Effects of measuring voltage with a multimeter
Emmett Ientilucci
Most people do not consider the effects a multimeter or digital multimeter (DMM) has in an electrical circuit.
For example, consider the circuit below:
R1
V
R2
Rm
The above schematic would be the actual schematic if you were trying to measure the voltage across R2. When
measuring voltage with a DMM, the meter acts as a resistor in parallel with the device you are trying to measure. As
long as the meters resistance is substantially larger than the resistance of the device you are trying to measure, the error
in the measured voltage will be minimal. The idea is to have the meters resistance be as large as possible (compared to
the device you are measuring). In this way, the meters resistance will not drastically alter the value of R2, the resistor
in question. Example, if R2 was 1 kΩ and Rm was 20 MΩ, the equivalent resistance would still be about 1 kΩ. The
voltage reading given by the meter would be for a 1 kΩ resistor. However, if R2 was 15 MΩ and Rm was 20 MΩ, the
equivalent resistance would now be 8.5 MΩ, not 15 MΩ! The voltage reading given by the meter would be for an
8.5 MΩ resistor not the 15 MΩ that is actually in the circuit!
Here is an example,
V total = 5v
R1 = 10 kΩ
R2 = 15 MΩ
Rm = 22 MΩ (typically)
Q: What is the voltage across R2?
First lets calculate the voltage across R2 with the meter out of the circuit. This is simply 2 resistors (R1, R2) in series.
V2 = I2 R2
What is I2?
I2 = I1 = I total
What is I total?
I total =
Vtotal
R total
R total = R1 + R2
V2 =
V total
R2
R1 + R2
V2 = V total
R2
R1 + R2
What is R total?
Now put it all together
or
This is commonly called a voltage divider
Plug in the numbers
V2 = (5v) [ 10MΩ / (100kΩ + 10MΩ) ]
V2 = 4.95 v
Ok, now put the meter in the circuit and have it measure this result. The meter should display a reading of 4.95v,
right? Not really......
Now there are 3 resistors to consider.
V2 = I2 R2
I2 = Itotal
What is I2?
Rm
R2 + Rm
... division of current or " current divider" What is I total?
Vtotal
R total
I total =
What is R total? We must consider Rm
R total = R1 + Req
V2 =
Req =
Rm
Vtotal
R2
R1 + Req R2 + Rm
V 2 = Vtotal
1
.....since R2 and Rm are in parallel
1 + 1
R2 Rm
The quanity
Rm R2
is simply Req
R2 + Rm
Req
R1 + Req
Now plug in the numbers.............
Re q =
1
1
1
10 MΩ + 22 MΩ
V 2 = 5v
= 6.97 MΩ (not 10 MΩ anymore!)
6.97 MΩ
100 kΩ + 6.97 MΩ
V 2 = 4.92 v
This should have been 4.95 v but the meters resistance in parallel with R2 induced a 30 mv difference. This is because
the value of R2 was close to Rm, the meters input impedance. This is sometime referred to as “loading” down the
circuit. In other words, the meter loaded down the circuit in test. This difference may not seem like much.....but it is
worth considering when measuring devices that have an input resistance close to the input impedance of the meter.