Visual walkthrough — Define energy (joules) vs power (watts)

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Step 1 — Two different questions: "how much" vs "how fast"

WHAT. Before any formula, notice there are two separate things we could ask about pouring water into a bucket: the total we poured and how quickly it flowed.

WHY. If these two were the same question, we would only need one word. They are not — so nature forces us to keep two ideas apart. We call the total energy and the rate power.

PICTURE. Two identical buckets end up equally full (same total), but one filled from a trickle over a long time and one from a firehose in a flash.

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Step 2 — Measuring "the total": the joule

WHAT. We need a ruler for energy. The chosen unit is the joule, written $\text{J}$.

WHY. To add up energy we must agree on one standard lump of it. Physics defines one joule as the work done pushing with a force of $1$ newton ($1\text{N}$) through a distance of $1$ metre ($1\text{m}$): $\underset{\text{one lump of energy}}{\underbrace{1\text{J}}}=\underset{\text{how hard you push}}{\underbrace{1\text{N}}}\times \underset{\text{how far it moves}}{\underbrace{1\text{m}}}$

PICTURE. A hand pushing a block one metre along the floor; the shaded "work" is the accent-red bar $1\text{N}\times 1\text{m}$.

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Step 3 — Measuring "how fast": rate = amount ÷ time

WHAT. A rate is always "amount per unit of time". Power is the rate of energy delivery, so: $P=\frac{\Delta E}{\Delta t}$

WHY. The triangle $\Delta$ (Greek "delta") just means "the change in" — how much of something happened. So $\Delta E$ = energy delivered, $\Delta t$ = time it took. Dividing gives "energy per unit time", which is exactly what "how fast" means. We divide (not multiply) because a bigger time for the same energy means a slower flow — the answer must shrink when $\Delta t$ grows, and division does that.

PICTURE. Two bars of equal energy $\Delta E$; the short-time bar makes a tall power spike, the long-time bar makes a low one — same numerator, different slope.

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Step 4 — From average to instant (why a slope, why the limit)

WHAT. $\frac{\Delta E}{\Delta t}$ is only the average rate over a stretch. To get the power at one exact moment we shrink the time window to nothing: $P={\lim }_{\Delta t\to 0}\frac{\Delta E}{\Delta t}=\frac{dE}{dt}$

WHY THIS TOOL. We use the derivative $\frac{dE}{dt}$ — not just plain division — because energy might flow unevenly (a flash starts slow, peaks, ends). "Average over a second" would hide the peak. The derivative answers a sharper question: "exactly how fast right now?" — the slope of the energy-vs-time curve at a single point.

PICTURE. A curved energy-vs-time graph. A long chord (average slope) is redrawn shorter and shorter until it becomes the red tangent line — the instantaneous power.

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Step 5 — Reverse it: energy is the area under power

WHAT. If power is the slope of energy, then energy is the accumulation of power. We add up every little "joules-per-second × tiny-second" slice: $E=\int Pdt\stackrel{P\text{ constant}}{\to }E=Pt$

WHY. The integral sign $\int$ means "sum up all the thin slices". Each slice is $P\times dt$ = (rate) × (tiny time) = a tiny lump of energy. Sum them and you recover the total. If $P$ never changes, all the slices stack into one clean rectangle, and a rectangle's area is just height × width: $E=\underset{\text{height = rate}}{\underbrace{P}}\times \underset{\text{width = duration}}{\underbrace{t}}$

PICTURE. A flat power line at height $P$; the red rectangle beneath it up to time $t$ is the energy, labelled "area = $P\times t$".

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Step 6 — Edge cases: what the picture says at the extremes

WHAT. Test the rectangle at its limits so no scenario surprises us.

WHY. A formula you trust must survive its degenerate inputs. We check three: zero time, zero power, and "same energy, opposite powers".

PICTURE. Three mini-rectangles: a zero-width sliver, a zero-height sliver, and two equal-area rectangles of very different shape.

• Zero time ($t\to 0$): width $0$ → area $0$ → $E=0$. No time, no energy delivered — even at huge power. A firehose that's on for zero seconds fills nothing.
• Zero power ($P=0$): height $0$ → area $0$ → $E=0$, for any time. The tap is off; wait forever, bucket stays empty.
• Same energy, different power: a tall-thin rectangle (high $P$, small $t$) and a short-wide one (low $P$, big $t$) can have the same area. This is the flash-vs-charger fact: equal energy, wildly different power. So a $40\text{W}$ bulb left on for hours can beat a $100\text{W}$ bulb on for a minute — see Kilowatt-hour and electricity billing.

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Step 7 — Plugging in electricity: deriving $P=IV$

WHAT. In a wire, energy comes from pushing charge through a voltage. We turn the mechanical picture into an electrical one and land on $P=IV$.

WHY. We need the electrical building blocks first:

• Charge $Q$, measured in coulombs — see Electric charge and the coulomb.
• Voltage $V$ = energy given to each coulomb of charge, i.e. $V=E/Q$, so $E=QV$. That is the very definition of voltage.
• Current $I=Q/t$ = charge flowing per second — the definition of current.

Now substitute step by step into $P=E/t$: $P=\frac{E}{t}=\frac{QV}{t}=\left(\underset{=I}{\underbrace{\frac{Q}{t}}}\right)V=IV$

Reading it left to right: energy per second $=$ (charge × voltage) per second $=$ (charge per second) × voltage $=$ current × voltage.

PICTURE. Charges (dots) marching through a voltage "hill" $V$; each dot carries energy $QV$; the flow rate $I=Q/t$ turns that into power $P=IV$.

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The one-picture summary

Everything above lives in one diagram: the energy-vs-time curve, whose slope is power ($P=dE/dt$) and whose area is energy ($E=\int Pdt$), with the constant-power case giving the clean rectangle $E=Pt$, and the electrical substitution turning $P$ into $IV$.

Recall Feynman: the whole walkthrough in plain words

Picture filling a bucket. The water in the bucket is energy (joules) — how much you have. How fast the tap runs is power (watts) — joules each second. If you draw a graph of total water going up as time goes across, the steepness of that line is the power (fast tap = steep line), and that steepness is what "derivative" means. Going the other way, the area underneath a steady tap-rate is just rate × time — a rectangle — and that area is the total water: $E=Pt$. Test the edges: zero time or zero rate means an empty bucket; a fast tap for a moment and a slow tap for ages can pour the same total. Finally, in a wire the "water" is charge: voltage says how much energy each drop carries ($E=QV$) and current says how many drops flow per second ($I=Q/t$), so the delivery rate is $P=IV$. One curve, one slope, one area — that's the entire idea.

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Connections

• Define energy (joules) vs power (watts)
• Voltage as energy per charge (V = J/C)
• Current as charge per second (I = Q/t)
• Ohm's Law and P = I²R = V²/R
• Electric charge and the coulomb
• Kilowatt-hour and electricity billing
• Work-energy theorem (mechanics)