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 VTH and RTH can be selected that will produce the same terminal properties as the "black box" in Figure 1.
For such a circuit, VTH = VOC where VOC is the open circuit voltage, the voltage at a-b when nothing is connected to it. Furthermore RTH = VOC/ISC where ISC 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:
-VTH + V1 + 20 = 0
By Ohm's Law:
V1 = I1RTH
Also by Ohm's Law:
I1 = V/R = 20/100 = .2 A
Plugging this into KVL:
(1) -VTH + .2RTH + 20 = 0
Similar analysis can be done for Measurement 2:
-VTH + V2 + 35 = 0
V2 = I2RTH
I2 = V/R = 35/350 = .1 A
(2) -VTH + .1RTH + 35 = 0
We now have two equations and two unknowns, which we can solve
simultaneously. Subtract Equation (2) from Equation (1):
.1RTH - 15 = 0
(3) RTH = 150 Ω
Now plug Equation (3) into Equation (1):
-VTH + .2(150) + 20 = 0
VTH = 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 VOC
Kirchhoff's Voltage Law around the dashed loop is:
-20 + V80 + V120 = 0
The Ohm's Law equations are:
V80 = 80I80
V120 = 120I120
These values can be plugged into the KVL equation:
-20 + 80I80 + 120I120 = 0
The Kirchhoff's Current Law equation at node C is:
-I80 - 1 + I120 = 0
Note that there is no current in the 22 Ω resistor because of the open circuit.
The KCL equation can be solved for I80:
I80 = I120 - 1
Now we plug this into the KVL equation:
-20 + 80(I120 - 1) + 120I120 = 0
Then we solve for I120:
-20 +80I120 - 80 + 120I120 = 0
I120 = (20 + 80)/(80 + 120) = 100/200 = .5 A
I120 can be used to solve for V120:
V120 = 120I120 = 120(.5) = 60 V
Now we'll find VOC using KVL:
-V120 + V22 + VOC = 0
VOC = V120 - V22
Since there is no current through the 22 Ω resistor,
there is no voltage across it:
VOC = V120 - 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 RTH
The 120 Ω resistor and the 80 Ω resistor are each connected between node B and node C;
therefore they are in parallel.
RBC = 80(120)/(80 + 120) = 48 Ω
RBC is in series with 22 Ω.
RTH = RBC + 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 RL, will be a maximum if RL = RTH.
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.
Consider V due to the 6 V source.
By voltage divider:
V6 = 6×2/(2 + 12) = 12/14 = 6/7 volts
Consider V due to the current source:
By current divider:
I4 = 4×12/(2 + 12) = 48/14 = 24/7 amps
V4 = 2I4 = 48/7 volts
Total V: V = V6 - V4 = 6/7 - 48/7 = -42/7 = -6 V
You can't use Superposition to find power. This is because, if V = Va + Vb, V2 is not Va2 + Vb2.
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.