A 12 V car battery dies not so much because its voltage drops but because chemic

Question

A 12 V car battery dies not so much because its voltage drops but because chemical reactions increase its internal resistance. A good battery connected with jumper cables can start the engine. Consider the automotive circuit in the figure below, where v = 8.00 V, R = 0.700 ohm and R_s = 0.0900 ohm. How much power could the good battery alone supply to the starter motor? How much power is the dead battery alone able to supply to the starter motor? With the jumper cables attached, how much power is supplied to the starter motor? With the jumper cables attached, how much current passes through the dead battery, and in which direction (charging or discharging the dead battery)?

Explanation / Answer

part a

means dead battery is disconnected from circuit

so current,i =12/(R5+0.01)=12/(0.1)=120 amp

power =vi=120*12=1440W=1.44KW

part b

Now good battery is disconnected from circuit

so

i=8/(0.7+0.09)=10.12658A

power=10.12658*81.012W

if jumper cable is attached

if two unidentical battery connected in parallel tah equavlent volatge

Eeq=12*R+8*0.01/(R+0.01)=11.9436V

1/req=1/0.01+1/0.7

req=0.009859ohm

now ieq=11.9436/(0.009859+0.09)=119.604A

P=VI=119.604*11.9436=1428.507A

part d

ieq=119.604A

now i in good battery=12/0.01=1200A

so it i in dead battery=1200-119.604=1080.369A

so mit is charging

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