Build and analyze a current divider: Worked examples

Before the table, one reminder of the two tools we use everywhere, so no symbol appears unearned:

Here $I_{T}$ is the total current arriving at the fork, $I_{k}$ is the current in branch $k$ , $R_{k}$ is that branch's resistance, and every branch feels the same voltage $V$ because they connect the same two nodes (see Kirchhoff's Current Law and Parallel resistance).

The scenario matrix

Every current-divider problem falls into one of these case classes. Each row is covered by at least one worked example below.

#	Case class	What's special	Example
A	Equal resistors	Perfect 50/50 split — the reference point	Ex 1
B	Unequal resistors	Normal asymmetric split; check the "swap"	Ex 2
C	One branch → ∞ (open circuit)	Degenerate: infinite resistance = broken wire	Ex 3
D	One branch → 0 (dead short)	Degenerate: zero resistance hogs everything	Ex 4
E	Extreme ratio (limiting behaviour)	1000:1 — where does almost all current go?	Ex 5
F	Three+ branches (conductance form)	Two-branch trick fails; must use $G$	Ex 6
G	Real-world word problem	LED + shunt resistor sizing	Ex 7
H	Exam twist (solve backwards)	Given a branch current, find $R$	Ex 8

Figure — Build and analyze a current divider

The figure above shows the whole family on one "resistance ratio" axis: from a dead short at the left (all current one way) through equal split in the middle to an open circuit at the right (all current the other way). Every example below is a point on this line.

Case A — Equal resistors (the 50/50 reference)

Case B — Unequal resistors (does the "swap" really hold?)

Worked example Ex 2 · Cell B

$I_{T} = 18 mA$ , $R_{1} = 2 k Ω$ , $R_{2} = 6 k Ω$ .

Forecast: $R_{1}$ is the easier (smaller) path, so it should grab the larger current. Since $R_{1}$ is 3× easier than $R_{2}$ , guess a 3:1 split → $13.5 mA$ and $4.5 mA$ .

Step 1. Current in the small resistor $R_{1}$ — put the other resistor $R_{2}$ on top. $I_{1} = 18 \cdot \frac{R _{2}}{R _{1} + R _{2}} = 18 \cdot \frac{6}{2 + 6} = 18 \cdot \frac{6}{8} = 13.5 mA$ Why this step? Remember $I_{1} = V / R_{1}$ and a smaller $R_{1}$ means a bigger $I_{1}$ . The formula encodes this by putting the big "other" resistor $R_{2}$ on top — bigger top ⇒ bigger share.

Step 2. Current in $R_{2}$ — now $R_{1}$ goes on top. $I_{2} = 18 \cdot \frac{R _{1}}{R _{1} + R _{2}} = 18 \cdot \frac{2}{8} = 4.5 mA$ Why this step? We cross over again: the current through $R_{2}$ depends on the other branch $R_{1}$ .

Verify: $13.5 + 4.5 = 18 = I_{T}$ ✅. Ratio $I_{1} : I_{2} = 13.5 : 4.5 = 3 : 1$ , exactly the inverse of $R_{1} : R_{2} = 2 : 6 = 1 : 3$ . Current ratio is the inverse of the resistance ratio — matches the forecast.

Case C — Open branch (one resistor → ∞)

Worked example Ex 3 · Cell C (degenerate)

$I_{T} = 7 mA$ , $R_{1} = 5 k Ω$ , and branch 2 is broken — an open wire, so $R_{2} \to \infty$ .

Forecast: A broken pipe carries no water. All $7 mA$ must go through branch 1.

Step 1. Push $R_{2} \to \infty$ inside the formula for $I_{2}$ . $I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} R_{2} \to \infty I_{T} \cdot \frac{R _{1}}{\infty} = 0 mA$ Why this step? We test a limit — not a normal number but "what happens as the resistance grows without bound." An infinite resistor is exactly a cut wire, so its current should collapse to zero; the algebra confirms it (finite top, infinite bottom → 0).

Step 2. Now $I_{1}$ . $I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} R_{2} \to \infty I_{T} \cdot 1 = 7 mA$ Why this step? As $R_{2} \to \infty$ , the fraction $\frac{R _{2}}{R _{1} + R _{2}} \to 1$ (the $R_{1}$ becomes negligible next to an infinite $R_{2}$ ). So branch 1 keeps everything.

Verify: $I_{1} + I_{2} = 7 + 0 = 7 = I_{T}$ ✅. Physical sanity: a divider with one branch open is just a single series path — no dividing happens.

Case D — Dead short (one resistor → 0)

Worked example Ex 4 · Cell D (degenerate)

$I_{T} = 7 mA$ , $R_{1} = 5 k Ω$ , and branch 2 is a dead short — a plain wire with $R_{2} = 0$ .

Forecast: A zero-resistance path is a superhighway with no friction. All current takes it; branch 1 gets nothing.

Step 1. Set $R_{2} = 0$ in the formula for $I_{1}$ . $I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} = 7 \cdot \frac{0}{5 + 0} = 0 mA$ Why this step? The share belonging to branch 1 is proportional to the other resistor $R_{2}$ . If $R_{2} = 0$ , branch 1's numerator is 0 — it gets no current.

Step 2. Then $I_{2}$ . $I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} = 7 \cdot \frac{5}{5 + 0} = 7 mA$ Why this step? The short swallows the full current. This is why a short across a component "starves" it — current abandons the resistive path.

Verify: $0 + 7 = 7 = I_{T}$ ✅. Note cases C and D are mirror images: infinity in one branch and zero in the other both send 100% one way — just opposite ways.

Case E — Extreme ratio (limiting behaviour, not quite degenerate)

Worked example Ex 5 · Cell E

$I_{T} = 100 mA$ , $R_{1} = 1 Ω$ , $R_{2} = 1000 Ω$ .

Forecast: $R_{1}$ is 1000× easier. Almost everything should go through $R_{1}$ , with a tiny trickle in $R_{2}$ . Guess $I_{1} \approx 100 mA$ , $I_{2} \approx 0.1 mA$ .

Step 1. $I_{1}$ with $R_{2} = 1000$ on top. $I_{1} = 100 \cdot \frac{1000}{1 + 1000} = 100 \cdot \frac{1000}{1001} \approx 99.9001 mA$ Why this step? When one resistor dwarfs the other, the fraction sits just below 1 — not exactly 1, so it's a limit approached, not reached. This is the difference between "extreme" (Ex 5) and "degenerate/open" (Ex 3).

Step 2. $I_{2}$ with $R_{1} = 1$ on top. $I_{2} = 100 \cdot \frac{1}{1001} \approx 0.0999 mA$ Why this step? The tiny numerator $R_{1} = 1$ over the big total gives a whisper of current — the physical "leakage" through the hard path.

Verify: $99.9001 + 0.0999 = 100 = I_{T}$ ✅. Lesson: even a huge resistor carries a little current unless it is truly infinite (open).

Case F — Three branches (the swap trick breaks, use conductance)

Worked example Ex 6 · Cell F

$I_{T} = 24 mA$ , $R_{1} = 10 Ω$ , $R_{2} = 30 Ω$ , $R_{3} = 60 Ω$ .

Forecast: $R_{1}$ is easiest, $R_{3}$ hardest → $I_{1} > I_{2} > I_{3}$ . The two-branch formula does not apply (there are three "other" resistors, not one).

Step 1. Convert each resistance to conductance $G = 1/ R$ (siemens, S). $G_{1} = \frac{1}{10} = 0.1, G_{2} = \frac{1}{30} \approx 0.0333, G_{3} = \frac{1}{60} \approx 0.0167 S$ Why this step? Conductance measures ease of flow, and current splits in direct proportion to ease — no confusing swap. With 3+ branches this is the only clean rule.

Step 2. Total conductance. $\sum G = 0.1 + 0.0333 + 0.0167 = 0.15 S$ Why this step? Parallel conductances add (they are all extra easy paths side by side). This $\sum G$ is the denominator every branch shares.

Step 3. Each branch current $I_{k} = I_{T} G_{k} / \sum G$ . $I_{1} = 24 \cdot \frac{0.1}{0.15} = 16 mA, I_{2} = 24 \cdot \frac{0.0333}{0.15} = 5.333 mA, I_{3} = 24 \cdot \frac{0.0167}{0.15} = 2.667 mA$ Why this step? Each fraction is "this branch's share of the total ease." $R_{1}$ owns $\frac{0.1}{0.15} = \frac{2}{3}$ of the ease, so it takes two-thirds of the current.

Verify: $16 + 5.333 + 2.667 = 24 = I_{T}$ ✅. Order $I_{1} > I_{2} > I_{3}$ matches the forecast, and it tracks $G_{1} > G_{2} > G_{3}$ .

Case G — Real-world word problem (LED shunt)

Worked example Ex 7 · Cell G

A driver supplies $I_{T} = 30 mA$ into a node. Branch 1 is an LED string modelled as $R_{1} = 100 Ω$ ; branch 2 is a shunt (bypass) resistor. You want exactly $20 mA$ through the LED. What $R_{2}$ do you pick?

Forecast: The LED should get two-thirds of the current, so its branch must be the easier one — the shunt $R_{2}$ should be bigger than $100 Ω$ so more current stays in the LED.

Step 1. Write the LED-branch current with the two-branch formula. $I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} \Rightarrow 20 = 30 \cdot \frac{R _{2}}{100 + R _{2}}$ Why this step? We know $I_{1}$ and want $R_{2}$ , so we set up the equation and solve backwards.

Step 2. Solve for $R_{2}$ . $\frac{20}{30} = \frac{R _{2}}{100 + R _{2}} \Rightarrow \frac{2}{3} (100 + R_{2}) = R_{2} \Rightarrow \frac{200}{3} = R_{2} - \frac{2}{3} R_{2} = \frac{1}{3} R_{2}$ $R_{2} = 200 Ω$ Why this step? Cross-multiplying isolates $R_{2}$ . The result being larger than $R_{1}$ matches the forecast — a bigger shunt "pushes" more current into the LED.

Verify: $I_{1} = 30 \cdot \frac{200}{100 + 200} = 30 \cdot \frac{200}{300} = 20 mA$ ✅. Then $I_{2} = 30 - 20 = 10 mA$ , and indeed $30 \cdot \frac{100}{300} = 10 mA$ ✅.

Case H — Exam twist (given a current, find total)

Worked example Ex 8 · Cell H

Exam question: "In a two-branch divider, $R_{1} = 4 Ω$ , $R_{2} = 12 Ω$ . A meter reads $I_{2} = 3 A$ . Find the total current $I_{T}$ and $I_{1}$ ."

Forecast: $R_{2}$ is the harder path (3× $R_{1}$ ), so $I_{2}$ should be the smaller current. Total will be bigger than $3 A$ ; $I_{1}$ should be about 3× larger than $I_{2}$ .

Step 1. Use the $I_{2}$ formula (the other resistor $R_{1}$ on top) and solve for $I_{T}$ . $I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} \Rightarrow 3 = I_{T} \cdot \frac{4}{4 + 12} = I_{T} \cdot \frac{4}{16} = \frac{I _{T}}{4}$ $I_{T} = 12 A$ Why this step? We have one branch current and the resistors, so invert the divider formula. The factor $\frac{4}{16} = \frac{1}{4}$ means branch 2 carries a quarter of the total — so the total is 4× the meter reading.

Step 2. Now find $I_{1}$ by KCL (fastest) or the formula. $I_{1} = I_{T} - I_{2} = 12 - 3 = 9 A$ Why this step? Kirchhoff's Current Law guarantees the branches sum to $I_{T}$ , so once one branch and the total are known, the other is free subtraction.

Verify: Formula check: $I_{1} = 12 \cdot \frac{R _{2}}{R _{1} + R _{2}} = 12 \cdot \frac{12}{16} = 9 A$ ✅. Ratio $I_{1} : I_{2} = 9 : 3 = 3 : 1$ , the inverse of $R_{1} : R_{2} = 4 : 12 = 1 : 3$ — matches the forecast.

Active recall

Connections

Build and analyze a current divider — the parent this page stress-tests.
Kirchhoff's Current Law — every "Verify" step uses $\sum I_{k} = I_{T}$ .
Voltage divider — contrast: series divides voltage, parallel divides current.
Parallel resistance — the $R_{∥}$ behind the derivation.
Ohm's Law — $I = V / R$ per branch, used in every case.
Conductance — the language for Cell F (3+ branches).