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Parallel and Series Voltage

  1. Feb 5, 2005 #1
    If three cells (1.5 volts each) are placed in series, the total yield is 4.5 volts. But if the same three cells are placed in parallel, the total yield is only 1.5 volts. What is the physical explanation for this?

    I realize that in both cases, power (I x V) remains the same so that there are changes in amperage but what is the physical explanation for the changes in total potential differences?
  2. jcsd
  3. Feb 5, 2005 #2
    The best physical explanation I can give is that you think of the Potential Difference like pressure in a system with water. When you have your cells in series it's like adding 3 pressures together, but when you have them in parallel you dont change the pressure you only change the total amount of water(current) that can be drawn out.
  4. Feb 5, 2005 #3
    Not too different from Erienion's explanation is mine...

    When an emf source is connected in parallel to more than 1 circuit element, the potential difference across each such element in parallel is equal because each element is exposed to the emf source in the same way...as far as this exposure is concerned there is no preference. But currents are different you say. But current drawn depends on the characteristics of the load in this case and even though the potential differences are same across each "exposure point" (strictly a pair of points) the difference in resistances accounts for the change.

    To the external world then, the parallel arrangement of cells is totally equivalent to a single emf source and no distinction (physical or mathematical) is possible in this case....in fact this is the core of Thevenin's Theorem for electric circuits. But I'll leave that for some other time...

    A mathematically rigorous argument follows from Kirchoff's Laws--which are essentially laws of energy (and charge) conservation. With some mathematical spadework you can actually show that the emf values you have mentioned hold. (But again that's not a physical explanation.)

  5. Feb 5, 2005 #4
    Thank you Erienion. Your explanation should work fine in explaining why there is more current when cells are place in parallel as opposed to series. I don’t see how I can use it to create a picture to explain why the pressures are different. Let’s see. Three pressures working one on top of each other on the same circuit (or water circuit) is the sum of those three pressures. However, if those three pressures act “side by side” on the same circuit the result is equal to only one pressure! The model seems to break down here.
  6. Feb 5, 2005 #5


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    Actually the voltage, or Potential difference, is more characteristic of a cell then current. Notice that AAA, AA, C, D are all 1.5V but have much different abilities sustain current. The chemical reaction in a cell determines the voltage of the cell, thus a battery will ALWAYS produce the same voltage. The trouble is as they age they use up the material to produce electrons so they lose the ability to produce current.

    When the cells are connected in series you combine the output of individual cells as measured from a common reference point. Each cell is producing its 1.5V but each successive cell starts at a higher over all potential, as established by the previous cells.

    A parallel set of cells, obviously must have the same output of a single cell since the common ends of all the cells are tied together.
  7. Feb 5, 2005 #6
    Yes, the problem seems to be easier with current

    I realize that the answer can be deduced mathematically, but I am trying to find a simple physical model (if possible).

    On the other hand, would you happen to know the formula for two or more cells of unequal emf placed in parallel? Example: cell1=1.5V, cell2=1.4V, and cell3=0.9V?
  8. Feb 5, 2005 #7


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    If the common ends of the cells (+ to +, - to -) are tied together by perfect wires you would see a significant current between the cells until the voltage equalizes. The end result may be damage to one or more of the cells or the wires connecting the cells. If you used something other then perfect wires then the ends of the cells would essentially be connected by resistors and you would measure a voltage drop across the resistor, until the cells equalized.
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