Measuring Voltages Across 10k, 100k, 1M & 4.7M Ohm Resistors

[balanto]

Homework Statement

We are supposed to figure out the voltages over every resistance when measuring with a voltmeter

Homework Equations

R1=R2=10k ohm
R3=R4=100k ohm
R5=R6=1M ohm
R7=R8=4.7M ohm
U=10V

The Attempt at a Solution

The thing I'm not quite understanding is whether the resistors are in series ( meaning R1 and R2 and etc) or/and parallel. I tried using voltage division over the resistors and realized they would have same voltage over them, 5V, which is wrong, I think.
V1=U*((R1)/(R1+R2))=U*(R/2R)=U/2=5V,

One second thing we should take account is there is an internal resistance when measuring the voltages with a voltmeter, which is parallel to the measuring object. The internal resistance is usually 10M ohm or higher. How should I do now?

 

[cnh1995]

l. I tried using voltage division over the resistors and realized they would have same voltage over them, 5V, which is wrong, I think.
V1=U*((R1)/(R1+R2))=U*(R/2R)=U/2=5V,

Since the resistance of voltmeter is not specified in the problem, I don't think it should be considered as 10MΩ. The voltmeter is ideal. So, your answer looks correct to me.

 

[Suraj M]

I second that
Go with your answer, unless specified, the voltage of the voltmeter, assume it to be infinity(ideal).

 

[balanto]

Thanks for your replies!

Lets assume that we knew that it wasn't an ideal voltmeter, how would you find the internal resistance? I think that is the part I'm not understanding

 

[Suraj M]

They should give you the internal resistance if it's not ideal.
There is no way of finding the internal resistance of the voltmeter, except of course working backwards- if they've given the potential difference.

 

[gneill]

Lets assume that we knew that it wasn't an ideal voltmeter, how would you find the internal resistance? I think that is the part I'm not understanding

If the circuit resistor values are known to a very good accuracy then you could work out an expression for what the voltmeter would read given that it has some fixed internal resistance. It's basic circuit analysis. Then with some algebra on the expression the meter resistance could be obtained from the actual reading of the voltage on the voltage divider.

Measure the power supply voltage first so that you have an accurate value to work with. Assume that the meter resistance is much higher than any internal resistance of the source, so won't influence its output. You may want to check the output at each voltage divider and choose one that provides an easily measured deviation from 1/2 U on the available voltage scales.