Imagine the following:

Electric motor power = 75 kW
Operates at max voltage of 420 Volts

Battery Pack size is 48 kWh
Cell specification: size 18650, 4.2 Max Voltage, 3.6 V(Nominal) and 3 Ah capacity.

Some basic math:
For 420 V (to operate the motor)i connect 100 cells in series with each having 4.2 Volts
Now 48 kWh = 48,000 Wh / 420 V = 115 Ah

Also 115 Ah / 3 Ah = 38 Cell in parallel
So total cells are 100 x 38 = 3800 cells making 48kWh battery pack.


When pack is fully charged the voltage = 420 and capacity = 115 Ah

For 75kW requirement ----- 75,000 W / 420 V = 179 amps required
Thus my max C-rate when battery pack is fully charged is = 179 amps / 115 Ah = 1.55 C rate

Now imagine the cell voltage has reduced to around 3 volts and capacity to around 0.4 Ah

Thus battery pack voltage is 300 volts and capacity is only 15 Ah

Now at this time if i need 75 kW of power from the pack than,

75,000 W / 300 V = 250 amps required

250 amps / 15 Ah = 16.7 C-rate !!!!!!!!!!!!!!!!!!!

Is this logic true ???????????????


William13 | 7 juin 2011

great question.

The confusion is between mA and mAh. The mA ability is combined with the voltage to define the c or current.

The mA actually stays fairly constant. You are thinking of the mAh which has declined with use.

Bhavin Patel | 7 juin 2011

This is what i think.....

mA or just Amps vary constantly......

because speed varies so power requirement vary....

Means kW vary now p = V * I

V has range of 4.2 V to lets say 3 volts
at battery pack 420 volts to 300 volts (for 100 cells i series)

in that case I will vary my friend

and current is time dependent so amps is amp-hr

which gradually decreases from 3Ah to 0.4Ah (or 115Ah to 15Ah)


we can limit the C-rate provided on has to bare the performance reduction as battery pack is discharged.....

To all Tesla-roadster users---------------

is your car giving the same performance like max velocity of 125 miles/hr and 0 to 60 in 4 sec--------

when your battery pack is almost discharged?

like if 56Kwh is fully charged state

what is the performance condition when 5Kwh of energy is left ?

William13 | 9 juin 2011

B Patel,
Lithium ion batteries are capable of delivering nearly constant amps until only 10% charge is left. The voltage drops gradually as you indicate.and then drops precipitously when close to 0% charge left. The maximum possible current does drop about 20%.

Back to your original issue, the current draw is continuously variable depending on acceleration. Elon says that the battery pack can charge from 10% to 90% over 45 minutes which for a 300 mile pack (90 kWh) equals a charge rate of 96 kW.

This compares with the motor which draws 300 kW max. This only occurs during maximal acceleration.

Again the current draw ability is not dependent on the state of charge for the batteries.

William13 | 9 juin 2011

By the way when talking about batteries, the c rate is 1 when a full charge is given or removed over one hour. Different batteries support different c rates without damage. For the Model S the max instantaneous discharge c will vary from about 3.5 to 6 depending on the size of the pack. Part of the damage potential is time dependent. Maximum acceleration would last 5.6 seconds if going to 60 mph. This is not all that long for batteries as demonstrated by the experience of Roadster owners.

lzlmewhut | 16 septembre 2015

Hi Bhavin Patel
I think your calculation is right. The problem is that may be 16.7 C-Rate discharge definitely is not continuous but just for a few seconds like for examples 5 seconds in acceleration. Since, the power from the motor is p=w*T(angular velocity*torque),thereby after acceleration, T will decrease, meanwhile, power decrease as well. And I also decrease with p.

Starter batteries in traditional cars can provider much higher C-Rate like pulse discharge at 50C for 10 seconds and 65C for 1 second.

johnse | 17 septembre 2015

Your calculation is confusing "current state of charge" with "total capacity". C definition is based on the total capacity.

Thus, if you are drawing consistent power from the battery, the C rate at which it is being drawn is the same whether you are at 30% or 100% state of charge.

lzlmewhut | 18 septembre 2015

Hi johnse,

Get it. Thank you.