Voltage Divider Calculator
Compute the unloaded and loaded output of a two-resistor divider, plus the current and the power in each resistor.
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About this tool
The circuit is Vin across R1 in series with R2, with the output taken across R2. The defaults are 5 volts, R1 of 10 kilohms, and R2 of 10 kilohms. The load box starts empty. Vin, R1, and R2 must be strictly positive or the report stops. The ideal output ignores the load and is Vin × R2 / (R1 + R2). With the defaults that is 5 × 10000 / 20000 = 2.5 V. The loaded output uses a bottom resistance that is R2 when the load is not a positive number, and the parallel combination of R2 and the load when it is. An empty load therefore makes the loaded line equal the ideal line.
Current is Vin divided by R1 plus that bottom resistance. Power in R1 is the current squared times R1. Power in the bottom is the current squared times the bottom resistance, so when a load is present that power is the power in the parallel pair, not a separate line for the load alone. With the defaults and no load, the current is 5/20000 amps, which prints as 250 µA, and each resistor dissipates 625 µW. Type a load of 10 kilohms, with the kilohm unit selected. The bottom becomes the parallel of 10 kΩ and 10 kΩ, which is 5 kΩ. Ideal output stays 2.5 V. Loaded output is 5 × 5000 / 15000 = 1.6666666666667 V. Current prints as 333.33333333333 µA.
A second familiar case is 9 volts, R1 of 2 kilohms, and R2 of 1 kilohm, with no load. Ideal and loaded output are both 3 V, and the current is 3 mA. The resistance menus default to kilohms, so the number 2 in R1 is 2000 ohms only while that menu says kilohms. Switching a menu to ohms without changing the number divides the resistance by 1000 and moves the output. The load menu also defaults to kilohms. A load that is blank or not positive is ignored. A load of 0 is not positive, so it does not short the output. The page will not model a short. You would need a positive load small enough to pull the voltage down. Source impedance is not a separate input. If the source has resistance, add it to R1 yourself. Resistor tolerance is not applied. The two power figures assume DC and the loaded current, and they do not check the wattage printed on a real part. The formatter writes 10000 ohms as 10 kΩ, 0.00025 amps as 250 µA, and 0.000625 watts as 625 µW. There is no AC impedance, no capacitor, and no Thevenin button. The ideal line is always present so you can see how far the load moved the output.
Example. The defaults, 5 V with R1 and R2 at 10 kΩ and no load, report an ideal output of 2.5 V, a loaded output of 2.5 V, current 250 µA, and 625 µW in each resistor. A 10 kΩ load drops the loaded output to 1.6666666666667 V.
Limit. Vin, R1, and R2 must be positive. A non-positive load is ignored. Tolerance, source impedance, and AC impedance are not in the formula.
What do the default 5 V, 10 kΩ, and 10 kΩ give?
Ideal output is 5 × R2 / (R1 + R2) = 2.5 V. With the load box empty, the loaded output is the same 2.5 V. Current is 250 µA, and each resistor shows 625 µW. The unit menus start on kilohms.
How does the load change the output?
A positive load is placed in parallel with R2. A 10 kΩ load across the default 10 kΩ bottom makes 5 kΩ. Loaded output becomes 1.6666666666667 V, while the ideal line stays 2.5 V. Current is then 333.33333333333 µA. A blank or zero load is ignored, so it does not short the divider.
Where is the source resistance?
There is no separate source-resistance field. Any resistance in series with the source has to be included in R1. The power lines use the loaded current and do not compare the result with a resistor's rated wattage.
Why do both output lines match when I leave the load empty?
The loaded path uses R2 itself unless the load parses as a positive number. Empty, zero, and negative loads all leave the bottom equal to R2, so the two voltages match. The ideal line is still printed so a later load has something to be compared with.