If the circuit shown in the "black box" in Figure 1 consists only of DC voltage sources, current sources, and resistors, then the black box can be emulated by the circuit shown in Figure 2.

Figure 1. Unknown Circuit

Figure 2. Thevenin Equivalent Circuit

This means that some V_{TH} and
R_{TH }can be selected that will produce the same
terminal properties as the "black box" in Figure 1.

For such a circuit, V_{TH }= V_{OC} where V_{OC}
is the open circuit voltage, the voltage at *a-b* when
nothing is connected to it. Furthermore R_{TH} = V_{OC}/I_{SC}
where I_{SC} is the short circuit current, the current
when a short is connected between *a *and *b*.

To find the Thevenin equivalent circuit of any unknown collection of sources and resistors, two measurements must be made. Assume a "black box" as in Figure 1 above. In this example we will measure the voltage across two different resistors connected to the terminals of the black box. When we connect a 100 Ω resistor to the terminals, we measure a voltage of 20 V, and when we connect a 350 Ω resistor, we get a voltage of 35 V, as shown in Figure 3 below.

Figure 3. Example Circuit for Thevenin Analysis

Kirchhoff's Voltage Law for Measurement 1 is:

-V_{TH} + V_{1} + 20 = 0

By Ohm's Law:

V_{1} = I_{1}R_{TH}

Also by Ohm's Law:

I_{1} = V/R = 20/100 = .2 A

Plugging this into KVL:

(1) -V_{TH} + .2R_{TH} + 20 = 0

Similar analysis can be done for Measurement 2:

-V_{TH} + V_{2} + 35 = 0

V_{2} = I_{2}R_{TH}

I_{2} = V/R = 35/350 = .1 A

(2) -V_{TH} + .1R_{TH} + 35 = 0

We now have two equations and two unknowns, which we can solve
simultaneously. Subtract Equation (2) from Equation (1):

.1R_{TH} - 15 = 0

(3) R_{TH} = 150 Ω

Now plug Equation (3) into Equation (1):

-V_{TH} + .2(150) + 20 = 0

V_{TH} = 50 V

The resulting Thevenin equivalent circuit is shown below in Figure 4.

Figure 4. Resulting Thevenin Equivalent Circuit

Besides being useful for characterizing unknown circuits, Thevenin's Theorem
is also good for simplifying known circuits. If we need to find the behavior
of a circuit between two points, Thevenin's Theorem can reduce the circuit to a
voltage source in series with a resistor. This is done in two steps:

(1) Find the open circuit voltage between the two points of interest.
This will be the Thevenin voltage.

(2) Short circuit all voltage sources, open circuit all current sources,
and find the resistance seen looking into the two points. This is the
Thevenin resistance.

Consider the example shown in Figure 5 below.

Figure 5. Example Circuit for Thevenin Reduction

The task is to find the Thevenin equivalent circuit between A and B. The first thing we will do is to find the open circuit voltage between A and B. Figure 6 below shows the analysis.

Figure 6. Analysis for Finding V_{OC}

Kirchhoff's Voltage Law around the dashed loop is:

-20 + V_{80} + V_{120} = 0

The Ohm's Law equations are:

V_{80} = 80I_{80}

V_{120} = 120I_{120}

These values can be plugged into the KVL equation:

-20 + 80I_{80} + 120I_{120} = 0

The Kirchhoff's Current Law equation at node C is:

-I_{80} - 1 + I_{120} = 0

Note that there is no current in the 22 Ω resistor because of the open circuit.

The KCL equation can be solved for I_{80}:

I_{80} = I_{120} - 1

Now we plug this into the KVL equation:

-20 + 80(I_{120} - 1) + 120I_{120} = 0

Then we solve for I_{120}:

-20 +80I_{120} - 80 + 120I_{120} = 0

I_{120} = (20 + 80)/(80 + 120) = 100/200 = .5 A

I_{120} can be used to solve for V_{120}:

V_{120} = 120I_{120} = 120(.5) = 60 V

Now we'll find V_{OC} using KVL:

-V_{120} + V_{22} + V_{OC} = 0

V_{OC} = V_{120} - V_{22}

Since there is no current through the 22 Ω resistor,
there is no voltage across it:

V_{OC} = V_{120} - 0 = 60 V

Now that we've found the open circuit voltage (which is the Thevenin voltage), we'll find the Thevenin resistance. Figure 7 shows the circuit with the voltage source replaced with a short circuit and the current source replaced with an open circuit.

Figure 7. Circuit for Finding R_{TH}

The 120 Ω resistor and the 80 Ω resistor are each connected between node B and node C;
therefore they are in parallel.

R_{BC} = 80(120)/(80 + 120) = 48 Ω

R_{BC} is in series with 22 Ω.

R_{TH} = R_{BC} + 22 = 48 + 22 = 70 Ω

The resulting Thevenin equivalent circuit is shown in Figure 8.

Figure 8. Result of Thevenin Circuit Reduction

Check out this Thevenin explanation and example: http://people.clarkson.edu/~svoboda/eta/cards/p12.html (Created by Dr. James A. Svoboda of Clarkson University).

Before going on, you should complete
Tutorial 6
which is a **simulated Thevenin laboaratory**.

In many applications in electrical engineering, it is desirable to draw maximum power from a circuit. It can be easily shown that this condition will exist if the load connected to the circuit is made equal to the Thevenin resistance. See Figure 9.

Figure 9. Maximum Power Transfer

P, the power in the load resistor R_{L}, will be a
maximum if R_{L }= R_{TH}.

Before going on, you should complete
Tutorial 6A
on **maximum power transfer**.

Whenever an "effect" is linearly related to its "cause", then the effect due to a combination of causes is the sum of the effects due to each cause acting alone, with all other causes removed.

To remove a voltage source, replace by a short circuit.

To remove a current source, replace by an open circuit.

Example

Consider V due to the 6 V source.

By voltage divider:

V_{6} = 6×2/(2 + 12) = 12/14 = 6/7 volts

Consider V due to the current source:

By current divider:

I_{4} = 4×12/(2 + 12) = 48/14 = 24/7 amps

V_{4} = 2I_{4} = 48/7 volts

Total V: V = V_{6} - V_{4} = 6/7 - 48/7 =
-42/7 = -6 V

You can't use Superposition to find power. This is because, if
V = Va + Vb, V^{2} is not Va^{2} + Vb^{2}.

Check out this superposition explanation and example: http://people.clarkson.edu/~svoboda/eta/ClickDevice/super.html (Created by Dr. James A. Svoboda of Clarkson University).

At this time you should complete
Tutorial 6B
on **superposition**.