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Taking let's say three ideal and identical batteries, why does the current flow increase when we add them in series compared to them being in parallel?

I can understand the height analogy for voltage but what would be a similar analogy for current which also explains the above question?

My question is based on this video.

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  • $\begingroup$ While voltage can correspond to a height difference, current (charges per second) can correspond to water flow (cubic metre per second). The battery can then be considered a pump that moves water up (increases voltage) $\endgroup$ – Steeven Oct 23 '17 at 22:43
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Using the height analogy, tripling the height of the water reservoir means that you will also get 3 times the pressure in the pipe. That increased pressure then increases the flow through the pipe.

Mathematically, for electrical circuits, we use Ohm's law: $I=V/R$

That means that the current increases when we increase the voltage. If we triple the voltage, and everything else stays the same, then the current will also triple.

If you put the batteries in parallel, the total voltage remains the same. This is the same as connecting 3 identical reservoirs to a pipe: you are not increasing the pressure. However, the parallel batteries (or reservoirs) will be able to supply more current (3 times, in the case of ideal batteries) when you reduce the resistance (make the pipe larger)

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  • $\begingroup$ The current won't be larger, since the voltage is the same. What will triple is the total charge released before the battery is drained. $\endgroup$ – Mark H Oct 23 '17 at 23:14
  • $\begingroup$ Note that I said "when you reduce the resistance" $\endgroup$ – hdhondt Oct 24 '17 at 1:21
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Intuitively, a parallel circuit requires the current to split in as many flows as branches, because the current must pass through all branches at almost the same time. The current arrives to a bifurcation in which it splits.

However, a series one doesn't need to split, the current is the same for all elements, because it goes in order, first through first element, and then all that current for the second one, and so on.

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I can understand the height analogy for voltage but what would be a similar analogy for current which also explains the above question?

I'll try a different approach.

Assume you have a resistor of resistance $1\, \Omega$ connected across an ideal cell that maintains $1\, \mathrm{V}$ across its terminals.

The voltage across the resistor is thus $1\, \mathrm{V}$ and so, the current through the resistor is, by Ohm's law, $1\, \mathrm{A}$.

Now, replace the $1\,\mathrm{V}$ cell with a $3\,\mathrm{V}$ cell.

The voltage across the resistor is thus $3\, \mathrm{V}$ and so, the current through the resistor is, by Ohm's law, $3\, \mathrm{A}$.


Assuming you have no problem with the above, take the next step and replace the $3\,\mathrm{V}$ cell with a $3\,\mathrm{V}$ battery consisting of three $1\,\mathrm{V}$ cells in series.

Since the voltage across the resistor is still $3\, \mathrm{V}$, the current through the resistor is still $3\, \mathrm{A}$.

So you see, the current increases when you place cells in series for the same reason the current increases when you use a cell with a larger voltage - the larger voltage across the same resistance produces a larger current.

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  • $\begingroup$ Thanks but I was searching more for an intuitive analogy since I know Ohm's law already. $\endgroup$ – philmcole Oct 25 '17 at 20:15
  • $\begingroup$ @philmcole, you're welcome but the point of this answer isn't Ohm's law, it's that one 3V source and (3) 1V sources (in series) are indistinguishable. Perhaps you don't fully appreciate this point now but maybe later you will. Cheers. $\endgroup$ – Alfred Centauri Oct 26 '17 at 0:35

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