If two cells are connected in series, say,


I understand that charge from cell 1 have a potential of 10 V would increase it's potential by 5V on moving through the second cell (neglecting internal resistance).

But since both are cells, doesn't the 5V cell have the capability to produce charge as well? The charge coming from that cell should only have a potential of 5V

I know this theory is definitely wrong but I want to know why. I think it's because of the way the cells are connected but I'm sure. In any case, even if I'm right, please elaborate.

  • $\begingroup$ Yeah, sorry. My mistake. I'll edit the question. $\endgroup$ – LeroyJD Aug 2 '16 at 5:27
  • $\begingroup$ That charge that went through only the 5 V cell started at a potential of 10 V. So its final potential is 15 V. (I'm assuming here that 0 V is at the far left of your diagram, and the polarity of the cells is positive to the right. Conventionally, the longer line is the positive side of the cell.) Am I not understanding the question? $\endgroup$ – garyp Aug 2 '16 at 11:18

Imagine connecting these two cells into a circuit, with some other components. Electrons will move through the wires of the circuit. When the electrons reach the positive terminal of a cell, new electrons with more energy will appear at the negative terminal of the same cell. This happens at both cells.

Each electron entering the 10V cell is replaced in the circuit by an electron with 10eV more energy; and each electron entering the 5V cell is replaced by an electron with 5eV more energy. All these electrons are indistinguishable from one another, except for the energy that they carry. And therefore, the overall effect is equivalent to a single cell where each electron is replaced by an electron with 15eV more energy - that is, a 15V cell.

Those 15eV of energy will then be used up by whatever additional components are in the circuit, including the internal resistances of both cells.

  • $\begingroup$ This would be perfect for current passing in loops after the initial loop. In the initial loop, the charges present in the conducting wire after the 10V cell only pass through the 5V cell don't they? So in the initial loop wouldn't 1C of charge have only 5J of expendable energy? $\endgroup$ – LeroyJD Aug 2 '16 at 5:37
  • $\begingroup$ I don't know what you mean by "initial loop". I'm visualising a circuit where you have these two cells plus a light bulb or something similar. All the charges involved travel through all three components. $\endgroup$ – Dawood ibn Kareem Aug 2 '16 at 5:38
  • $\begingroup$ Say you add a switch to the circuit. The initial loop is when the circuit is closed and charges move through one loop of the circuit and return back to their original position. In that loop, charges initially located in front of the 10V cell would only pass through the 5V cell $\endgroup$ – LeroyJD Aug 2 '16 at 5:45
  • $\begingroup$ perfect explanation, well done!! $\endgroup$ – don_Gunner94 Aug 2 '16 at 6:42
  • 1
    $\begingroup$ I don't follow. The voltage of a cell measures the difference in the energy of the electrons at each end of the cell; not the total amount of energy of the electrons at the negative terminal. $\endgroup$ – Dawood ibn Kareem Aug 2 '16 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.