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Difficulty: HardMultimeter: Voltage and Current Measurement

An electrical technician troubleshooting a 24 V DC24\text{ V DC} control panel accidentally connects a digital multimeter (configured for DC voltage measurement with test leads plugged into the COM\text{COM} and V/Ω\text{V}/\Omega jacks) in series with an electromechanical relay coil instead of in parallel across it. When the 24 V DC24\text{ V DC} circuit is energized, which of the following describes the resulting behavior of the relay and the multimeter display?

  1. A
    The internal multimeter fuse blows immediately due to excessive current, and the meter display goes blank.
  2. B
    The relay energizes normally, and the multimeter displays 0.0 V0.0\text{ V} because the meter behaves as a low-resistance path.
  3. The relay remains de-energized, and the multimeter displays approximately 24 V24\text{ V}.Answer
  4. D
    The relay chatters rapidly on and off while the multimeter displays an alternating current ripple error.

Answer

The relay remains de-energized, and the multimeter displays approximately 24 V24\text{ V}.
A digital multimeter configured for voltage measurement has a very high internal input impedance (typically 10 MΩ10\text{ M}\Omega). When placed in series with a load like a relay coil, it acts like an open circuit, limiting the current flow to a few microamperes. This current is insufficient to actuate the relay. Because the meter's resistance is overwhelmingly larger than the coil resistance, the voltage divider effect causes virtually the entire 24 V24\text{ V} source voltage to appear across the multimeter terminals.

Step-by-Step Solution

1
Determine the internal electrical characteristics of the digital multimeter in DC voltage mode.
The meter presents an extremely high input impedance, typically 10 MΩ10\text{ M}\Omega (10000000 Ω10{}000{}000\text{ }\Omega).
Voltmeters are designed with high internal resistance to avoid drawing significant current or loading down circuits when placed in parallel.
2
Calculate the total series resistance and the resulting circuit current.
Total circuit resistance is Rtotal=Rmeter+Rcoil≈10 MΩR_{\text{total}} = R_{\text{meter}} + R_{\text{coil}} \approx 10\text{ M}\Omega, yielding a current of approximately I=24 V10 MΩ=2.4 μAI = \frac{24\text{ V}}{10\text{ M}\Omega} = 2.4\text{ }\mu\text{A}.
According to Ohm's law, this microampere-level current is far below the tens of milliamperes required to energize the relay coil.
3
Determine the voltage distribution across the series combination.
Nearly all 24 V24\text{ V} drops across the 10 MΩ10\text{ M}\Omega multimeter, while less than 1 mV1\text{ mV} drops across the relay coil.
By the voltage divider principle, the component with the dominant resistance drops the vast majority of the supply voltage, causing the meter to read approximately 24 V24\text{ V}.

Key Concept

Internal input impedance of a voltmeter and the circuit loading effect of series voltmeter insertion
Estimated Time:1m 30s
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Topics

Multimeter: Voltage and Current Measurement

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