Timeline for Potential of cells in series
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2016 at 14:08 | comment | added | Dawood ibn Kareem | 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. | |
| Aug 2, 2016 at 14:04 | comment | added | Sharpfawkes | Again I understand this answer perfectly, even better than before thanks to @RedGrittyBrick and DavidWallace 's improved answer. But I still have the same doubt. If you had a switch in the circuit, and closed it, the initial loop would be when first couple of electrons produced at the negative electrode of the 5V cell reached the positive of the 10V. However, in this loop, there is a conducting wire (having electrons) between the two cells. The electrons in this wire, only reach the positive electrode of the 5V cell. Sorry if I seem extremely ignorant, I just want to clarify this. | |
| Aug 2, 2016 at 11:34 | comment | added | Dawood ibn Kareem | Thank you, @RedGrittyBrick, for your help with this answer. I think it's better now. I now see that talking about a "charge moving through a cell" is kind of ambiguous; and I certainly didn't want to paint a false impression of electrons passing all the way through a cell. If you think my answer is still inadequate, I'd really appreciate it if you could comment again, and I will delete it. Thanks again for helping. | |
| Aug 2, 2016 at 11:32 | history | edited | Dawood ibn Kareem | CC BY-SA 3.0 |
Worded in terms of charged particles, instead of charges
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| Aug 2, 2016 at 11:08 | comment | added | Dawood ibn Kareem | @RedGrittyBrick I see what you mean. I am guilty of playing fast-and-loose with the word "charge". I think I need to re-word this to clarify that when current moves through a dry cell, the electrons are chemically absorbed at one end, and new electrons chemically produced at the other. If I can't work out a better way of wording what I've written, I shall delete this answer. | |
| Aug 2, 2016 at 10:56 | comment | added | RedGrittyBrick | This explanation makes me uneasy. For example in "The charges would all still have to go through both cells to get back to their starting position." contradicts the idea that the charges never return to their starting position because, in a primary cell, the negative charges (which travel through the metal wires of a typical circuit) are produced at an anode and permanently consumed at the cathode and never reach the anode ever again. The circuit is completed by totally different and separate positive charges flowing in the opposite direction. | |
| Aug 2, 2016 at 6:42 | comment | added | don_Gunner94 | perfect explanation, well done!! | |
| Aug 2, 2016 at 5:45 | comment | added | Sharpfawkes | 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 | |
| Aug 2, 2016 at 5:38 | comment | added | Dawood ibn Kareem | 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. | |
| Aug 2, 2016 at 5:37 | comment | added | Sharpfawkes | 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? | |
| Aug 2, 2016 at 5:34 | history | answered | Dawood ibn Kareem | CC BY-SA 3.0 |