The short answer

The familiar divider formula assumes its output is effectively unloaded. A load to ground sits in parallel with the lower resistor, reduces its effective resistance, and pulls the output down.

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Ten-kilohm divider with a ten-kilohm loadA five-volt source feeds a ten-kilohm upper resistor. The output node connects to a ten-kilohm lower resistor and a ten-kilohm load in parallel; both return to source negative. The ideal loaded output is about 1.67 volts, rather than the unloaded 2.5 volts.+−Vin = 5 VUpper 10 kΩLower 10 kΩLoad 10 kΩVout ≈ 1.67 V
Ten-kilohm divider with a ten-kilohm load

Locate the midpoint and return

Put an upper resistor between a 5 V source and an output node, and a lower resistor between that node and source negative. Measure Vout from the midpoint to the same negative reference. The usual ratio assumes an ideal regulated source, ohmic resistors, and no significant current leaving the midpoint. With no load, one current flows through both resistors; their voltage drops share the source voltage. An input connected to Vout may look passive, but if it draws current it changes the network being calculated.

Unloaded: Vout = Vs × Rlower / (Rupper + Rlower)

Worked example: 5 V with two 10 kΩ resistors

With Rupper = Rlower = 10 kΩ and an ideal 5 V source, the total series resistance is 20 kΩ. Current is 5 V / 20 kΩ = 0.25 mA. Each resistor drops 2.5 V, so the midpoint sits at 2.5 V relative to source negative. These are ideal nominal numbers. Real resistor tolerances, source variation, and the measuring instrument's finite input resistance affect a real result. A sufficiently high-impedance voltmeter usually perturbs the divider less than a low-resistance load, but no physical meter is perfectly invisible.

Iunloaded = 5 V / 20 kΩ = 0.25 mA; Vout = 2.5 V

Add 10 kΩ from midpoint to ground

A 10 kΩ load between Vout and source negative is in parallel with the 10 kΩ lower resistor. Their combined resistance is 5 kΩ. The divider is now 10 kΩ above 5 kΩ, so Vout = 5 V × 5 kΩ / 15 kΩ ≈ 1.6667 V, not 2.5 V. The upper resistor carries 5 V / 15 kΩ ≈ 0.333 mA. Since both lower branches have 1.6667 V across 10 kΩ, the lower resistor carries about 0.167 mA and the load also carries about 0.167 mA. The two lower branch currents add to the upper current, allowing for rounding.

10 kΩ ∥ 10 kΩ = 5 kΩ; Vout ≈ 1.6667 V; Iupper ≈ 0.333 mA; Ilower ≈ Iload ≈ 0.167 mA

High input impedance is a design condition

A voltmeter or sensing input draws less current when its input resistance is much larger than the divider's effective output resistance. For this 10 kΩ/10 kΩ pair, the ideal source set to zero leaves 10 kΩ ∥ 10 kΩ = 5 kΩ looking into the midpoint. A 10 kΩ load is only twice that resistance, so a large voltage shift is unsurprising. Check a device's specified input impedance, leakage, sampling behavior, and permitted voltage range rather than assuming every digital or analog input is an open circuit. Resistor tolerance and temperature can add errors even when loading is small.

Ideal divider output resistance = Rupper ∥ Rlower = 5 kΩ here

If the measured output is unexpectedly low

Disconnect power before moving leads. Confirm both resistor values and that the output really is between them rather than at one end. Check the return connection and rail continuity. Then measure source voltage and Vout with a correctly rated voltmeter across the intended two nodes; if Vout rises when the load is disconnected, loading is a strong suspect. An accidental short to ground, a faulty component, or a sagging source can also pull the output down. Measure across each resistor to distinguish these possibilities. Do not connect a current-mode meter across the midpoint and ground.

  1. Predict the unloaded midpoint from measured supply and resistor values.
  2. Identify everything attached to the midpoint, including the measuring input.
  3. Recalculate with each load in parallel with the lower branch, then compare safe voltage readings.

Choose a proper supply when current matters

A passive divider cannot hold its output fixed while its load current varies. Decreasing both resistors can reduce loading error for a given load, but raises standing current and heat; check resistor power and source capacity. For an actual low-voltage load supply, use an appropriately rated regulator or buffer designed for the required current and voltage. Divider calculations here describe steady DC with ideal resistors and an approximately constant 5 V source, not startup transients, nonlinear loads, or a guarantee of safe device operation. Keep the example on an isolated low-voltage supply.

Sources and further reading

Diagrams are simplified teaching illustrations, not measured results or construction plans. See how these guides are prepared.