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DC Electricity Intermediate 8 min read Nov 2026

Current Divider

A current divider splits a total current between parallel branches. Learn the formula, why the smaller resistance takes more current, and how to verify with KCL.

Definition

A current divider is a parallel connection in which a total current splits between branches in inverse proportion to their resistances. For two branches, I₁ = I_total × R₂/(R₁ + R₂) and I₂ = I_total × R₁/(R₁ + R₂).

Learning objectives

  • Apply the two-branch current divider formula.
  • Explain why the smaller resistance carries more current.
  • Verify branch currents with Kirchhoff’s current law.
  • Recognise when source resistance or AC reactance invalidates the simple formula.
Current Divider diagram Resistors in parallel R₁ R₂ same voltage, currents add
Current Divider — schematic illustration
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Progressive explanation

Start with the plain-language idea, then move into the engineering detail. Each level stands on its own, so you can stop at the depth you need.

Beginner The core idea
A current divider splits current between parallel paths. Because each branch has the same voltage, the branch with the smaller resistance carries the larger current. If one branch has twice the resistance of the other, it carries half the current.

This is the opposite of a voltage divider, where the larger resistance takes the larger share of voltage. Current division explains why a short circuit takes most of the current, why parallel lamps share current according to their power ratings, and why adding a low-resistance branch increases the total current.
Intermediate How it behaves in a circuit
For two parallel resistors, the total current is I_total and the branch currents are I₁ = I_total × R₂/(R₁ + R₂) and I₂ = I_total × R₁/(R₁ + R₂). Notice that the opposite resistance appears in each numerator: the smaller resistance carries the larger current. The branch voltages are equal: I₁R₁ = I₂R₂ = V.

For more than two branches, use conductances: Iₙ = I_total × Gₙ/ΣG, where G = 1/R. Kirchhoff’s current law provides the check that the branch currents sum to the total. The formula assumes the total current is fixed; if the source has significant internal resistance, adding or changing a branch changes the total current too.
Advanced Limits, assumptions, engineering detail
The current divider formula is a special case of the general parallel-network relation. Because each branch is defined by its admittance, current division in AC circuits uses admittances and phasors, and the split depends on frequency when reactance is present. In a two-branch AC divider, the branch currents are not necessarily in phase with the total current, so arithmetic addition of RMS values is invalid.

Practical current sharing is affected by wire resistance, contact resistance and component tolerance. Precision current shunts and current-sense resistors are chosen with low temperature coefficient because their resistance changes with self-heating. In power electronics, parallel switching devices need balancing networks or active control to share current, because a small difference in on-resistance can cause a large imbalance at high current.

Formulas and variables

Current divider for two branches
I₁ = I_total × R₂ / (R₁ + R₂)
I₁ Current in branch 1 · A I_total Total current entering the parallel group · A R₁ Resistance of branch 1 · Ω R₂ Resistance of branch 2 · Ω
6 A entering 10 Ω and 15 Ω gives I₁ = 6 × 15/25 = 3.6 A.
Second branch
I₂ = I_total × R₁ / (R₁ + R₂)
I₂ Current in branch 2 · A I_total Total current · A R₁, R₂ Branch resistances · Ω
With 6 A and 10 Ω/15 Ω, I₂ = 6 × 10/25 = 2.4 A.
General conductance form
Iₙ = I_total × Gₙ / ΣG
Iₙ Current in branch n · A Gₙ Conductance of branch n (1/Rₙ) · S ΣG Sum of branch conductances · S
For 10 Ω and 15 Ω, G = 0.10 S and 0.0667 S; the 10 Ω branch takes 0.10/0.1667 of the total.

Units of measurement

QuantitySymbolUnit
Current I ampere (A)
Resistance R ohm (Ω)
Conductance G siemens (S)
Voltage V volt (V)

Worked examples

Worked example

Current split between two resistors

Problem. A total current of 6.0 A enters a parallel pair of 10 Ω and 15 Ω. Find each branch current.

Solution. I₁₀ = 6.0 × 15/(10 + 15) = 90/25 = 3.6 A. I₁₅ = 6.0 × 10/25 = 60/25 = 2.4 A. Check: 3.6 + 2.4 = 6.0 A, and both branches have 3.6 × 10 = 36 V and 2.4 × 15 = 36 V.

I₁₀ = 3.6 A; I₁₅ = 2.4 A; branch voltage = 36 V.
Worked example

Three-way current division using conductances

Problem. A total current of 3.0 A splits between 4 Ω, 12 Ω and 6 Ω in parallel. Find the current in the 4 Ω branch and the 12 Ω branch.

Solution. Conductances: G₁ = 1/4 = 0.25 S, G₂ = 1/12 ≈ 0.0833 S, G₃ = 1/6 ≈ 0.1667 S; ΣG ≈ 0.50 S. I₄ = 3.0 × 0.25/0.50 = 1.5 A. I₁₂ = 3.0 × 0.0833/0.50 = 0.50 A. The remaining 1.0 A flows in the 6 Ω branch. Common voltage: 1.5 × 4 = 0.5 × 12 = 1.0 × 6 = 6.0 V.

I₄ = 1.5 A; I₁₂ = 0.50 A; I₆ = 1.0 A.

Practical applications

  • Parallel load current sharing
  • Current-sense shunt networks
  • Parallel battery and converter branches
  • Biasing and signal-splitting networks

Automotive applications

Current division explains why a shorted branch in a vehicle circuit can starve the intended load and blow a fuse: the low-resistance fault takes most of the current. It also helps interpret parallel loads, such as two lamps of different power ratings sharing a supply. When adding accessories, the new parallel branch changes the current distribution and can overload a shared feed.

Industrial applications

Industrial current sharing appears in parallel feeders, parallel battery strings and parallel power converters. Cables and converters are matched and derated to share current evenly; a small resistance difference can cause a significant imbalance at high current. In instrumentation, current dividers are used to extend the range of a current-measuring circuit, but the shunt resistance and its temperature coefficient then determine the accuracy.

Measurement procedure

  1. Measure the total current and each branch current separately, then verify that the branch currents add to the total.
  2. Measure the voltage across the parallel group; each branch current should equal V/R for that branch.
  3. Use a clamp meter on individual conductors; clamping several parallel conductors together reads the net current.
  4. For low-resistance shunts, use a four-wire measurement to avoid lead resistance affecting the result.

Common mistakes

  • Using the voltage divider formula for a current divider.
  • Assigning the larger current to the larger resistance.
  • Forgetting to verify the branch currents sum to the total current.
  • Applying the simple formula when the source resistance is significant and changes the total current.

Safety considerations

  • A low-resistance parallel branch can draw a large current and overheat wiring; ensure protection is correctly rated.
  • Current-sense shunts can be at a hazardous potential; use isolated measurement equipment.
  • Do not open a branch of a current-carrying parallel circuit unless the resulting redistribution is safe and understood.

Frequently asked questions

Why does the smaller resistance take more current?
Both branches have the same voltage, and I = V/R, so a smaller resistance produces a larger current.
Can I add the branch currents arithmetically?
Yes for DC or for AC with purely resistive branches. For reactive AC branches the currents may be out of phase and must be added as phasors.
What is the difference between a current divider and a voltage divider?
A voltage divider uses series resistors and divides voltage in proportion to resistance. A current divider uses parallel branches and divides current in inverse proportion to resistance.
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