Current Divider Calculator

Basic Current Divider Calculator

Current Divider Formula:

I₁ = I_total × R₂ / (R₁ + R₂) I₂ = I_total × R₁ / (R₁ + R₂)

I = 10mA

I = 100mA

R1 = 10kΩ, R2 = 10kΩ

R1 = 1kΩ, R2 = 10kΩ

Total Current (I_total):

Current through R1 (I₁):

Current through R2 (I₂):

Resistance 1 (R₁):

Resistance 2 (R₂):

Enter any three values and click Calculate.

Note: The circuit diagram updates automatically as you enter values.

Auto-calculate as you type

Multi-Resistor Current Divider

Multi-Resistor Current Divider Formula:

I_n = I_total × (1/R_n) / Σ(1/R_i) where Σ(1/R_i) is the sum of reciprocals of all resistances

Total Current (I_total):

Resistor 1 (R₁):

Resistor 2 (R₂):

Add resistors and enter total current to calculate individual currents.

Note: You can add up to 10 resistors in parallel.

AC/Impedance Current Divider

AC Current Divider Formula:

I₁ = I_total × Z₂ / (Z₁ + Z₂) where Z = R + jX (impedance)

Current Input

Total Current Magnitude:

Phase Angle (degrees):

Frequency (Hz):

Impedance 1

Resistance (R₁):

Inductance (L₁):

Capacitance (C₁):

Impedance 2

Resistance (R₂):

Inductance (L₂):

Capacitance (C₂):

Enter impedance values and current to calculate AC current division.

Tolerance Analysis

Tolerance Analysis:

Analyzes worst-case scenarios considering resistor tolerances

Total Current:

Resistance 1:

R1 Tolerance (%):

Resistance 2:

R2 Tolerance (%):

Enter resistor values and tolerances to analyze current variations.

Tolerance Analysis Details

This analysis shows how resistor tolerances affect current distribution.

Standard Resistor Calculator

Total Current:

Desired Current through R1:

Resistor Series:

Enter total current and desired branch current to find standard resistors.

Current Divider Examples & Applications

LED Current Control

Given: I_total=20mA, R₁=470Ω, R₂=220Ω

I₁ = 20mA × 220Ω / (470Ω + 220Ω) = 6.4mA
I₂ = 20mA × 470Ω / (470Ω + 220Ω) = 13.6mA
Used for controlling LED brightness in parallel

Current Sensing Application

Given: I_total=500mA, R_shunt=0.1Ω, R_sense=1kΩ

I_shunt = 500mA × 1000Ω / (0.1Ω + 1000Ω) ≈ 500mA
I_sense = 500mA × 0.1Ω / (0.1Ω + 1000Ω) ≈ 0.05mA
Used for high-current measurement with minimal loss

Signal Processing Example

Given: I_total=10mA, R₁=10kΩ, R₂=10kΩ

I₁ = 10mA × 10kΩ / (10kΩ + 10kΩ) = 5mA
I₂ = 10mA × 10kΩ / (10kΩ + 10kΩ) = 5mA
Equal current division for balanced amplifier inputs