How to calculate original voltage drop across resistor if circuit is not loaded

Thread Starter

darkc0der

Joined Dec 20, 2023
14
Suppose my dmm loads the circuit when i try to measure the potential difference across a resistor. Is there a way to know what the original voltage drop would have been if the multimeter did not load the circuit? Is there an equation that i can plug in the new voltage to find the original?
 

k1ng 1337

Joined Sep 11, 2020
960
Find out the resistance of the probes then ask yourself if a voltage divider is formed during a voltage measurement. Next, use Kirchhoff's laws to calculate how much the probes load the circuit.

Keep in mind some meters are passive and others are active. Regardless of the method, the circuit under test is affected due to coupling. Spend some time researching the different methods for voltage measurement because the answer depends on your test apparatus.
 
Last edited:

WBahn

Joined Mar 31, 2012
30,072
Suppose my dmm loads the circuit when i try to measure the potential difference across a resistor. Is there a way to know what the original voltage drop would have been if the multimeter did not load the circuit? Is there an equation that i can plug in the new voltage to find the original?
Most DMMs appear as a high resistance to most circuits, particularly if you are measuring a DC voltage, which is probably the case if you are using a DMM in the first place.

To make the adjustment, you need to know not only the effective input resistance of the meter, but also the resistance that you are taking the measurement across (the load resistor) as well as the resistance of the circuit driving it, as seen by the load resistor.

If either the load resistor or the effective resistance of the circuit driving it is much smaller than the input resistance of the meter, the adjustment will be very small and likely less than the tolerance of the DMM's voltage measurement capabilities.

Most DMMs have an input resistance of 10 MΩ, so unless both of the other resistances are greater than, say 100 kΩ, it's probably not worth making the effort.
 

Thread Starter

darkc0der

Joined Dec 20, 2023
14
Find out the resistance of the probes then ask yourself if a voltage divider is formed during a voltage measurement. Next, use Kirchhoff's laws to calculate how much the probes load the circuit.

Keep in mind some meters are passive and others are active. Regardless of the method, the circuit under test is affected due to coupling. Spend some time researching the different methods for voltage measurement because the answer depends on your test apparatus.
Kirchoff's law you mean i have to measure the current entering the test point?
 

WBahn

Joined Mar 31, 2012
30,072
Kirchoff's law you mean i have to measure the current entering the test point?
No. Doing that would add yet another measurement taken in a way such that making that measurement affects that measurement -- and probably in a much more significant way.

What is it you are trying to do? Why are you trying to do it? Provide details. What is the resistor you are measuring the voltage across? What is providing the voltage to that resistor? What is the DMM that you are using? How accurate does your measurement need to be?
 

crutschow

Joined Mar 14, 2008
34,459
You need to determine the equivalent Thevenin resistance of the node you are measuring, to determines how much the reading is affected by the meter input resistance.
 
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