Use Thévenin’s theorem to determine 
.

Solution
Lets break the circuit at the 
 load as shown in Fig. (1-27-2).
You may also watch the video of solving the problem below:
Also pdf version of the solution is available here.

Now, we should find an equivalent circuit that contains only an independent voltage source in series with a resistor, as shown in Fig. (1-27-3).

Unknowns are 
 and 
. 
 is the open circuit voltage 
 shown in Fig. (1-27-2).
It is trivial that the current of 
 resistor is equal to the current of the current source, i.e. 
. Therefore, 
. The Thévenin theorem says that 
. Please note that it is not saying that 
 is the voltage across the load in the original circuit (Fig. (1-27-1)).
To find the other unknown, 
, we turn off independent sources and find the equivalent resistance seen from the port, as this is an easy way to find 
 for circuits without dependent sources. Recall that in turning independent sources off, voltage sources should be replace with short circuits and current sources with open circuits. By turning sources off, we reach at the circuit shown in Fig. (1-27-4).

The 
 resistor is short circuited and the 
 one is open. Therefore, their currents are zero and 
.
Now that we have found 
 and 
, we can calculate 
 in the original circuit shown in Fig. (1-27-1) using the Thévenin equivalent circuit depicted in Fig. (1-27-3). It is trivial that
.
We used the Thévenin Theorem to solve this circuit. A much more easier way to find 
 here is to use the current devision rule. The current of the current source is divided between 
 and 
 resistors. Therefore,![]()
Now, replace the current source with a 
 voltage source as shown below and solve the problem. The answers are 
, 
 and 
. Please let me know how it goes and leave me a comment if you need help 🙂

Leave a Reply