What electrical power measures

After thiswhat you will be able to doCalculate a component's power and transferred energy from its voltage, current, and elapsed time, and track where that energy goes.

Questionwhat this lesson answersVoltage tells us energy per charge and current tells us charge per time. What rate of energy transfer appears when both act together, and how does it become heat or stored energy?

Not coveredwhat this lesson leaves outWe calculate power and energy for ideal direct-current components and resistive heating. We do not model battery chemistry, alternating-current power factor, or thermal design.

Voltage and current describe two different ratios:

V=ΔEq,I=ΔqΔt.V = \frac{\Delta E}{q}, \qquad I = \frac{\Delta q}{\Delta t}.

Multiply them and the charge cancels:

VI=ΔEqΔqΔt=ΔEΔt.VI = \frac{\Delta E}{q}\frac{\Delta q}{\Delta t} = \frac{\Delta E}{\Delta t}.

That rate of energy transfer is power:

P=VI.P = VI.

One watt is one joule transferred each second:

1W=1J/1s.\SI{1}{W} = \SI{1}{J}/\SI{1}{s}.
Step 1 of 3

Multiply the two rates

Voltage supplies joules per coulomb. Current supplies coulombs per second. Their product is joules per second, or power.

Energy per time

Power is the speed of energy transfer

Change the voltage, current, or time. The stream shows a rate, while the ledger records the amount accumulated over the chosen interval.

+-9 Vloadsourceenergy changes form here0.18 W = 9 V x 0.02 A
Voltage
9 V
Current
20 mA
Energy
10.8 J in 60 s
Ideal source and load account
QuantityInOutBalance
Source transfer1.08 × 101 J0 J1.08 × 101 J
Load received0 J1.08 × 101 J1.08 × 101 J
Total1.08 × 101 J1.08 × 101 JBalanced 0 J

0.18 W runs for 60 s, so 10.8 J is transferred. The ideal account balances because the source and load exchange the same energy.

The formula is not a memorised multiplication with no story underneath. A voltage says how much energy each coulomb transfers; a current says how many coulombs pass each second. Together they say how many joules transfer each second.

Energy is power over time

If a component draws a steady power, the energy transferred over a time interval is

ΔE=PΔt.\Delta E = P\Delta t.

A 9V{\SI{9}{V}} source driving 20mA{\SI{20}{mA}} through a resistor transfers energy at

P=9V×0.020A=0.18W.P = \SI{9}{V}\times\SI{0.020}{A} = \SI{0.18}{W}.

After 60s{\SI{60}{s}}, the transferred energy is

ΔE=0.18W×60s=10.8J.\Delta E = \SI{0.18}{W}\times\SI{60}{s} = \SI{10.8}{J}.

Power is the rate; energy is the accumulated amount. A small power left on for a long time can transfer more energy than a large power used briefly. Confusing the two is the electrical version of confusing speed with distance.

Three equivalent resistor formulas

For a resistor, V=IR{V = IR} lets us rewrite P=VI{P = VI} in two other forms:

P=VI=I2R=V2R.P = VI = I^2R = \frac{V^2}{R}.

Choose the form that uses the values already known. If a resistor carries 20mA{\SI{20}{mA}} and has resistance 450Ω{\SI{450}{\Omega}}, then

P=I2R=(0.020A)2×450Ω=0.18W.P = I^2R = (\SI{0.020}{A})^2\times\SI{450}{\Omega} = \SI{0.18}{W}.

The same answer comes from V=IR=9V{V = IR = \SI{9}{V}} and P=V2/R{P = V^2/R}. The equality is a consistency check: changing the algebraic form does not change the physical transfer rate.

In an ordinary resistor, the energy leaves the organised electrical description and becomes heat. The moving charge is not used up, and the current does not disappear. Charge exits the resistor at the same rate it entered; the energy per charge is lower because the resistor has transferred energy to its material and surroundings. A lamp can transfer some of that energy into light, a motor into motion, and a heater mostly into thermal motion.

The source and load share the account

An ideal source supplies the same power that the rest of the circuit receives. If a circuit draws 0.18W{\SI{0.18}{W}}, the source transfers 0.18J{\SI{0.18}{J}} each second to its components. A real battery also transfers some energy inside its own internal resistance, so its chemical energy decreases faster than the useful load power alone suggests.

This accounting viewpoint is useful because it ties the four central quantities together:

The final lesson applies those definitions to instruments. A meter does not merely read a number from outside the circuit. It has to join the circuit in a particular way, and that connection can change the voltage, current, and power being measured.

Doorswhat to read next, and why

Symbolswhat each one means, and whether we defined it, measured it, or just started there

PStatus: defined
power, the rate at which energy is transferred
EStatus: defined
energy transferred to or from a component
tStatus: defined
the time interval over which energy transfer is counted
VStatus: defined
energy transferred per unit charge across the component
IStatus: defined
charge transferred per unit time through the component
heatStatus: empirical
energy transferred into disordered microscopic motion in a material
What these classifications mean
defined
circular by construction, true because we chose it
empirical
a measured claim about the world that could have come out otherwise
bottoms out
a primitive of the model, with nothing under it here
door
used here, explained elsewhere