Visual walkthrough — State and apply Ohm's Law (V = IR)

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Step 1 — Draw the thing we are pushing charge through

WHAT. Start with the simplest possible object: a single piece of wire (or a resistor) with two ends. We connect a battery to it. Nothing more.

WHY. Every idea in this topic lives inside one object. Before we name any quantity, we must have a picture to pin those names onto — otherwise the letters $V$, $I$, $R$ float free and mean nothing.

PICTURE. In Figure s01 the amber block is our component. The two white terminals are labelled A (higher) and B (lower). The cyan arrow shows the direction we choose to call positive for charge flow: from A, through the block, to B. Look at that arrow — every sign in this page is measured against it.

Figure s01 — our single block with end A (higher) and end B (lower); the cyan arrow is the agreed positive direction of current, the amber bracket marks the push measured across A-to-B.

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Step 2 — What "push" ($V$) really counts

WHAT. We define potential difference $V$: the energy given to each unit of charge as it drops from A down to B.

WHY. We need a number for "push", not a feeling. The honest way to measure a push is: how much energy does each passing charge gain from it? Big energy-per-charge = big push.

PICTURE. In Figure s02, think of a hill. A is the top, B is the bottom. $V$ is the height of the hill — how far each charge falls. Watch the cyan double-headed arrow: it measures that height, and a taller hill hands each charge more energy.

Figure s02 — the hill picture: end A sits at the top, end B at the bottom, and the cyan double arrow is the height $V$ (energy handed to each unit of charge as it falls).

$V=\underset{\begin{array}{c}\text{joules per coulomb}\\ =\text{ one volt}\end{array}}{\underbrace{\frac{\text{energy given (joules, J)}}{\text{charge (coulombs, C)}}}}[\text{unit: volt, }\text{V}]$

• The top of the fraction is energy in joules — the push's "strength."
• The bottom is the amount of charge in coulombs sharing that energy.
• The whole ratio is per-charge, so it does not depend on how much charge you send — only on how tall the hill is. See Potential Difference (Voltage).

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Step 3 — What "flow" ($I$) really counts

WHAT. We define current $I$: how much charge passes a point in the wire each second.

WHY. "Flow" also needs a number. The natural measure of flow is a rate: charge per second, exactly like litres-per-second for water.

PICTURE. In Figure s03, imagine standing at the amber dashed line on the wire with a counter. Every charge that zips past in the arrow direction, you click. $I$ is your clicks per second.

Figure s03 — cyan charges stream along the wire in the positive (arrow) direction; the amber dashed line is your counting point, and $I$ is how much charge crosses it each second.

$I=\underset{\begin{array}{c}\text{coulombs per second}\\ =\text{ one ampere}\end{array}}{\underbrace{\frac{\text{charge passing (coulombs, C)}}{\text{time (seconds, s)}}}}[\text{unit: ampere, }\text{A}]$

• Top: the amount of charge that went by, in coulombs.
• Bottom: the time it took, in seconds.
• Big $I$ = a thick, fast stream. Tiny $I$ = a trickle. See Electric Current.

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Step 4 — Do the experiment: change the push, watch the flow

WHAT. Keep the same block. Turn a knob to raise $V$ step by step. At each setting, read off $I$. Write the pairs down: $(V,I)$.

WHY. We do not yet know how push and flow are related. Maybe doubling the push doubles the flow — maybe it quadruples it, maybe it does something weird. The only way to find out is to measure. This is exactly what Georg Ohm actually did.

PICTURE. In Figure s04, each experiment becomes one amber dot on a graph: $V$ (volts) up the side, $I$ (amperes) along the bottom. Watch where the dots land.

Figure s04 — measured $(V,I)$ pairs plotted as amber dots; they fall exactly on a straight cyan line that passes through the white origin dot at $(0,0)$.

For a plain metal wire at steady temperature, the dots do something remarkable: they land on a perfectly straight line that passes through the corner (0, 0). Push zero → flow zero, and every step up in $V$ steps $I$ up by the same amount.

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Step 5 — Read the meaning of that straight line

WHAT. A straight line through the origin means one thing in mathematics: $V$ and $I$ are directly proportional. Their ratio $V/I$ is the same at every point on the line.

WHY use the ratio here? Because "straight line through origin" and "constant ratio" are literally the same fact. For any point on that line, if you divide its height ($V$) by its across-distance ($I$), you get the same number every time — that number is the line's steepness.

PICTURE. In Figure s05, pick three different dots on the line. Draw the little right-angled step ("rise over run", the white base and amber upright) to each. All three steps have the same slope. That shared slope is one fixed number.

Figure s05 — three "rise over run" steps drawn at different points on the same cyan line; the amber uprights show that $V/I$ is identical everywhere — one locked number.

$\underset{\begin{array}{c}\text{push}\\ ÷\text{ flow}\end{array}}{\underbrace{\frac{V}{I}}}=\underset{\text{at every point on the line}}{\underbrace{\text{the same constant}}}$

• If this ratio changed as you moved along, the dots would curve, not stay straight.
• Straight ⇒ ratio locked ⇒ we are allowed to give that ratio a permanent name.

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Step 6 — Name the constant: that is Resistance

WHAT. Define resistance as that locked ratio: $R\equiv \frac{V}{I}$

WHY this definition and not another? "Opposition to flow" should be big when a big push $V$ produces only a small flow $I$. Look at the fraction: big top, small bottom → big $R$. It behaves exactly like the word "opposition" demands. The symbol $\equiv$ means "is defined to be" — we are choosing this, not deriving it.

PICTURE. In Figure s06, the same line from Step 5, now with its slope labelled $R$. Compare the two lines: the steep cyan line (lots of push needed for a little flow) = large $R$; the shallow amber line = small $R$.

Figure s06 — two blocks compared: the steep cyan line has large $R$ (much $V$ for little $I$), the shallow amber line has small $R$. The slope of each line IS its resistance.

$R=\frac{V}{I}=\underset{\text{steeper}\Rightarrow \text{more resistance}}{\underbrace{\text{slope of the }V\text{-vs-}I\text{ line}}}[\text{unit: ohm, }\Omega ]$

Because $V$ is in volts and $I$ in amperes, $R$ comes out in volts per ampere, and that gets its own name: the ohm ($\Omega$). See Resistance for what physically sets this slope (material, length, thickness).

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Step 7 — Rearrange into the famous form $V=IR$

WHAT. We have the defined fact $R=V/I$. To turn it into a push-predictor, multiply both sides by $I$: $R\times I=\frac{V}{I}\times I⟹V=IR$ The $I$ on the right cancels the $I$ underneath, leaving $V$ alone — that is the entire "trick", ordinary algebra, no magic.

WHY rearrange at all? The form $R=V/I$ defines R. The form $V=IR$ predicts: give me the flow $I$ and the block's resistance $R$, and I hand back the push $V$ needed. Same truth, more useful shape. There are three shapes because there are three unknowns you might be missing — the algebra, not a gimmick, gives all three:

$V=\underset{\text{flow (A)}}{\underbrace{I}}\times \underset{\text{opposition (}\Omega \text{)}}{\underbrace{R}}I=\frac{V}{R}R=\frac{V}{I}$

PICTURE. Figure s07 is just a memory aid for those three shapes we already derived — not a new law. Place $V$ on top, $I$ and $R$ side by side underneath. Whichever letter you cover, the picture shows the arrangement you get from the algebra above: covering $V$ leaves $I$ and $R$ side by side (multiply), covering $I$ or $R$ leaves $V$ sitting over the other (divide). If you ever doubt it, re-derive it from $R=V/I$ in one line — the triangle is only there to save you the re-derivation.

Figure s07 — the VIR memory triangle: $V$ on top, $I$ and $R$ beneath. Covering a letter reveals the rearrangement already proven by algebra above — it stores the result, it does not replace the reasoning.

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Step 8 — The degenerate cases (never get caught out)

WHAT. Check what the formula says at the extreme, "silly" inputs. A rule you trust must survive its own edges.

WHY. Real circuits hit zeros. If you have never watched $V=IR$ at $V=0$ or $R=0$, those cases will surprise you at exactly the wrong moment.

PICTURE. In Figure s08, the straight line now runs through the origin in both directions, with the corner cases flagged. Note how reversing the push (going bottom-left of the origin) reverses the current too — consistent with our sign convention from Step 1.

Figure s08 — the ohmic line extended through the origin into negative $V$ and $I$; amber markers flag the zero case, the $R=0$ flat line, and the reversed-push quadrant where both $V$ and $I$ go negative together.

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Step 9 — The one case where the picture bends (non-ohmic)

WHAT. Not every block gives a straight line. Heat one up (a lamp filament) and re-run Step 4: the dots curve.

WHY it curves. As current flows, the filament heats, its atoms vibrate harder, electrons collide more — so $R$ grows as you push harder. The slope $V/I$ is no longer one fixed number, so the "constant ratio" of Step 5 fails, and with it the proportionality.

PICTURE. In Figure s09, two lines share one graph: the straight cyan ohmic block, and the amber filament that gets steeper as it heats. Watch the amber curve bend upward — each extra amp needs more than its fair share of extra volts.

Figure s09 — cyan straight line (ohmic, fixed $R$) versus amber upward-bending curve (filament, $R$ rising with heat). The bending means $V/I$ is no longer a single number.

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The one-picture summary

Figure s10 compresses everything: the hill (push $V$) drives a stream (flow $I$) through a block whose slope on the $V$–$I$ graph is its opposition $R$ — and the whole law is just that one line's equation.

Figure s10 — left to right: the hill of height $V$ (joules per coulomb), the stream $I$ (coulombs per second) through the amber block of resistance $R$ (volts per ampere), and the mini graph whose slope is $R$. All bound by $V=IR$.

Recall Feynman retelling — the whole walkthrough in plain words

We drew one block with a top end A and a bottom end B, and agreed once and for all: the push $V$ is how much higher A sits than B, and current counts as positive when it flows the way our arrow points. Between the ends is a hill of height $V$ — that height is measured in volts, meaning joules of energy handed to each coulomb of charge. Along the wire runs a stream $I$ — coulombs of charge counted each second, measured in amperes. Then we did the honest thing: we cranked the push up bit by bit and wrote down the flow each time. The dots drew a straight line from the corner. A straight line from the corner means the push and the flow always divide to the same number — so we named that number $R$, the block's stubbornness, in ohms (volts per amp). Written as a recipe instead of a definition, "$R$ equals push over flow" becomes, by multiplying both sides by $I$, "push equals flow times stubbornness," $V=IR$ — and the little triangle is just a note-to-self for that one line of algebra. We checked the silly ends: no push, no flow; zero stubbornness, runaway flow; infinite stubbornness, dead; reverse the push and both signs flip together. And we met the odd block — a lamp — whose line bends because pushing it makes it hot and hotter means stubborner. There the single-number trick fails, and only the straight-line kind keeps Ohm's Law simple.

Recall Q: In three symbols, why is the

$V$–$I$ graph a straight line through the origin? Because $R=V/I$ is constant, so $V=IR$ is $V=(\text{constant})\times I$ — the equation of a straight line through $(0,0)$ with slope $R$.

Recall Q: What are the SI units of

$V$, $I$ and $R$ in terms of joules, coulombs and seconds? $V$ = joules per coulomb (volt), $I$ = coulombs per second (ampere), $R$ = volts per ampere (ohm).

Recall Q: State the sign convention used on this page.

$V=V_A-V_B$ (higher end minus lower), and $I$ is positive in the chosen arrow direction (A → B); reverse the push and both go negative together.

Recall Q: What happens to

$I$ as $R\to 0$ and as $R\to \infty$ (fixed $V$)? $R\to 0\Rightarrow I\to \infty$ (short circuit); $R\to \infty \Rightarrow I\to 0$ (insulator).

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Connections

• Ohm's Law (parent) — the note this deep dive expands.
• Potential Difference (Voltage) — Step 2, the hill-height $V$.
• Electric Current — Step 3, the counted stream $I$.
• Resistance — Steps 5–6, the slope $R$.
• Series and Parallel Circuits — where "across vs through" matters.
• Power in Circuits (P = VI) — the next thing $V$ and $I$ combine to give.
• Non-ohmic Components — Step 9, the bending curves.