Define resistance and the ohm: Visual walkthrough — Hardware deep dive

This is the picture-first companion to the parent note. If any word below is new, we build it here — we assume nothing.

Step 1 — Draw the flow of charge (this is current)

WHAT. Picture a wire as a hallway. Little packets of electric charge march down it. If we stand at one spot and count how much charge walks past us each second, that count is the current.

WHY. Before we can talk about "opposing the flow," we must first agree on what "the flow" even is. Flow means amount per second — a rate. That is exactly what current measures, so it is our starting brick.

PICTURE. Below, the blue dots are charge. The dashed line is our counting gate. In one second, some number of dots cross it.

Why divide by $t$ and not something else? Because "flow" is inherently per second. A garden hose that delivers 10 litres is meaningless until you say "per minute." Same here: charge only becomes current once we pin it to time.

Step 2 — Draw the push (this is voltage)

WHAT. The dots do not move on their own — something has to shove them. That shove is the voltage. But voltage is not "force on one dot"; it is energy handed to each coulomb of charge to get it through.

WHY. We need a number for "how hard we are pushing" so we can later compare push to flow. Measuring energy per coulomb (not total energy) means the push is a property of the source, not of how much charge happens to be around.

PICTURE. Think of a hill. A coulomb of charge sits at the top with energy; rolling down, it spends that energy passing through the component. The height of the hill is the voltage.

See Voltage and Potential Difference for the full story of the "hill." For us it is simply the push in joules per coulomb.

Step 3 — Do the experiment: push harder, watch the flow

WHAT. Take one fixed metal wire, kept at the same temperature. Set a small push and read the flow. Double the push, read again. Triple it, read again. Write each pair $(V, I)$ on the board.

WHY. We are not allowed to assume how push relates to flow — we must look. Nature has to tell us the pattern; our job is only to notice it.

PICTURE. Each experiment is one point. Plot push $V$ sideways, flow $I$ upward. The points do not scatter randomly — they fall on a straight line that passes through the corner where both are zero.

Two things the picture is screaming at us:

The line goes through the origin: zero push gives zero flow. (No shove, no march — obvious, but the graph confirms it.)
The line is straight: every time we add the same push, we get the same extra flow.

This observed straightness is Ohm's Law. It is a fact about this material, not a law of the universe — some components curve. We handle those in Step 7.

Step 4 — Turn "straight line through origin" into an equation

WHAT. A straight line that passes through the origin has one, and only one, freedom: its steepness. So the whole relationship is captured by a single number — the slope.

WHY. Any straight line through the origin obeys $y = k x$ : output equals a constant times input. Here the input is push $V$ and the output is flow $I$ . Matching them:

$flow (up) I = steepness k push (across) V$

PICTURE. The slope $k$ is "rise over run" — how much extra flow you get per extra volt. A steep line: a little more push buys a lot more flow. A shallow line: even big pushes barely raise the flow.

Step 5 — Flip the slope and name it: this is resistance

WHAT. The slope $k = I / V$ measures how easily charge flows. But we usually care about the opposite question — how strongly the material fights back. So we flip it over and give the flipped number a name: resistance $R$ .

WHY. "Opposition" should get bigger when flow gets harder. But the slope gets smaller when flow gets harder (a flat line = choked flow). Flipping fixes the direction: $R = 1/ k$ grows exactly when the material resists more.

$R = \frac{1}{k} = \frac{1}{I / V} = push per unit of flow \frac{V}{I}$

PICTURE. Same line, two readings: the slope $I / V$ is the "ease" arrow (chalk blue), and its reciprocal $V / I$ is the "resistance" we read off. Steep line ⇒ small $R$ ; flat line ⇒ big $R$ .

Step 6 — Where does the unit $Ω$ come from?

WHAT. The ohm was not decreed; it falls out of $R = V / I$ once you carry the units along.

WHY. A derived quantity inherits its unit from the quantities that built it. We already know $V$ is volts and $I$ is amperes, so:

$[R] = \frac{[ V ]}{[ I ]} = \frac{volt}{ampere} = ohm = Ω$

PICTURE. A "unit factory": volts go in the top, amperes come in the side, an ohm rolls out. And if we substitute the base units from Steps 1–2:

$1 Ω = \frac{1 V}{1 A} = \frac{1 J/C}{1 C/s} = 1 \frac{J \cdot s}{C ^{2}}$

Step 7 — The edge cases (never leave the reader stranded)

Real components are not always the tidy straight line of Step 3. We must show every scenario.

Case A — Zero push ( $V = 0$ ). No shove, no flow: $I = 0$ . On the graph this is the origin. Resistance is still defined by the slope of the line, so $R$ keeps its value even here — a resistor sitting unpowered still has its resistance.

Case B — Zero flow with nonzero push ( $I = 0, V \neq = 0$ ). You push but nothing gets through. Then $R = V /0 \to \infty$ : infinite resistance. This is a perfect insulator or a broken (open) wire. See Conductors and Insulators.

Case C — Huge flow from tiny push ( $V \to 0, I$ large). $R = V / I \to 0$ : a perfect conductor, offering essentially no opposition.

Case D — A curved $I$ – $V$ line (non-ohmic). A lamp filament heats up, gets harder to push through, and the line bends. Then there is no single slope. We must read $R = V / I$ at one chosen point — and that $R$ now changes with the operating point. The straight-line law of Step 3 was a special (though very common) case.

PICTURE. All four cases on one board: the vertical "insulator" line, the horizontal "conductor" line, the tidy ohmic diagonal, and the bending filament curve.

The one-picture summary

Everything above, compressed: charge flow gives $I$ (Step 1), the push gives $V$ (Step 2), experiment shows a straight line (Step 3), its slope becomes an equation (Step 4), flipping the slope names resistance (Step 5), and the unit falls out as the ohm (Step 6).

Recall Feynman retelling — say the whole walk in plain words

Imagine charge marching down a hallway. Count how many march past you each second — that's current. Now measure how hard you're shoving them, in energy handed to each coulomb — that's voltage. Do an experiment: shove twice as hard, twice as many march past. Plot shove sideways and flow upward, and the dots line up dead straight through the corner. A straight line through the corner is just "flow = some fixed number times shove." That fixed number tells you how easily charge flows — so flip it over to describe how hard the hallway fights back, and call that resistance, $R = V / I$ . Carry the units along and out pops a brand-new unit, the ohm: one ohm is when one volt of shove buys exactly one ampere of march. And at the extremes — an unpluggable wall is infinite ohms, a perfect wire is zero ohms, and a lamp that heats up bends the line so its resistance changes as it warms.

Recall Quick self-check

What does the slope of the $I$ – $V$ line equal? ::: $1/ R$ (steeper = lower resistance). Why do we flip the slope to define resistance? ::: So the number grows when flow gets harder; the raw slope shrinks when flow gets harder. What is $R$ for a perfect insulator? ::: Infinite ( $I = 0$ with $V \neq = 0$ ). A 12 V push gives 4 A. Read $R$ off the definition. ::: $R = 12/4 = 3 Ω$ .

Connections

Define resistance and the ohm — the parent note this walkthrough illustrates.
Electric Current — the $I$ built in Step 1.
Voltage and Potential Difference — the $V$ built in Step 2.
Ohm's Law — the straight line observed in Step 3.
Conductors and Insulators — the zero- and infinite- $R$ edge cases.
Resistivity and Resistance of a Wire — what sets the slope in the first place.
Power Dissipation in Resistors — the next step: energy lost across $R$ .