Kirchhoff's Rules: Solve 9V Battery Circuit

[kaspis245]

Homework Statement

A circuit contains 9V battery ,which internal resistance is unknown, an ammeter and a voltmeter, whose resistances are also unknown. When a resistor was connected to the voltmeter in parallel, the voltmeter's value was reduced 2 times and the ammeter's value doubled. What did the voltmeter show before the resistor was connected?

Homework Equations

Kirchhoff's Rules

The Attempt at a Solution

Here's my drawing:

I know that:
$I_2=2I_1$
$V_1=2V_2$

From first drawing I get:
$V_1=I(r_1+r)$
$I_1=I-9/r$

If I could express the same values (V₁ and I₁) maybe I could equate them and get the answer, but I am not sure how to do this.

 

[BvU]

Hello Kaspis,

You confuse yourself by using i, i₁ and i₂ in both pictures. They are not the same !

Also, you want to introduce the resistances of the meters R_vm and R_am (which are the same in both pictures).

 

[phinds]

Draw the voltmeter with it's internal resistance shown and the same thing with the ammeter, then analyze the circuit. Don't just write thinks down without drawing the whole circuit you are trying to analyze.

 

[kaspis245]

I've redrawn the circuit with internal resistances shown:

 

[phinds]

(1) why do you show two resistors associated with the voltmeter?

(2) where is the analysis that you are supposed to do based on the diagram (once you are sure what diagram you are using) ?

 

[kaspis245]

r₁ is voltmeter's internal resistance and R is the resistance of the newly connected resistor. Are these circuits drawn correctly?

 

[phinds]

r₁ is voltmeter's internal resistance and R is the resistance of the newly connected resistor. Are these circuits drawn correctly?

Ah yes, I forgot about that.

Yeah they are ready for analysis, so do it.

 

[kaspis245]

I'm a bit new to this, but here's what I did:

From the left circuit:
$\frac{V_1}{r}-\frac{V_2-V_1}{r_1}-\frac{V_3-V_2}{r_2}=0$

From the right circuit:
$\frac{V_1}{r}-\frac{V_2-V_1}{\frac{Rr_1}{R+r_1}}-\frac{V_3-V_2}{r_2}=0$

Is it correct?

 

[BvU]

What about the 9V from the battery ? (construct the simplest possible case if in doubt -- just an ideal battery and 1 resistor).

 

[kaspis245]

I think I've corrected it:

From the left circuit:
$\frac{V_1-9}{r}-\frac{V_2-V_1}{r_1}-\frac{V_3-V_2}{r_2}=0$

From the right circuit:
$\frac{V_1-9}{r}-\frac{V_2-V_1}{\frac{Rr_1}{R+r_1}}-\frac{V_3-V_2}{r_2}=0$

 

[kaspis245]

I can also note that $\frac{V_3-V_2}{r_2}$ from the first equation is equal to $\frac{V_3-V_2}{2r_2}$ from the second equation since the ammeter's value doubled when resistor was connected. Also $\frac{V_2-V_1}{2}$ from the first equation is equal to $V_2-V_1$ from the second equation since the voltmeter's value was reduced 2 times.
I've tried to equate the first equation with the second, but I don't get a clear answer. Is there another way to solve it or am I doing something wrong?

 

[ehild]

I'm a bit new to this, but here's what I did:

From the left circuit:
$\frac{V_1}{r}-\frac{V_2-V_1}{r_1}-\frac{V_3-V_2}{r_2}=0$

From the right circuit:
$\frac{V_1}{r}-\frac{V_2-V_1}{\frac{Rr_1}{R+r_1}}-\frac{V_3-V_2}{r_2}=0$

Is it correct?

No. It is wrong.

An ideal voltmeter has infinite internal resistance. The internal resistance of the real voltmeter is supposed to be connected parallel with an ideal voltmeter.
In your circuit on the left, no current flows.
Also notice, that everything is connected in series in the left circuit. The currents through r1 and r2 do not add up.

 

[kaspis245]

So my drawing of the left circuit is incorrect, but I don't think this affects the equations. Even if I connect voltmeter's internal resistance parallel to the voltmeter, total resistance of that part of the circuit remains the same $r_1$.

 

[ehild]

So my drawing of the left circuit is incorrect, but I don't think this affects the equations. Even if I connect voltmeter's internal resistance parallel to the voltmeter, total resistance of that part of the circuit remains the same $r_1$.

The total resistance of which part of the circuit?

 

[kaspis245]

The total resistance between V1 and V2 would remain the same regardless if I connect V and r₁ parallely:

 

[ehild]

The total resistance between V1 and V2 would remain the same regardless if I connect V and r₁ parallely:

If you connect an ideal voltmeter parallel with resistance r1, The resistance between V1 and V2 is r1. If you connect r1 in series with the voltmeter, the total resistance is infinite. The circuit on the right is still wrong.
What do you know about the currents flowing through each resistor?

 

[kaspis245]

I've redrawn the circuit but again, the total resistance between V1 and V2 does not change:

Current passing through r: $\frac{V_1-9}{r}$
Current passing through r₁: $\frac{V_2-V_1}{\frac{Rr_1}{R+r_1}}$ (the voltage of this part is 2 times bigger in the left circuit)
Current passing through r₂: $\frac{V_3-V_2}{r_2}$ (this current is 2 times bigger in the right circuit)