I'm told at school that the Electromotive Force (e.m.f) of a battery equals the potential difference between the terminals of the battery when there is no current.
How is that possible? How can there be a potential difference with no charge flowing?
[Physics] How to there be a voltage when there is no current
batterieselectric-currentelectricityvoltage
Related Solutions
What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So your question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"?
The answer to this is that we assume all potential energy lost, is again lost due to inner collisions with other atoms, and that's why materials heat up! Also, this implies a steady current because the drift velocity will not be changing.
Edit:
It can be shown that $F\Delta x= \Delta KE$ this means that a force acting on some distance will produce a change in kinetic energy. You can imagine an electric field where the work done by it is simply $W= qE\Delta x$. By the way notice that if I were to divide that expression by the charge $q$ I would obtain the voltage across the thing i'm concerned with. Namely $V=E\Delta x$.
So in summary, if I do some work, then I have some change in kinetic energy. If there is a voltage drop, it follows that positive work was done and kinetic energy increased, which means a velocity increment.
Across a resistor there is a voltage drop. So why isn't it that charged particles are going faster and then I can measure a current increase? Well that's due to the explanation I gave above my edit portion. Namely, that all that increase in kinetic energy is absorbed due to collisions with the neighboring atoms.
By the way, if you're curious enough to visit this website, I suggest you look up a video on Work-Energy theorem on youtube. This concept is pretty straightforward and I'm sure you can understand it.
Real chemical cells in batteries produce a potential difference which is determined by the chemistry of the cell. A lead/sulfuric acid cell will have an chemical potential ($\epsilon$, the emf), of slightly greater than 2.0 V. Couple six of those cells in series (like a car battery) and you get about 12 V.
Those cells, however, have an effective internal resistance ($r_i$). If current $I$ flows through the cell, the resistance drops the total potential difference to $\epsilon - Ir_i$. If one can minimize the current by having an extremely high external resistance (a good quality voltmeter), the reading of the meter will approach $\epsilon$.
$$V_{meter}=\epsilon-Ir_i$$ $$I=\frac{\epsilon}{r_{meter}+r_i}$$
If $r_{meter}\to\infty$, you can see what happens to $V_{meter}.$
Best Answer
The analogy of electricity to flowing water may come in handy here. In this analogy, a potential difference is like a difference in height. One lake on top of a mountain and another in a valley, for example, might represent the two terminals of the battery, which are at different potentials. If you think about that situation, it's clear that no water flows from the upper lake to the lower one because there's no path for it to get there. The same goes for current: when there's no path from the negative terminal of the battery to the positive terminal, current won't flow.