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If a multimeter has a resistance (1M ohm, say) when measuring voltages how do I take that into account in my error?

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By treating the multimeter as a resistor in parallel with the rest of the circuit, connected at the contact points. When you measure the potential difference across those points, you can compute the potential difference in the absence of the multimeter by comparing the resistance of the circuit to the resistance of the circuit-in-parallel-with-multimeter. Of course, if you have a constant voltage power supply, this will make no difference. If you have a constant current power supply, then this might be a worthwhile exercise. However, unless you know the effective resistance of the circuit very precisely, or the circuit has a high effective resistance (> 10k$\Omega$), the effect of the multimeter in parallel is likely to be much smaller than the uncertainty in the resistance of the circuit. If you are dealing with a circuit with a high effective resistance, then you should find a multimeter suited to dealing with these systems (one with an even larger resistance).

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Thank you, it helped a lot. –  General Stubbs Feb 21 '13 at 2:45

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