Problem 15-3    Source:   Question 2 - Test P.1 - 2012 - UFRGS.

In the circuit shown in Figure 15-03.1:

a) Obtain the Thévenin equivalent with respect to the terminals a-b of the circuit, determining the open circuit voltage and the short-circuit current.

b) Calculate the resistance of Thévenin by eliminating independent sources. Compare your result with Thévenin's resistance found in the previous item.

Solution of the Problem 15-3   -    Thévenin/Norton Method

Item a

To transform sources we must decide where to start. A good start is we make an explosion of the X node. See in the Figure 15-03.2 the new configuration of the circuit.

By transforming the 9 volt source and the 20 ohm resistor which is in series with it, in a source of current in parallel with the resistor, we obtain the circuit shown in the Figure 15-03.3. Note that the negative pole of the 9 volt source is in contact with the ground . Therefore, we must be careful in making the transformation so that we do not put the current source and the 20 ohms resistor in parallel with the 60 ohms which would completely change the circuit.

From the above circuit, we see that the two current sources provide a total current from 2.25 A to the a node. Thus, to calculate the current through the resistor of 60 ohms , just make a current divider. Note that the 10 ohms resistor must be added to 60 ohms , obtaining 70 ohms . Soon:

Therefore, we can calculate the voltage that appears in the terminals a - b for open circuit, that is, the tension of Thevenin.

Since we know the tension of Thevenin, let's calculate the short-circuit current. To do so, we have the circuit shown in Figure 15-03.4:

Note that the 60 ohms resistor is out of the circuit due to the short between the points a - b. Therefore, the 10 ohms resistor is in parallel with that of 20 ohms . Then, the short-circuit current is the current that will cross the resistor of 10 ohms . Again, making a current divider, we have:

Now we can calculate the resistance of Thevenin, given by:

In the Figure 15-03.5 we can see the equivalent circuit of Thévenin.

Item b

To calculate Thévenin resistance, one must proceed in such a way that the voltage sources are short-circuited and the current sources are circuits. So, we have the circuit seen in the Figure 15-03.6.

Note that with the short circuit in the voltage source, the two resistors, the 5 ohms and the 25 ohms , were shorted as well. So they're out of the loop. The resistor remained of 10 ohms in series with the resistor of 20 ohms , totaling 30 ohms . As can be seen in the circuit, this resistor is in parallel with the resistor of 60 ohms . Now, by calculating this parallel, we find R_th.

Therefore, we conclude that by any method we find the same value for R_th.