How to calculate a voltmeter reading with internal resistance?

Hey!

I have this little problem with calculating the reading on a voltmeter when it has internal resistance.

Say you have a 500Ohms and a 2000Ohms resistors placed in series with a 60V supply, and there is a voltmeter with internal resistance of 2000Ohms being used to read the PD across each. What would be the reading on the voltmeter when a) placed across the 500Ohm resistor and b) placed across the 2000Ohm resistor?

My first instinct is to use V = IR. R (total) being 2000 + 2000 + 500 = 4500, so I = 60/4500. Then I would say that V = (60/4500) * 500 for the 500ohm resistor. This gives 20/3V. Then to account for the lost volts I would say V = (60/4500) * 2000 which gives 80/3V. I then would do 20/3 - 80/3, but this gives 20V and the answer sheet says 10V.

What have I done wrong?

It's been a long time since I did this, so I may be a little rusty, but surely you need to use the fact that the voltmeter and the $500 \Omega$ resistor are in parallel before you use $V=IR$ to find the current in the circuit. Then I guess you can deduce from the ratio of resistances of the parallel part and the remaining $2000 \Omega$ resistor to deduce what the potential difference the voltmeter measures is.

Thank you so much! Can't believe I forgot the parallel bit.

The total resistance when the voltmeter is in parallel with the 500 ohm resistor is 2400 ohms around the whole circuit. I therefore is 0.025A, so the ratio of the resistances is 4:20 (parallel bit to fixed 2000Ohm resistor), so 60/24 * 4 = 10V.

So, if I wanted to work out the current through each component, for curiosity, I = V/R, so 10/500 and 10/2000?