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it's a theoretical question. is there any voltage drop when measuring a 12v battery with a multimeter(voltmeter mode)? so if we get 12 Volts reading near the battery, does this changes if we extends the cables of multimeter to 1 km ??

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    \$\begingroup\$ Yes. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 25 '14 at 23:46
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    \$\begingroup\$ It really depends on what you mean by change. My point is that the answer also depends on the precision of your voltmeter. Of course there is some, typically minute, voltage drop across the leads but, again typically, this will not be reflected as a change in the reading on your voltmeter. \$\endgroup\$ – Alfred Centauri Oct 26 '14 at 0:22
  • \$\begingroup\$ I'm Talking about theory, change can be in microVolts. \$\endgroup\$ – Ali Mkahal Oct 28 '14 at 12:17
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A typical DMM has a very high (but not infinite) input impedance, typically ~10Mohm or bigger.

Now suppose you have very long leads. This will also have some finite resistance. Forming a voltage divider (I'm moving all of the resistance due to leads to above the multimeter. Mathematically this is equivalent to having two leads with 1/2 the length on each side):

\begin{equation} V_{out} = \frac{R_{DMM}}{R_{DMM} + R_{leads}} V_{in} \end{equation}

Computing the equivalent lead resistance for 1km 24AWG wire on each side, we get \$R_{leads} = 166 \Omega\$.

Then with a 10Mohm dmm resistance, \begin{equation} \frac{V_{out}}{V_{in}} = 0.9999834 \end{equation} Or an error of 0.00166%. You'd be pretty hard pressed even measuring this error with most multimeters, and errors from other sources will swamp any errors due to the voltage drop in the leads.

There is slightly more error if you add in the battery's internal resistance, but still not significant.

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    \$\begingroup\$ Just for completeness : the question doesn't specify a DMM. Analog multimeters often used a 50uA meter movement, giving an impedance labelled 20kohm/volt. (Slightly confusing, this refers to the max volts on that meter range' not the actual voltage being measured). So if you use the 20V range for your measurement, the impedance would be 400kilohms. Leading to 25x as much error ... or about 0.03%. Still not a lot. \$\endgroup\$ – Brian Drummond Oct 26 '14 at 10:34
  • \$\begingroup\$ Some meters also have a select-able "high impedance" or "high Z" mode which are on the order of Giga ohms (ex.: Keysight TrueVolt meters have a selectable >10Giga ohms for the 10V range). The error here is now ~1.66ppm. Probably will have to start characterizing leakage currents well before any of this... \$\endgroup\$ – helloworld922 Oct 26 '14 at 20:27
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Generally, voltage drop through the cables is proportional to the resistance of the cables and the current drawn by the multimeter. Most voltmeters have very high input impedance, but it wouldn't take much loss per meter to add up to a lot of resistance. The voltage across the meter will be the battery voltage minus two times the input current of the meter times the loss per cable (once in each direction).

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