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  1. Home
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  4. /Current Divider Calculator

Current Divider Calculator

Calculator

Results

Current through R1

8

A

Current through R2

2

A

Current through R3

0

A

Voltage across Branches

800

V

Equivalent Parallel Resistance

80

Ω

Total Power

8,000

W

Power in R1

6,400

W

Power in R2

1,600

W

Power in R3

0

W

R1 Current Share

80

%

R2 Current Share

20

%

R3 Current Share

0

%

Results

Current through R1

8

A

Current through R2

2

A

Current through R3

0

A

Voltage across Branches

800

V

Equivalent Parallel Resistance

80

Ω

Total Power

8,000

W

Power in R1

6,400

W

Power in R2

1,600

W

Power in R3

0

W

R1 Current Share

80

%

R2 Current Share

20

%

R3 Current Share

0

%

The current divider is the dual of the voltage divider: just as a voltage divider splits voltage across series resistors in proportion to resistance values, a current divider splits current among parallel resistors in inverse proportion to their resistance values. More specifically, the current through each parallel branch equals the total current multiplied by the ratio of total parallel conductance to that branch's conductance: I_n = I_total × G_n / G_total = I_total × (1/R_n) / (1/R_total).

For two parallel resistors, the simplified current divider formula is: I1 = I_total × R2/(R1+R2) and I2 = I_total × R1/(R1+R2). Notice that I1 depends on R2 (the other resistor), not R1 — the branch with lower resistance carries more current, but the formula uses the opposite resistor in the numerator. This is the key distinction from the voltage divider, where Vout = Vin × R2/(R1+R2) depends on R2 itself.

Current dividers appear throughout circuit design. In current mirror circuits — the backbone of analog IC design — a reference current is mirrored (copied) to multiple output branches by sizing transistors rather than resistors, but the underlying principle is current division. In instrumentation amplifiers, current flow through matched resistor networks must be precisely controlled. In power distribution, current sharing among parallel conductors or fuses depends on their relative resistance.

Practical current sharing challenges arise when parallel components have even small resistance mismatches. In parallel power MOSFETs carrying high load current, a MOSFET with slightly lower R_DS(on) carries more current, heats up more, and if the temperature coefficient is positive (R_DS(on) increases with temperature), self-stabilizes. This 'positive temperature coefficient' is why MOSFETs parallel better than bipolar transistors, which have negative temperature coefficient that causes thermal runaway in parallel configurations.

Battery cells in parallel are another current divider application. Each cell has internal resistance, and current divides inversely to internal resistance. Cells with lower internal resistance (healthier cells) contribute more current, while aged, high-internal-resistance cells contribute less. This self-balancing effect is benign up to a point, but if cell internal resistance differs greatly, the low-resistance cell may be forced to supply most of the current, accelerating its degradation.

This calculator provides branch currents for two or three parallel resistors, the common voltage across the parallel combination, and the total equivalent resistance — a complete current divider analysis toolkit.

Visual Analysis

How It Works

I_n = I_total × G_n / G_total where G_n = 1/R_n and G_total = 1/R1 + 1/R2 + (1/R3 if R3 > 0). Parallel voltage V = I_total × R_total = I_total / G_total. Total resistance R_total = 1/G_total. Verify: I1 + I2 + I3 = I_total (KCL). Set R3 = 0 to use only two branches.

Understanding Your Results

Branch with lowest resistance carries highest current (inverse relationship). Branch currents sum exactly to I_total (Kirchhoff's Current Law check). The voltage across all branches is equal: V = I_total × R_parallel. For equal resistors: each branch carries I_total/n equally. Current imbalance = (I_max − I_min)/I_total × 100%.

Worked Examples

Parallel LED Strings

Inputs

i total0.3
r1100
r2100
r3100

Results

i10.1
i20.1
i30.1
v parallel10
r total33.333

Three equal 100 Ω resistors in parallel divide 300 mA equally — 100 mA per branch, 10 V across each (same voltage). This equal splitting requires matched resistor values for equal LED brightness.

Ammeter Shunt Design

Inputs

i total1
r11
r2999
r30

Results

i10.999
i20.001
i30
v parallel0.999
r total0.999

A 1 Ω shunt in parallel with a 999 Ω meter movement: 99.9% of current (999 mA) goes through the shunt, only 1 mA through the meter. This extends a 1 mA full-scale meter to 1 A range.

Frequently Asked Questions

I1 = I_total × R2 / (R1 + R2) and I2 = I_total × R1 / (R1 + R2). Note the crossover: I1 uses R2, I2 uses R1. Lower resistance branch gets more current. Sum: I1 + I2 = I_total × (R2+R1)/(R1+R2) = I_total ✓. This formula applies only to two branches; for more, use the conductance formula.

Ammeters use a low-resistance shunt (R_shunt) in parallel with the meter movement (R_meter). The shunt diverts most of the measured current: I_meter = I_total × R_shunt/(R_shunt+R_meter). For a 1 mA full-scale movement (R_meter = 100 Ω) to read 10 A: R_shunt = (0.001 × 100) / (10 − 0.001) ≈ 0.01 Ω. The shunt must dissipate I²R power accurately.

Parallel MOSFETs share current inversely proportional to their R_DS(on). For equal sharing, use devices from the same lot with matched R_DS(on). Small source resistors (0.01–0.1 Ω) in series with each MOSFET degeneration resistors add balancing resistance that dominates over R_DS(on) mismatch, improving current sharing to within ±5% per source resistor voltage drop.

A short circuit (R → 0) in one parallel branch creates zero voltage across the entire parallel combination (ideal case). With zero voltage across remaining branches, their currents also drop to zero. In practice, the source has finite resistance, creating a voltage equal to I_sc × R_source — still very small. This is why fuses and breakers must interrupt short-circuit current in parallel branch circuits.

Norton's theorem states that any linear two-terminal network can be replaced by an ideal current source I_N in parallel with resistance R_N. The Norton current I_N is the short-circuit current at the terminals; R_N equals the Thevenin resistance. When a load is connected, current divides between R_N and R_load per the current divider formula: I_load = I_N × R_N/(R_N + R_load).

In AC circuits, replace R with complex impedance Z. The current divider formula becomes I_n = I_total × Z_parallel / Z_n, where Z_parallel is the parallel combination impedance. At resonance in parallel LC circuits, the tank impedance becomes very high (Z_parallel → ∞ for ideal LC), and source current is nearly zero while circulating tank currents can be Q times the source current.

Sources & Methodology

Nilsson & Riedel, Electric Circuits. Boylestad, Introductory Circuit Analysis. Sedra & Smith, Microelectronic Circuits. IEEE Std 100.
R

Roboculator Team

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