Calculate electrical power (P = VI, P = I²R): Visual walkthrough

Step 1 — What is charge, and what is a "drop"?

WHAT. Picture a bucket of tiny beads. Each bead is a chunk of electric charge. We measure a pile of charge in coulombs (symbol $Q$ , unit C). One coulomb is just "this many beads."

WHY start here. Every energy idea in a circuit is really "beads falling down a hill." Before we talk about speed or power, we need the two actors: the beads (charge) and the hill (the voltage drop). No hill, no fall, no energy.

PICTURE. The hill on the left is the voltage drop; the pile of beads at the top is the charge $Q$ waiting to slide down.

Step 2 — Voltage = energy handed to each bead

WHAT. Voltage $V$ is the height of the hill: how much energy one coulomb gains by falling down it. Its unit, the volt, is defined as joules-per-coulomb: $V = energy per one coulomb \frac{E}{Q}$

Here $E$ is total energy in joules (J), $Q$ is the charge in coulombs.

WHY this definition. We could describe the hill by "how steep" or "how long," but what a circuit cares about is: how much energy does each unit of charge pick up? That is exactly $E / Q$ . A taller hill (bigger $V$ ) means each bead arrives at the bottom carrying more energy.

PICTURE. Two hills side by side — a short one (small $V$ ) and a tall one (big $V$ ). The same bead drops further on the tall hill, so it delivers more energy. See Voltage and Potential Difference.

Step 3 — Current = how many beads fall per second

WHAT. Current $I$ is the flow rate of charge: how much charge passes a point each second. Its unit, the amp, is coulombs-per-second: $I = charge per second \frac{Q}{t}$

$t$ is time in seconds (s).

WHY this definition. Energy alone isn't power. A slow trickle of beads and a roaring flood might drop the same beads down the same hill — but one does it slowly and one does it fast. To capture "fast," we need a rate, and $Q / t$ is that rate. See Electric Current.

PICTURE. A pipe with beads streaming past a checkpoint. Count how many cross the dashed line in one second — that count is $I$ .

Step 4 — What is power? A rate of energy

WHAT. Power $P$ is energy delivered per second: $P = joules per second \frac{E}{t}$

Its unit is the watt (W), and $1 W = 1 J/s$ .

WHY. Component ratings, heat, battery drain — they all ask "how fast is energy moving right now?" not "how much total." That's a rate, so we divide energy by time. See Energy and the Joule.

PICTURE. Two devices delivering the same 100 J: one over 10 s (low power), one over 1 s (high power). Same energy, very different power.

Step 5 — Snap the pieces together: $P = V I$

WHAT. We now have three facts, each a little Lego brick:

From Step 2: $E = V Q$
From Step 3: $Q = I t$
From Step 4: $P = \frac{E}{t}$

Substitute in order. First put $E = V Q$ into $P = E / t$ : $P = \frac{E}{t} = \frac{V Q}{t}$ Now put $Q = I t$ in place of $Q$ : $P = \frac{V ( I t )}{t} = V I \frac{t}{t} = V I$

WHY the $t$ cancels. The $t$ in "beads per second" and the $t$ in "energy per second" are the same clock. When both appear top and bottom, they divide out — leaving a formula with no time in it at all. Power depends only on the hill height $V$ and the flow rate $I$ .

PICTURE. The three bricks flowing into one box: the $t$ 's meet and cancel (struck out in red), leaving the clean result.

Step 6 — Feed in Ohm's law to get the other two forms

WHAT. Ohm's Law (V = IR) says the drop across a resistor is $V = I R$ , where $R$ is resistance in ohms ( $Ω$ ). Slot it into $P = V I$ two different ways:

Replace $V$ with $I R$ : $P = V I = (I R) I = I^{2} R$ Replace $I$ with $V / R$ (since $I = V / R$ ): $P = V I = V \cdot \frac{V}{R} = \frac{V ^{2}}{R}$

WHY. Sometimes you don't know both $V$ and $I$ . Ohm's law lets you rewrite the same power using whichever pair you actually measured. All three describe the identical wattage.

PICTURE. One triangle ( $V$ , $I$ , $R$ ) with three doors out — each door drops you into one of the three power formulas.

Step 7 — The degenerate cases (never let the reader get stuck)

WHAT. Check the extremes so nothing surprises you:

PICTURE. Four dials, each pinned to an edge case, with the resulting power labelled.

No current ( $I = 0$ ): an open switch, a broken wire. $P = V \times 0 = 0$ . There is a hill (voltage present), but no beads fall — so zero power. Voltage alone burns nothing.
No voltage ( $V = 0$ ): two points at the same height, a dead short with no drop. $P = 0 \times I = 0$ . Beads may drift, but they fall nowhere — zero energy delivered.
Zero resistance ( $R \to 0$ ): using $P = V^{2} / R$ , dividing by a shrinking $R$ makes $P \to \infty$ . A real short circuit tries to dump huge power — this is why fuses exist. See Resistors and Power Ratings.
Huge resistance ( $R \to \infty$ ): an insulator. $I = V / R \to 0$ , so $P = I^{2} R \to 0$ . Almost no current flows, almost no power. (Careful: $V^{2} / R \to 0$ too — same answer, no contradiction.)

The one-picture summary

Everything on one canvas: charge and hill (Steps 1–2) → flow rate (Step 3) → rate of energy (Step 4) → the $t$ 's cancel into $P = V I$ (Step 5) → Ohm's law fans it into the trio (Step 6).

Recall Feynman retelling of the whole walkthrough

Imagine beads at the top of a hill. How tall the hill is = voltage $V$ (energy each bead gets). How many beads slide down per second = current $I$ . Multiply "energy per bead" by "beads per second" and the per-bead and per-second stitch together into energy per second = power. That's $P = V I$ — and notice the clock cancelled, so power doesn't care how long you wait, only how tall the hill and how fast the flood. If the hill is rough (resistance $R$ ), Ohm's law ( $V = I R$ ) lets you rewrite the same power as $I^{2} R$ or $V^{2} / R$ depending on what you measured. The square in $I^{2} R$ warns you: crank the current and the heat rockets up fourfold. Turn off the flow and power vanishes; short the hill flat and it vanishes too. Same one idea, drawn every way.

Recall Quick self-check

Why does the $t$ cancel in the $P = V I$ derivation? ::: Because $Q = I t$ carries a $t$ on top, and $P = E / t$ has a $t$ on the bottom — they are the same clock, so they divide out, leaving no time in the formula. With $I = 0$ but $V = 5 V$ , what is the power? ::: $P = V I = 5 \times 0 = 0 W$ — voltage alone delivers nothing. Doubling the current in a fixed resistor multiplies power by how much? ::: By 4, because $P = I^{2} R$ and $2^{2} = 4$ .

Connections

Parent topic — full note
Voltage and Potential Difference — the "hill height," source of $E = V Q$ .
Electric Current — the "flow rate," source of $Q = I t$ .
Energy and the Joule — what $E$ and the watt (J/s) actually measure.
Ohm's Law (V = IR) — the bridge to $I^{2} R$ and $V^{2} / R$ .
Resistors and Power Ratings — why the $R \to 0$ short-circuit case matters.
Heat Dissipation and Cooling — the real-world cost of the $I^{2} R$ square.