Use Thévenin’s theorem to determine .
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 🙂