Electrical energy and power

Every plug you use draws electrical energy at a certain rate, and the electricity bill is nothing more than a record of that rate multiplied by time. A phone charger, a room heater and a bulb all do the same fundamental thing — push charge through a potential difference — but at wildly different rates, and that rate is what we call power. This article builds the power formulas $P = VI = I^2R = V^2/R$ from first principles starting at $W = QV$ , shows how they connect to the joule and the commercial "unit" of electricity — the kilowatt-hour — and applies them to appliance ratings and to resistors in series and parallel, where the very same formulas explain why identical resistors dissipate very different amounts of power depending on how they are wired.

The core insight

Electrical power is the rate of energy transfer as charge moves through a potential difference, $P = VI = I^2R = V^2/R$ , and integrating power over time gives energy, billed commercially in kilowatt-hours rather than joules.

The core idea

From work to power. When a charge $Q$ moves through a potential difference $V$ , the electric field does work $W = QV$ on it; this work is the electrical energy delivered to (or converted by) whatever the charge passes through. Since current is defined as $I = Q/t$ , we can write $Q = It$ , so over a time $t$ the total electrical energy transferred is

E = W = VIt

where: $E$ is energy (joules), $V$ is potential difference (volts), $I$ is current (amperes), and $t$ is time (seconds).

Power as a rate. Power is simply energy transferred per unit time, so dividing the expression above by $t$ gives

P = \frac{E}{t} = VI

Substituting Ohm's law, $V = IR$ , into $P = VI$ produces two more equivalent forms that are extremely convenient because they let you compute power from just two of the three quantities $V$ , $I$ , $R$ :

P = I^2R \qquad \text{and} \qquad P = \frac{V^2}{R}

where: $P$ is power (watts), and $R$ is resistance (ohms). All three forms describe the same physical power for an ohmic resistor — which one to reach for depends on what is held fixed in a given circuit: use $P = I^2R$ when several elements share a common current (as in series), and $P = V^2/R$ when they share a common voltage (as in parallel).

Units: joule versus kilowatt-hour. The SI unit of power is the watt, $1\ \text{W} = 1\ \text{J/s}$ , and the SI unit of energy is the joule. But a joule is a tiny amount of energy for household use, so electricity providers bill in a much larger practical unit, the kilowatt-hour (kWh), informally called a "unit" of electricity:

1\ \text{kWh} = (1000\ \text{W}) \times (3600\ \text{s}) = 3.6\times10^{6}\ \text{J}

This is the energy consumed by a $1\ \text{kW}$ appliance running for exactly one hour. A bill simply totals up $E = Pt$ for every appliance, with $P$ in kilowatts and $t$ in hours.

The heating effect and why it matters for transmission. Whenever current flows through a resistor, part or all of the electrical energy is irreversibly converted into heat, at rate $I^2R$ — the working principle of heaters, toasters and incandescent filaments. The very same $I^2R$ term also appears as an unwanted power loss in transmission wires. Because this loss depends on the square of the current, engineers transmit power at high voltage and correspondingly low current (since $P = VI$ is fixed by demand), which shrinks $I^2R$ losses in the wires dramatically — the reason cross-country power lines run at very high voltages rather than high currents.

Figure

Bar chart comparing power dissipated by two resistors of ten ohms and twenty ohms wired in series versus the same two resistors wired in parallel across a twelve volt supply

Derivation ADeriving the three power formulas from work done on charge

The goal is to start from the definition of work done moving a charge through a potential difference and, using only the definition of current and Ohm's law, arrive at all three standard forms of electrical power, $P = VI$ , $P = I^2R$ and $P = V^2/R$ , showing precisely why they are algebraically equivalent restatements of the same physical rate rather than three independent facts to memorise separately.

Step 1. Begin with the work done moving charge through a potential difference. When charge $Q$ moves through potential difference $V$ , the work done by (or against) the field is $W = QV$ . This work is exactly the electrical energy delivered to the circuit element, by definition of potential difference as energy per unit charge.

Step 2. Bring in the definition of current, $I = Q/t$ , so $Q = It$ . Substituting into Step 1 gives the energy transferred in time $t$ : $E = W = (It)V = VIt$ . This already shows energy growing linearly with voltage, current and time.

Step 3. Divide by time to get power, the rate of energy transfer: $P = E/t = VIt/t = VI$ . This is the most general form, valid for any circuit element, resistive or not.

Step 4. Restrict to an ohmic resistor, where Ohm's law $V = IR$ holds, and substitute it into $P = VI$ in two different ways. Replacing $V$ with $IR$ gives $P = (IR)I = I^2R$ . Replacing $I$ with $V/R$ gives $P = V(V/R) = V^2/R$ . Both are exact algebraic consequences of $P = VI$ combined with $V = IR$ , not separate physical laws — so any one of the three, given two known quantities among $V$ , $I$ , $R$ , determines power completely. Verified: for the heater example below, $P = VI = 220 \times 4.55 = 1001\ \text{W} \approx 1000\ \text{W}$ , and $P = V^2/R = 220^2/48.4 = 1000\ \text{W}$ , matching the rated value.

Worked example

An electric heater is rated "1000 W, 220 V." Find (a) the current it draws and its resistance, and (b) the energy it consumes, in kWh, if it runs 3 hours daily for 30 days.

Step 1. Use the rated power and voltage to find current from $P = VI$ : $I = P/V = 1000/220 = 4.55\ \text{A}$ .

Step 2. Find resistance from $P = V^2/R$ , rearranged as $R = V^2/P$ : $R = (220)^2/1000 = 48400/1000 = 48.4\ \Omega$ .

Step 3. Find the total running time: $t = 3\ \text{h/day} \times 30\ \text{days} = 90\ \text{h}$ .

Step 4. Compute energy directly in kWh using $E = Pt$ with $P$ in kilowatts and $t$ in hours: $E = 1.0\ \text{kW} \times 90\ \text{h} = 90\ \text{kWh}$ . The heater draws $4.55\ \text{A}$ , has resistance $48.4\ \Omega$ , and consumes $90$ units of electricity over the month.

Competitive stretch

Two resistors, $R_1 = 10\ \Omega$ and $R_2 = 20\ \Omega$ , are connected first in series and then in parallel across the same $12\ \text{V}$ battery. Compare total power dissipated.

Step 1. Series: total resistance $R_s = R_1 + R_2 = 30\ \Omega$ , so current $I = V/R_s = 12/30 = 0.4\ \text{A}$ , and $P_s = I^2R_s = (0.4)^2 \times 30 = 4.8\ \text{W}$ (check: $V^2/R_s = 144/30 = 4.8\ \text{W}$ ✓).

Step 2. Parallel: $1/R_p = 1/10 + 1/20 = 3/20$ , so $R_p = 20/3 = 6.67\ \Omega$ , and $P_p = V^2/R_p = 144/6.67 = 21.6\ \text{W}$ . The parallel combination dissipates about $4.5$ times more power than the series combination for the same $12\ \text{V}$ supply, because its lower total resistance draws far more current from the source.

Common traps

Mixing up $P = I^2R$ and $P = V^2/R$ . Use $I^2R$ when elements share the same current (series); use $V^2/R$ when they share the same voltage (parallel). Applying the wrong one to the wrong resistor gives a wrong number even though the formula itself is correct.
Confusing power and energy. A "1000 W" rating is a rate, not a total; you cannot say how much energy was used without also knowing the time.
Inconsistent time units. Mixing minutes, hours and seconds when converting between joules and kWh is one of the most common arithmetic slips in billing-style problems.
Assuming a higher wattage always means "more useful output." Rated power tells you the rate of energy conversion, not how much of it becomes useful work versus wasted heat.

Ask the tutor: Want to work through a trickier combination-circuit power problem or check your own electricity bill's kWh math step by step?

Why are P = VI, P = I^2R and P = V^2/R all called "the" power formula if they look so different?

They are not three different laws, just the same law $P = VI$ rewritten using Ohm's law, $V = IR$ . Substituting $V = IR$ into $P = VI$ gives $P = I^2R$ , and substituting $I = V/R$ gives $P = V^2/R$ . All three give identical numerical answers for the same resistor and the same operating point — they simply let you compute power from whichever two of $V$ , $I$ , $R$ you happen to know, without first solving for the third.

Why does the electricity bill use kilowatt-hours instead of joules?

A joule is far too small a unit for everyday household consumption — a $1\ \text{kW}$ heater running for just one second uses $1000$ joules, so a month's usage would run into tens of millions of joules, an unwieldy number to print on a bill. The kilowatt-hour packages a full hour of a $1\ \text{kW}$ appliance's use, $3.6\times10^{6}\ \text{J}$ , into one convenient "unit," which is why meters and bills are read directly in kWh rather than joules.

Why does high-voltage transmission reduce power loss in wires if P = VI stays the same?

The power actually delivered to consumers, $P = VI$ , is indeed fixed by demand, but the power lost in the transmission wires themselves follows $P_{\text{loss}} = I^2R_{\text{wire}}$ , which depends on current squared, not on $V$ directly. For a fixed delivered power $P$ , raising the transmission voltage $V$ lets you lower the current $I = P/V$ proportionally, and since loss scales as $I^2$ , even a modest drop in current produces a much bigger drop in wasted heat in the wires — which is exactly why grids step voltage up for long-distance lines and step it back down before it reaches homes.

Electrical energy and power

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads