Problem 15-11    Source:    Problem 68 of the circuit list Electrical Circuits I, School of Engineering, UFRGS, Prof. Dr. Valner Brusamarello.

In the circuit shown in Figure 15-11.1, determine the value read by the ammeter A when the S key is in position 2, knowing that when the S key is in position 1 the circuit contained and highlighted by the orange rectangle in the figure, transfers the maximum possible power.

Solution of the problem using Method of Transforming Sources   Click here!

Solution of the Problem 15-11   -    Thévenin/Norton Method

First we must remember that a circuit like the one highlighted in an orange rectangle in the figure above can transfer the maximum power to the load (in this case, load = 6 ohms) if the load value is equal to the internal resistance of the circuit. From the graph shown to the right of the figure, we realize that the value of the voltage source of the circuit A is 126 volts. It is not possible to know the value of the internal resistance (which is in series with the voltage source). But based on what has been said above, we can calculate, since the parallel of this with the resistance of 9 ohms must be equal to 6 ohms for there to be the maximum power transfer . Therefore, we can write:

Performing the calculation:

With this value we can find the Thévenin equivalent of the whole circuit that is inside of the orange rectangle. The Thévenin voltage (using a voltage divider) will be:

And the Thévenin resistance will be the parallel of the two resistors, or:

In the circuit of the Figure 15-11.2, we must calculate its equivalent Thévenin. For this, a voltage source of 10 volts is inserted between the terminals a-b. This voltage value is convenient, because based on the circuit, we conclude that i = 1 A. Assuming this value for i and making the mesh through of the two 2 ohms resistors, we have:

From this equation we can easily find the value of I, because i = 1 A. Then:

With this value we can calculate the Thévenin resistance, or:

This circuit has only dependent sources, and the Thévenin equivalent is a single resistor, since Thévenin's voltage is zero. With this data we can simplify the original circuit for the scheme shown in the Figure 15-11.3.

Note that the circuit was reduced to one voltage source and three resistors in series. Soon, easily calculate the value that the ammeter measures, or:

In the Figure 15-11.4 we see the complete circuit with the indication of the currents in the various components that form the circuit.