Power: How Components Do Work

So far we've talked about voltage (the push), current (the flow), and resistance (the friction). Now comes the question that connects everything to what circuits are actually for: how much work is being done?

The answer is power — the rate at which energy is being delivered, converted, or consumed. Measured in watts (W), where one watt is one joule of energy per second.

Power is the universal "what's happening here?" measurement for any electrical component:

• A resistor turning electrical energy into heat at 5 watts is heating up like a 5 W heater.
• An LED producing light at 0.04 watts is working at the same rate as a small flashlight bulb.
• A motor producing motion at 100 watts is delivering enough mechanical work to lift a 10 kg weight one meter every second.

Different components convert power to different kinds of energy, but the watt measures the rate equally well for all of them. This makes power the most useful number for comparing what very different components are doing in the same circuit.

The next section ties power to the V and I you already know. We begin with steady DC circuits and the formula P = V × I; AC needs one more distinction.

For the steady DC circuits modeled by the slider below, electrical power is:

$$P = V \times I$$

Here V is the voltage across the component and I is the current through it. Their product is the rate of energy transfer, measured in watts. For a resistor, that electrical energy becomes heat; an LED also produces light.

Principle — measure voltage across, current through. Voltage is a difference between two points; current passes through a component. In this steady DC model, multiplying them gives the component's power. A component can also return energy to a source, so the direction of energy transfer matters.

The units show why the relationship works: one volt is one joule per coulomb, and one ampere is one coulomb per second. Therefore:

$$\frac{\text{joules}}{\text{coulomb}} \times \frac{\text{coulombs}}{\text{second}} = \frac{\text{joules}}{\text{second}} = \text{watts}$$

For changing voltage and current, the more general statement is instantaneous power: $p(t) = v(t)i(t)$. With AC, multiplying the RMS voltage and RMS current gives apparent power in volt-amperes (VA), which need not equal the average real power in watts. Real power is $P = V_{\mathrm{rms}} I_{\mathrm{rms}} \times \text{power factor}$; for sinusoidal voltage and current, the power factor is $\cos\phi$, where $\phi$ is their phase difference. We will use that distinction in the AC lesson.^[1]

Try the slider: within its independent, steady DC input model, doubling either V or I doubles P, and doubling both quadruples P. This slider does not predict whether contact with a real circuit is safe. Shock severity depends on the current path through the body and the duration of contact, among other conditions; a person can become part of a current path when touching energized parts. Treat potentially energized conductors as hazardous and use the applicable safe work procedures.^[2]

Power isn't consumed in the everyday sense — energy is conserved, always. What's actually happening is that each component converts the electrical power flowing through it into some other form of energy. The form depends on what the component is built to do:

Whatever the component, the watt is the same. A 60 W incandescent bulb dissipates the same amount of power as a 60 W heater — the bulb just happens to convert some of that power to a few percent of light along with the heat. (LEDs are much more efficient: a modern LED bulb produces the same light output as a 60 W incandescent for around 8 W of input power.)

This is why "how many watts" is the universal first question when sizing wires, fuses, breakers, supplies, or batteries: the watt tells you how much energy per second has to be delivered, dissipated, or stored, regardless of what the component does with it.

Principle — energy is always conserved. Power flowing INTO a component must equal power flowing OUT of it (as heat, light, motion, sound, etc.) plus power being stored (in a battery's chemistry, in a capacitor's electric field, in an inductor's magnetic field). For a plain resistor, "out" is essentially all heat; for an LED, it's mostly light; for a motor, mostly motion. But none of it disappears — it always shows up somewhere.

Power is the rate — energy per second. Energy is the amount — power × time.

$$\text{Energy} = \text{Power} \times \text{Time}$$

Same idea as distance vs. speed: 60 km/h is a rate; 60 km is the amount you've covered after one hour at that rate. A 60 W bulb is a rate; 60 watt-hours (Wh) is the energy it consumes after one hour of being on.

This is the unit your electricity bill is in, just bigger: kilowatt-hours (kWh) = 1000 Wh. A house running an average of 500 W around the clock for 30 days uses 0.5 kW × 24 h × 30 = 360 kWh — typical of a small Japanese household per month.

It's also the unit hidden inside battery capacity ratings, though batteries are usually labeled in milliamp-hours (mAh) at a specific voltage. To get the actual energy in watt-hours, multiply the mAh by the battery's voltage and divide by 1000:

$$\text{Wh} = \frac{\text{mAh} \times \text{V}}{1000}$$

A few examples to anchor:

So if you want to know "how long will this run my LED?" — divide energy by power. A 15 Wh phone battery powering a 1 W LED would last 15 / 1 = 15 hours (in theory; real efficiency losses make it less).

This is also why a "fast charger" matters: charging is just delivering energy, and a 30 W charger fills a 15 Wh battery in 15 / 30 = 0.5 h = 30 min (in theory).

Remember the LED-without-a-resistor question from earlier in the course? It now has its proper answer in P = V × I.

Here's the chain again with the power numbers filled in:

The LED isn't "killed by too much voltage" in any direct sense — it's killed by too much V × I. Power is what physically heats the semiconductor junction; once the junction temperature exceeds its limit, the semiconductor crystal damages permanently and the LED stops working (or stays stuck in some weird half-broken state).

This generalises. Almost every "what burns out?" question in electronics is really a power question. A wire melts when I²R (the wire's own dissipation) exceeds the rate it can shed heat to its surroundings. A resistor "burns out" when its V·I exceeds its watt rating (¼ W is typical for through-hole resistors; ⅛ W for many surface-mount). A power supply trips its overcurrent protection when V_out × I_out passes a threshold designed to keep the internal pass transistor or transformer safe. A breaker trips when I × the line voltage approaches the rate at which the wire would start a fire.

Working principle for every component you'll ever specify: check the watts. The component's datasheet has a maximum power rating; your job is to keep V × I below that rating in every operating condition. Voltage limits and current limits are usually just convenient simplifications of the underlying power limit.

Some power values to anchor your intuition for the rest of the course:

For a simplified calculation, treat a 1500 W kettle's heating element as a resistive load at 100 V, so its power factor is approximately one. Its nominal current is then 1500 / 100 = 15 A. If other loads share a branch circuit, their currents add. Actual current, protective-device behavior, and permissible loading depend on the equipment and installation; this estimate does not establish that a particular outlet or circuit can safely supply the kettle.^[1]

For 第二種電気工事士 study, start by identifying the type of load and the assumptions in the question. For a resistive load at its rated voltage, I ≈ P / V is a useful estimate; other AC loads may have a power factor below one. Breaker, conductor, and receptacle selection requires the applicable installation rules and equipment ratings, not this estimate alone.