Current Divider Calculator

Use the current divider calculator to estimate how the current divides across each branch of different parallel circuits: resistive, inductive, or capacitive.

As a current divider, through R131.973% of the totalnote the ratio is the OPPOSITE way round from the voltage divider — current prefers the lower resistance

Output voltage3.8367 V12 V × 4,700 ÷ (10,000 + 4,700)

Division ratio0.31972831.973% of the input

Current through the divider816.3265 µAthe quiescent current this divider wastes continuously

Power wasted in the divider9.7959 mW

Dropped across R18.1633 V

Thévenin resistance3.1973 kΩR1 in parallel with R2 — the output impedance, and what decides how much a load disturbs the divider

As a current divider, through R268.027% of the total

Second arm output3.8367 V

Bridge output (difference)0 Vbalanced — R1/R2 equals R3/R4, so the bridge reads zero regardless of the supply voltage

Balance condition2.12766 against 2.12766a bridge balances when these ratios match, which is why it measures a resistance ratio without needing an accurate supply

The formula

V_out = V_in × R₂ ÷ (R₁+R₂); a current divider uses the opposite ratio

The ratio flips between voltage and current

A voltage divider gives the lower resistor's share: V_out = V_in × R₂/(R₁+R₂). A current divider gives the opposite ratio, because current takes the path of least resistance — the branch with R₁ carries the fraction R₂/(R₁+R₂). Both are shown above precisely because swapping them is such an easy mistake.

The Thévenin resistance is the figure that decides whether a divider is usable. It is R₁ in parallel with R₂, and any load comparable to it will drag the output down noticeably. A rule of thumb is that the load should be at least ten times the Thévenin resistance, or the divider needs a buffer amplifier.

There is a genuine trade-off in choosing the resistor values. Low values give a stiff output that tolerates loading but waste current continuously; high values waste almost nothing but cannot drive anything and pick up noise. A Wheatstone bridge sidesteps the supply-accuracy problem entirely: it balances on a ratio, so the reading is independent of the supply voltage.