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## Homework Statement

An automobile battery has a terminal voltage of 12.8 V with no load. When the starting motor (which draws 90A) is running, the terminal voltage drops to 11.0 V. What is the internal resistance of the battery?

## Homework Equations

V= EMF- Ir

## The Attempt at a Solution

I think I'm just completely over thinking the problem and maybe missing something. In order to find EMF:

EMF= I(R+r) and in order to find this I need the R of the load in the circuit. This would be the Motor. The voltage drop is 1.8V so Ohm's Law would give V/I= (1.8)/(90A)=.02 Ohms, which seems way to low.

What am I missing here?