Calculating electromotive for, and internal resistance of a battery.

[shyronnie]

From a battery a current of 2.0 A, with a voltage of 114.2 V. When Amperage is increased to 8.0 A, the voltage drops to 112.6 V.

Find electromotive power, and the internal resistance of the battery.

How do i do this? Simultaneous equations with the formula r_i=(E-U)/I or by differentiation, or??

Thanks in advance.

 

[(*steve*)]

Think of a battery as being a perfect voltage source in series with a resistor.

In this case you don't know the actual battery voltage, OR the value of the resistor.

But you do know how much extra voltage the resistor drops when the current changes.

You can use that to determine the value of the resistor.

You could also solve it as a pair of simultaneous equations, but this way is easier if you are familiar with Ohm's Law.

 

[shyronnie]

Why did my thread get deleted?

My thread got deleted. Why?

 

[(*steve*)]

I must have had your post selected when I deleted some other posts. Sorry. I've put it back.

 

[shyronnie]

I must have had your post selected when I deleted some other posts. Sorry. I've put it back.

Steve: Which formula du i have to use for the simultaneous equations. I can't figure this one out.

 

[(*steve*)]

You need an equation that has as unknowns the battery voltage and ESR as unknowns, and where the current and measured output voltage are the knowns.

You will have two equations with two unknowns and you can then solve for them.

 

[shyronnie]

You need an equation that has as unknowns the battery voltage and ESR as unknowns, and where the current and measured output voltage are the knowns.

You will have two equations with two unknowns and you can then solve for them.

I found out. thanks.

 

[john monks]

The change in battery voltage is small. Therefore you can come up with a close approximation of the internal battery resistance by ohm's law. You take the change in battery voltage and divide it by the change in current. You get [the answer -- which we don't provide]