# Current output by batteries is ....?

Discussion in 'Homework Help' started by Mod91, Sep 4, 2018.

1. ### Mod91 Thread Starter New Member

Sep 4, 2018
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Two batteries connected in series feed a 0.16 Ohm resistor with 80 Watts of power at 3.58 Volts.
Each 4 Volts battery is capable of supplying up to 20 Amps of continuous current.

Ohm's Law calculation gives 22.37 Amps of current draw for the entire circuit but .....

Question 1: How much current is drawn from each battery ?
Question 2: Will the current draw per battery be divided because Wattage output per battery is divided (80 Watts/2 batteries) ?

2. ### dl324 AAC Fanatic!

Mar 30, 2015
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Welcome to AAC!

Your current calculation is wrong. Current will be 8V/0.16 ohms = 50A.
1. The same current will be drawn from each battery.
2. The drain from each battery will be the same; assuming the batteries are matched.
Where do you get 3.58V?

3. ### oz93666 Senior Member

Sep 7, 2010
282
56
Your calculation is correct (almost ) current is 22.36A ( dl324 has forgotten about the internal resistance of each cell , and resistance of fuse , switch and connecting wires)

1. current running through each cell is the same 22.36A

2. only if the voltage under load is the same for each cell will the power drawn from each cell be the same

4. ### dl324 AAC Fanatic!

Mar 30, 2015
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Without that information being given, I assumed 0 ohms of parasitic resistance.

5. ### Ylli Member

Nov 13, 2015
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Your numbers do not make sense. Two 4 volt batteries in series gives you 8 volts, not 3.58 volts. 8 volts across 0.16 ohms creates a dissipation of 400 watts.

6. ### oz93666 Senior Member

Sep 7, 2010
282
56
The question is not presented very clearly but it deals with a very realistic problem , two 18650 lithium cells , these typically have 20A max discharge ,have a no load voltage of 4V , and have an internal resistance of aprox. 40 to 100 m Ohms each ...two of these wired as shown will give a voltage of around 3.58V under a load of 22A ...

Last edited: Sep 4, 2018
7. ### Mod91 Thread Starter New Member

Sep 4, 2018
11
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Thank you guys for replies.

I should have clarified it. It is a circuit with a power regulator with matching batteries that draws only 3.58 Volts collectively to fulfill 80 Watts of power demand. Lets assume the circuit has 100% power efficiency for this calculation (which won't be the case in practice, 75% - 90% is more realistic).

Does it mean each battery will still output 80 Watts of power at 1.79 Volts drawing 22.36 Amps ?

My goal is to reduce continuous current load per battery in order to reduce heating (or over-heating) of batteries or to seek for batteries that have higher continuous current rating.

8. ### dl324 AAC Fanatic!

Mar 30, 2015
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Two of these hypothetical batteries are going to be more like 7.16V. 7.16V/0.16 ohms = 44.75A. Still closer to 50A than 22A...

9. ### oz93666 Senior Member

Sep 7, 2010
282
56
Each cell will produce about 80W , but half of this will heat the cell , half is available to be used .... that's 40W heating each cell !!! ...

These are 18650 cells ?? they will be flat in 2 mins and will not have a long life running at such high currents ...

As you say , much better to run these cells at a few Amps , get more if space permits.

10. ### Mod91 Thread Starter New Member

Sep 4, 2018
11
0
Yes, these are 18650s 4.2V (or 3.6V nominal), Samsung 25R to be more precise.
Samsung 25R is rated at 20 Amps continuous current discharge according to Samsung's data sheet.

11. ### crutschow Expert

Mar 14, 2008
20,494
5,803
If you have a buck switching regulator with 100% efficiency, then each (matched) battery will be delivering 40W of power.

That would be 11.1A/battery at 3.6V.

12. ### oz93666 Senior Member

Sep 7, 2010
282
56
Yes ...2500mA ... each cell ideally can hold 9 W Hrs .... but running at 20A half is only useable so 2 cells will give you about 8 W Hrs of useable power , enough to run for 5 mins theoretically , but the voltage will drop during that time .... Is 5 mins ( realistically 3 mins ) run time enough ??

If so make sure cells are well ventilated to keep them cool .... performance and life will be reduced running at 20 A.

Keep connecting wires thick to reduce circuit resistance ... Is a fuse necessary? it will have a significant resistance and reduce current .

13. ### mvas New Member

Jun 19, 2017
1
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If your goal is to reduce the current load (amps) per battery, then connect 10 batteries in parallel using a Star type connection,
where each battery has its own (+) wire and (-) wire directly to the main power distribution block.
Then connect your load to the main power distribution block with heavy gauge wire.
Now, each one of the 10 batteries would supply 2 amps of the total 20 amp load.
You could fuse each battery at 5 amps.
Voltages of all batteries must be nearly equal, BEFORE connecting in parallel.