Current Dividers
A current divider is the parallel-circuit counterpart of the voltage divider. Where a voltage divider splits a voltage between series resistors, a current divider splits a current between parallel branches. The mathematics is elegantly dual: in a voltage divider the output voltage is proportional to the chosen resistor's fraction of the total; in a current divider the current through one branch is proportional to the other branch's resistance — because the lower-resistance branch attracts more current. This counter-intuitive twist catches many beginners, so this lesson works carefully through the physical reason, the formula, and its derivation before building up to more complex cases.
- The Physical WHY: Conductance and the Path of Least Resistance
- The Two-Branch Current Divider Formula
- Step-by-Step Derivation
- Three or More Branches: The Conductance Method
- Worked Examples
- Current Dividers and Voltage Dividers: The Dual Pair
- Fault Diagnosis: Open and Short Circuits
- Ammeter Shunts: A Practical Application
- Current Divider Calculator
- Experiment: Measure Current Division
- Current Dividers in Ham Radio
A current divider: total current splits between two parallel branches in inverse proportion to their resistances. The lower-resistance branch carries more current.
View LargerThe Physical WHY: Conductance and the Path of Least Resistance
Before memorising any formula, it is worth understanding exactly why current splits the way it does — and why the formula contains the "opposite" resistor in the numerator.
Think of water flowing into a pipe junction that splits into two branches. A wide pipe (low resistance to flow) lets water through easily; a narrow pipe (high resistance) restricts it. If both branches connect the same two points, the pressure difference across both branches is identical. The wide pipe carries more flow because the same pressure drives more current through less restriction. The narrow pipe carries less — not zero, but a fraction determined by how much narrower it is relative to the wide pipe.
Electricity behaves the same way. In a parallel circuit, all branches share the same voltage across them (this is the definition of a parallel connection). Each branch carries a current I = V/R, where V is the common voltage. A larger resistance means a smaller current; a smaller resistance means a larger current.
The clearest way to state the current divider principle is:
Conductance G = 1/R (measured in siemens, symbol S).
A branch with conductance Gn carries a fraction Gn/Gtotal of the total current.
A high-conductance (low-resistance) branch carries a large fraction. A low-conductance (high-resistance) branch carries a small fraction.
This conductance framing is the intuitive one. The traditional two-branch formula (with R2 in the numerator of I1) is simply this same relationship rewritten for two resistors — and the cross-over of resistor labels is a direct consequence of G = 1/R. Let us see exactly why.
The Two-Branch Current Divider Formula
For two resistors R1 and R2 in parallel carrying a total current Itotal:
I2 = Itotal × R1 / (R1 + R2)
Notice: I1 contains R2 in the numerator, and I2 contains R1. The formula always uses the other branch's resistance. Two quick consistency checks help you remember this:
- If R2 is very large (approaches infinity), I1 approaches Itotal — almost all current goes through R1 when R2 is near-open. This makes sense physically.
- If R1 = R2, then I1 = I2 = Itotal/2 — equal resistances split the current equally. This also makes sense.
Step-by-Step Derivation
Here is exactly how the formula comes from first principles.
(1) Both branches share the same voltage V across them (parallel connection).
(2) The parallel resistance is Rparallel = R1 × R2 / (R1 + R2).
(3) The common voltage is V = Itotal × Rparallel.
Current through R1:
I1 = V / R1 = [Itotal × R1 × R2 / (R1 + R2)] / R1
The R1 in the numerator of Rparallel cancels with the R1 in the denominator:
I1 = Itotal × R2 / (R1 + R2)
Current through R2:
I2 = V / R2 = [Itotal × R1 × R2 / (R1 + R2)] / R2
The R2 cancels:
I2 = Itotal × R1 / (R1 + R2)
The key insight: When you compute V/R1, the R1 in the product R1 × R2 cancels with the R1 in the denominator. What is left in the numerator is R2 — the other branch. This is why the formula "crosses over": each branch's current formula contains the other branch's resistance.
Alternatively, using conductances directly (this avoids the cross-over entirely):
I1 = Itotal × G1 / (G1 + G2)
I2 = Itotal × G2 / (G1 + G2)
This form is directly intuitive: each branch carries a fraction of the total current equal to its share of the total conductance. No cross-over — the branch with G1 in the formula is the branch that carries I1.
Three or More Branches: The Conductance Method
The two-branch formula does not generalise cleanly to three or more branches when expressed in terms of resistance — the algebra becomes messy. The conductance form extends trivially:
In = Itotal × Gn / (G1 + G2 + … + GN)
where Gn = 1/Rn (siemens)
Procedure for any number of branches:
- Convert each resistance to a conductance: Gn = 1/Rn.
- Sum the conductances: Gtotal = G1 + G2 + … + GN.
- Each branch current: In = Itotal × Gn / Gtotal.
- Verify: sum all branch currents — they must equal Itotal.
Three-branch current divider solved using conductances. Each branch current is proportional to that branch's conductance — the higher-conductance (lower-resistance) branch always carries the most current.
View LargerWorked Examples
Example 1 — Two branches (basic)
Using the two-branch formula:
IR100 = 60 × 150 / (100 + 150) = 60 × 150 / 250 = 36 mA
IR150 = 60 × 100 / (100 + 150) = 60 × 100 / 250 = 24 mA
Sanity check 1 — KCL: 36 + 24 = 60 mA ✓
Sanity check 2 — via common voltage:
Rparallel = (100 × 150)/(100 + 150) = 15000/250 = 60 Ω
V = 60 mA × 60 Ω = 3.6 V
IR100 = 3.6/100 = 36 mA ✓ IR150 = 3.6/150 = 24 mA ✓
Ratio check: IR100/IR150 = 36/24 = 1.5 = R2/R1 = 150/100 = 1.5 ✓
The current ratio always equals the inverse resistance ratio.
Example 2 — Three branches (conductance method)
Step 1 — Compute conductances:
G1 = 1/100 = 10.000 mS
G2 = 1/220 = 4.545 mS
G3 = 1/470 = 2.128 mS
Gtotal = 10.000 + 4.545 + 2.128 = 16.673 mS
Step 2 — Branch currents:
I1 = 120 × 10.000/16.673 = 71.97 mA
I2 = 120 × 4.545/16.673 = 32.71 mA
I3 = 120 × 2.128/16.673 = 15.31 mA
Verification: 71.97 + 32.71 + 15.31 = 119.99 mA ≈ 120 mA ✓ (rounding)
Verify via node voltage:
Rparallel = 1/Gtotal = 1/0.016673 = 59.97 Ω
Vnode = 120 mA × 59.97 Ω = 7.197 V
I1 = 7.197/100 = 72.0 mA ✓ I2 = 7.197/220 = 32.7 mA ✓ I3 = 7.197/470 = 15.3 mA ✓
Example 3 — Reverse problem: find unknown R given branch currents
Step 1 — Total current:
Itotal = 32 + 48 = 80 mA
Step 2 — The 48 mA branch is R2. Use the current divider formula for R1:
I1 = Itotal × R2/(R1 + R2)
32 = 80 × R2/(150 + R2)
32(150 + R2) = 80 R2
4800 + 32 R2 = 80 R2
4800 = 48 R2
R2 = 100 Ω
Step 3 — Node voltage:
V = I1 × R1 = 32 mA × 150 Ω = 4.8 V
Verify: I2 = 4.8/100 = 48 mA ✓ Itotal = 32 + 48 = 80 mA ✓
Example 4 — Find unknown R such that a branch carries a given percentage
Setting up the equation:
I1 = 0.20 × Itotal = Itotal × R2/(R1 + R2)
0.20 = R2/(470 + R2)
0.20(470 + R2) = R2
94 + 0.20 R2 = R2
94 = 0.80 R2
R2 = 94/0.80 = 117.5 Ω → use nearest standard value: 120 Ω
Verify with 120 Ω:
I1 = Itotal × 120/(470 + 120) = Itotal × 120/590 = 0.2034 × Itotal ≈ 20.3% ✓ (close enough given standard resistor tolerance)
Current Dividers and Voltage Dividers: The Dual Pair
Voltage dividers and current dividers are electrical duals — every quantity in one has an exact counterpart in the other, with voltage and current swapped, and series and parallel connections swapped. Understanding this duality helps you remember both formulas and predict behaviour from one by analogy with the other.
| Property | Voltage Divider | Current Divider |
|---|---|---|
| Resistor configuration | Series | Parallel |
| Source type | Voltage source (Vin) | Current source (Itotal) |
| Quantity split | Voltage (same current in all) | Current (same voltage across all) |
| Output proportional to | Resistance (R/Rtotal) | Conductance (G/Gtotal = (1/R)/(1/Rtotal)) |
| Larger R → | More voltage across it | Less current through it |
| Two-element formula | Vout = Vin × Rbottom/(Rtop+Rbottom) | I1 = Itotal × R2/(R1+R2) [opposite R] |
| General formula | Vn = Vin × Rn/Rtotal | In = Itotal × Gn/Gtotal |
| Law applied | KVL (voltages sum to Vin) | KCL (currents sum to Itotal) |
| Loading effect | Load in parallel with R reduces Vout | Load in series with branch reduces I through that branch |
| Practical application | Bias voltage reference, sensor conditioning | Ammeter range extension, RF current monitoring |
The deepest insight from this table: in a voltage divider, the element with more resistance gets more voltage. In a current divider, the element with more resistance gets less current. These are dual statements — both follow from V = IR applied in their respective topologies.
Fault Diagnosis: Open and Short Circuits
Understanding how a parallel circuit fails — and what measurements reveal which fault is present — is essential for practical troubleshooting.
Open-circuit fault in one branch
If one branch wire breaks or a component inside it fails open (infinite resistance), that branch carries zero current. In a circuit driven by a voltage source, the remaining branches are unaffected — they still see the same supply voltage and carry the same current as before. Only the total current drawn from the supply decreases.
- Symptom: One branch measures zero current; all others measure normal. Total supply current is lower than expected.
- Voltage measurement: Measure voltage across each branch — all should still read supply voltage (the open branch has voltage across it but carries no current).
- How to find it: Measure current in each branch individually, or measure the resistance of each branch with the power off. The open branch reads infinite (or very high) resistance.
In a current-source-driven divider (where Itotal is fixed regardless of the load), an open branch forces all that current through the remaining branches — they carry more current than intended and may overheat. The voltage across the combination also rises.
Short-circuit fault in one branch
If one branch fails short (zero resistance across it), the voltage across the entire parallel combination drops to zero. A short-circuit sets V = 0 across itself and, because all branches share the same voltage, all other branches also see V = 0 and carry zero current. All the current flows through the short.
- Symptom: Voltage across the combination is zero (or very low). Only the short branch carries any measurable current. All other branches read zero current. A fuse in the supply may blow.
- How to find it: With power off, measure resistance across each branch. The shorted branch reads near zero ohms.
- In ham equipment: A shorted bypass capacitor across a parallel resistor creates exactly this fault — it drags the node voltage to ground and starves all other branches. The symptom is an amplifier stage with no DC bias voltage at the node.
| Fault type | Voltage across combination | Affected branch current | Other branch currents |
|---|---|---|---|
| Normal operation | V = Itotal × Rparallel | Normal (I = V/R) | Normal |
| One branch open-circuit | Unchanged (voltage source) or rises (current source) | Zero | Unchanged (voltage source) or increases (current source) |
| One branch short-circuit | Drops to zero | All current flows through short | All drop to zero |
Ammeter Shunts: A Practical Application
A moving-coil ammeter movement (galvanometer) deflects full-scale at a small current — typically 50 µA to 1 mA — and has a small internal resistance Rm. To measure larger currents, a shunt resistor Rsh is placed in parallel with the movement. Most of the measured current bypasses the meter through Rsh, and only the tiny fraction that produces full-scale deflection passes through the movement.
Rsh = Rm × Im / (Imax − Im)
Where Im is the full-scale meter current and Imax is the maximum current to be measured. This is a direct current divider: the shunt carries (Imax − Im) and the meter carries Im. Both see the same voltage at full-scale: Vfs = Im × Rm.
Meter: 1 mA full-scale, Rm = 50 Ω.
Rsh = 50 × 1 / (100 − 1) = 50/99 = 0.505 Ω
At 100 mA: Ish = 99 mA, Vfs = 1 mA × 50 Ω = 50 mV = 99 mA × 0.505 Ω = 50 mV ✓
Multiplication factor: 100:1
Rsh = 50 × 1 / (10 − 1) = 50/9 = 5.56 Ω
At 10 mA: Ish = 9 mA, Vfs = 1 mA × 50 Ω = 50 mV = 9 mA × 5.56 Ω = 50 mV ✓
Multiplication factor: 10:1 — notice Rsh is 11× larger for 10× smaller range.
Rsh = 50 × 1 / (1000 − 1) = 50/999 = 0.0500 Ω
This is a very low resistance — in practice it would be a short length of copper or manganin wire of precise cross-section. The shunt voltage is still 50 mV at full scale.
Notice: range increased 10× (from 100 mA to 1 A) → Rsh decreased by approximately 10× (from 0.505 Ω to 0.050 Ω). This is the expected pattern.
A key practical point: the shunt resistor must be stable over temperature and time. Manganin alloy (copper-manganese-nickel) is the standard choice for precision shunts because its resistance changes very little with temperature.
Current Divider Calculator
Current Divider Calculator
Enter the total current and up to four parallel branch resistances. The calculator finds the current through each branch, the node voltage, and power dissipated in each branch.
Ammeter Shunt Designer
Find the shunt resistor value to extend a meter movement to a larger current range.
⚖ Experiment: Measure Current Division Between Two Branches
Directly verify the current divider formula by measuring branch currents in a parallel circuit and comparing to predictions.
- 9 V battery and clip connector
- R1: 100 Ω; R2: 220 Ω in parallel
- Breadboard and jumper wires
- Digital multimeter
- Use the calculator above to predict: total current, I through 100 Ω, and I through 220 Ω for a 9 V supply.
- Connect 100 Ω and 220 Ω in parallel across the 9 V battery on a breadboard.
- Measure total supply current: break the positive lead from the battery and insert the multimeter in series (current measurement mode).
- Measure branch currents individually: break one end of the 100 Ω resistor and insert the meter in series with it. Record IR100. Repeat for the 220 Ω branch.
- Verify: IR100 + IR220 ≈ Itotal.
- Compare the measured ratio IR100/IR220 to the predicted ratio R2/R1 = 220/100 = 2.2.
- Simulate an open-circuit fault: disconnect the 100 Ω branch. Measure Itotal and IR220. Compare to the prediction: IR220 = 9/220 = 40.9 mA, unchanged from before.
Predicted values: V = 9 V; Rparallel = (100 × 220)/(100 + 220) ≈ 68.75 Ω; Itotal ≈ 130.9 mA; IR100 = 9/100 = 90.0 mA; IR220 = 9/220 = 40.9 mA. The ratio IR100/IR220 ≈ 2.2 = R2/R1. When the 100 Ω branch is removed, Itotal drops to 40.9 mA — only the 220 Ω branch remains — while IR220 stays at 40.9 mA, unchanged (the voltage source keeps it at 9/220 regardless of the other branch).
Current Dividers in Ham Radio
Ammeter shunts in station meters
Analogue panel meters measuring supply current in older transceivers and linear amplifiers use shunt resistors as current dividers. The meter movement — typically 50–100 µA full-scale — has a low-value shunt in parallel that carries almost all the measured current. A 20 A supply meter might use a shunt of just a few milliohms; the shunt is designed so the voltage across it (usually 50–75 mV at full scale) is exactly what drives the meter movement to full deflection. The shunt material — usually manganin — must have very low temperature coefficient to maintain calibration across the temperature range of a shack environment.
RF current monitors
In antenna current monitors used to verify consistent field strength, a current transformer samples a fraction of the antenna current. The transformer secondary is loaded with a resistor, and the meter reads the voltage proportional to secondary current. The primary-to-secondary current ratio follows the transformer turns ratio — a generalisation of the current divider concept to magnetically coupled inductors. The current in the secondary is a known fraction of the antenna current, giving an indirect measurement of radiated power.
Parallel transistor stages and current sharing
In high-power RF amplifiers, two or more transistors may be connected in parallel to share the total collector current. If the transistors are perfectly matched, each carries half the total current. In practice, slight differences in transconductance and VBE cause one transistor to carry more current — it heats up, its transconductance increases (in some devices), and it takes even more current, leading to thermal runaway. Small-value emitter resistors (typically a few ohms) in series with each transistor act as equalising elements: the transistor carrying more current develops more voltage across its emitter resistor, reducing its effective VBE and forcing a more equal split. This negative-feedback mechanism is the engineering solution to the current-divider inequality problem in parallel transistor circuits.
Bias current injection and ALC circuits
In some automatic level control (ALC) circuits, a variable DC current is injected at a node that also connects to the main signal path. KCL applies at this node: the signal path current and the injected bias current must both be accounted for. Careful current divider analysis is needed to ensure the injected current shifts the operating point by the intended amount without disturbing the signal path impedance. If the ALC injection resistor is too small relative to the signal path impedance, it will also divert signal current, reducing gain more than intended.
Frequently Asked Questions
Why does the current divider formula use the opposite resistor in the numerator?
Because both branches see the same voltage. The current through R1 is V/R1, and V = Itotal × Rparallel = Itotal × (R1 × R2)/(R1 + R2). Dividing by R1 to get I1, the R1 in the numerator of Rparallel cancels with the R1 in the denominator, leaving R2 in the numerator. In conductance terms, this is simply I1 = Itotal × G1/(G1+G2) — no cross-over, naturally proportional to that branch's own conductance.
Does a current divider work the same way for AC circuits?
Yes, but the "resistances" become impedances (complex numbers combining resistance and reactance). The formula becomes I1 = Itotal × Z2/(Z1 + Z2), where Z1 and Z2 are complex impedances. Because impedances are complex, the branch currents may be out of phase with each other and with the total current. The magnitudes and phases must be handled as phasors. For purely resistive circuits at DC or low-frequency AC, the standard DC formula applies directly.
What is the dual relationship between voltage dividers and current dividers?
They are electrical duals: series/parallel topology swaps, voltage/current swaps, resistance/conductance swaps, and KVL/KCL swaps. In a voltage divider, Vout = Vin × Rn/Rtotal (proportional to resistance). In a current divider, In = Itotal × Gn/Gtotal (proportional to conductance). A larger resistance gets more voltage in a series circuit; a larger resistance gets less current in a parallel circuit.
How do I design a shunt for a digital multimeter?
Most digital multimeters do not use a classical moving-coil shunt. Instead, they use a precision low-value shunt resistor in series with the measurement path and measure the voltage across it using the high-accuracy internal voltmeter. The displayed current is Vshunt/Rshunt. The ammeter shunt formula above is relevant to analogue meter movements and external shunts for calibrating clamp meters or extending the range of analogue panel meters — not to modern digital bench instruments.
What happens to the other branches when one branch short-circuits?
A short circuit sets the voltage across the entire parallel combination to zero. Since all branches share the same voltage (by definition of parallel connection), all other branches see 0 V and carry 0 A. Every amp of the original total current flows through the short. This is why a shorted component in a parallel circuit can be so destructive — the short can carry much more current than it was designed for. In practice, a fuse or current-limited supply provides protection, but the short must be found and removed to restore normal operation.
What is a current mirror and how is it related to a current divider?
A current mirror is an active circuit (using transistors) that forces a second branch to carry the same current as a reference branch, regardless of the load resistance. It is more powerful than a passive current divider because it maintains its current ratio even if the supply voltage or load changes significantly. Current mirrors are used in integrated circuit biasing, active loads in amplifiers, and precision current sources. They can be thought of as "programmable current dividers" — the ratio is set by transistor geometry rather than by a passive resistor ratio.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.