# Potential difference when two emf sources are connected in a simple loop circuit

I am having difficulty understanding the equation below for the potential difference between the two terminals of an electromotive source when two sources are connected from Sear and Zemansky's University Physics.

So suppose we have two emf sources connected in a simple loop circuit with the lower source having larger emf, e.g. a battery connected to a car's alternator as in Figure 25.23. Let $$\mathcal{E}$$ and $$r$$ be the emf and internal resistance of the battery. If $$a$$ is the positive terminal and $$b$$ is the negative terminal (and we assume the wire connected has zero resistance) then the book states that the current flows from the positive to the negative terminal in the battery and the potential difference $$V_{ab}=\mathcal{E}+Ir$$ instead of $$V_{ab}=\mathcal{E}-Ir$$ as in the case when the battery is the sole emf source.

But I can't understand why the potential difference is given like this. Isn't $$V_{ab}$$ determined by the emf and internal resistance of the lower source? My understanding is that in this setup the lower source serves the role as the emf of the circuit and the upper source becomes the resistor. So then if $$\mathcal{D}$$ and $$r'$$ are the emf and internal resistance of the lower source, I think we should have $$V_{ab}=\mathcal{D}-Ir'$$ if the current flows from $$b$$ to $$a$$ in the lower source. I would greatly appreciate if anyone could help me understand what is going on here and why we get the equation given.

What you have to remember is that for a resistor the current always flows from one node to another at a lower potential.

Thus when the current trough a cell is reversed (the difference between a cell discharging and a cell being charged) the potential difference across the internal resistance within the cell reverses.

Discharging on the left and charging on the right - note the current directions.

Below the circuit diagram are graphs of potential against position to show the difference in the voltage across the terminals of a cell, discharging $$V_1=\mathcal E-ir$$ and charging $$V_2=\mathcal E+ir$$

• What I still don't understand is so if the lower emf is the energy source of the current, then shouldn't the potential difference $V_{ab}$ equal to the emf minus the current*internal resistance of the lower emf? So say the upper source has emf $10V$ and lower source has emf $20V$ and the internal resistances are small, then the equation says $V_{ab}=10V$, the voltage of the upper source, not equal to the energy source $20V$. Commented Jun 20 at 17:49
• $\dots$ if the lower emf is the energy source of the current, then shouldn't the potential difference Vab equal to the emf minus the current $\times$ internal resistance of the lower emf? is what happens when the cell is discharging, ie supplying energy to the external circuit which is the situation in my left-hand diagram and the in diagram $25.22$. Commented Jun 20 at 21:11
• The book uses $\mathcal{E}$ and $r$ for the emf and internal resistance of the upper emf always. So if I understand correctly when two emf are connected, and the lower source has greater emf, then the lower cell is discharging, ie supplying energy to the upper cell. Using $\mathcal{E}_1, r_1, \mathcal{E}_2, r_2$ for the emf and internal resistances where $1$ stands for the upper cell and $2$ for the lower cell, then we should have $\mathcal{E}_2 - ir_2=\mathcal{E}_1+ir_1$ right? Commented Jun 21 at 4:54
• Because the terminal voltage of both cells should be equal. My question was if $\mathcal{E}_2$ is much greater than $\mathcal{E}_1$ and $r_1, r_2$ are very small or zero, how does this hold? Commented Jun 21 at 4:55
• If you try and charge a battery with too large a charging current the battery will heat up and possibly explode. Some charger actually have a temperature probe which can be placed on the outside of the battery being charged to make sure the temperature of the battery does not rise to a dangerous level. Commented Jun 21 at 8:39