Calculating voltage which voltmeter is showing

[akaliuseheal]

Homework Statement

Schematic of circuit is not given, only the text which I translated into English.

Using a voltmeter with internal resistance of 6k ohm, voltage between two points, 1 and 2 of a circuit of constant current, is measured to be 8v.
Then, using a voltmeter with internal resistance of 10k ohm, between the same points, voltage is measured to be 12v.
What voltage will the voltmeter with internal resistance of 15k ohm measure?

Homework Equations

I=U/R

The Attempt at a Solution

I tried to solve it by calculating currents getting 1.3mA and 1.2mA in first and second case, but wasn't sure what to do next.
These voltmeters should be represented as resistors and the answer is 16V.

 

[gneill]

You're told that the current in the circuit is constant, which I take to mean that the current leaving node 2 is the same as the current entering node 1:

So $I$ is an unknown constant and $R$ is some unknown resistance lying between nodes 1 and 2. That's two unknowns. Fortunately you were given two cases where the meter resistance and the measured voltage are given, so you can construct two equations in two unknowns.

 

[akaliuseheal]

I would like to thank you for looking into this but I am unsure on how to do that. I mean, how would system of equations look like?

 

[gneill]

Start symbolically: don't plug in any numbers, just use variables. See if you can write an expression for the voltage $V$ in terms of $I$, $R$ and $R_m$. Or, write an expression for $I$ in terms of $V$, $R$ and $R_m$. Either way is fine (although the latter may be more straight forward).

 

[akaliuseheal]

So like voltage/current divider?
V = (Rm || R) * I

 

[gneill]

So like voltage/current divider?
V = (Rm || R) * I

Sure, that would work. Or since the voltage is the same across both resistors it's easy to write the sum of the currents:

$I = \frac{V}{R} + \frac{V}{R_m}$

Whatever you are more comfortable with.

 

[akaliuseheal]

So it's like this.

R=30k ohm
I = 0,0016A

Replacing those values gets me the voltage of 16V.
Thanks, was struggling with this trivial problem.

 

[gneill]

Glad I could help.