Physics · Electrical energy and power

Electrical energy and work at a terminal

E=UIt assumes constant voltage and current over the interval. Generally integrate the simultaneous product; separate averages cannot generally be multiplied. The sign follows the chosen passive orientation.

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Formula

Calculate

E_{el} = P t = U I t

E_el in J; P in W; t in s; U in V; I in A. 1 kWh=3.6 MJ.

Variables & units

Variables & units
Symbol	Meaning	Unit
$E_{\mathrm{el}}$	Electrical energy transferred	J
$P$	Electrical power	W
$t$	Duration	s
$U$	Voltage	V
$I$	Electric current	A

LaTeX

E_\mathrm{el}=P\,t=UIt

QUANTA STUDYquanta-study.de

Power-time plot: the first rectangle represents 5184 J, the second 2016 J. Widths have ratio 3 to 7, heights 6 to 1. The 12 W mean line has the same total area. — Three minutes at 28.8 W and seven minutes at 4.8 W give 7200 J. Duration-weighted mean power is 12 W.

Electrical energy calculator

Find electrical energy from power and time, from voltage, current and time, or from voltage and charge, with values that stay constant.

Calculation model

Find

Power P

Power taken in, constant or as an average, e.g. 2 kW for a kettle; negative means the device delivers power.

Time t

How long the device runs, e.g. 3 min.

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

Working · rearrange for E, substitute, evaluate

1. Rearrange for E
$E=P\Delta t$
2. Substitute values (in SI units)
- P = 2 kW = 2000 W
- t = 3 min = 180 s
$2000\cdot 180$
3. Result
E = 0.1 kWh

Result: Electrical energy E

E = 0.1 kWh

With P = 2 kW and t = 3 min, E = 0.1 kWh.

Quick tips

Unit check: kW times h gives kWh, and 1 kWh = 3,600,000 J. 2 kW for 3 min, i.e. 0.05 h, is 0.1 kWh.
Twice the power or twice the time means twice the energy. A 2 kW kettle running for 3 min uses as much as a 10 W lamp for 10 h.

Assumptions and limits

Passive terminal orientation and nonnegative duration. P is constant or an already correctly determined mean over this exact interval. Positive energy means absorption, negative means delivery. No arbitrary time profile, automatic useful energy or tariff calculation.

Derivation & background

Specify which component absorbs energy and direct its current into the positive voltage terminal. For a charge increment, $\mathrm dE=u\,\mathrm dq$ . Using $i=\frac{\mathrm dq}{\mathrm dt}$ gives $\mathrm dE=u(t)i(t)\,\mathrm dt$ . Integration over the interval gives $E=\int_{t_0}^{t_1}u(t)i(t)\,\mathrm dt$ . For constant voltage and current, $E=UI\Delta t=P\Delta t$ . For multiple operating phases, add their energies. Check joules, watt-hours and kilowatt-hours. Cost uses input energy; efficiency separately describes the useful fraction.

Exam blueprint

Validity range

Derivation steps

Transferred energy is the time integral of instantaneous electrical power.

1Specify which component absorbs energy and direct its current into the positive voltage terminal.
2For a charge increment, $\mathrm dE=u\,\mathrm dq$ .
3Using $i=\frac{\mathrm dq}{\mathrm dt}$ gives $\mathrm dE=u(t)i(t)\,\mathrm dt$ .
4Integration over the interval gives $E=\int_{t_0}^{t_1}u(t)i(t)\,\mathrm dt$ .
5For constant voltage and current, $E=UI\Delta t=P\Delta t$ . For multiple operating phases, add their energies.
6Check joules, watt-hours and kilowatt-hours. Cost uses input energy; efficiency separately describes the useful fraction.

Rearrangements

Find the duration

t=\frac{E_\mathrm{el}}{UI}

U and I must stay constant over the interval.

Task variant

A 12 V device draws 2 A for 60 s. How much electrical energy is transferred?

E=12·2·60=1,440 J.
Check using power: P=24 W and 24·60=1,440 J.

Common mistakes

Calling a kilowatt-hour a power unit or multiplying watts by minutes without conversion.

A kilowatt-hour is energy; convert time consistently and check that P·t gives joules.

Exam context

Energy use of devices, batteries, kilowatt-hours, and electricity costs.

Formula cluster

Electricity and energy

Voltage times current gives power; over time it gives energy.

P = U·I

Electric Power

U = R·I

Ohm's Law

η = Enutz/Ezu = P̄nutz/P̄zu

Efficiency of an Energy Conversion

Worked example

A 12 V device draws 2 A for 60 s: E=12·2·60=1,440 J. Check using power: P=24 W and 24 W·60 s=1,440 J.

Applications

Electrical devices, batteries, energy use, electricity costs, and quantitative energy balances.

How do you calculate with Electrical Energy at Constant Voltage and Current?

Here is how to work through a typical Electrical Energy at Constant Voltage and Current (Eel = P·t = U·I·t) task step by step:

1
Task
A 12 V device draws 2 A for 60 s. How much electrical energy is transferred?
Solution path
1. E=12·2·60=1,440 J.
2. Check using power: P=24 W and 24·60=1,440 J.

Understand and apply the equation

High power for a short time can transfer the same energy as low power over a longer period. Electrical energy counts the complete transfer during the interval considered. This requires the actual time profile and a clearly chosen system boundary. Examples progress from a constant load through multiple operating phases to transfer that may temporarily be negative.

The interval belongs to the energy quantity

A watt is a rate; a watt-hour is an amount of energy. The distinction resembles speed versus travelled distance: a rate alone does not give the total. Start by specifying the beginning and end of the interval and the device’s input port. Cost and efficiency can then be calculated in separate, explicitly identified steps.[1]

\begin{aligned}E_{\mathrm{abs}}&=\int_{t_0}^{t_1}u(t)i(t)\,\mathrm dt\\{}E&=UIt=Pt\end{aligned}

Passive sign convention: current enters the positive voltage terminal; t here denotes duration.[1]

Positive E is electrical energy absorbed across the boundary; negative E is net delivery. An ideal storage element can do both at different times. Device electrical consumption means input transfer, not automatically useful heat or mechanical work. A nameplate power alone is not an energy amount; interval and actual operating profile are missing.[1]

Electrical power gives the instantaneous transfer rate at the same oriented terminal. Energy then follows from its time area, including any negative contributions. A different boundary or interval changes the balance even if the power reading has the same units.[1]

Combine transported charge with energy per charge

With constant power, the area under the power curve is a rectangle. For multiple operating phases, add rectangles with their respective widths. A continuously changing response requires an integral. Under passive port orientation, negative area represents return of energy; do not replace it by its magnitude when calculating net absorption.[1]

\begin{aligned}\mathrm dE&=u\,\mathrm dq\\{}\mathrm dq&=i\,\mathrm dt\\{}E&=\int u\,\mathrm dq\end{aligned}

At constant U this gives E=UΔq even when current varies over the interval.[1]

Only constant I gives Δq=IΔt and hence UIt with one current value. Average current also works when U is constant. If both voltage and current vary, multiplying their averages is generally invalid: their time correlation matters. The integral of their product gives the correct terminal balance.[1][2]

Joules and kilowatt-hours share the same dimension

\begin{aligned}1\,\mathrm{Wh}&=3600\,\mathrm J\\{}1\,\mathrm{kWh}&=3.6\cdot10^6\,\mathrm J\end{aligned}

Watts times hours; k contributes a factor 1000 without turning this into a power unit.[1]

P in watts and t in seconds gives joules. P in kilowatts and t in hours gives kilowatt-hours. Ampere-hours instead measure charge: 1 Ah=3600 C. A battery Ah specification becomes electrical energy only with suitable voltage or its profile. Nominal voltage times Ah is a model estimate, not automatically usable energy at every load.[1]

Constant load, operating phases and return

A DC device runs for seven minutes

Given: Constant $U=18\,\mathrm V$ , $I=0.60\,\mathrm A$ , duration $7.0\,\mathrm{min}$ . Model tariff: 0.32 euros per kilowatt-hour.
Find: Find input energy and usage-only cost.

Approach: Integrate constant power over duration, then convert the energy unit.
$\begin{aligned}E&=UIt\\{}E_{\mathrm{kWh}}&=\frac{E_{\mathrm J}}{3.6\cdot10^6}\end{aligned}$
Convert duration to seconds and find power.
$\begin{aligned}t&=420\,\mathrm s\\{}P&=18\cdot0.60=10.8\,\mathrm W\end{aligned}$
Calculate energy and express the same amount in kWh.
$E=10.8\cdot420=4536\,\mathrm J=0.00126\,\mathrm{kWh}$
Apply the explicitly supplied model tariff.
$K=0.00126\cdot0.32=0.0004032\,\mathrm{EUR}$

Result: 4.536 kJ; model usage cost 0.04032 cents, excluding fixed fees.

Check: Check using charge: $\Delta q=0.60\,\mathrm A\cdot420\,\mathrm s=252\,\mathrm C$ . Multiplying $18\,\mathrm V\cdot252\,\mathrm C$ again gives 4536 J.

Count active and standby phases separately

Given: Constant $U=24\,\mathrm V$ ; $3\,\mathrm{min}$ at $1.2\,\mathrm A$ , followed by $7\,\mathrm{min}$ at $0.20\,\mathrm A$ .
Find: Find total energy and mean power over ten minutes.

Approach: Weight each constant operating phase by its own duration.
$\begin{aligned}E&=\sum_jUI_j\Delta t_j\\{}\overline P&=\frac{E}{\sum_j\Delta t_j}\end{aligned}$
Use each phase’s own time and power.
$\begin{aligned}E_1&=24\cdot1.2\cdot180=5184\,\mathrm J\\{}E_2&=24\cdot0.20\cdot420=2016\,\mathrm J\end{aligned}$
Add energies before averaging.
$\begin{aligned}E&=7200\,\mathrm J=0.00200\,\mathrm{kWh}\\{}\overline P&=\frac{7200}{600}=12\,\mathrm W\end{aligned}$
Calculate time-weighted current as an independent check.
$\begin{aligned}\overline I&=\frac{1.2\cdot180+0.20\cdot420}{600}=0.50\,\mathrm A\\{}U\overline I&=12\,\mathrm W\end{aligned}$

Result: 7.2 kJ and 12 W mean input; a simple average of the two current values would be wrong.

Check: Active operation absorbs 28.8 W, standby 4.8 W. The 12 W mean lies between them and includes the longer standby interval.

Rectangular operating-phase areas were a simple special case of the integral. A straight power curve can be evaluated just as clearly using triangle areas. The sign determines whether each area describes absorption or return.[1]

A continuous power profile changes direction

Given: At an ideal store’s passively oriented port, the supplied profile from 0 to 8 s is $p(t)=12\,\mathrm W-(3\,\mathrm{W/s})t$ . There are no other energy flows; initial stored energy is 10 J.
Find: Find energy absorbed, returned and stored at the end.

Approach: the zero separates absorption from delivery. Integrate each area separately and add with signs for the net balance.
$\begin{aligned}t_*&=\frac{12\,\mathrm W}{3\,\mathrm{W/s}}=4\,\mathrm s\\E_{\mathrm{abs}}&=\int_0^{4\,\mathrm s}p(t)\,\mathrm dt\\E_{\mathrm{return}}&=-\int_{4\,\mathrm s}^{8\,\mathrm s}p(t)\,\mathrm dt\end{aligned}$
The two triangles have equal areas and opposite signs.
$\begin{aligned}E_{\mathrm{abs}}&=\frac12\cdot4\,\mathrm s\cdot12\,\mathrm W=24\,\mathrm J\\E_{\mathrm{return}}&=24\,\mathrm J\\\Delta E&=24-24=0\,\mathrm J\end{aligned}$
Distinguish final energy from total transfer magnitude.
$\begin{aligned}E_f&=10\,\mathrm J+\Delta E=10\,\mathrm J\\E_{\mathrm{throughput}}&=24+24=48\,\mathrm J\end{aligned}$

Result: The store absorbs 24 J and returns 24 J. Final energy remains 10 J although transfer magnitudes total 48 J.

Check: An independent net check is the trapezoidal area: mean power from endpoints +12 W and −12 W is zero. Stored energy first rises to 34 J then falls to 10 J; it is never negative.

Equal areas: absorption and return

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Energy transfer counts positively above the time axis and negatively below it. Net area is zero, while the magnitudes sum to 48 J.[1]

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Explore further: Account synchronously for changing voltage

Consider two equal 10 s phases: first 10 V and 2 A, then 20 V and 1 A. Each phase supplies 20 W, giving 400 J total. Separate averages 15 V and 1.5 A instead give 22.5 W and 450 J. Here the product of averages overestimates by 12.5%. The correct average is that of the product, not the product of independently averaged quantities.[1]

\overline P=\overline{ui}

Sinusoidal AC requires the power factor; the more general integral remains valid.[2]

Distinguish input, storage and useful energy

\begin{aligned}E_{\mathrm{useful}}&=\eta E_{\mathrm{in}}\\{}E_C&=\frac{1}{2}CU_C^2\end{aligned}

Specified constant overall efficiency for the first relation; an ideal linear capacitor for the second.[1][3]

Efficiency multiplies the same clearly defined input-energy amount; UIt does not establish it. Capacitor voltage changes during charging, so stored energy is not charge times final voltage but ∫ $U_C$ dq= $\frac{Q^2}{2C}$ . Electrical energy may be dissipated, stored or returned later. The chosen accounting boundary determines which transfer is called consumption.[3][1]

Efficiency of an energy conversion compares the defined useful output with complete input. Apply it to the same energy boundary and interval; whole-installation efficiency cannot be taken unchecked from one motor’s value. Electrical terminal energy remains a separately determined transfer quantity.[1]

Five exercises with duration and model limits

Calculate · Foundation

A constant 24 W load runs for 15 min. Find E in J and Wh.

Show hint

15 min=900 s=0.25 h.

Learning goal: Convert power-time energy to joules and Wh.

Show solution and calculation

Approach: load power is constant; use the same duration for both energy units.
Transferred energy is 21,600 J, corresponding to 6 Wh or 0.006 kWh.

Check: each watt-hour equals 3600 J.

E=24\cdot900=21600\,\mathrm J

Transfer · Application

An ideal constant 9 V source transfers 2.0 Ah of charge. Find model energy.

Show hint

Convert Ah to C or directly V×Ah to Wh.

Learning goal: Combine constant voltage and charge transfer to obtain energy.

Show solution and calculation

Approach: with constant voltage, transferred charge directly determines energy.
Transferred charge is 7200 C.
At constant 9 V this gives 64,800 J or 18 Wh.
An actually changing battery voltage would instead require evaluating $\int u\,\mathrm dq$ .

E=9\cdot(2\cdot3600)=64800\,\mathrm J

Transfer · Application

Input energy 500 J, specified efficiency 0.72. Find useful energy and the remaining balance.

Show hint

Do not equate electrical input with useful energy.

Learning goal: Separate useful energy from remaining balance terms.

Show solution and calculation

Approach: the supplied efficiency refers to the stated input energy over the same interval.
Useful energy 360 J; remainder 140 J.
Only if other storage or delivery is excluded is that 140 J entirely waste heat.

E_{\mathrm{useful}}=0.72\cdot500=360\,\mathrm J

Interpret · Application

Under passive sign convention, an ideal store absorbs 120 W for 30 s, then delivers 60 W for 20 s. Find absorbed, returned and net stored energy if no other energy flows occur.

Show hint

The second power is negative in the absorption balance.

Learning goal: Separate absorbed, returned and net transferred energy.

Show solution and calculation

Approach: time integrals give +3600 J and −1200 J.
Net absorption and stored-energy increase are 2400 J.
The sum of transfer magnitudes is 4800 J, not storage increase.

Check: 3600 J=1200 J+2400 J closes the lossless balance. A real store with losses would require additional flows.

\begin{aligned}E_{\mathrm{net}}&=\int p\,dt=120\cdot30-60\cdot20=2400\,\mathrm J\\E_{\mathrm{throughput}}&=3600+1200=4800\,\mathrm J\end{aligned}

[1]

Error analysis · Application

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as $E=U_fQ_f=10\,\mathrm{mJ}$ . Correct the calculation; source consumption or resistor losses are not being asked.

Show hint

Capacitor voltage increases with q during charging.

Learning goal: Account for changing voltage when charging a capacitor.

Show solution and calculation

Approach: stored energy is $\int U(q)\, dq$ with $U=\frac qC$ , yielding $\frac12 CU^2$ =5 mJ.
The mistaken calculation applies final voltage to all transferred charge and doubles the area.

Check: Qfinal=1 mC and charge-averaged capacitor voltage is 5 V. Depending on the charging circuit, a source may deliver more energy; that is a different boundary.

E_C=\int_0^{Q}\frac qC\,dq=\frac{Q^2}{2C}=\frac12CU^2=5\,\mathrm{mJ}

[3]

State time profiles and tariff conditions

The introductory 12 V/2 A/60 s case gives 1440 J with P=24 W. Example prices are expressly chosen model tariffs, not current market prices. Usage costs are energy in kWh times price per kWh; fixed fees and other tariff conditions are separate. For varying loads, switched devices and AC measurements, time resolution must capture the actual profile. Nominal voltage times peak current times full duration often overestimates intermittent operation.[1][2]

In brief

Specify the interval and accounting boundary. Integrate simultaneous voltage times current or add the appropriate phase energies. Then distinguish net absorption, stored energy, useful energy and cost; they are not automatically the same quantity.

Sources for this explanation

Quanta · Updated October 2, 2026

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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.

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Change history

October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
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. Electrical Energy at Constant Voltage and Current. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrische-energie

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Frequently asked questions about Electrical Energy at Constant Voltage and Current

How do I calculate electrical energy from voltage and current?+

If voltage and current remain constant over the entire interval, transferred electrical energy is their product times duration. Use volts, amps and seconds to obtain joules. Values must refer to the same two-terminal element; passive orientation makes positive energy absorption and negative energy delivery. The retained 12 V/2 A/60 s case gives 1440 J. With fixed voltage and varying current, total transferred charge may also be used. When both vary, instead integrate their simultaneous product. Useful energy is an additional accounting quantity and may differ from electrical input because of losses or storage.

Is a kilowatt-hour power or energy?+

A kilowatt-hour is energy. It is one kilowatt of power multiplied by one hour, corresponding to 3.6 million joules. A kilowatt alone instead measures power. The same energy consumption can result from different combinations of power and duration. In the original 18 V device example, 10.8 W for seven minutes gives 4536 J or 0.00126 kWh. Ampere-hours are charge, not energy: suitable voltage or its profile is needed for an energy calculation. For real batteries, nominal voltage times ampere-hours is therefore a model estimate, not guaranteed usable energy under arbitrary load.

What assumptions are built into E=UIt?+

The short form with one numerical value assumes constant voltage and current over the interval. Alternatively, constant voltage together with a correctly time-averaged current is sufficient. Multiplying separately averaged varying voltage and current is generally invalid. In the original counterexample, 10 V at 2 A followed by 20 V at 1 A gives 20 W in both phases. Two equal ten-second phases transfer 400 J. Separate averages would instead predict 450 J. For sinusoidal AC, mean real power includes power factor. Source and load orientation must also be defined to give energy transfer an unambiguous sign.

How do I calculate electrical energy from power?+

For constant power, multiply by duration. With different operating phases, calculate each energy separately and add them; a continuous profile instead uses the integral of power. The original 24 V device runs three minutes at 1.2 A and seven minutes at 0.20 A. Phase energies 5184 J and 2016 J total 7200 J. Dividing by 600 seconds gives 12 W mean input. An unweighted average of the two current values would be wrong because phase durations differ. Areas under a power-time diagram show the weighting directly; one peak power value alone does not determine total energy.

How are electrical energy and electricity costs related?+

Usage-only cost is electrical input energy in kilowatt-hours multiplied by the explicitly stated price per kilowatt-hour. An example tariff is a calculation input, not a current price claim. The original 18 V/0.60 A/7 min case gives 0.00126 kWh. At the supplied €0.32/kWh, this costs €0.0004032 or 0.04032 cents. The small energy quantity explains the small charge. Fixed fees and other tariff items are separate. Time-variable pricing requires matching energy and price in each phase. Useful-heat calculations with efficiency also remain separate: billing does not automatically count only the useful part of electrical input.

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Electrical energy and work at a terminal

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Symbol

Meaning

Unit

E_{\mathrm{el}}

Electrical energy transferred

P

Electrical power

t

Duration

U

Voltage

I

Electric current

Derivation & background

Exam blueprint

Validity range

Derivation steps

Transferred energy is the time integral of instantaneous electrical power.

1Specify which component absorbs energy and direct its current into the positive voltage terminal.
2For a charge increment, $\mathrm dE=u\,\mathrm dq$ .
3Using $i=\frac{\mathrm dq}{\mathrm dt}$ gives $\mathrm dE=u(t)i(t)\,\mathrm dt$ .
4Integration over the interval gives $E=\int_{t_0}^{t_1}u(t)i(t)\,\mathrm dt$ .
5For constant voltage and current, $E=UI\Delta t=P\Delta t$ . For multiple operating phases, add their energies.
6Check joules, watt-hours and kilowatt-hours. Cost uses input energy; efficiency separately describes the useful fraction.

Understand and apply the equation

The interval belongs to the energy quantity

\begin{aligned}E_{\mathrm{abs}}&=\int_{t_0}^{t_1}u(t)i(t)\,\mathrm dt\\{}E&=UIt=Pt\end{aligned}

Passive sign convention: current enters the positive voltage terminal; t here denotes duration.[1]

Combine transported charge with energy per charge

\begin{aligned}\mathrm dE&=u\,\mathrm dq\\{}\mathrm dq&=i\,\mathrm dt\\{}E&=\int u\,\mathrm dq\end{aligned}

At constant U this gives E=UΔq even when current varies over the interval.[1]

Joules and kilowatt-hours share the same dimension

\begin{aligned}1\,\mathrm{Wh}&=3600\,\mathrm J\\{}1\,\mathrm{kWh}&=3.6\cdot10^6\,\mathrm J\end{aligned}

Watts times hours; k contributes a factor 1000 without turning this into a power unit.[1]

Constant load, operating phases and return

A DC device runs for seven minutes

Given: Constant $U=18\,\mathrm V$ , $I=0.60\,\mathrm A$ , duration $7.0\,\mathrm{min}$ . Model tariff: 0.32 euros per kilowatt-hour.
Find: Find input energy and usage-only cost.

Approach: Integrate constant power over duration, then convert the energy unit.
$\begin{aligned}E&=UIt\\{}E_{\mathrm{kWh}}&=\frac{E_{\mathrm J}}{3.6\cdot10^6}\end{aligned}$
Convert duration to seconds and find power.
$\begin{aligned}t&=420\,\mathrm s\\{}P&=18\cdot0.60=10.8\,\mathrm W\end{aligned}$
Calculate energy and express the same amount in kWh.
$E=10.8\cdot420=4536\,\mathrm J=0.00126\,\mathrm{kWh}$
Apply the explicitly supplied model tariff.
$K=0.00126\cdot0.32=0.0004032\,\mathrm{EUR}$

Result: 4.536 kJ; model usage cost 0.04032 cents, excluding fixed fees.

Check: Check using charge: $\Delta q=0.60\,\mathrm A\cdot420\,\mathrm s=252\,\mathrm C$ . Multiplying $18\,\mathrm V\cdot252\,\mathrm C$ again gives 4536 J.

Count active and standby phases separately

Given: Constant $U=24\,\mathrm V$ ; $3\,\mathrm{min}$ at $1.2\,\mathrm A$ , followed by $7\,\mathrm{min}$ at $0.20\,\mathrm A$ .
Find: Find total energy and mean power over ten minutes.

Approach: Weight each constant operating phase by its own duration.
$\begin{aligned}E&=\sum_jUI_j\Delta t_j\\{}\overline P&=\frac{E}{\sum_j\Delta t_j}\end{aligned}$
Use each phase’s own time and power.
$\begin{aligned}E_1&=24\cdot1.2\cdot180=5184\,\mathrm J\\{}E_2&=24\cdot0.20\cdot420=2016\,\mathrm J\end{aligned}$
Add energies before averaging.
$\begin{aligned}E&=7200\,\mathrm J=0.00200\,\mathrm{kWh}\\{}\overline P&=\frac{7200}{600}=12\,\mathrm W\end{aligned}$
Calculate time-weighted current as an independent check.
$\begin{aligned}\overline I&=\frac{1.2\cdot180+0.20\cdot420}{600}=0.50\,\mathrm A\\{}U\overline I&=12\,\mathrm W\end{aligned}$

Result: 7.2 kJ and 12 W mean input; a simple average of the two current values would be wrong.

Check: Active operation absorbs 28.8 W, standby 4.8 W. The 12 W mean lies between them and includes the longer standby interval.

A continuous power profile changes direction

Given: At an ideal store’s passively oriented port, the supplied profile from 0 to 8 s is $p(t)=12\,\mathrm W-(3\,\mathrm{W/s})t$ . There are no other energy flows; initial stored energy is 10 J.
Find: Find energy absorbed, returned and stored at the end.

Approach: the zero separates absorption from delivery. Integrate each area separately and add with signs for the net balance.
$\begin{aligned}t_*&=\frac{12\,\mathrm W}{3\,\mathrm{W/s}}=4\,\mathrm s\\E_{\mathrm{abs}}&=\int_0^{4\,\mathrm s}p(t)\,\mathrm dt\\E_{\mathrm{return}}&=-\int_{4\,\mathrm s}^{8\,\mathrm s}p(t)\,\mathrm dt\end{aligned}$
The two triangles have equal areas and opposite signs.
$\begin{aligned}E_{\mathrm{abs}}&=\frac12\cdot4\,\mathrm s\cdot12\,\mathrm W=24\,\mathrm J\\E_{\mathrm{return}}&=24\,\mathrm J\\\Delta E&=24-24=0\,\mathrm J\end{aligned}$
Distinguish final energy from total transfer magnitude.
$\begin{aligned}E_f&=10\,\mathrm J+\Delta E=10\,\mathrm J\\E_{\mathrm{throughput}}&=24+24=48\,\mathrm J\end{aligned}$

Result: The store absorbs 24 J and returns 24 J. Final energy remains 10 J although transfer magnitudes total 48 J.

Check: An independent net check is the trapezoidal area: mean power from endpoints +12 W and −12 W is zero. Stored energy first rises to 34 J then falls to 10 J; it is never negative.

Equal areas: absorption and return

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Energy transfer counts positively above the time axis and negatively below it. Net area is zero, while the magnitudes sum to 48 J.[1]

Free with a Quanta account. Your download stays ready after registration.

Explore further: Account synchronously for changing voltage

\overline P=\overline{ui}

Sinusoidal AC requires the power factor; the more general integral remains valid.[2]

Distinguish input, storage and useful energy

\begin{aligned}E_{\mathrm{useful}}&=\eta E_{\mathrm{in}}\\{}E_C&=\frac{1}{2}CU_C^2\end{aligned}

Specified constant overall efficiency for the first relation; an ideal linear capacitor for the second.[1][3]

Five exercises with duration and model limits

Calculate · Foundation

A constant 24 W load runs for 15 min. Find E in J and Wh.

Show hint

15 min=900 s=0.25 h.

Learning goal: Convert power-time energy to joules and Wh.

Show solution and calculation

Approach: load power is constant; use the same duration for both energy units.
Transferred energy is 21,600 J, corresponding to 6 Wh or 0.006 kWh.

Check: each watt-hour equals 3600 J.

E=24\cdot900=21600\,\mathrm J

Transfer · Application

An ideal constant 9 V source transfers 2.0 Ah of charge. Find model energy.

Show hint

Convert Ah to C or directly V×Ah to Wh.

Learning goal: Combine constant voltage and charge transfer to obtain energy.

Show solution and calculation

Approach: with constant voltage, transferred charge directly determines energy.
Transferred charge is 7200 C.
At constant 9 V this gives 64,800 J or 18 Wh.
An actually changing battery voltage would instead require evaluating $\int u\,\mathrm dq$ .

E=9\cdot(2\cdot3600)=64800\,\mathrm J

Transfer · Application

Input energy 500 J, specified efficiency 0.72. Find useful energy and the remaining balance.

Show hint

Do not equate electrical input with useful energy.

Learning goal: Separate useful energy from remaining balance terms.

Show solution and calculation

Approach: the supplied efficiency refers to the stated input energy over the same interval.
Useful energy 360 J; remainder 140 J.
Only if other storage or delivery is excluded is that 140 J entirely waste heat.

E_{\mathrm{useful}}=0.72\cdot500=360\,\mathrm J

Interpret · Application

Under passive sign convention, an ideal store absorbs 120 W for 30 s, then delivers 60 W for 20 s. Find absorbed, returned and net stored energy if no other energy flows occur.

Show hint

The second power is negative in the absorption balance.

Learning goal: Separate absorbed, returned and net transferred energy.

Show solution and calculation

Approach: time integrals give +3600 J and −1200 J.
Net absorption and stored-energy increase are 2400 J.
The sum of transfer magnitudes is 4800 J, not storage increase.

Check: 3600 J=1200 J+2400 J closes the lossless balance. A real store with losses would require additional flows.

\begin{aligned}E_{\mathrm{net}}&=\int p\,dt=120\cdot30-60\cdot20=2400\,\mathrm J\\E_{\mathrm{throughput}}&=3600+1200=4800\,\mathrm J\end{aligned}

[1]

Error analysis · Application

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as $E=U_fQ_f=10\,\mathrm{mJ}$ . Correct the calculation; source consumption or resistor losses are not being asked.

Show hint

Capacitor voltage increases with q during charging.

Learning goal: Account for changing voltage when charging a capacitor.

Show solution and calculation

Approach: stored energy is $\int U(q)\, dq$ with $U=\frac qC$ , yielding $\frac12 CU^2$ =5 mJ.
The mistaken calculation applies final voltage to all transferred charge and doubles the area.

Check: Qfinal=1 mC and charge-averaged capacitor voltage is 5 V. Depending on the charging circuit, a source may deliver more energy; that is a different boundary.

E_C=\int_0^{Q}\frac qC\,dq=\frac{Q^2}{2C}=\frac12CU^2=5\,\mathrm{mJ}

[3]

State time profiles and tariff conditions

In brief

Sources for this explanation

Quanta · Updated October 2, 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 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
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. Electrical Energy at Constant Voltage and Current. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/elektrische-energie

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

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Frequently asked questions about Electrical Energy at Constant Voltage and Current

How do I calculate electrical energy from voltage and current?+

Is a kilowatt-hour power or energy?+

What assumptions are built into E=UIt?+

How do I calculate electrical energy from power?+

How are electrical energy and electricity costs related?+

Retain Electrical Energy at Constant Voltage and Current for exams

Create a curated FSRS exam set for Eel = P·t = U·I·t: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Electrical energy and work at a terminal

Variables & units

Electrical energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electricity and energy

Worked example

Applications

How do you calculate with Electrical Energy at Constant Voltage and Current?

The interval belongs to the energy quantity

Combine transported charge with energy per charge

Joules and kilowatt-hours share the same dimension

Constant load, operating phases and return

A DC device runs for seven minutes

Count active and standby phases separately

A continuous power profile changes direction

Equal areas: absorption and return

Explore further: Account synchronously for changing voltage

Distinguish input, storage and useful energy

Five exercises with duration and model limits

A constant 24 W load runs for 15 min. Find E in J and Wh.

An ideal constant 9 V source transfers 2.0 Ah of charge. Find model energy.

Input energy 500 J, specified efficiency 0.72. Find useful energy and the remaining balance.

Under passive sign convention, an ideal store absorbs 120 W for 30 s, then delivers 60 W for 20 s. Find absorbed, returned and net stored energy if no other energy flows occur.

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as E=UfQf=10 mJE=U_fQ_f=10\,\mathrm{mJ}E=Uf​Qf​=10mJ. Correct the calculation; source consumption or resistor losses are not being asked.

State time profiles and tariff conditions

In brief

Sources for this explanation

Frequently asked questions about Electrical Energy at Constant Voltage and Current

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Electrical Energy at Constant Voltage and Current for exams

Electrical energy and work at a terminal

Variables & units

Electrical energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electricity and energy

Worked example

Applications

How do you calculate with Electrical Energy at Constant Voltage and Current?

The interval belongs to the energy quantity

Combine transported charge with energy per charge

Joules and kilowatt-hours share the same dimension

Constant load, operating phases and return

A DC device runs for seven minutes

Count active and standby phases separately

A continuous power profile changes direction

Equal areas: absorption and return

Explore further: Account synchronously for changing voltage

Distinguish input, storage and useful energy

Five exercises with duration and model limits

A constant 24 W load runs for 15 min. Find E in J and Wh.

An ideal constant 9 V source transfers 2.0 Ah of charge. Find model energy.

Input energy 500 J, specified efficiency 0.72. Find useful energy and the remaining balance.

Under passive sign convention, an ideal store absorbs 120 W for 30 s, then delivers 60 W for 20 s. Find absorbed, returned and net stored energy if no other energy flows occur.

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as E=UfQf=10 mJE=U_fQ_f=10\,\mathrm{mJ}E=Uf​Qf​=10mJ. Correct the calculation; source consumption or resistor losses are not being asked.

State time profiles and tariff conditions

In brief

Sources for this explanation

Frequently asked questions about Electrical Energy at Constant Voltage and Current

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Electrical Energy at Constant Voltage and Current for exams

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as $E=U_fQ_f=10\,\mathrm{mJ}$ . Correct the calculation; source consumption or resistor losses are not being asked.

An ideal 100 μF capacitor starts uncharged and reaches 10 V. A solution gives its stored energy as $E=U_fQ_f=10\,\mathrm{mJ}$ . Correct the calculation; source consumption or resistor losses are not being asked.