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An unloaded voltage divider provides a fraction of its input voltage across the lower resistor R2.
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Formula
U in V; R in Ω
| Symbol | Meaning | Unit |
|---|---|---|
| Uout | Output voltage across lower resistor R2 | V |
| Total input voltage across the series pair | V | |
| R1, R2 | Series resistors; output is measured across R2 | Ω |
U_\mathrm{out}=U_\mathrm{in}\frac{R_2}{R_1+R_2}The equation follows from the same current through both series resistors and Ohm’s law. It assumes an unloaded output; a connected load lies in parallel with R2 and changes the division. At high resistance values, input and leakage currents can also matter. State polarity and output tap clearly.
The equation follows from the same current through both series resistors and Ohm’s law. It assumes an unloaded output; a connected load lies in parallel with R2 and changes the division. At high resistance values, input and leakage currents can also matter. State polarity and output tap clearly.
An unloaded voltage divider provides a fraction of its input voltage across the lower resistor R2.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Voltage divider with two series resistors” and check its unit and magnitude.
For Uin=12 V, R1=2 kΩ and R2=1 kΩ, Uout=12·1/(2+1)=4 V.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: U in V; R in Ω.
Formula cluster
An unloaded voltage divider provides a fraction of its input voltage across the lower resistor R2.
For Uin=12 V, R1=2 kΩ and R2=1 kΩ, Uout=12·1/(2+1)=4 V.
Create reference and measurement voltages from a known supply.
Here is how to work through a typical Voltage divider with two series resistors (Uout = Uin·R₂/(R₁+R₂)) task step by step:
Task
Work through the numerical example for “Voltage divider with two series resistors” and check its unit and magnitude.
Solution path
For Uin=12 V, R1=2 kΩ and R2=1 kΩ, Uout=12·1/(2+1)=4 V.
The chosen tap is between R1 and R2, with R2 connected to the reference node. Voltage across R2 is therefore the lower share of total series voltage. Across R1, the numerator would instead be R1. Always state the output tap and reference direction. The simple ratio assumes two series resistors and a high-impedance output measurement. Output is across R2. A real load changes the effective lower resistance, while tolerance and source resistance shift the result; include them when accuracy matters. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
A load RL at the output is in parallel with R2. Replacing R2 by R2∥RL lowers the effective lower resistance and usually reduces Uout. The unloaded formula is a good approximation only when RL is much larger than R2. Measurement-device impedance matters for accurate dividers. The simple ratio assumes two series resistors and a high-impedance output measurement. Output is across R2. A real load changes the effective lower resistance, while tolerance and source resistance shift the result; include them when accuracy matters. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
For target ratio r=Uout/Uin, r=R2/(R1+R2). Many resistor pairs give the same ratio but draw different current and load the source differently. Smaller resistors improve load robustness but increase power loss. Choose values based on source impedance, load and power too. The simple ratio assumes two series resistors and a high-impedance output measurement. Output is across R2. A real load changes the effective lower resistance, while tolerance and source resistance shift the result; include them when accuracy matters. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
Write the quantity and unit from the variable list next to every value: U in V; R in Ω. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For Uin=12 V, R1=2 kΩ and R2=1 kΩ, Uout=12·1/(2+1)=4 V. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. The simple ratio assumes two series resistors and a high-impedance output measurement. Output is across R2. A real load changes the effective lower resistance, while tolerance and source resistance shift the result; include them when accuracy matters. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
The background states this starting condition: The equation follows from the same current through both series resistors and Ohm’s law. It assumes an unloaded. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. The simple ratio assumes two series resistors and a high-impedance output measurement. Output is across R2. A real load changes the effective lower resistance, while tolerance and source resistance shift the result; include them when accuracy matters. Finally check units, assumptions, measurement conditions and reference quantities against the problem before using the result.
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How do you rearrange Voltage divider with two series resistors for Base equation for the target quantity?
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Create a curated FSRS exam set for Uout = Uin·R₂/(R₁+R₂): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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