Current flow with batteries in series vs parallel

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, $3, mathrm{A}$.

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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.

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.