Build and analyze a current divider: Visual walkthrough

Step 0 — The four symbols we are allowed to use

Before drawing anything, let's earn every letter so nothing sneaks up on you later.

The one law tying them together is Ohm's Law:

$flow I = \frac{V push}{fight R}$

Read it in plain words: more push → more flow; more fight → less flow. That single sentence powers the entire derivation.

Figure — Build and analyze a current divider

Look at the figure: one incoming wire (the fat blue arrow, $I_{T}$ ) reaches a junction — a single dot where wires meet — and faces two roads, $R_{1}$ and $R_{2}$ . Our whole job is to find how much flow, $I_{1}$ and $I_{2}$ , takes each road.

Step 1 — Both roads feel the SAME push

WHAT: The two resistors are wired between the same two dots — the top junction and the bottom junction. So they share one identical voltage, which we call $V$ .

WHY: Voltage is a difference between two points. If both resistors are pinned to the same top point and the same bottom point, there is only one difference to speak of. This is exactly the idea behind Parallel resistance — "parallel" means "sharing both endpoints."

PICTURE:

The green band marks the top node (all one voltage) and the gray band marks the bottom node. Every point on the green band is at the same push; every point on the gray band too. The gap between them is our shared $V$ — the same for $R_{1}$ and for $R_{2}$ .

Step 2 — Nothing piles up: the currents must sum to $I_{T}$

WHAT: Whatever flows in must flow out. So $I_{1} + I_{2} = I_{T}$ .

WHY: Charge can't accumulate at a dot — there's no bucket there. This is Kirchhoff's Current Law (KCL): current in = current out. It is bookkeeping, not physics you have to feel — think of a road fork: every car entering must leave down one of the two roads.

PICTURE:

The fat arrow $I_{T}$ enters; two thinner arrows $I_{1}$ , $I_{2}$ leave. The figure shades the arrow widths to hint that the flows may be unequal — but their total width equals the incoming width.

We now have two true statements (Step 1's shared $V$ , Step 2's KCL). Two equations, three unknowns ( $I_{1}, I_{2}, V$ ). We need to eliminate $V$ .

Step 3 — Replace the two roads by one "equivalent" road

WHAT: Pretend the whole two-resistor bundle is a single resistor, $R_{∥}$ , carrying the full $I_{T}$ at the same shared $V$ .

WHY: If the bundle behaves like one resistor, we can write $V = I_{T} R_{∥}$ and later swap it into Step 1. This gives us a handle on $V$ in terms of things we know ( $I_{T}$ and the $R$ 's).

PICTURE:

On the left: the two real resistors. On the right: one gray box labelled $R_{∥}$ . Same terminals, same $V$ , same total flow — indistinguishable from outside.

Sanity of $R_{∥}$ : two equal roads $R_{1} = R_{2} = R$ give $R_{∥} = \frac{R ^{2}}{2 R} = \frac{R}{2}$ — half the resistance, because two roads carry twice the flow. Good.

Step 4 — Solve for $I_{1}$ and watch $R_{1}$ vanish

WHAT: Use $I_{1} = V / R_{1}$ (Step 1) and substitute $V = I_{T} R_{∥}$ (Step 3).

WHY: This kills the unknown $V$ and leaves $I_{1}$ in terms of the currents and resistances we actually know.

PICTURE:

The figure lays the algebra out as a chain, colouring the $R_{1}$ that cancels in red so you see it disappear.

I_{1} = \frac{V}{R _{1}} = \frac{I _{T} R _{∥}}{R _{1}} = \frac{I _{T}}{R _{1}} \cdot \frac{R _{1} R _{2}}{R _{1} + R _{2}}

$\frac{V}{R _{1}}$ — Ohm's law on branch 1: its flow is its push over its fight.
$\frac{I _{T} R _{∥}}{R _{1}}$ — we swapped in the shared push $V = I_{T} R_{∥}$ .
The red $R_{1}$ on top (inside $R_{∥}$ ) cancels the red $R_{1}$ on the bottom.

WHY the swap makes physical sense: $I_{1} = V / R_{1}$ , so a bigger $R_{1}$ gives a smaller $I_{1}$ . The formula must go down as $R_{1}$ grows — and indeed $R_{1}$ only appears in the denominator. The numerator $R_{2}$ tells you how easy the escape route is; a fat $R_{2}$ (hard to escape) keeps flow in your own branch.

Step 5 — Does it obey KCL? Add the two currents

WHAT: Check that $I_{1} + I_{2}$ comes back to $I_{T}$ .

WHY: Any derivation must satisfy the law it started from (Step 2). If it didn't, we'd have made an algebra slip.

I_{1} + I_{2} = I_{T} \frac{R _{2}}{R _{1} + R _{2}} + I_{T} \frac{R _{1}}{R _{1} + R _{2}} = I_{T} \frac{R _{2} + R _{1}}{R _{1} + R _{2}} = I_{T} ✓

The tops add to exactly the bottom, so the fraction is $1$ . Bookkeeping closes.

Step 6 — Every case, so nothing surprises you

The formula must behave sensibly at the extremes. Let's walk each corner.

The figure is a strip of four mini-circuits; read them alongside the rows below.

Formula Limiting cases of

$I_{1} = I_{T} \frac{R _{2}}{R _{1} + R _{2}}$

Case	What happens to $I_{1}$	Why (in words)
$R_{1} = R_{2}$	$I_{1} = I_{T} \cdot \frac{R}{2 R} = \frac{I _{T}}{2}$	Equal roads → equal split.
$R_{2} \to \infty$ (branch 2 broken/open)	$\frac{R _{2}}{R _{1} + R _{2}} \to 1$ , so $I_{1} \to I_{T}$	Escape route sealed; all flow must use $R_{1}$ .
$R_{2} \to 0$ (branch 2 a bare wire, "short")	$\frac{R _{2}}{R _{1} + R _{2}} \to 0$ , so $I_{1} \to 0$	A zero-resistance road hogs everything; $R_{1}$ gets none.
$R_{1} \to 0$ (branch 1 shorted)	$\frac{R _{2}}{0 + R _{2}} = 1$ , so $I_{1} \to I_{T}$	Now your branch is the free road; it takes all of $I_{T}$ .

The one-picture summary

One frame carries the whole story: $I_{T}$ enters the node → the green shared-voltage band forces $V = I_{1} R_{1} = I_{2} R_{2}$ → KCL forces $I_{1} + I_{2} = I_{T}$ → collapsing to $R_{∥}$ eliminates $V$ → $R_{1}$ cancels → out drops $I_{1} = I_{T} \frac{R _{2}}{R _{1} + R _{2}}$ , with the other resistor on top.

Recall Feynman retelling — say it to a 12-year-old

Water reaches a fork with two pipes. Both pipes start and end at the same two puddles, so they feel the same pressure — that's our $V$ . No water vanishes at the fork, so the two out-flows must add back up to what came in — that's KCL. Now here's the punchline: the flow down your pipe is pressure ÷ how skinny your pipe is. When we write the pressure in terms of the total flow, the "your pipe skinniness" on the bottom partly cancels, and what's left on top is the other pipe's skinniness. That's why the far resistor sits on top: a fat, hard-to-use other pipe keeps more water in yours. Seal the other pipe → you get everything; make the other pipe a free highway → you get nothing. Every case, one picture.

Active recall

Connections

Build and analyze a current divider — the parent topic this walkthrough visualises.
Kirchhoff's Current Law — Step 2's conservation of current.
Ohm's Law — $I = V / R$ , used in every step.
Parallel resistance — where $R_{∥}$ (Step 3) comes from.
Conductance — the cleaner language for the $N$ -branch generalisation.
Voltage divider — the series dual; contrast the "swap."