Battery Thevenin Equivalent

1. Sep 29, 2010

Trentonx

1. The problem statement, all variables and given/known data
An automobile battery when connected to a car radio provides 12.5 V to the radio. When connected to the headlights, it provides 11.7 V. Assume the radio is modeled as 6.25 ohm resistor and the headlights are .65 ohms. Find the Thevenin Equivalent for the battery.

2. Relevant equations
V=IR

3. The attempt at a solution
The battery is modeled as an ideal voltage source, Vth, with an internal resistance of Rth, and the Rl is either the radio or headlights are the load across the terminals.
I looked at the loop equations
Vth-RthI-RLI=0
for both resistors, then tried to solve both simultaneously. I'm thinking my setup isn't reflecting the problem. Does the battery voltage matter? It seemed to cancel out. Help?

2. Sep 30, 2010

d-rock

I will tell you that the battery voltage shouldn't matter IF you are solving for the Thevenin resistance, Rth.

In some Thevenin equivalent problems you can usually do the zero-source method, but this only works if there ISN'T a dependent source.

This method involves turning 'off' all of the independent sources. Batteries become a short, and current sources become open circuit wires. Then you can calculate the Thevenin resistance from this method.

3. Sep 30, 2010

vk6kro

Regard the battery as an EMF voltage E, and a series resistance R.

You need to find both of these.

Current in radio = 12.5 volts / 6.25 ohms =

Current in headlights = 11.7 volts / 0.65 ohms =

Now you can work out the equations for E and R:

Work out the value of the internal resistance in terms of the voltage across it and the current in it:
(E - 12.5 )/ radio current = R
(E - 11.7) / headlight current = R

Solve for E and R.

4. Sep 30, 2010

Trentonx

I see my error know. I was treating the current as the same for both of the circuits, leading to an incorrect cancellation. Thanks for putting my on the right track.