Voltage, current and power at low frequency
An analog voltmeter of 20 kΩ/V measuring a 100 kΩ divider reads 4 V where there are 5. It is not out of calibration: connecting it changed the circuit.
01Every instrument modifies what it measures
To measure a voltage, the voltmeter is connected in parallel and draws some current; to measure a current, the ammeter is inserted in series and adds resistance. In both cases the circuit changes when the instrument is connected, and what is measured is the modified circuit. This loading effect is a systematic error that can be calculated and, almost always, made negligible by choosing the instrument well.
A divider of two equal resistors fed with 10 V: at the midpoint there are 5 V, with a Thévenin resistance equal to half of each resistor. Four voltmeters of different input resistance measure that point.
The ammeter has its own effect: the burden voltage. On the 200 mA range, a typical multimeter has about 1 Ω between the shunt and the fuse. In a 5 V circuit with a load of 25 Ω, that lowers the current from 200 mA to 192 mA: a 3.8% error that no instrument accuracy can compensate.
02Analog instruments
| Mechanism | What it responds to | Use |
|---|---|---|
| Moving coil and permanent magnet | Average value of the current | DC; with a rectifier, sinusoidal AC |
| Moving iron | RMS value, up to a few hundred Hz | 50 Hz AC switchboards |
| Electrodynamic | Product of two currents: RMS value or power | Laboratory wattmeters |
The moving-coil voltmeter is specified by its sensitivity in ohms per volt: a 20 kΩ/V one on the 10 V range has 200 kΩ of input resistance, and on the 1 V range, only 20 kΩ. The lower the range, the greater the loading effect.
03Digital instruments
A digital multimeter converts with a dual-slope or sigma-delta integrating ADC, which averages over a whole number of mains cycles and thus rejects the 50 Hz hum. Its DC input resistance is 10 MΩ on all ranges, and some offer more than 10 GΩ on the lowest ones. In AC there are two classes: average-responding meters calibrated for a sine wave, and true-RMS meters, which measure any waveform correctly within their bandwidth and crest factor. The difference is analyzed in non-sinusoidal signals.
The uncertainty of the reading is evaluated as seen in errors and uncertainty.
04Power measurement
In DC, \( P = V I \), but the connection matters: if the voltmeter is placed directly across the load, the ammeter also measures the voltmeter’s current; if it is placed before the ammeter, it also measures the drop across it. The first error is relative to \( R_L / R_v \) and the second to \( R_a / R_L \): for low-resistance loads the voltmeter should go across the load, and for high ones, before the ammeter. The boundary is at \( R_L = \sqrt{R_a R_v} \).
In AC, the active power is the average of the instantaneous product, which a wattmeter calculates directly:
In three-phase three-wire systems two wattmeters are enough (the Aron method): the sum of their readings is the total power, with any load. If the load is balanced, moreover, \( \tan\varphi = \sqrt{3}\,(W_1 - W_2)/(W_1 + W_2) \).
05Measuring current without breaking the circuit
| Sensor | Principle | Measures |
|---|---|---|
| Shunt | Drop across a known resistance, of milliohms | DC and AC, requires breaking the circuit |
| Current-transformer clamp | The conductor is the primary of a current transformer | AC only |
| Hall-effect clamp | A Hall sensor in the core gap measures the field | DC and AC |
| Rogowski coil | A coreless coil measures the derivative of the current, which is integrated | High-current, high-frequency AC |
06In the laboratory
Measure the midpoint of 10 kΩ, 1 MΩ and 10 MΩ dividers with an analog and a digital multimeter. Compare with the lab calculation.
Measure the power in a 10 Ω load and in a 100 kΩ one with the two possible connections of voltmeter and ammeter, and explain which is better in each case.
Measure the power of a three-phase motor with two wattmeters and calculate the power factor.
07Common mistakes
- Measuring high-impedance circuits with an analog meter. The voltmeter becomes part of the circuit.
- Forgetting the ammeter’s burden voltage. In low-voltage circuits it can be the largest source of error.
- Calculating AC power as V times I. That is the apparent power; the power factor is missing.
- Leaving the multimeter on the current range and measuring voltage. It is a short circuit through the shunt; the fuse, if it is the right one, saves it.
08Self-assessment
What does a 200 kΩ voltmeter read at a 5 V point with 50 kΩ Thévenin resistance?
\( 5 \cdot 200 / 250 = 4 \) V: 20% less.
What is the input resistance of a 20 kΩ/V voltmeter on the 50 V range?
1 MΩ.
With \( R_a = 0.1\ \Omega \) and \( R_v = 10\ \text{M}\Omega \), which connection is better for a 10 Ω load?
\( \sqrt{0.1 \cdot 10^7} = 1000\ \Omega \): the load is smaller, so the voltmeter should go directly across it.
Two wattmeters read 3 kW and 1 kW on a balanced three-phase load. What is the power factor?
\( \tan\varphi = \sqrt 3 \cdot 2/4 = 0.866 \), \( \varphi = 40.9° \), \( \cos\varphi \approx 0.76 \).
09Further reading
- Albert D. Helfrick and William D. Cooper. Modern Electronic Instrumentation and Measurement Techniques (Spanish edition: “Instrumentación electrónica moderna y técnicas de medición”). Prentice Hall, 1990. The classic text: analog and digital instruments, bridges and oscilloscopes.
- Keithley Instruments. Low Level Measurements Handbook. 7th ed., Keithley, 2014. Loading effect, leakage and noise when measuring very small voltages and currents.
- Alan S. Morris and Reza Langari. Measurement and Instrumentation: Theory and Application. 3rd ed., Academic Press, 2020. A current overview of instruments and sensors, with their errors.