Voltage dividers: why a connected load changes the output
Compare an unloaded resistor divider with a loaded output using equivalent resistance.
Method
The basic voltage-divider formula assumes no current leaves the midpoint except through the lower resistor. If R1 connects the supply to the midpoint and R2 connects the midpoint to the return, output voltage is Vin × R2 / (R1 + R2). A load connected between the midpoint and return is in parallel with R2. Replace R2 with that parallel equivalent before using the same divider formula. A measuring device also acts as a load, even when its input resistance is large.
Worked example
Use the resistor values and supply shown below. Equal resistors give half the supply voltage when unloaded. Connecting a load equal to the lower resistor halves that lower branch's equivalent resistance, so the midpoint voltage drops. The supply current also increases. To reproduce the loaded calculation, find the parallel equivalent first, then enter it as the lower resistance in Voltage divider. Check the direction of the change: this positive-resistance load cannot raise the ideal midpoint voltage.
Vin = 12 V; R1 = 1000 Ω; R2 = 1000 Ω
Vout = 12 × 1000 / (1000 + 1000) = 6 V
RL = 1000 Ω; R2 ∥ RL = 500 Ω
Vout = 12 × 500 / (1000 + 500) = 4 V
Checks and limits
The calculation assumes a steady ideal supply and fixed resistances. Component tolerance, heating and source resistance can change a real measurement. A divider is suitable for some signal and reference tasks but does not hold a power supply voltage constant as current demand changes. Check each resistor's dissipated power and ratings separately. Active loads, capacitors and frequency-dependent impedances require an appropriate circuit model; this resistor-only calculation does not establish electrical safety or approve a practical design.