Battery internal resistance voltage divider

[Jeanguypataterubberboot]

Having a bit of trouble with this question: "A storage battery of emf 34 V and internal resistance of 0.1 ohm is to be charged at a rate of 20 A from a source of 110 volts. What series resistance is needed in the circuit?"

How do you solve this. The text book I am using doesn't explain this well.

Thanks.

Pierre.

 

[KlausST]

Hi,

first please post a drawing including all the known values. Then we can discuss about it.

Is the 110V source pure DC? Without internal resistance?

Klaus

 

[rmanalo]

Well since it's only an exercise, The direction of the current is from a higher potential (source) to a lower potential (battery). so the charging current is equalt to the potential difference divided by the sum of the internal resistance and the series resistance. All values except for the series resistance is given and therefore can be calculated.

In reality however, you'll need a constant current source since the battery is a nonlinear element.

 

[Jeanguypataterubberboot]

Hi,

first please post a drawing including all the known values. Then we can discuss about it.

Is the 110V source pure DC? Without internal resistance?

Klaus

It's a question in an electronic math textbook. Unfortunately they didn't give anymore details.

Pierre

 

[KlausST]

Hi,

It's a question in an electronic math textbook. Unfortunately they didn't give anymore details.

Yes, I guessed that.

But believe me, it´s a good idea to use a pencil and paper. ;-) For experienced persons and even more for not that experienced persons.
Sadly our brain is better in "reading" pictures than in reading words.

Klaus

 

[Jeanguypataterubberboot]

I agree 100%

 

[BradtheRad]

To make it simple we could start out saying the overall voltage operating in the circuit is 110-34. THen if we set current to 20A we want a safety resistor =(110-34)/20.

But then the exercise states 0.1 ohm internal resistance in the battery. Perhaps this is information we ought to take into account. This creates additional 2V as it carries 20A (20 * 0.1). So (as we try to make a well-reasoned answer), raw math says the safety resistor should be (110-34-2)/20.

In real life it's hard to predict whether internal resistance stays fixed at 0.1 ohm during charging, or whether battery chemistry adds greater resistance with greater charge rates.