Terminal voltage of car battery being charged at 62.5 A

[lando45]

Hey,

I got this question in a homework from a teacher of mine:

"A car battery with a 15.0 V emf and an internal resistance of 0.0430 is being charged with a current of 62.5 A. What is the potential difference across its terminals?"

I've tried nearly every formula I know on it. EMF = IR + Ir gives me:

15.0 = (62.5 x R) + (62.5 x 0.0430)
15.0 = (62.5 x R) + 2.6875
62.5 x R = 12.3125
R = 0.197

If I then use that value in V = IR I get:

V = 62.5 x 0.197 = 12.3125

I tried that answer but it's incorrect...please, this is driving me nuts! Can anyone help me out? Does this seem like an incredibly hard question for a 16 year old?!

 

[Chi Meson]

Your answer would be correct if it was the BATTERY that was providing the current. But here the battery is being charged. The voltage source has to be at LEAST equal to the battery's EMF in order to charge it up.

About how many volts more than the battery's EMF do you think this charger needs to be (minimally) to charge THIS battery? (think: the chargers "push" needs to "outpush" the battery plus get over some internal something.)

 

[lando45]

Ah...so would it just be EMF of the battery (15.0 V) divided by the internal resistance of the battery (0.0430 ohms) ?

 

[lando45]

Wait I just realized that can't be right...

 

[Chi Meson]

No.

You have already found the quantity 2.7 volts. This is the amount of potential that is dropped across the internal resistance with this particular charging current.

The charging voltage must be at least a little more than the sum of (EMF of battery) + (whatever voltage is dropped outside or inside the battery).

This solution involves a little trick I like to call : "addition"

 

[lando45]

Ha Ok thanks a lot for the help man, I think I understand now ! Thanks again