Use Thévenin’s theorem to determine  .
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Solution
Lets break the circuit at the  load as shown in Fig. (1-27-2).
 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
 and  .
.  is the open circuit voltage
 is the open circuit voltage  shown in Fig. (1-27-2).
 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.
 resistor is equal to the current of the current source, i.e.  . Therefore,
. Therefore,  . The Thévenin theorem says that
. The Thévenin theorem says that  . Please note that it is not saying that
. Please note that it is not saying that  is the voltage across the load in the original circuit (Fig. (1-27-1)).
 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
, 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).
 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
 resistor is short circuited and the  one is open. Therefore, their currents are zero and
 one is open. Therefore, their currents are zero and  .
.
Now that we have found  and
 and  , we can calculate
, 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
 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
 here is to use the current devision rule. The current of the current source is divided between  and
 and  resistors. Therefore,
 resistors. Therefore,
Now, replace the current source with a  voltage source as shown below and solve the problem. The answers are
 voltage source as shown below and solve the problem. The answers are  ,
,  and
 and  . Please let me know how it goes and leave me a comment if you need help 🙂
. Please let me know how it goes and leave me a comment if you need help 🙂

 resistor is in parallel with  3V voltage source. the series combination of
 resistor is in parallel with  3V voltage source. the series combination of  and
 and  resistors is connected to the series combination of 3V and -1 volt voltage sources. Therefore,
 resistors is connected to the series combination of 3V and -1 volt voltage sources. Therefore,  .
.  is in fact the voltage across the
 is in fact the voltage across the  that can be found by voltage division rule:
 that can be found by voltage division rule: 
 and
 and  . To find
. To find  (or
 (or  , if you are using a current source). I am planning to add an example soon.
, if you are using a current source). I am planning to add an example soon. . Let me know if you still need help.
. Let me know if you still need help. but for the solved problem (Fig. (1-27-1)) it is
 but for the solved problem (Fig. (1-27-1)) it is  .
. for the homework.
 for the homework. . No, it won’t be the same.
. No, it won’t be the same. resistor because it is short circuited and no current would pass though it.
 resistor because it is short circuited and no current would pass though it. . For a current source, the internal resistance is infinite. This is because there can be a situation (short circuit) where the voltage across a current source is zero while the current is a constant number
. For a current source, the internal resistance is infinite. This is because there can be a situation (short circuit) where the voltage across a current source is zero while the current is a constant number  .
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