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Suppose two cells of emf and internal resistance e1, r1 and e2, r2 respectively are connected in series. The negative electrode of 1st cell is connected to negative electrode of the second cell. Why is the voltage drop across the second cell (-e2-Ir2) (I is the current) but not (-e2+Ir2)?

In my intuitive sense of speaking, for example, if there was just one cell with current I being drawn from it, then the terminal voltage is (e - Ir) as the current emerges from the positive electrode and the voltage drop across internal resistance is negative as internally the cell current emerges from negative electrode.

The negative e2 in the series combination makes sense as the current emerges from the negative electrode of the cell, but why is the potential drop across internal resistance still negative as even though the cell's internal current emerges from the positive electrode this time?

The only time I know this makes sense is when it is charging, but even with charging of cell the problem shifts to the sign of emf instead. I think I'm going wrong somewhere with this interpretation, and if so, is there a better way to interpret this intuitively? Attached photo is for reference of the setup I'm concerned for

A peculiarly simple series cell combination

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3 Answers 3

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Great confusion can arise in circuit analysis when the analyst confuses (1) the current flow direction that is arbitrarily (but immutably) associated with a positive value with (2) the current flow direction that's intuitively felt to be "right" (such as current leaving a positive electrode).

Direction (1) must be applied universally and consistently, and the appeal of using direction (2) must be ignored. (Indeed, the intuitive direction (2) is often wrong anyway.)

If the current is found to be a negative number, then fine; the direction of (1) is opposite that of (2). Regardless, that detail is taken care of by the sign anyway.

For your diagram, if a voltage drop is assumed to be positive if the potential is higher on the left than on the right, then there's a voltage drop of $-Ir_1$ and $-Ir_2$ over components 1 and 2, respectively, from Ohm's law. In addition, the cells provide a voltage drop of $e_1$ and $-e_2$ because of their orientation. The sum provides the total voltage drop over component 2: $-e_2-Ir_2$.

Other models of single cells (e.g., that give $e-Ir$ consistently) may incorporate a current direction assumed to leave the positive electrode (e.g., direction (2) as discussed above) and are therefore incompatible and unusable with the present analysis because we can have only one current convention: direction (1) as discussed above.

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  • $\begingroup$ Thank you! This partly alleviated my confusion. So did I go wrong where my direction of current assumption was not lining up with the actual current direction, at least inside of the cell? $\endgroup$
    – Dkmg2k
    Mar 3, 2023 at 22:26
  • $\begingroup$ No—as I wrote above, that's never a problem; it may produce a negative answer, but that negative current value is consistent. The problem arose because you used two contradictory current conventions: the one labeled on the sheet and the one that says positive current always exits a cell at its positive terminal. $\endgroup$ Mar 3, 2023 at 22:38
  • $\begingroup$ I see now, thank you! $\endgroup$
    – Dkmg2k
    Mar 3, 2023 at 22:55
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Why is the voltage drop across the second cell (-e2-Ir2) (I is the current) but not (-e2+Ir2)?

It all depends on the direction you chose (assume) to be the direction of the current.

The direction of the current $I$ that you chose makes the change in potential across $r_2$ going from right to left a voltage drop, or $-Ir_2$, and not a voltage rise. The polarity of the emf of cell 2, on the other hand, does not depend on the direction of the current. Going from right to left, no matter which direction you choose for the current, it is $-e_{2}$.

On the other hand, if you chose the direction of the current $I$ opposite to that shown, then you would have a voltage rise of $+Ir_2$ going from right to left coupled with the emf voltage drop of $-e_2$.

I think I wrongly assumed the emf current convention is the same as with the convention of potential drop/rise across internal resistance.

There is no emf current convention. The emf is what it is. The potential drop/rise across the internal resistance depends solely on the assumed direction of current. That assumed direction could turn out to be wrong once you applied KVL to the circuit. If the analysis shows the value to be negative, it simply means the direction is opposite to that assumed.

Hope this helps.

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  • $\begingroup$ Thank you, this helps a lot. I think I wrongly assumed the emf current convention is the same as with the convention of potential drop/rise across internal resistance. $\endgroup$
    – Dkmg2k
    Mar 3, 2023 at 23:02
  • $\begingroup$ @Dkmg2k There is no emf current convention. The emf is what it is. The potential drop/rise across the internal resistance depends solely on the assumed direction of current. That assumed direction could turn out to be wrong once you applied KVL to the circuit. $\endgroup$
    – Bob D
    Mar 3, 2023 at 23:06
  • $\begingroup$ @Dkmg2k I've edited my answer to include your comment and my response. $\endgroup$
    – Bob D
    Mar 3, 2023 at 23:14
  • $\begingroup$ This was so confusing! I meant to say I assumed emf would apparently change signs with the direction of current, which is not so, as you pointed it out; as there is no convention of changing signs of emf with direction of current $\endgroup$
    – Dkmg2k
    Mar 3, 2023 at 23:17
  • $\begingroup$ @Dkmg2k Then now you know a battery emf does not change signs with the direction of current. You should be a able to move on now. $\endgroup$
    – Bob D
    Mar 4, 2023 at 0:05
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There are two things to remember.

When an ideal call is said to have an emf $\mathcal E$ it means that the potential difference across it is always $\mathcal E$ or put another way on passing through the cell from the negative terminal to the positive terminal the potential increases by $\mathcal E$.

When a current flows through a resistor it always flows from one node to a node at a lower potential.

A graph of potential against position may help you understand what is going on?

enter image description here

I have labelled the nodes $A,\,B,\,C,\,D$ and $E$, and made node $C$ the reference node.

Current $I$ flowing from left to right - gray graph
$\bf A \to B$ - potential drops by $\mathcal E_1$
$\bf B \to C$ - potential drops by $IR_1$
$\bf C \to D$ - potential drops by $IR_2$
$\bf D \to E$ - potential rises by $\mathcal E_0$
Thus the potential of node $A$ relative to node $C$ is $\mathcal E_1 \color{red}+ IR_1$,
the potential of node $E$ relative to node $C$ is $\mathcal E_2 \color{red}- IR_2$ and
the potential of node $A$ relative to node $E$ is $\mathcal E_1 + IR_1+IR_2 - \mathcal E_2$.

Current $I$ flowing from right to left - orange graph
$\bf E \to D$ - potential drops by $\mathcal E_2$
$\bf D \to C$ - potential drops by $IR_2$
$\bf C \to B$ - potential drops by $IR_1$
$\bf B \to A$ - potential rises by $\mathcal E_1$
Thus the potential of node $A$ relative to node $C$ is $\mathcal E_1 \color{red}- IR_1$,
the potential of node $E$ relative to node $C$ is $\mathcal E_2 \color{red}+ IR_2$ and
the potential of node $E$ relative to node $A$ is $\mathcal E_2 + IR_2+IR_1 - \mathcal E_1$.

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