How Do You Calculate Thevenin Equivalent for a Battery with Different Loads?

[Trentonx]

Homework Statement

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.

Homework Equations

V=IR

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
V_th-R_thI-R_LI=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?

 

[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.

 

[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.

 

[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.

 

[DeeNos]

I would approach this problem by first understanding the concept of Thevenin Equivalent. The Thevenin Equivalent is a simplified circuit that represents the behavior of a more complex circuit. In this case, we are trying to find the equivalent circuit for the battery when connected to either the car radio or the headlights.

To find the Thevenin Equivalent, we need to find the Thevenin voltage (Vth) and the Thevenin resistance (Rth). The Thevenin voltage is the open-circuit voltage, which is the voltage across the terminals when there is no load connected. The Thevenin resistance is the equivalent resistance seen from the load terminals when all the sources are removed.

In this problem, we are given the voltages across the radio and the headlights when connected to the battery. This means that we can calculate the Thevenin voltage for both cases. For the radio, Vth = 12.5 V and for the headlights, Vth = 11.7 V. However, we do not have enough information to calculate the Thevenin resistance.

To find the Thevenin resistance, we need to find the equivalent resistance seen from the load terminals. This can be done by first finding the total resistance of the circuit when the load is connected. In this case, the total resistance is the sum of the resistance of the radio and the headlights, which is 6.25 ohms + 0.65 ohms = 6.9 ohms.

Next, we need to find the equivalent resistance seen from the load terminals. This can be done by using Ohm's law, V = IR, where V is the voltage across the load, I is the current flowing through the load, and R is the equivalent resistance. In this case, we know the voltage across the load (12.5 V for the radio and 11.7 V for the headlights) and the total resistance (6.9 ohms). We can rearrange the equation to solve for the equivalent resistance, which gives us R = V/I. Plugging in the values, we get R = 12.5 V/6.25 ohms = 2 ohms for the radio and R = 11.7 V/0.65 ohms = 18 ohms for the headlights.

Therefore, the Thevenin Equivalent for the battery when connected to the radio is a 12.5 V ideal voltage source with