# How much current is drawn? Absolutely confused

• Sid55
In summary, two batteries connected in series feed a 0.16 Ohm resistor with 80 Watts of power at 3.85 Volts. Each 4 Volts battery is capable of supplying up to 20 Amps of continuous current.
Sid55
Moved from a technical forum, no template.
Two batteries connected in series feed a 0.16 Ohm resistor with 80 Watts of power at 3.85 Volts. Each 4 Volts battery is capable of supplying up to 20 Amps of continuous current.

Ohm's Law calculation gives 22.36 Amps of current draw for the entire circuit but ...

Question 1: How much current is drawn from each battery ?
Question 2: Will each battery still give out 22.36 Amps of current or will the current draw per battery be divided because Wattage output is divided between two batteries (80 Watts/2 batteries) ?

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Sid55 said:
Ohm's Law calculation gives 22.36 Amps of current draw for the entire circuit but ...
Ohm would not agree with you.

phinds said:
Ohm would not agree with you.
My mistake. 3.58 Volts not 3.85 Volts being drawn by the circuit.

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This is homework. Our rules easy that you must show your work before they can help.

So tell us your thoughts on those two questions.

Have you studied series and parallel connections yet ?

It's in the definitions of current and power. .

Always apply your basics - that's how you come to trust them.

Is that really the diagram from your question? It seems to show 3.58 V going ‘towards’ the resistor. Voltage doesn’t flow like that - it’s measured across two points - potential difference.

Current, however, does flow. An ammeter must be spliced into the wire to measure the flow. Imagine splicing in an ammeter at any point in the circuit. Would it measure different values at different points?

Sid55 said:
My mistake. 3.58 Volts not 3.85 Volts being drawn by the circuit.
You don't seem to understand the difference between voltage and current. As guineafowl already pointed out, voltage doesn't "flow", current does. Also, you say the voltage source are in series and you have drawn them in series, BUT ... your "3.58volts" is MUCH more likely to be correct if they are in parallel

I think you seriously need to go back and study the very basics.

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phinds said:
You don't seem to understand the difference between voltage and current. As guineafowl already pointed out, voltage doesn't "flow", current does. Also, you say the voltage source are in series and you have drawn them in series, BUT ... your "3.58volts" is MUCH more likely to be correct if they are in parallel

I think you seriously need to go back and study the very basics.
Speaking about basics, my question is very basic.
How can I reduce continuous current load per battery below 17 Amps to give it a headroom for energy efficiency loss in this circuit (of about 15%) ?

Design needs :
#
Need at least 5(plus) Volts at times to achieve the same power level so wiring 2 batteries in parallel won't cut it. Don't like noticeable voltage sags.
# Can't let batteries run too hot because of higher continuous current discharge per battery than that rated by the battery manufacturer, in this case Samsung.

Design options:
#
Less feasible option - Go with parallel-series configuration with 4 instead of 2 batteries. Requires a wholesale hardware reconfiguration.
# More feasible option - Stay with the current series configuration of 2 batteries but go with higher rated continuous current discharge batteries. Though this option will reduce battery run-time by as much as 40%.

Sid55 said:
peaking about basics, my question is very basic.
But your misunderstandings are even more basic. Items in series carry the same current. There is no split of the current.

Load is not voltage, not current, but V*I.

If you want to split the load between two batteries, they must be in parallel. But in this case, that gives you only 4V, not 5V.

What kind of battery gives 4V? That is an unusual number. Are you sure it is correct?

Sid55 said:
Speaking about basics, my question is very basic.
But you don't UNDERSTAND the basics. You really need to do that before we can give you much help. We'll happily give you all the help we can in your working towards understanding circuits in general so that you can figure these things out yourself, but we're not here to design your circuit for you.

anorlunda said:
What kind of battery gives 4V? That is an unusual number. Are you sure it is correct?

Fully charged LiPo is 4.2 V, but it drops down while discharging. Safe open circuit final voltage is typically listed as 3.3 V.

anorlunda
One thought at a time
Sid55 said:
How can I reduce continuous current load per battery below 17 Amps
?
Connect the batteries in parallel as shown in @Babadag 's post. Each will carry half the current.

Sid55 said:
to give it a headroom for energy efficiency loss in this circuit (of about 15%) ?

"energy efficiency loss" ? That's gibberish. Do you mean voltage headroom ?
Where did 15% come from ?

Sid55 said:
# Need at least 5(plus) Volts at times to achieve the same power level so wiring 2 batteries in parallel won't cut it.
5 volts from 4 volt batteries ? Boost regulator or bite the bullet and more batteries.

phinds said:
But you don't UNDERSTAND the basics. You really need to do that before we can give you much help. We'll happily give you all the help we can in your working towards understanding circuits in general so that you can figure these things out yourself, but we're not here to design your circuit for you.

I'm trying to get batteries or replace ones that I already have with suitable continuous current discharge batteries for a particular application by knowing how much current will be pulled form each battery.

The device in question has a power regulator. Meaning, when I set it at a particular power level (Watts) the power regulator adjusts current draw and voltage based on resistor's resistance reading. The resistor module is also replaceable for achieving different resistance levels. And I repeat, it uses 2 batteries in series.

An example of a low capacity-very high continuous discharge battery :
18650 LG HB6 4.2V (or 3.6V nominal / 1500mAh) rated at 30 Amps continuous discharge

Examples of medium capacity-high continuous discharge batteries :
18650 Sony VTC5A 4.2V (or 3.6V nominal / 2600mAh) rated at 25 Amps continuous discharge
18650 Samsung 25R 4.2V (or 3.6V nominal / 2500mAh) rated at 20 Amps continuous discharge

Examples of high capacity-low continuous discharge batteries :
18650 Sanyo NCRGA 4.2V (or 3.6V nominal / 3500mAh) rated at 10 Amps continuous discharge
18650 LG MJ1 4.2V (or 3.6V nominal / 3400mAh) rated at 8 Amps continuous discharge

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This looks like an e-cig. These have boost regulators that can adjust duty cycle to modulate output power.

If you have two batteries in series, each will experience the same current draw. The only way to share current would be to parallel them. But bear in mind that max continuous current ratings may not apply to intermittent ‘draws’, as when you take a puff. You may get away with short-term overcurrents.

Guineafowl said:
This looks like an e-cig. These have boost regulators that can adjust duty cycle to modulate output power.

If you have two batteries in series, each will experience the same current draw. The only way to share current would be to parallel them. But bear in mind that max continuous current ratings may not apply to intermittent ‘draws’, as when you take a puff. You may get away with short-term overcurrents.

Thank you Guineafowl for your two cents.

A pair of 2500mAh batteries rated at 20 Amps continuous discharge offers remarkably good battery run-time with absolutely no voltage sag at all and battery heating issues are very minimal when 0.4 Ohm resistor is used.

However, at the same rate of usage, battery run-time plummets and battery heating issues are very noticeable when 0.16 Ohm resistor is used. The energy efficiency loss is enormous at 0.16 Ohm, I believe.

Sid55 said:
A pair of 2500mAh batteries rated at 20 Amps continuous discharge offers remarkably good battery run-time with absolutely no voltage sag at all and battery heating issues are very minimal when 0.4 Ohm resistor is used.
Well, of course.
From the very well-known Ohm's Law (one form of it), I = E/R. If you have two 4-volt batteries in series, you get a total of 8 volts, ignoring for the sake of simplicity the internal resistance of the batteries.
With a .4 Ohm resistor,
I = E/R = 8/.4 = 20 A
P = I * E = 20 * 8 = 160 Watts
With a .16 Ohm resistor,
I = E/R = 8/.16 = 50 A! That's a lot of current!
P = I * E = 50 * 8 = 400 W
Sid55 said:
However, at the same rate of usage, battery run-time plummets and battery heating issues are very noticeable when 0.16 Ohm resistor is used. The energy efficiency loss is enormous at 0.16 Ohm, I believe.

If the goal is 80 watts into 0.16 ohms
then why does one need 5 volts?

P = E2/R
so E = √(P X R)

with 0.4Ω
E = √(80 X.0.4) = √32 = 5.66 volts
and that's a lot to ask of a 4 volt battery.

with 0.16Ω
E=√80 X 0.16) = √12.8 = 3.56 volts
which those Samsung 25R's ought to make at 20 amps, per your post #14

Can i assume the current , 5.66V /0.4Ω = 14.15 amps, isn't overworking the battery ? You said it works okay with 0.4 ohm.

for 80 watts into 0.16 ohm

3.56 volts / 0.16 amps = 22.4 amps

and that's only 11.2 amps per battery in parallel
which sure won't overwork them if they're accustomed to 14.2 in series..

Check my arithmetic. I rounded to three figures.

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jim hardy said:
"energy efficiency loss" ? That's gibberish.
I'll have to disagree here. Look in just about any battery spec sheet and you will find that the available amp-hours decreases with increased load current.

Look at the upper set of curves here.
For a 3.5V cutoff voltage at 2A load you will get 2000mAH, or 1 hour of life.
For a 10A load you will get 1000mAH or 1/10 of an hour, a factor of 5 for load results in a factor of 10 in run time.

Cheers,
Tom

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Tom.G said:
I'll have to disagree here. Look in just about any battery spec sheet and you will find that the available amp-hours decreases with increased load current.

Of course they do.
It's the phrase "energy efficiency loss" that doesn't parse for me.
What he loses at high discharge rates is capacity

The performance curves for the cell he pictured are here
https://www.powerstream.com/p/INR18650-25R-datasheet.pdf
and energy isn't drastically affected by discharge rate.

what does affect energy is how many of the available amp-hours he uses
and that's a strong function of his 'cut off voltage'

So he has to bite the bullet
and either boost, rearrange batteries, or stick with 0.4 ohms.

old jim

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Sid55 said:
Question 1: How much current is drawn from each battery ?

Sid55 said:
I'm trying to get batteries or replace ones that I already have with suitable continuous current discharge batteries for a particular application by knowing how much current will be pulled form each battery.

That question is easy to answer...

The batteries are in series so the same current flows through both batteries...

If the current in the load resistor is 10A then the current in each battery is 10A.
If the current in the load resistor is 20A then the current in each battery is 20A.
etc

Sid55 said:
Question 2: Will each battery still give out 22.36 Amps of current or will the current draw per battery be divided because Wattage output is divided between two batteries (80 Watts/2 batteries) ?

That question can also be answered (in the context of just changing the batteries).

Short answer: The current drawn from each battery will still be equal to the current in the load resistor because the batteries are still in series.

Longer answer: The load resistor in your OP has quite a small value. This means that the internal resistance of the battery might be significant enough to effect the current in the load resistor. If you change the batteries to a different make/size/type the new ones may have a different internal resistance which may cause current in the load resistor might change.

# More feasible option - Stay with the current series configuration of 2 batteries but go with higher rated continuous current discharge batteries. Though this option will reduce battery run-time by as much as 40%.

That'll work.

BTW - Nice Job @CWatters, of answering the original question.

old jim

Sid55 said:
A pair of 2500mAh batteries rated at 20 Amps continuous discharge offers remarkably good battery run-time with absolutely no voltage sag at all and battery heating issues are very minimal when 0.4 Ohm resistor is used.

However, at the same rate of usage, battery run-time plummets and battery heating issues are very noticeable when 0.16 Ohm resistor is used. The energy efficiency loss is enormous at 0.16 Ohm, I believe.

The data sheet for the Samsung INR18650-25R...

https://www.powerstream.com/p/INR18650-25R-datasheet.pdf

..says the cells have a typical internal resistance of 22mOhms. You have two cells in series so the total Internal resistance should be around 44mOhms or 0.044 Ohms.

So with a 0.16 Ohm load resistor the total resistance in the circuit is around 0.16+0.044 = 0.204 Ohms. If you connected that to an 8V source (two 4V sources in series) the current draw would be around 40 Amps! 8/0.2 = 40A. This is way too high for the Samsung INR18650-25R. The data sheet says the max continuous current is 20A. You are overloading the cells when you use a 0.16Ohm load resistor. No wonder they get hot.

Be aware that Li cells can be dangerous things. Especially if you abuse them like this.

jim hardy
jim hardy said:
If the goal is 80 watts into 0.16 ohms
then why does one need 5 volts?

P = E2/R
so E = √(P X R)

with 0.4Ω
E = √(80 X.0.4) = √32 = 5.66 volts
and that's a lot to ask of a 4 volt battery.

with 0.16Ω
E=√80 X 0.16) = √12.8 = 3.56 volts
which those Samsung 25R's ought to make at 20 amps, per your post #14

Can i assume the current , 5.66V /0.4Ω = 14.15 amps, isn't overworking the battery ? You said it works okay with 0.4 ohm.

for 80 watts into 0.16 ohm

3.56 volts / 0.16 amps = 22.4 amps

and that's only 11.2 amps per battery in parallel
which sure won't overwork them if they're accustomed to 14.2 in series..

Check my arithmetic. I rounded to three figures.
Thanks a lot for crunching those numbers. I'll be getting 25 Amp continuous discharge batteries.

## 1. How is current measured?

Current is measured in units called amperes (A) using a device called an ammeter. The ammeter is connected in series with the circuit and measures the flow of electric charge through the circuit.

## 2. What factors affect the amount of current drawn?

The amount of current drawn depends on the resistance of the circuit and the voltage applied to it. Ohm's law (I = V/R) states that the current is directly proportional to the voltage and inversely proportional to the resistance.

## 3. How do I calculate the current drawn in a circuit?

To calculate the current drawn in a circuit, you need to know the voltage and resistance of the circuit. You can use Ohm's law (I = V/R) to calculate the current, or you can use the power formula (P = VI) and divide the power by the voltage to get the current.

## 4. Why is it important to know the amount of current drawn in a circuit?

It is important to know the amount of current drawn in a circuit because it can help determine the health and efficiency of the circuit. If the current drawn is too high, it can cause overheating and potentially damage the components. Additionally, knowing the current drawn can help with troubleshooting and identifying any issues in the circuit.

## 5. How can I reduce the amount of current drawn in a circuit?

The amount of current drawn in a circuit can be reduced by increasing the resistance or decreasing the voltage. This can be achieved by using resistors in the circuit or adjusting the power source. It is important to note that changing the current may also affect the overall functionality of the circuit.

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