Electric Circuit Connected to Ground

[Ricky2357]

Homework Statement

Given the attached scheme, if $$R_{1}$$=$$R_{2}$$=$$R_{3}$$=10 Ω and $$\epsilon_{1}=20 V$$ , $$\epsilon_{2}=10 V$$ determine the potentials at the points A,B,C,D,E. The sources of emf have no internal resistance.

Homework Equations

My question is: Why is it that no current exists along the branch BE? If one made this particular apparatus I am convinced he would observe no current. But in theory, why does this happen? Shouldn't the source of emf $$\epsilon_{2}$$ produce some current?
And even if $$\epsilon_{2}$$ , $$R_{3}$$ did not exist, why would the current ''choose'' to move around the loop instead of going towards the ground?

The Attempt at a Solution

Assuming no current exists at the branch BE, we assign zero potential at points D,E. Because of $$\epsilon_{2}$$ we have $$V_{B}=-10 V$$.
Due to our assumption, current exists only within the loop. We easily find its value:
$$I=1 A$$. Using the mathematical expression of Ohm's law we may now easily obtain the wanted potentials:
$$V_{A}=0 V$$ , $$V_{C}=-20 V$$

 

[kamerling]

There is no return path for the current. If there was a current from B to E, the electrons would either pile up somewhere in the loop or the loop would get depleted of electrons.

What would happen if you connected the loop to point B is that a current would flow for a very short time, until a very small negative charge is distributed around the loop. This charge would prevent any further electrons going through the loop through BE. Their electric field would be the source of the overall negative potential of the loop.
(It's much easier to forget about this and just go with: "no closed loop - no current")

 

[Moetasim]

, V_{D}=0 V , V_{E}=0 V.

It is a fundamental principle in electric circuits that current will always take the path of least resistance. In this case, the path through the loop has a lower resistance than the branch BE, so the majority of the current will flow through the loop. This is why no current exists at the branch BE. Additionally, the source of emf \epsilon_{2} will only produce current if there is a complete circuit for the current to flow through. In this case, the branch BE is not part of the circuit, so no current will flow through it. As for why the current "chooses" to move around the loop instead of going towards the ground, it is simply due to the path of least resistance. The current will always flow through the path with the lowest resistance, and in this case, that is the loop.