Voltmeter Loading Error Calculator
How much a meter's finite input resistance pulls down the voltage it is trying to measure.
Inputs
Results
Formula breakdown
Reading accuracy versus meter resistance
The curve shows what percentage of the true voltage the meter displays. The shaded band is the reading range allowed by your loading-error limit.
Equivalent circuit
Conversion tables
| Rm/Rs | Loading error | Reading for Vt = 10 V |
|---|
| Target error | Min Rm/Rs | Min Rm for Rs = 10 kΩ |
|---|
Typical meter input resistances
| Instrument | Input resistance |
|---|
Key formulas and references
- Horowitz & Hill, The Art of Electronics, Cambridge University Press: source and load impedance.
- Nilsson & Riedel, Electric Circuits, Pearson: Thévenin equivalent circuits.
- Wolf & Smith, Student Reference Manual for Electronic Instrumentation Laboratories, Prentice Hall: meter loading.
- Morris & Langari, Measurement and Instrumentation, Academic Press.
- Tektronix, The XYZs of Oscilloscopes primer: probe and input loading.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| V_measured | Measured voltage (with loading effect) | V |
| V_true | True (open-circuit) voltage | V |
| R_m | Voltmeter internal resistance | Ω |
| R_s | Source (Thevenin) resistance | Ω |
What it means
When a voltmeter is connected to measure a voltage, it draws current from the circuit, affecting the measured value. The loading error is due to the meter’s internal resistance R_m forming a voltage divider with the source resistance R_s. The measured voltage is V_measured = V_true · (R_m / (R_m + R_s)). To minimize loading error, the voltmeter should have a very high resistance compared to the source resistance. This is why digital voltmeters have input impedances of 10MΩ or more. Example: A source with V_true=10V and R_s=100kΩ. A voltmeter with R_m=1MΩ reads V_measured = 10 * (1e6/(1e6+1e5)) = 10 * (1e6/1.1e6) = 9.09V, giving a 9.1% error. A meter with 10MΩ would give 9.9V (1% error).
Worked example
Voltmeter Loading Error – Practical Example
Real‑World| Parameter | Value |
|---|---|
| Vtrue | 5 V |
| Rs | 1 MΩ |
| Rm | 1 MΩ |
| Formula | Vm = Vtrue · Rm/(Rm+Rs) |
Common mistakes
Watch unit consistency and the assumptions behind the formula; misapplying it outside its valid conditions is the most frequent error.Applications
Voltmeter loading error occurs because the meter draws current, forming a voltage divider with the source resistance. V_measured = V_true · (R_m/(R_m+R_s)). Engineers use this to understand why high‑impedance voltmeters are needed for low‑power circuits. By selecting a meter with high input impedance, they minimise this error. This is a practical consideration in circuit measurements.
- Measurement error analysis in voltage measurements
- Selection of voltmeters with adequate input impedance
- Design of buffer amplifiers for measurement isolation
- Educational understanding of loading effects
Related calculators · Instrumentation & Measurement
Recently updated formulas
Suggest an improvement
Found a mistake? Have a better example? Let us know.
You may send up to 3 suggestions per 24 hours.