Library / Electrical / ELE-02
Electrical
Electrical work done
Energy transferred by a current flowing through a potential difference for a given time.
- Ref
- ELE-02
- Rev
- v1
- Field
- Electrical
- Result
- W [J]
W in joules (J), I in amperes, t in seconds, V in volts. Since Q = It, this is the same as W = QV.
Each coulomb that passes carries V joules across the potential difference, and It coulombs pass in time t. Multiply them together to get the energy transferred, in joules.
Calculator
- Power P = IV36 W
- Charge Q = It360 C
- Energy in kWh0.0012 kWh
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The three inputs multiply, so a 10% change in any one of them changes W by 10%. The ranking in the driver chart mostly reflects how wide each typical range is.
- Energy bills use kWh: 1 kWh = 3.6 MJ. A 1 kW load running for one hour uses one kWh.
- Dividing W by t gives power: power is the rate of energy transfer.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V Potential difference | 1 to 400 V | 360 | 144,000 | 143,600 | +10% | +10 % |
| t Time | 1 to 3,600 s | 36 | 129,600 | 129,600 | +10% | +10 % |
| I Current | 0.1 to 20 A | 144 | 28,800 | 28,660 | +10% | +10 % |
Sweep of Time
| Time (s) | Energy transferred (J) |
|---|---|
| 1 | 36 |
| 2.783 | 100.2 |
| 7.746 | 278.9 |
| 21.56 | 776.1 |
| 60 | 2,160 |
| 167 | 6,012 |
| 464.8 | 16,730 |
| 1,293 | 46,570 |
| 3,600 | 129,600 |
Sweep one variable
Variables and how each one contributes
Energy transferred
What the equation gives you. Everything else feeds it.
Current
Steady current through the component. Directly proportional: at a fixed voltage and time, twice the current transfers twice the energy.
Time
How long the current flows. Directly proportional: a device left on twice as long transfers twice the energy.
Potential difference
Voltage across the component, not necessarily the supply voltage. Directly proportional: each coulomb carries more energy across a larger potential difference.
Worked example with the values above
- Charge moved: Q = It = 3 A × 120 s = 360 C.
- Energy: W = QV = 360 C × 12 V = 4,320 J.
- In kilowatt-hours that is 0.0012 kWh (1 kWh = 3.6 MJ).
Assumptions and limits
- Constant current and voltage for the whole time.
- V is the voltage across the component you are interested in.
- Direct current. For AC use RMS values and include the power factor.
Common mistakes
- Entering time in minutes or hours. t must be in seconds to get joules.
- Confusing joules (energy) with watts (power).
- Using the supply voltage instead of the voltage across the component.
Statistics and uncertainty
The formula is deterministic. For independent measurement errors the relative errors of I, t and V combine in quadrature. If voltage and current vary during the interval, W is the integral of v(t)·i(t) dt, which differs from the product of the average values when the two vary together.
Where it is used
- Energy use of a device over a mission, and battery sizing.
- Heat dissipated in a resistor over time.
- Energy per pulse in pulsed loads.
Maths you need
- Energy and power: power is energy per unit time, P = W / t
- Units: J, Wh and kWh: 1 Wh = 3600 J