Voltages and current in a circuit: incorrect signal?

[Granger]

Homework Statement

Consider the circuit in the figure and calculate V1, V2 and IA. Notice that the symbols V and A represent, respectively, an ideal voltmeter and an ideal ammeter.

Homework Equations

Ohm's Law
$$V=RI$$

KVL
The sum of the voltages of a mesh in a circuit equals zero. We apply a positive sign to the voltages following the direction of the mesh and a negative sign otherwise.

The Attempt at a Solution

So first because we are in the presence of an ideal voltmeter (zero internal resistance) and an ideal ammeter (infinite internal resistance), I switched them with an open circuit and a short circuit, respectively.
Then I applied the KVL to the only mesh in the circuit and obtained:

$$V_2 + V_1 + 15V =0$$

Because $$V_1 = 10 \Omega \times I_A$$ and $$V_2= 5 \Omega \times I_A$$

Substituting in the first equation we get to $$I_A=-1A$$.

Substituting the found current in the other two equations we get to $$V_1 = -10 V$$ and $$V_2= -5 V$$.

However the exercise solution gives me the same results but with the positive sign. What have I done wrong?

Note: I used passive convention in all components.

Thanks!

 

[phinds]

However the exercise solution gives me the same results but with the positive sign. What have I done wrong?

Your mistake is in believing that they have it right and you have it wrong. It's the other way 'round.

 

[Merlin3189]

$V_2 + V_1 + 15V =0$

Because V₁ = 10 Ω x I_A
and V₂= 5 Ω x I_A

Substituting in the first equation we get to $$I_A=-1A$$.

Substituting the found current in the other two equations we get to $$V_1 = -10 V$$ and $$V_2= -5 V$$.

Seems to me that V1 is across the 5Ω and V2 across the 10Ω, but I agree with your signs.

 

[phinds]

Seems to me that V1 is across the 5Ω and V2 across the 10Ω, but I agree with your signs.

Oh, right. I missed that because it was so obvious that his signs were correct that I thought that was the only mistake.

 

[Windadct]

Right away I picked up this: " ideal voltmeter (zero internal resistance) and an ideal ammeter (infinite internal resistance)" -- Ideal voltmeter / Ammeter but thinking about ideal sources! - Certainly would through off the results...

 

[phinds]

Right away I picked up this: " ideal voltmeter (zero internal resistance) and an ideal ammeter (infinite internal resistance)" -- Ideal voltmeter / Ammeter but thinking about ideal sources! - Certainly would through off the results...

Huh? NOT having ideal everything would throw off the results, or is that what you meant?

OOPS: I see now what you meant. I missed it.

 

[Granger]

Thank you everyone for the help.

The result if correct bud I did indeed commit two mistakes:
- first I switched V1 and V2
- second, I did a conceptual mistake by saying "So first because we are in the presence of an ideal voltmeter (zero internal resistance) and an ideal ammeter (infinite internal resistance)". It's the other way around: ideal voltmeters have infinite internal resistance (and that's why we switch them to open circuits) and ammeters have zero internal resistance (and that's why we switch them to short circuits). I made a confusion with sources of voltage and current and how we shut them in the superposition theorem: we shut down a voltage source so that we have a zero tension (switch to a short circuit) and we shut down a current source so that we have a zero current (switch to an open circuit).

I think everything is fine now.
Thanks again!

 

[CWatters]

Deleted. My mistake.

 

[phinds]

Deleted. My mistake.

See post 6. I made exactly the same mistake.