product: $\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ . Joule heating in an ohmic conductor follows $P=I^2R$ ; consumption energy can be converted using $1\,\mathrm{kWh}=3.6\cdot10^6\,\mathrm J$ .

Exam blueprint

Validity range

$P=UI$ applies to constant DC quantities; $p=ui$ is the more general instantaneous relation. Resistor forms require ohmic behaviour. Sinusoidal AC rms values require the power factor.

Derivation steps

Voltage assigns energy to transferred charge; charge current turns it into the energy-transfer rate.

1Direct current into the positive voltage terminal of the component. Positive p then denotes absorbed power.
2Voltage describes energy per transferred charge: $\mathrm dE=u\,\mathrm dq$ .
3Divide by the corresponding time interval and use $i=\frac{\mathrm dq}{\mathrm dt}$ .
4Thus $p(t)=u(t)i(t)$ instantaneously. Constant DC quantities give $P=UI$ .
5For an ohmic resistor, inserting $U=RI$ gives $P=I^2R=\frac{U^2}{R}$ .
6For sinusoidal AC, average the simultaneous product: $\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ .

Rearrangements

Current from power and voltage

I = \frac{P}{U}

For constant DC values or a purely resistive AC load with matching rms values. Calculated operating current alone does not establish fuse or wiring design.

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 modelled as purely resistive absorbs 2300 W at 230 V rms. Find its rms current.

Approach: the purely resistive load gives $I_{\mathrm{rms}}=\frac P{U_{\mathrm{rms}}}$ .
Substitution gives $\frac{2300\,\mathrm W}{230\,\mathrm V}=10\,\mathrm A$ .
Reverse calculation restores 2300 W.
A 16 A rating alone does not establish suitable protection or wiring design; additional operating and design conditions are missing.

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

$E = P\cdot{}t = 100 W \times{} 5 h = 500 Wh = 0.5 kWh$ .

Common mistakes

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

Power is the rate; energy = power × time.

Applying $P =\frac{ U^{2}}{R}$ 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’s law in circuit analysis.

Formula cluster

Energy in circuits

Connects circuit quantities with energy and cost considerations.

U = R·I

Ohm's Law

P = W/t

Mechanical Power

E = ½CU²

Energy Stored in a Capacitor

Worked example

The retained kettle case has $U_{\mathrm{rms}}=230\,\mathrm V$ and $I_{\mathrm{rms}}=8.7\,\mathrm A$ . For approximately resistive input, $P=230\cdot8.7\,\mathrm W=2001\,\mathrm W\approx2.0\,\mathrm{kW}$ .

Applications

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

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 modelled as purely resistive absorbs 2300 W at 230 V rms. Find its rms current.
Solution path
1. Approach: the purely resistive load gives $I_{\mathrm{rms}}=\frac P{U_{\mathrm{rms}}}$ .
2. Substitution gives $\frac{2300\,\mathrm W}{230\,\mathrm V}=10\,\mathrm A$ .
3. Reverse calculation restores 2300 W.
4. A 16 A rating alone does not establish suitable protection or wiring design; additional operating and design conditions are missing.
2
Task
A 100 W lamp runs for 5 hours. How much energy does it use in kWh?
Solution path
$E = P\cdot{}t = 100 W \times{} 5 h = 500 Wh = 0.5 kWh$ .

Understand and apply the equation

Two devices can use the same energy yet absorb it at different rates. Electrical power describes that rate of energy transfer. Voltage and current must refer to the same port at the same instant. This simple idea leads to DC calculations, signed power balances and mean power for changing signals.

How do you calculate electrical power with direct current?

For constant DC quantities, $P=UI$ ; in general, $p(t)=u(t)i(t)$ uses simultaneous values at the same two-terminal element. A watt is a joule per second. A constant 12 W therefore transfers 12 J each second. Before calculating, specify the component being considered. Whether energy enters or leaves follows from a directed current and labelled voltage terminals, not from the device’s name alone.[1]

\begin{aligned}p(t)&=u(t)i(t)\\{}P&=UI\end{aligned}

The current arrow enters the terminal marked +: the passive sign convention.[1]

With this convention, p>0 means electrical energy absorption and p<0 delivery. A resistor absorbs; a battery can deliver while discharging and absorb while charging. Positive motor input power is not entirely useful mechanical power. Heat, light and stored field energy are distinct destinations of the same electrical transfer.[1]

How does power follow from energy per unit time?

For a charge increment, voltage describes energy transferred per charge. Current counts how much charge passes per time. Their product combines these statements. Only afterwards do we average or multiply by duration. This keeps instantaneous power, mean power and total energy distinct.[1]

\begin{aligned}\mathrm dE_{\mathrm{abs}}&=u\,\mathrm dq\\{}i&=\frac{\mathrm dq}{\mathrm dt}\\{}p&=\frac{\mathrm dE_{\mathrm{abs}}}{\mathrm dt}=ui\end{aligned}

Voltage is transferred energy per charge; directed charge flow sets the rate.[1]

U and I must refer to the same two-terminal element and instant. Multiplying supply voltage by an unrelated branch current does not account for a defined load. The resulting rate is generally time dependent. Energy equals one power value times duration only when that product is constant; varying quantities require instantaneous power or its correct average.[1]

When do the resistor power formulas apply?

Both forms follow from Ohm’s law for the same purely ohmic resistor. Use the current through it or voltage across its own terminals; its positive resistance stays constant within the linear model regime.[2][1]

\begin{aligned}p&=i^2R=\frac{u^2}{R}\\{}&R>0\end{aligned}

For an ohmic resistor at the operating point; constant R defines the linear model regime.[2][1]

I²R and $\frac{U^2}{R}$ describe resistive dissipation. Using winding resistance in $\frac{U^2}{R}$ does not automatically give motor input: back emf and other energy conversion belong to the machine. At fixed current, a larger resistance raises dissipation; at fixed voltage it lowers it. These statements hold different quantities constant.[1][2]

Explore further: AC: an average rather than arbitrary meter readings

\overline P=\frac{1}{T}\int_0^T u(t)i(t)\,\mathrm dt=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi

The last equality applies to steady sinusoidal voltage and current at the same frequency.[3]

For a purely resistive load the power factor is one, so the product of rms voltage and rms current gives mean real power. An ideal capacitor may alternately absorb and return energy without net real-power consumption. For distorted current, cosφ alone is not the complete power factor. The general time integral remains valid; the product of rms values is initially apparent power in VA.[3]

Calculate DC power, phase offset and pulsed voltage

Two DC loads across the same source

Given: Ideal source with $U=12\,\mathrm V$ ; constant parallel resistors $R_1=18\,\Omega$ and $R_2=36\,\Omega$ .
Find: Determine branch and total power.

Approach: Use Ohm for each parallel branch, then add input powers.
$\begin{aligned}I_j&=\frac{U}{R_j}\\{}P_j&=UI_j\end{aligned}$
Each branch has 12 V; calculate its current separately.
$\begin{aligned}I_1&=\frac{12\,\mathrm V}{18\,\Omega}=\frac{2}{3}\,\mathrm A\\{}I_2&=\frac{12\,\mathrm V}{36\,\Omega}=\frac{1}{3}\,\mathrm A\end{aligned}$
Add branch rates and use source current.
$\begin{aligned}P_1&=8\,\mathrm W\\{}P_2&=4\,\mathrm W\\{}P_{\mathrm{tot}}&=12(\frac{2}{3}+\frac{1}{3})=12\,\mathrm W\end{aligned}$

Result: Loads absorb 12 W in total; the ideal source supplies 12 W.

Check: In passive orientation the source has −12 W; all absorbed powers sum to $-12+8+4=0$ .

Phase offset changes real power

Given: Sinusoidal quantities: $U_{\mathrm{rms}}=48\,\mathrm V$ , $I_{\mathrm{rms}}=2.0\,\mathrm A$ , $\varphi=60^\circ$ .
Find: Distinguish real and apparent power.

Approach: Separate the rms product from the sinusoidal power factor.
$\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$
First form the rms product.
$S=48\cdot2=96\,\mathrm{VA}$
Only the in-phase contribution sets the average.
$\overline P=96\cos60^\circ=48\,\mathrm W$

Result: Mean electrical input is 48 W, not 96 W.

Check: At φ=0 the same rms values would give 96 W; at ±90° the ideal mean would be zero.

The rms definition averages the square of a quantity, so it can also be used for the following rectangular profile. Power here follows directly from constant resistance and on-time fraction; no sinusoidal phase formula is assumed.[1][3]

Pulsed voltage across a pure resistor

Given: An ideal 12 Ω resistor periodically receives 12 V for 1 ms and 0 V for the next 3 ms. Switching edges are ideal; no other stores or losses.
Find: Find mean power, energy per period and rms values.

Approach: find instantaneous power in each phase and then time-weight it. Voltage and current are proportional for this pure resistor.
$\begin{aligned}D&=\frac{t_{\mathrm{on}}}{T}\\P_{\mathrm{on}}&=\frac{U_{\mathrm{on}}^2}{R}\\\overline P&=DP_{\mathrm{on}}\\U_{\mathrm{rms}}&=U_{\mathrm{on}}\sqrt D\end{aligned}$
The on-time fraction is one quarter.
$\begin{aligned}D&=\frac{1}{4}\\I_{\mathrm{on}}&=\frac{12\,\mathrm V}{12\,\Omega}=1\,\mathrm A\\P_{\mathrm{on}}&=12\,\mathrm W\\\overline P&=3\,\mathrm W\end{aligned}$
Evaluate energy and quadratically averaged quantities separately.
$\begin{aligned}E_T&=3\,\mathrm W\cdot0.004\,\mathrm s=0.012\,\mathrm J\\U_{\mathrm{rms}}&=6\,\mathrm V\\I_{\mathrm{rms}}&=0.50\,\mathrm A\\\overline U\,\overline I&=3\,\mathrm V\cdot0.25\,\mathrm A=0.75\,\mathrm W\end{aligned}$

Result: Mean power is 3 W and energy per period is 12 mJ. Multiplying the ordinary means gives a different value here.

Check: The area of a 12 W pulse over 1 ms is 12 mJ. Also $\frac{U_{\mathrm{rms}}^2}{R}=3\,\mathrm W$ gives the same mean. This rms calculation does not require a sinusoid, but does require the assumed constant pure resistance.

Power pulses and their mean

QUANTA STUDYquanta-study.de

Area per period is 12 mJ. The dashed mean-power line has the same area over that period. Ideal edges are part of the supplied model.[1][3]

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Watts and kilowatt-hours: rearrange power and calculate energy

\begin{aligned}[UI]&=\mathrm{V\,A}=\frac{\mathrm J}{\mathrm C}\frac{\mathrm C}{\mathrm s}=\mathrm W\\{}I&=\frac{P}{U}\\{}U&=\frac{P}{I}\end{aligned}

Divide only by a nonzero reference quantity; these are constant DC quantities.[1]

1 kW is 1000 J/s; 1 kWh is 3.6 MJ. A watt is a rate, a kilowatt-hour an energy quantity. The introductory kettle case of 230 V and 8.7 A gives 2001 W, about 2.0 kW: on AC this explicitly assumes rms values and approximately resistive input. Fuse suitability additionally depends on wiring, inrush current and trip characteristics; this power calculation alone does not establish it. For constant absorbed power, electrical energy follows from $E=Pt$ . For kWh, use power in kW and time in hours; consumption costs additionally require a specified price per kWh.[1][3]

Five solved exercises

Calculate · Foundation

An ohmic 24 Ω resistor has 6 V across it. Find current and power.

Show hint

Use Ohm first or $\frac{U^2}{R}$ directly.

Learning goal: Calculate ohmic current and power absorption.

Show solution and calculation

Approach: find current in the same resistor, then calculate power.
$I=\frac{6}{24}\,\mathrm A=0.25\,\mathrm A$ ; P=6×0.25=1.5 W.
Check I²R=0.0625×24=1.5 W.

\begin{aligned}I&=\frac UR=\frac{6\,\mathrm V}{24\,\Omega}=0.25\,\mathrm A\\P&=UI=1.5\,\mathrm W\\I^2R&=0.25^2\cdot24\,\mathrm W=1.5\,\mathrm W\end{aligned}

Interpret · Application

In passive orientation u=9 V and i=−0.20 A. Interpret the power.

Show hint

Keep the sign.

Learning goal: Interpret negative passive power as delivery.

Show solution and calculation

Approach: passive port orientation determines what the sign means. p=−1.8 W: the element delivers 1.8 J each second.
Negative power is not negative resistive heat.

p=9(-0.20)=-1.8\,\mathrm W

Transfer · Application

A constant 75 W load runs for 40 min; model tariff €0.28/kWh. Find energy and usage cost.

Show hint

Convert minutes to hours for kWh.

Learning goal: Translate electrical power into energy and model costs.

Show solution and calculation

Approach: first integrate constant input power over time, then apply the supplied model tariff.
$E=0.075\,\mathrm{kW}\cdot(\frac{2}{3})\,\mathrm h=0.050\,\mathrm{kWh}$ =180 kJ; cost €0.014=1.4 cents, excluding fixed fees.

E=75\cdot2400=180000\,\mathrm J

Interpret · Application

An ideal two-terminal element has u=10 V, i=0 A in the first half of a 2 s period, then u=0 V, i=2 A in the second. Transition times are neglected. Find average power; can mean voltage and mean current simply be multiplied?

Show hint

Form the simultaneous product in each half first.

Learning goal: Distinguish mean product from product of means.

Show solution and calculation

Approach: both instantaneous products are zero, so average power is zero.
Mean values 5 V and 1 A would incorrectly give 5 W.

Check: integrating ui over the period gives no net absorption. The ideal model omits fast-transition losses; any real contributions would require separate measurement. Timing correlation between measurements is essential.

\overline P=\frac{10\cdot0\cdot1+0\cdot2\cdot1}{2}\,\mathrm W=0\ne\overline u\,\overline i=5\,\mathrm W

[1]

Error analysis · Application

A sinusoidal load has Urms=100 V, Irms=2 A and phase angle 60°. A solution gives 200 W and calls it ohmic. Which claims are wrong, and what real power is absorbed?

Show hint

Power factor is cosφ; a purely ohmic load has φ=0.

Learning goal: Distinguish apparent and real power in sinusoidal AC.

Show solution and calculation

Approach: the RMS product is apparent power, 200 VA.
Multiplying by cos60°= $\frac12$ gives real power 100 W.
The supplied phase angle contradicts a purely ohmic model.

Check: over a complete period $T$ , absorbed energy is $E_T=\overline P T=(100\,\mathrm W)T$ . Exact energy for an arbitrary one-second interval is not determined without frequency and starting phase. This expression assumes the stated sinusoids; nonsinusoidal loads require more general averaging.

\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi=100\cdot2\cdot\frac12=100\,\mathrm W

[3][2]

What the terminal model establishes

The relation accounts for electrical transfer across the chosen boundary. It does not establish ideal efficiency, constant temperature, rating or protection. Switched-load measurements require synchronous u and i with appropriate bandwidth. Spatially extended high-frequency lines and radiation require additional field and transmission-line models; one resistor does not capture them.[1][3]

In brief

Calculate the simultaneous product with a clear port orientation. Resistor forms apply to the appropriate ohmic branch; for changing signals the timing relationship between voltage and current matters. Energy comes from integrating power over time.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Distinguished DC, instantaneous and real power more clearly; retained pulse and phase calculations and directed energy and kWh questions to the energy page without an additional full calculation.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

Cite this reference page

Quanta. Electric Power. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/elektrische-leistung

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Electric Power

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

For constant DC voltage and current, $P=UI$ . Both quantities must describe the same load. With current directed into the positive voltage terminal, positive power means absorption and negative power delivery. An ohmic resistor also permits current squared times resistance or voltage squared divided by resistance. AC instead requires the average of the simultaneous product. For sinusoidal quantities, rms values and power factor give real power. The retained kettle case 230 V and 8.7 A yields about 2 kW if its input is approximately resistive. Electrical input and useful mechanical or thermal output must be distinguished for real devices.

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

A watt measures energy transfer rate: one watt is one joule per second. A kilowatt-hour instead measures energy, specifically 3.6 million joules. A constant 75 W load running for 40 minutes absorbs 0.050 kWh or 180 kJ. Without duration, 75 W does not determine consumption. Conversely, one energy quantity does not define a unique power profile: short high-power and long low-power operation can transfer the same energy. For varying loads, add energy from operating phases or integrate instantaneous power. Electricity usage is ordinarily billed as energy; a fixed fee is a separate tariff component.

When do you use

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

and when

P = I^{2}\cdot{}R

Both variants follow for an ohmic resistor from $P=UI$ and its component relation. If current is known, current squared times resistance is often direct, for example when calculating line heat loss. If voltage across that exact resistor is known, use voltage squared divided by resistance. At fixed current dissipation increases with R; at fixed voltage it decreases. $\frac{A 6 V}{24} \Omega{}$ model gives 0.25 A and 1.5 W by both methods. These variants do not describe a motor simply through winding resistance: back emf and useful work additionally belong in its complete energy balance.

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

Electrical operating current can be derived from absorbed real power in the appropriate DC model. Sinusoidal AC additionally requires the power factor; multiplying rms values initially gives apparent power. This calculation identifies electrical load but does not establish suitability of a particular fuse. Inrush, trip characteristics, conductor conditions and coordinated protection are additional requirements. The kettle example gives about 8.7 A in a purely resistive 230 V model. That is not a general fuse approval. Several parallel loads may jointly draw substantially more current even if each individual device has only moderate power.

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

First determine actual absorbed electrical energy. A constant load uses power times duration; kilowatt-hours require kilowatts and hours. Multiply by the explicitly stated price per kilowatt-hour. The original $\frac{75 W}{40} \min{}$ case gives 0.050 kWh; at the chosen model tariff € $\frac{0.28}{kWh}$ this costs €0.014 in usage charges. That tariff is a calculation input, not a current market quotation. Fixed charges, time-variable tariffs and other contract terms need separate treatment. Nameplate power times total on-time may overestimate a cycling device; use its actual operating phases or an appropriate energy measurement.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §9.5 Electrical Power

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Electrical power

P=UI

applies to constant DC quantities;

p=ui

is the more general instantaneous relation. Resistor forms require ohmic behaviour. Sinusoidal AC rms values require the power factor.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

P

Electric power

U

Electric voltage

I

Electric current

Derivation & background

Exam blueprint

Validity range

$P=UI$ applies to constant DC quantities; $p=ui$ is the more general instantaneous relation. Resistor forms require ohmic behaviour. Sinusoidal AC rms values require the power factor.

Derivation steps

Voltage assigns energy to transferred charge; charge current turns it into the energy-transfer rate.

1Direct current into the positive voltage terminal of the component. Positive p then denotes absorbed power.
2Voltage describes energy per transferred charge: $\mathrm dE=u\,\mathrm dq$ .
3Divide by the corresponding time interval and use $i=\frac{\mathrm dq}{\mathrm dt}$ .
4Thus $p(t)=u(t)i(t)$ instantaneously. Constant DC quantities give $P=UI$ .
5For an ohmic resistor, inserting $U=RI$ gives $P=I^2R=\frac{U^2}{R}$ .
6For sinusoidal AC, average the simultaneous product: $\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$ .

How do you calculate with Electric Power?

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

Task

A fan heater modelled as purely resistive absorbs 2300 W at 230 V rms. Find its rms current.

Solution path

Approach: the purely resistive load gives $I_{\mathrm{rms}}=\frac P{U_{\mathrm{rms}}}$ .
Substitution gives $\frac{2300\,\mathrm W}{230\,\mathrm V}=10\,\mathrm A$ .
Reverse calculation restores 2300 W.
A 16 A rating alone does not establish suitable protection or wiring design; additional operating and design conditions are missing.

Task

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

Solution path

$E = P\cdot{}t = 100 W \times{} 5 h = 500 Wh = 0.5 kWh$ .

How do you calculate electrical power with direct current?

\begin{aligned}p(t)&=u(t)i(t)\\{}P&=UI\end{aligned}

The current arrow enters the terminal marked +: the passive sign convention.[1]

How does power follow from energy per unit time?

\begin{aligned}\mathrm dE_{\mathrm{abs}}&=u\,\mathrm dq\\{}i&=\frac{\mathrm dq}{\mathrm dt}\\{}p&=\frac{\mathrm dE_{\mathrm{abs}}}{\mathrm dt}=ui\end{aligned}

Voltage is transferred energy per charge; directed charge flow sets the rate.[1]

When do the resistor power formulas apply?

\begin{aligned}p&=i^2R=\frac{u^2}{R}\\{}&R>0\end{aligned}

For an ohmic resistor at the operating point; constant R defines the linear model regime.[2][1]

Explore further: AC: an average rather than arbitrary meter readings

\overline P=\frac{1}{T}\int_0^T u(t)i(t)\,\mathrm dt=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi

The last equality applies to steady sinusoidal voltage and current at the same frequency.[3]

Calculate DC power, phase offset and pulsed voltage

Two DC loads across the same source

Given: Ideal source with $U=12\,\mathrm V$ ; constant parallel resistors $R_1=18\,\Omega$ and $R_2=36\,\Omega$ .
Find: Determine branch and total power.

Approach: Use Ohm for each parallel branch, then add input powers.
$\begin{aligned}I_j&=\frac{U}{R_j}\\{}P_j&=UI_j\end{aligned}$
Each branch has 12 V; calculate its current separately.
$\begin{aligned}I_1&=\frac{12\,\mathrm V}{18\,\Omega}=\frac{2}{3}\,\mathrm A\\{}I_2&=\frac{12\,\mathrm V}{36\,\Omega}=\frac{1}{3}\,\mathrm A\end{aligned}$
Add branch rates and use source current.
$\begin{aligned}P_1&=8\,\mathrm W\\{}P_2&=4\,\mathrm W\\{}P_{\mathrm{tot}}&=12(\frac{2}{3}+\frac{1}{3})=12\,\mathrm W\end{aligned}$

Result: Loads absorb 12 W in total; the ideal source supplies 12 W.

Check: In passive orientation the source has −12 W; all absorbed powers sum to $-12+8+4=0$ .

Phase offset changes real power

Given: Sinusoidal quantities: $U_{\mathrm{rms}}=48\,\mathrm V$ , $I_{\mathrm{rms}}=2.0\,\mathrm A$ , $\varphi=60^\circ$ .
Find: Distinguish real and apparent power.

Approach: Separate the rms product from the sinusoidal power factor.
$\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi$
First form the rms product.
$S=48\cdot2=96\,\mathrm{VA}$
Only the in-phase contribution sets the average.
$\overline P=96\cos60^\circ=48\,\mathrm W$

Result: Mean electrical input is 48 W, not 96 W.

Check: At φ=0 the same rms values would give 96 W; at ±90° the ideal mean would be zero.

Pulsed voltage across a pure resistor

Given: An ideal 12 Ω resistor periodically receives 12 V for 1 ms and 0 V for the next 3 ms. Switching edges are ideal; no other stores or losses.
Find: Find mean power, energy per period and rms values.

Approach: find instantaneous power in each phase and then time-weight it. Voltage and current are proportional for this pure resistor.
$\begin{aligned}D&=\frac{t_{\mathrm{on}}}{T}\\P_{\mathrm{on}}&=\frac{U_{\mathrm{on}}^2}{R}\\\overline P&=DP_{\mathrm{on}}\\U_{\mathrm{rms}}&=U_{\mathrm{on}}\sqrt D\end{aligned}$
The on-time fraction is one quarter.
$\begin{aligned}D&=\frac{1}{4}\\I_{\mathrm{on}}&=\frac{12\,\mathrm V}{12\,\Omega}=1\,\mathrm A\\P_{\mathrm{on}}&=12\,\mathrm W\\\overline P&=3\,\mathrm W\end{aligned}$
Evaluate energy and quadratically averaged quantities separately.
$\begin{aligned}E_T&=3\,\mathrm W\cdot0.004\,\mathrm s=0.012\,\mathrm J\\U_{\mathrm{rms}}&=6\,\mathrm V\\I_{\mathrm{rms}}&=0.50\,\mathrm A\\\overline U\,\overline I&=3\,\mathrm V\cdot0.25\,\mathrm A=0.75\,\mathrm W\end{aligned}$

Result: Mean power is 3 W and energy per period is 12 mJ. Multiplying the ordinary means gives a different value here.

Power pulses and their mean

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Area per period is 12 mJ. The dashed mean-power line has the same area over that period. Ideal edges are part of the supplied model.[1][3]

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Watts and kilowatt-hours: rearrange power and calculate energy

\begin{aligned}[UI]&=\mathrm{V\,A}=\frac{\mathrm J}{\mathrm C}\frac{\mathrm C}{\mathrm s}=\mathrm W\\{}I&=\frac{P}{U}\\{}U&=\frac{P}{I}\end{aligned}

Divide only by a nonzero reference quantity; these are constant DC quantities.[1]

Five solved exercises

Calculate · Foundation

An ohmic 24 Ω resistor has 6 V across it. Find current and power.

Show hint

Use Ohm first or $\frac{U^2}{R}$ directly.

Learning goal: Calculate ohmic current and power absorption.

Show solution and calculation

Approach: find current in the same resistor, then calculate power.
$I=\frac{6}{24}\,\mathrm A=0.25\,\mathrm A$ ; P=6×0.25=1.5 W.
Check I²R=0.0625×24=1.5 W.

\begin{aligned}I&=\frac UR=\frac{6\,\mathrm V}{24\,\Omega}=0.25\,\mathrm A\\P&=UI=1.5\,\mathrm W\\I^2R&=0.25^2\cdot24\,\mathrm W=1.5\,\mathrm W\end{aligned}

Interpret · Application

In passive orientation u=9 V and i=−0.20 A. Interpret the power.

Show hint

Keep the sign.

Learning goal: Interpret negative passive power as delivery.

Show solution and calculation

Approach: passive port orientation determines what the sign means. p=−1.8 W: the element delivers 1.8 J each second.
Negative power is not negative resistive heat.

p=9(-0.20)=-1.8\,\mathrm W

Transfer · Application

A constant 75 W load runs for 40 min; model tariff €0.28/kWh. Find energy and usage cost.

Show hint

Convert minutes to hours for kWh.

Learning goal: Translate electrical power into energy and model costs.

Show solution and calculation

Approach: first integrate constant input power over time, then apply the supplied model tariff.
$E=0.075\,\mathrm{kW}\cdot(\frac{2}{3})\,\mathrm h=0.050\,\mathrm{kWh}$ =180 kJ; cost €0.014=1.4 cents, excluding fixed fees.

E=75\cdot2400=180000\,\mathrm J

Interpret · Application

An ideal two-terminal element has u=10 V, i=0 A in the first half of a 2 s period, then u=0 V, i=2 A in the second. Transition times are neglected. Find average power; can mean voltage and mean current simply be multiplied?

Show hint

Form the simultaneous product in each half first.

Learning goal: Distinguish mean product from product of means.

Show solution and calculation

Approach: both instantaneous products are zero, so average power is zero.
Mean values 5 V and 1 A would incorrectly give 5 W.

\overline P=\frac{10\cdot0\cdot1+0\cdot2\cdot1}{2}\,\mathrm W=0\ne\overline u\,\overline i=5\,\mathrm W

[1]

Error analysis · Application

A sinusoidal load has Urms=100 V, Irms=2 A and phase angle 60°. A solution gives 200 W and calls it ohmic. Which claims are wrong, and what real power is absorbed?

Show hint

Power factor is cosφ; a purely ohmic load has φ=0.

Learning goal: Distinguish apparent and real power in sinusoidal AC.

Show solution and calculation

Approach: the RMS product is apparent power, 200 VA.
Multiplying by cos60°= $\frac12$ gives real power 100 W.
The supplied phase angle contradicts a purely ohmic model.

\overline P=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi=100\cdot2\cdot\frac12=100\,\mathrm W

[3][2]

What the terminal model establishes

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Distinguished DC, instantaneous and real power more clearly; retained pulse and phase calculations and directed energy and kWh questions to the energy page without an additional full calculation.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Electric Power. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/elektrische-leistung

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

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