Question

2. (3 points) A Tesla Roadster uses 6,831 Panasonic 3.7 V Lithium ion cells (these are about twice the size of a AA battery) in its battery pack. These are arranged in 69 parallel rows, with each row containing 99 of the batteries in series (so the pack voltage is 366.3 Volts). Internal resistance of a battery is larger when it is discharged, but for now let's assume it remains constant. The internal resistance of these batteries varies from 0.25 Ohms each when mostly discharged to 0.05 Ohms when mostly charged. For now though assume it is a constant 0.15 Ohms. If the charging system is charging at a constant 10 kW, what percent of the energy coming from the charger is lost to heat over the internal resistance?

Answer 1

99 batteries in series for each row

internal resistance of each row R = 0.15*99 = 14.85 ohm

charging voltage of the series V = 3.7 *99 = 366.3 V

Charging power P = 10kW

= IV

Total charging current I = 10kW/366.3 = 27.3 A

All 69 parallel rows are equal and hence current through each row is same

current through each row = 27.3/69 = 0.396 A

Power dissipated in internal resistance = I²R = 0.396²*14.85 = 2.325 W - each row

Total power dissipated (69 rows) = 2.325*69 = 160.4 W

percent power lost = 160.4/10000 *100 = 1.6 %