Solving for Internal Resistance: Thenevin Equivalent Circuit Explained

[phrygian]

Homework Statement

An automobile battery has a terminal voltage of 12.8 V with no load. When the starter motor, which draws 90 A, is running, the terminal voltage drops to 11 V. What’s the internal resistance of the battery?

Homework Equations

I think the solution must involve finding a Thenevin equivalent circuit?

The Attempt at a Solution

This is a problem from a homework written by my teacher, and from the reading in the book I have absolutely no idea where to start this one. Can anyone help point me in the right direction and help me see what the circuit is even supposed to look like?

Thanks for the help

 

[Doc Al]

I think the solution must involve finding a Thenevin equivalent circuit?

No need for anything like that.

Hint: Think of the battery as being a voltage source in series with an internal resistance. When no current flows, the voltage drop across that resistance is 0. What's the voltage drop when 90 A flows through it?

 

[phrygian]

So the answer is as simple as 1.8 V / 90 A? Is the starter motor a resistor or what do you write it as when you draw the circuit?

 

[Doc Al]

So the answer is as simple as 1.8 V / 90 A?

Yep.

Is the starter motor a resistor or what do you write it as when you draw the circuit?

It doesn't matter what you write for the starter motor, since it's outside the terminals of the battery. I guess you could show it as a resistor if you want.

 

[Doc Al]

You might find this helpful: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dcex6.html"