Build and analyze a current divider: Exercises — Hardware deep dive

Before we start, one figure fixes the picture in your head. Every symbol below points at something in it.

Figure — Build and analyze a current divider

Level 1 — Recognition

Goal: spot when a current-divider applies and read the formula, no heavy arithmetic.

Recall Solution L1-1

No. In series there is only one path, so the same current flows through both resistors — nothing splits. What divides in series is the voltage (that's a Voltage divider). A current divider needs parallel branches that share one voltage $V$ . See Kirchhoff's Current Law: splitting only happens where a wire physically forks.

Recall Solution L1-2

$R_{1}$ (the smaller resistance) carries more. Current takes the easy road — low resistance = high Conductance = bigger share. No numbers needed; the smaller- $R$ branch always wins.

Recall Solution L1-3

$R_{1}$ goes on top: $I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}}$ Why the other resistor? Look at figure s01: both branches sit across the same $V$ , so by Ohm's Law $I_{2} = V / R_{2}$ and $I_{1} = V / R_{1}$ . The whole bundle carries $I_{T} = V / R_{∥}$ with $R_{∥} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ . Substituting $V = I_{T} R_{∥}$ into $I_{2} = V / R_{2}$ : $I_{2} = \frac{I _{T} R _{∥}}{R _{2}} = \frac{I _{T}}{R _{2}} \cdot \frac{R _{1} R _{2}}{R _{1} + R _{2}} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} .$ The $R_{2}$ cancels, leaving the other resistor $R_{1}$ on top. That cancellation is the whole story — hence the mnemonic "CROSS to find your current."

Level 2 — Application

Goal: plug numbers into the two-branch formula and verify with KCL.

Recall Solution L2-1

$I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} = 20 \cdot \frac{1}{3 + 1} = 20 \cdot \frac{1}{4} = 5 mA$ $I_{2} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}} = 20 \cdot \frac{3}{4} = 15 mA$ Check (KCL): $5 + 15 = 20 mA = I_{T}$ ✓. Sensible: $R_{2}$ (1 k) is the easy path, so it hogs $15$ mA.

Recall Solution L2-2

Rearrange $I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}}$ for $I_{T}$ : $I_{T} = I_{1} \cdot \frac{R _{1} + R _{2}}{R _{2}} = 9 \cdot \frac{2 + 6}{6} = 9 \cdot \frac{8}{6} = 12 mA$ Check: then $I_{2} = 12 - 9 = 3 mA$ , and $I_{1} / I_{2} = 9/3 = 3 = R_{2} / R_{1}$ ✓ (the easy branch $R_{1}$ carries $3 \times$ more).

Recall Solution L2-3

First, why the current ratio equals the swapped resistance ratio. Both branches share the one voltage $V$ (they're parallel, figure s01), so by Ohm's Law: $I_{1} = \frac{V}{R _{1}}, I_{2} = \frac{V}{R _{2}} .$ Divide the two — the common $V$ cancels, and the resistors flip because they're in the denominators: $\frac{I _{1}}{I _{2}} = \frac{V / R _{1}}{V / R _{2}} = \frac{R _{2}}{R _{1}} \Rightarrow R_{2} = R_{1} \cdot \frac{I _{1}}{I _{2}} .$ Here $I_{2} = I_{T} - I_{1} = 10 - 4 = 6 mA$ , so $R_{2} = 6 k Ω \cdot \frac{4}{6} = 4 k Ω$ Check: $I_{1} = 10 \cdot \frac{4}{6 + 4} = 10 \cdot \frac{4}{10} = 4 mA$ ✓.

Level 3 — Analysis

Goal: many branches, conductance form, and interpreting the split.

Recall Solution L3-1

The two-resistor swap does not work for three branches — use conductances (see Parallel resistance). $G_{1} = \frac{1}{4} = 0.25, G_{2} = \frac{1}{6} \approx 0.1667, G_{3} = \frac{1}{12} \approx 0.0833 S$ $\sum_{j} G_{j} = 0.25 + 0.1667 + 0.0833 = 0.5 S (j = 1, 2, 3)$ $I_{1} = 24 \cdot \frac{0.25}{0.5} = 12 mA, I_{2} = 24 \cdot \frac{0.1667}{0.5} = 8 mA, I_{3} = 24 \cdot \frac{0.0833}{0.5} = 4 mA$ Check: $12 + 8 + 4 = 24 mA$ ✓. Note $R_{1} : R_{2} : R_{3} = 4 : 6 : 12$ gives currents $12 : 8 : 4 = 3 : 2 : 1$ — inversely proportional to resistance, as expected.

Recall Solution L3-2

$I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} = I_{T} \cdot \frac{1}{\frac{R _{1}}{R _{2}} + 1}$ As $R_{2} \to \infty$ , the term $R_{1} / R_{2} \to 0$ , so $I_{1} \to I_{T}$ . All the current flows through $R_{1}$ . Physical meaning: an infinite-resistance branch is an open circuit — no current can enter it, so branch 1 receives everything. Degenerate case handled ✓.

Recall Solution L3-3

$I_{1} = I_{T} \cdot \frac{R _{2}}{R _{1} + R _{2}} R_{2} \to 0 I_{T} \cdot \frac{0}{R _{1} + 0} = 0$ $I_{1} \to 0$ and $I_{2} \to I_{T}$ . Physical meaning: a zero-resistance path is a perfect short — current gladly dumps everything through the free path and avoids $R_{1}$ entirely. This is why a short "steals" current from parallel components.

Level 4 — Synthesis

Goal: combine the divider with Ohm's law, series/parallel reduction, and multi-step reasoning.

Recall Solution L4-1

Step 1 — collapse branch B to one resistor. Series adds: $R_{B} = 3 + 3 = 6 k Ω$ . Now it's a clean two-branch divider $R_{A} = 2 k Ω ∥ R_{B} = 6 k Ω$ . Step 2 — current into branch B (put the other resistor $R_{A}$ on top): $I_{B} = I_{T} \cdot \frac{R _{A}}{R _{A} + R _{B}} = 16 \cdot \frac{2}{2 + 6} = 16 \cdot \frac{1}{4} = 4 mA$ Step 3 — voltage across $R_{B 2}$ by Ohm's Law (the same $4$ mA flows through both series parts): $V_{B 2} = I_{B} \cdot R_{B 2} = 4 mA \times 3 k Ω = 12 V$ Check: branch A gets $I_{A} = 16 - 4 = 12 mA$ ; shared voltage $V = I_{A} R_{A} = 12 mA \times 2 k Ω = 24 V$ , and $I_{B} R_{B} = 4 \times 6 = 24 V$ ✓ — same voltage, as required.

Recall Solution L4-2

The LED current from the two-branch formula (its "other" resistor is $R_{b}$ ): $I_{L E D} = I_{T} \cdot \frac{R _{b}}{R _{L E D} + R _{b}}$ Solve for $R_{b}$ . Let $f = I_{L E D} / I_{T} = 6/30 = 0.2$ . Then $f (R_{L E D} + R_{b}) = R_{b}$ , so: $f R_{L E D} = R_{b} (1 - f) \Rightarrow R_{b} = R_{L E D} \cdot \frac{f}{1 - f} = 2 k Ω \cdot \frac{0.2}{0.8} = 0.5 k Ω = 500 Ω$ Check: $I_{L E D} = 30 \cdot \frac{0.5}{2 + 0.5} = 30 \cdot \frac{0.5}{2.5} = 6 mA$ ✓. The bypass ( $500 Ω$ , low resistance) soaks up the excess $24 mA$ so the LED stays safe.

Level 5 — Mastery

Goal: design and prove, including tricky sign/limit reasoning.

Recall Solution L5-1

Key idea: currents split in proportion to conductance $G_{k}$ , so the current ratio equals the conductance ratio: $G_{1} : G_{2} : G_{3} = 3 : 2 : 1$ Find the ratio unit. $R_{1} = 2 k Ω$ gives $G_{1} = 1/ R_{1} = 1/ (2 k Ω) = 0.5 mS$ . This is the " $3$ -parts" entry, so one part $= G_{1} /3 = 0.5/3 = \frac{1}{6} mS$ . Scale to the other branches: $G_{2} = 2 parts = 2 \cdot \frac{1}{6} = \frac{1}{3} mS, G_{3} = 1 part = 1 \cdot \frac{1}{6} = \frac{1}{6} mS$ Convert back to resistance ( $R = 1/ G$ ): $R_{2} = 1/ G_{2} = 3 k Ω, R_{3} = 1/ G_{3} = 6 k Ω$ Currents (fractions $\frac{3}{6}, \frac{2}{6}, \frac{1}{6}$ of $I_{T}$ since $3 + 2 + 1 = 6$ ): $I_{1} = 18 \cdot \frac{3}{6} = 9 mA, I_{2} = 18 \cdot \frac{2}{6} = 6 mA, I_{3} = 18 \cdot \frac{1}{6} = 3 mA$ Check: $9 + 6 + 3 = 18$ ✓ and $9 : 6 : 3 = 3 : 2 : 1$ ✓.

Recall Solution L5-2

Substitute $G_{1} = 1/ R_{1}$ , $G_{2} = 1/ R_{2}$ : $I_{1} = I_{T} \cdot \frac{1/ R _{1}}{1/ R _{1} + 1/ R _{2}}$ Multiply top and bottom by $R_{1} R_{2}$ (the common denominator — this clears the nested fractions): $I_{1} = I_{T} \cdot \frac{( 1/ R _{1} ) R _{1} R _{2}}{( 1/ R _{1} ) R _{1} R _{2} + ( 1/ R _{2} ) R _{1} R _{2}} = I_{T} \cdot \frac{R _{2}}{R _{2} + R _{1}}$ So the conductance form on top for branch 1, $G_{1}$ , turns into the other resistor $R_{2}$ — that's why the swap appears. ∎

Recall Solution L5-3

The fraction is $\frac{I _{1}}{I _{T}} = \frac{R _{2}}{R _{1} + R _{2}} .$ Trace the curve in figure s02 (drawn with $R_{1}$ fixed at $1$ ):

At $R_{2} = 0$ : fraction $= 0$ (short across branch 2 steals everything — L3-3). This is the amber dot at the origin.
At $R_{2} = R_{1}$ : fraction $= \frac{1}{2}$ (equal resistors ⇒ equal split) — the amber dot at mid-height.
As $R_{2} \to \infty$ : fraction $\to 1$ (open branch 2 ⇒ all current in branch 1 — L3-2), the dashed ceiling the curve approaches. It rises monotonically from $0$ toward $1$ and never reaches or exceeds either bound, because $R_{2} < R_{1} + R_{2}$ always (so the fraction $< 1$ ) and $R_{2} \geq 0$ (so $\geq 0$ ). A branch can never carry negative current or more than $I_{T}$ — exactly what physical conservation demands.

The steepest climb is near $R_{2} = 0$ : when branch 2 is still very "easy," small changes in $R_{2}$ shift a lot of current — read that off the sharp initial rise of the cyan curve.

Active recall

Does current divide in series or parallel branches?

Parallel — they share one voltage, so resistance alone sets each branch current.

For

I_{2}

with two resistors, which goes on top?

The other resistor,

R_{1}

I_{2} = I_{T} R_{1} / (R_{1} + R_{2})

As one parallel branch

R \to \infty

, its current goes to…?

Zero — an open branch carries no current; everything flows elsewhere.

To split currents

3 : 2 : 1

, put resistances in what ratio?

2 : 3 : 6

(conductances in

3 : 2 : 1

) — inverse of the current ratio.

Why does the two-resistor "swap" trick fail for 3+ branches?

It's algebra special to two branches; for

N \geq 3

use

I_{k} = I_{T} G_{k} / \sum_{j} G_{j}

Connections

Build and analyze a current divider — the parent note these exercises drill.
Kirchhoff's Current Law — every "sum of branch currents $= I_{T}$ " check.
Ohm's Law — used to turn branch currents into branch voltages (L4).
Parallel resistance — reducing whole branches before dividing.
Conductance — the correct language for multi-branch and design problems.
Voltage divider — the series dual you must not confuse with this.