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Say there is a circuit with two 1.5V cells, and a 100 ohm resistor.

If you connect two cells in series, then the total emf is 3V. And the current will be 3/100 = 0.03 A. (Using V = IR):
enter image description here

If you have the cells in parallel, then the total emf is 1.5V, as the terminals of the cells are electrically the same point. So the current will only be 0.015 A:
enter image description here

But if you just had one 1.5V cell, so the total emf is again 1.5V. The current will still be 0.015 A.
enter image description here

So what benefit does adding a second cell have? The emf and the current is the same no matter if you use 2 cells in parallel or just one cell.

I'm assuming that all the cells are identical, and internal resistance is negligible.

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When do the batteries in the various circuits you draw run out of juice? What is the maximum current that each battery configuration can supply? In both cases remember that you have to use real--not idealized--batteries in your real projects. –  dmckee May 16 '12 at 19:48
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up vote 5 down vote accepted

Real batteries have a finite energy storage capacity. Adding additional cells adds additional capacity (this is why I would add them, you haven't really specified any context so it's hard to say what you are looking for).

Also, it's worth noting:

I'm assuming that all the cells are identical, and internal resistance is negligible.

This is an important assumption. In a real circuit, you would lose some power ($I^2R$) to the internal resistance of the battery. You can reduce this power loss by adding cells, reducing the current each cell provides.

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So by having the cells in parallel, and having less current though each cell, you minimise the power wastage by the internal resistance? –  Jonathan. May 16 '12 at 20:01
    
I don't understand what you mean by energy capacity? Do you mean it will take longer for two cells in parallel to run out, compared to 2 cells in series? –  Jonathan. May 16 '12 at 20:02
    
@Jonathan. Yes, less current from each cell will reduce the amount of power lost to the internal resistance. If you halve the current, you lose 1/4 as much power (assuming linear internal resistance approximation). With 2 cells, this means the power loss (in both cells together) is 1/2 as much as one cell. –  tmac May 16 '12 at 20:09
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@Jonathan. By energy capacity, I just mean that the batteries will run out. Adding cells in parallel keeps the terminal voltage the same (as in your analysis) and adds energy storage capacity. I am comparing two cells in parallel to one cell. –  tmac May 16 '12 at 20:12
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