Thevenin Print

8/3/2019 Thevenin Print

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Thevenin: Two Loop ProblemTo apply Thevenin's Theorem to the solution of the two loop problem, consider the current

through resistor R 2 below. Replacing the network to the left of R 2 by its Thevenin equivalent

simplifies the determination of I2 .

Top of Form

For R 1 =

23

Ω, R 2 =

12

Ω, R 3 =

34

Ω,

and voltages V1 =

220

V andV2 =

220

V,

the Thevenin voltage is=

131.23

V

since R 1 and R 3 form a simple voltage divider.

The Thevenin resistance is=

13.72

Ω.

This reduces the circuit to a single loop for which the calculated current is

=

3.45

A

Bottom of Form



Note: To avoid dealing with so many short circuits, any resistor with value zero will default to 1when a voltage is changed. It can be changed back to a zero value if you wish to explore theeffects of short circuits. Ohms and amperes are the default units, but if you put in resistor valuesin kilohms, then the currents will be milliamperes.

Thévenin’s Theorem – Circuit with Two Independent Sources

Fig. (1-27-1) - Circuit with two independent sources

Solution

Lets break the circuit at the load as shown in Fig. (1-27-2).

Fig. (1-27-2) - Breaking circuit at the load

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



Fig. (1-27-3) - The Thevenin equivalent circuit

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 theoriginal 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 dependentsources. Recall that in turning independent sources off, voltage sources should be replace withshort circuits and current sources with open circuits. By turning sources off, we reach at thecircuit shown in Fig. (1-27-4).

Fig. (1-27-4) - Turning off the sources to find Rth

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 andresistors. 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

Fig (1-27-5) - Homework

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Thevenin Print