# When potential difference is equal to the emf, isn't the current 0?

I have been told that the $emf$ is equal to the potential difference across the terminals of a cell when no current is flowing.

Does that mean that the current is zero ($I=0$)?

• If no current is flowing, than $I = 0$. This is sometimes called the open-circuit voltage (depending on context). Commented Jun 22, 2015 at 14:00
• Note that questions of the form Is this correct are not good formats for this Q&A site because the answer (Yes|No) is too short to be a valid answer. It would be best if you wrote the question to be about clarifying your understanding than asking if it is correct. Commented Mar 17, 2016 at 10:11

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 $V_{meter}.$

When the battery is set up a chemical reaction which moves electrons from one terminal (which becomes the positive terminal) to the other (which becomes the negative terminal).
Thus an electric field is set up inside the battery which applies a force on the electrons which are being moved by the chemical reaction.
Eventually the electric field is large enough to stop any electrons moving from the positive terminal to the negative terminal (no current is flowing) and now the potential difference across terminal is constant and is equal to the emf of the battery $\mathcal{E}$.
If an external circuit is added then a current $I$ will flow and the chemical reaction will move electrons from the positive terminal to the negative terminal. However in doing that there will be some resistance to the motion of the electrons (internal resistance) and so there will be a drop in potential inside the battery $V_\text{internal}$ so the potential difference across the terminals drops.

$$V_{\text{terminal}}= \mathcal{E} - V_\text{internal}$$

Yes, potential difference being equal to the emf suggests an open circuit(no current being drawn)
When a circuit is open(open key or infinite resistance), the battery is still doing work per unit charge but is only building up an imbalance inside the battery until equilibrium which is when electrostatic forces successfully cancel out the battery. At this point the battery's force cancel's the electromagnetic force without the flow of current