Voltage of open circuit

Question

A battery with emf $\varepsilon$ and internal resistance $r$ is connected with a resistor $R$ in the following open circuit. What is the voltage $V_{ab}=V_a-V_b$?

The answer is $- \varepsilon$. "No current. There is no voltage change across R and r.". But I don't really understand why ... I was thinking intuitively it should be $0$? Then thinking of how to get 0, I was thinking ... $V_a = - \varepsilon$ since its on the negative terminal, the $V_b = + \varepsilon$ since its on the positive terminal. But $V_a - V_b = -2 \varepsilon$ ... how do I make sense of this?

Answer 1

Think of the open as infinite resistance. Your current will be $I = E/(R+\infty) = 0$. Now, voltage dropping across every resistor is proportional to the current and resistance: $V(r) = I * r$. Since infinite resistance is infinitely larger than R, all $E$ will drop across $\infty$ and 0v is left to R. Particluarly, $V_R = 0\mathrm{v} * R = 0\mathrm{v}$. Meantime, $V_\infty = 0\mathrm{v} * \infty = E$. The only way to get a finite value with one of multipliers 0 is to multiply it by $\infty$. Actually, $0 * \infty$ is uncertainty. But here we know that it must be E since $V_\infty$ is the only place where all voltage, generated by supply, must be dropped.

Answer 2

But Va−Vb=−2ε ... how do I make sense of this?

It's incorrect to write $V_a = \varepsilon$. The voltage $\varepsilon$ is across the battery.

Try this: place a ground symbol on the wire between the battery and the $a$ terminal; this is your zero node or the place you put the black lead of your voltmeter.

Now, if you place the red lead on the terminal of the battery connected to the resistor, you'll measure $\varepsilon$ volts.

If you place the red lead on the other side of the resistor, you'll measure $\varepsilon + V_R$ volts but $V_R = 0V$ so you still measure $\varepsilon$ volts.

But note that now you're measuring the voltage $V_{ba}$ since the read lead is connected to terminal b and the black lead is on terminal a.

So, $V_{ba} = \varepsilon = -V_{ab}$

Answer 3

Ok , Potential is work done by electrostatic force as you move from A to B per Couloumb. So transfer 1 coulomb from A to B and as you move through the battery, E is directed from +ve plate to -ve plate . So work done by $\vec{E}$ is ($V_-) - (V_+$) , which means $-EMF$.

And yes that is an assumption that charge must reach the positive plate with the same velocity with which it entered the negative plate .