# Paralleling battery cells

If I have a 10 A load with 10 V. If I connect 5 identical cells of 10 V each with 1 ohm internal resistance in parallel, then, their Thevenin equivalent will be 10 V in series with 0.2 ohm internal resistance.

The I^2*R losses is

• 5 cells * 2^2 A * 1 ohm = 20 W

Whereas, if I use one cell of 10 V with 1 ohm internal resistance, then,

• 10^2 A * 1 ohm = 100 W

A huge difference!

But if I connect five identical cells in series where each cell is 2 V and each with an internal impedance of 1 ohm, the heat losses would be:

5 cells * 2^2 A * 1 ohm = 20 W

But it's Thevenin equivalent would be

10 V with 5 ohm resistance

The heat losses would be

10^2 * 5 ohm = 500 W

If the above is correct, does this mean parallel connection is preferred because it gives more reliability due to redundancy and it has less heat stress on the cells?

• why should the reliability increase? Could you explain that claim? Jul 11, 2021 at 13:49
• (your loss calculation is incorrect. The individual resistor doesn't see the sum voltage. Add a schematic to your question, and this will become clear.) Jul 11, 2021 at 13:49
• Are the five identical batteries in the first paragraph in parallel? How did you get 2 V by parallel connection of 10 V batteries? For series connection how did five 1 ohm cells give a total resistance of 1 ohm? Jul 11, 2021 at 13:51
• 10 volts in parallel with 10 volts doesn't make 5 volts. Jul 11, 2021 at 14:22
• OMAR, It is unrealistic to assume the current will stay constant when you change the battery voltage or series resistance so you example is purely hypothetical and cannot be used to justify a practical use , even if its true Jul 12, 2021 at 6:22

If I connect 5 identical batteries of 10 V each with 1 Ohm internal resistance, then, their Thevenin equivalent will be 2 V in series with 0.2 Ohms internal resistance.

• The I^2*R losses is 5 cells * 2^2 A * 0.2 Ohms = 4 W
• 10^2 A * 1 Ohms = 20 W
• A huge difference!

If you connect five identical batteries of 10 V each with 1 ohm (lowercase 'o') internal resistance in parallel then their Thevenin equivalent will be 10 V in series with 0.2 Ω. For a 2 A load the I2R losses would be $$\ I^2 R = 2^2 \times 0.2 = 0.8 \ \text W\$$.

I think you mean to connect five 2 V cells in parallel in which case you would still get 0.8 W lost in the battery of cells.

But if I connect five identical cells in series where each cell has an internal impedance of 1 Ohms, the heat losses would be:

• 5 cells * 2^2 A * 1 Ohms = 20 W

Five cells in series will give you 10 V, 5 Ω. If you draw 2 A from this then you will have $$\ P = I^2 R = 2^2 \times 5 = 20 \ \text W\$$ lost in the battery.

If the above is correct, does this mean parallel connection is preferred because it gives more reliability due to redundancy and it has less heat stress on the cells?

No. What you have missed is that

• your 2 V parallel system is delivering a power to the load of (approx.) 2 V × 2 A = 4 W. Losses are 0.8 W = 20%.
• To deliver the same power from your 10 V system you only need $$\ I = \frac P V = \frac 4 {10} = 0.4 \ \text A\$$. Power delivered to the load is still (approx.) 4 W and power lost to heat is $$\ I^2R = 0.4^2\times 5 = 0.8 \ \text W\$$. Losses are 0.8 W = 20%.

It's the same either way as you would intuitively expect.

• I corrected the question it was wrong. Sorry.
– OMAR
Jul 12, 2021 at 5:30

If I connect 5 identical batteries of 10 V each with 1 ohm internal resistance,... But if I connect five identical cells in series where each cell has an internal impedance of 1 ohm, the heat losses would be...

...different, because you are are not considering equivalent batteries. Five 2 V cells in series, each with an impedance of 1 Ω, makes a 10 V battery with 5 Ω impedance. That's rather different from a 10 V battery with 1 Ω impedance, and even more different from 5 of those batteries in parallel.

does this mean parallel connection is preferred because it gives more reliability due to redundancy and it has less heat stress on the cells?

No. If you want 10 V then you must have cells in series, even if they are inside a single battery case. For some applications (eg. large battery installations) it is often better to use individual cells in series because they can be individually maintained and replaced if they fail. However it may also be useful to have two or more battery banks in parallel, so one bank can be taken offline for maintenance without disrupting service.

• I corrected the question it was wrong. Sorry.
– OMAR
Jul 12, 2021 at 5:30

1st assumption is incorrect.

The Thevenin Equiv. for parallel will be 10V @ 1/5 ohm since the 10V is a 0 ohm assumption with an ESR of 1 ohm for each of the 5 batteries in parallel. So the voltage does not drop with no load, which is how Thevenin equiv voltages are computed.

Thus a 10A load @ 10V = 1 ohm equiv will see a voltage drop of 0.2/(1+0.2)*10V = 10/6= 1.67V from 10V.

The total battery load now is 1.2 ohms , so the current drops to 10/1.2= 8.33A or 83.3 W in total, where the battery losses are only 1/6th of the total resistance and power dissipation shared by 5 batteries or 83.3W /(6*5=)30 = 2.78W

• I corrected the question it was wrong. Sorry.
– OMAR
Jul 12, 2021 at 5:31