I have a very simple series circuit which I originally built hoping to measure voltage from an Arduino, but while calibrating realized an irregularity, so stripped it down to its basic parts.


simulate this circuit – Schematic created using CircuitLab

The circuit now consists of a lithium-ion battery (unknown Ri), a 10Mohm resistor, and a 1Mohm resistor. When measuring the voltage drops across these resistors, the voltage does not add up.

Unlike in previously asked questions, I am not referring to a small margin of error around 5-10%. Instead, the voltage when measured across both resistors is 3.68V. R1 has a measured voltage of 1.78V across it, and R2 has a voltage of 0.17V across it. This does not add up by a very large margin, and I have no idea what could be causing it.

I have considered possible bad connections across the wires and in the breadboard, and have not been able to find any significant voltage drops to make up for the 1.7V difference. I'm at a loss for what could be causing what appears to be a violation of KVL and would appreciate any explanation for this.

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    \$\begingroup\$ What do you use to measure the voltage? Arduino? \$\endgroup\$ – Simus994 Jul 23 '17 at 17:16
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    \$\begingroup\$ Did you include your multimeter internal resistance? The cheap multimeter will have 1 Meg resistance and better one will have 10Meg. \$\endgroup\$ – G36 Jul 23 '17 at 17:17
  • \$\begingroup\$ (1) Your voltmeter's finite resistance is "modifying" the circuit when you use it (likely); or, (2) another path is being added when you take a measurement (unlikely.) \$\endgroup\$ – jonk Jul 23 '17 at 17:17
  • \$\begingroup\$ I measured the voltages with a Greenlee DM-20 multimeter, which apparently does have only a 1M internal resistance. I didn't realize it would be so low. Thanks! \$\endgroup\$ – Nathan5802 Jul 23 '17 at 17:23
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    \$\begingroup\$ Is it 100 or 10 Mohm? It's a pretty important part of your question. \$\endgroup\$ – pipe Jul 23 '17 at 22:07

Your meter is affecting the measurement

Your voltmeter is also connected to the circuit, and in a different position for each of your two measurements. A "perfect" voltmeter would have an infinite resistance, but any real voltmeter has a non-infinite one. So some current flows through it, and that affects your measurement. So your actual measurements look like this:


simulate this circuit – Schematic created using CircuitLab

The first one is fine, the meter still measures the battery voltage. In the second one, the meter is in parallel with R1. This reduces the effective resistance of the top half of the potential divider, and reduces the voltage across it. In the third case, the resistor meter is in parallel with the bottom half, and reduces it. So the total comes to less than 3.68V.

Bonus: You have enough information here to calculate the resistance of the meter. Can you work out what it is?

  • \$\begingroup\$ The bonus is an excellent addition .. +1! \$\endgroup\$ – donjuedo Jul 24 '17 at 17:12

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