Understanding example in Wikipedia entry for open circuit voltage

[zenterix]

Homework Statement

I was trying to understand what an open-circuit voltage is so I looked it up on Wikipedia.

Relevant Equations

There is an example there that is the following.

Consider the circuit (Wikipedia, Open Circuit Voltage)

If we want to find the open-circuit voltage across the 5$\Omega$ resistor, first disconnect it from the circuit:

Find the equivalent resistance in loop 1 to find the current in loop 1. Use Ohm's law with that current to find the potential drop across the resistance C. Note that since no current is flowing through resistor B, there is no potential drop across it, so it does not affect the open-circuit voltage.

The open-circuit voltage is the potential drop across the resistance C, which is: $\frac{C}{C+A}100V_{~}$

I am having a bit of difficulty understanding the steps and concepts here.

I redrew the circuit as follows

The red square is the same piece of the circuit with the dotted lines around it in the original diagram (not sure what this is for though).

It is not clear to me why there is no current through resistor B. This is my main question.

Assuming that this is true, then the current will be the same everywhere in loop 1 and resistors A and C are in series so

$$R_{eq} = R_A+R_C$$

and we obtain the current as

$$i=\frac{V}{R_{eq}}=\frac{V}{R_A+R_C}$$

We can find the potential drop through C from

$$V-R_Ai-R_Ci=0$$

$$R_Ci=V-R_Ai=V-\frac{R_AV}{R_A+R_C}=\frac{VR_C}{R_A+R_C}$$

which agrees with Wikipedia.

 

[kuruman]

Homework Statement: I was trying to understand what an open-circuit voltage is so I looked it up on Wikipedia.
Relevant Equations: There is an example there that is the following.

which agrees with Wikipedia, except for the sign.

That's because you made an algebraic error. The second equation does not follow from the first.
$$V-R_Ai-R_Ci=0$$ $$R_Ci=R_Ai-V$$

 

[zenterix]

A little after writing out my original question I realized the following. "Open circuit" means that the terminals between which we are calculating potential are disconnected from anything. Any potential difference in this circuit is coming from the AC voltage source.

Current doesn't flow through B because it has nowhere to go after that (though this seems intuitively ok, is there a more technical explanation in terms of the math involved?). My guess is that there is no electric field in that section and therefore no potential difference.

Anyways, if no current flows through B, then it will flow through the closed circuit which is loop 1.

When we close the circuit (with the 5$\Omega$) resistor, then we have a potential difference. Now, I believe this potential difference will be different from the open circuit potential difference because of the presence of resistor B. Is this correct?

 

[kuruman]

You got the general idea. However, your guess

that there is no electric field in that section and therefore no potential difference.

is not a good one. There is an electric field between the open points and a potential difference $V_L$ between them. If you place a charged particle in that region, it will experience a force and move from a region of high potential energy to a region of low potential energy. If you connect the two points with a resistor, there will be a current. The situation (minus resistors A, B and C) is similar to any of the electrical outlets that you have at home.