Formula & Calculator
Voltmeter Loading Error
Describes how a voltmeter's finite input resistance R_m loads the circuit and causes a measurement error.
Interpretation
Voltmeter loading error: the measured voltage is reduced because the meter draws current, forming a voltage divider with the source resistance.
V_measured = V_true · (R_m/(R_m+R_s)).
Example: V_true=10V, R_m=1MΩ, R_s=100kΩ → V_measured = 10 × (1e6/1.1e6) ≈ 9.09V.
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
Frequently Asked Questions
When a voltmeter with input resistance R_m is connected across a source with source resistance R_s, the measured voltage is V_measured = V_true · (R_m / (R_m + R_s)). The error is due to the voltmeter drawing current.
It causes the measured voltage to be lower than the true open‑circuit voltage. The error is more significant when the source resistance is high or the voltmeter input resistance is low.
Use a voltmeter with a very high input resistance (e.g., digital multimeter with 10 MΩ or more). For very high‑impedance sources, use a buffer amplifier.
Common errors: 1) using the wrong resistance values, 2) forgetting to include the source resistance, 3) applying the formula to DC only (AC has capacitance effects), 4) ignoring the effect of the voltmeter's capacitance.
V_measured = V_true · (10 MΩ / (10 MΩ + 0.1 MΩ)) ≈ 0.99 V_true. The error is about 1%.
Selecting the right voltmeter for high‑impedance circuits, and understanding measurement limitations.
At high frequencies, the input capacitance can further load the circuit, causing additional error and phase shift. This is important in RF measurements.
Low‑voltage measurements are more susceptible to loading errors because the error is a fraction of the true value, independent of voltage level.