I am having trouble understanding this concept. I know that a voltmeter has an internal resistance of its own. When using a voltmeter to measure the voltage drop across a resistor, does the measurement shown on the voltmeter INCLUDE the voltage drop caused by the internal resistance of the voltmeter itself?
the voltmeter is insert in parallel with the load that you are trying to measure, as a result, the v drop across the internal resistance will be same as the drop across the load. the only different when compare to the case with no voltmeter is that the circuit will draw a different current...that's why you want it to be as large as possible so that this difference would be small..... since adding something in parallel changes the overall resistance. now, to measure voltage drop the voltmeter rely on a device called galvanometer (ie. a current reader) which is supposedly very sensitive to tiny current flow. since the internal resistance of a voltmeter is not infinite, some leakage current will flow through it, and therefore by carefully calibrating the current to voltage ratio, one can work out what is the voltage drop.
The voltmeters resistance is assumed to be infinate relative to the resistor being measured - this way non of the current flows through it. But in reality it's still finate so there may be very small differences between the voltage measured and the actual voltage. Since this difference is close to constant it can be factored into the result.