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Finding Thevenin’s Equivalent

Thevenin’s Theorem

Thevenin’s theorem states that a two terminal circuit containing voltage sources, current sources, and resistors can be modeled as a voltage source in series with a resistor.  The benefit of using a Thevenin equivalent is that it makes analyzing how a circuit interacts with other circuits a much simpler process.  Consider the circuit below.  Suppose you want to know the loaded voltage of the circuit (VL) for three different loads connected to nodes a and b.  The three loads are 200 Ω,  2 kΩ, and 20 kΩ.  How fast could you find each Vab?


The best method for quickly assessing the performance of this circuit at a variety of loads is to find the circuit’s Thevenin equivalent.


Once we have the equivalent circuit, the problem can be solved using three simple voltage dividers.


V_{L1} = V_{th} (\frac{R_{L1}}{R_{L1} + R_{th}}) = 2 (\frac{200}{200 + 6k}) = 0.065V       V_{L2} = 0.5V    V_{L3} = 1.54V

Finding Thevenin Equivalents in Practice

By using the Thevenin equivalent model, the problem was solved without the need to perform a full circuit analysis each time the load changed.  However, you must learn how to acquire Thevenin equivalents before you can take advantage of them.  In practice, finding the Thevenin equivalent of a circuit is simple.

  • Find Vth by measuring the open circuit voltage with a multimeter.
  • Find Rth by connecting a current meter to the two terminals and dividing Vth by the measured current (called a short-circuit current).  Note that this isn’t the safest method and should never be used in practice! A safer and more proper method for finding Rth is outlined in Lab 3.  That is, find a load resistance that produces a noticeable drop in the load voltage.  Then find Rth by applying the voltage divider formula.

V_{L} = V_{th} (\frac{R_{L}}{R_{L} + R_{th}}) \rightarrow R_{th} = R_{L} (\frac{V_{th} - V_{L}}{V_{L}})

Thevenin Equivalent Resistance

In theory, finding the Thevenin equivalent is more difficult because we can’t rely on lab equipment to do the work for us.  On paper, the Thevenin equivalent resistance (Rth) is easier to find.


1. If there is a load, remove it – Remember that Thevenin’s theorem applies to two terminal circuits.  This implies that the circuit is unloaded (see Figure 4).  After all, Vth is equal to the open circuit voltage, as you saw above.

2. Short out any voltage sources and open up any current sources – All sources must be deactivated to find the Thevenin equivalent resistance.


3. Find the equivalent resistance looking into the two nodes – In Figure 4, the two nodes would be a and b.  The easiest way to find the equivalent resistance is to start at node a and end at node b.  If there are multiple paths from a to b, then there are parallel resistors somewhere in the circuit.

Example 1: Find the Thevenin equivalent resistance with respect to nodes a and b for the circuit in Figure 5.

Solution: Follow the steps listed above.  There is no load to remove because nodes a and b are already open.  Next, deactivate the sources by shorting the voltage source and opening the current source.  Shorting the voltage source effectively shorts out the 40 Ω and 15 Ω resistors, meaning they have no bearing on the value of Rth.  The last circuit in Figure 6 shows the two paths current can flow from a to b.  Rth is given by the equation below.

R_{th} = (4 + 26) || 10 = 7.5 \Omega


Thevenin Equivalent Voltage

Any of the circuit analysis techniques learned so far can be used to find the Thevenin equivalent voltage (Vth).  Voltage dividers, branch currents, node voltage, superposition, and source transformations (the last two will be covered soon) are all legitimate methods for finding Vth.  Of the two Thevenin equivalent values, Vth tends to be the harder one to find because there is no one best way to work all problems.  Node voltage may work well on one circuit, but may be weighed down by multiple variables on another.  Voltage division isn’t always easy, either.

Until you can recognize which method to use, perhaps the best approach is to try one method and see if it works.  If the math becomes too overwhelming, move on to a different method.  Also, realize that the equivalent voltage is measured across two nodes, as shown in Figure 7.  Be certain Vb is connected to ground before declaring that Va is equal to Vth. In this case Vth is equal to Va – Vb. Finally, if there is a load, remember that you must remove it before working the problem! Once again, a Thevenin equivalent is derived from an unloaded circuit.  A loaded circuit alters the equivalent values.

Example 2: Find the Thevenin equivalent voltage with respect to nodes a and b for the circuit in Figure 5.

Solution: The circuit has no ground marked.  This must be placed first.  In general, it’s most convenient to place the ground at the negative terminal of a voltage source.  Placing the ground at the negative terminal of the 17.4 V makes the voltage at the node connected to the source’s positive terminal 17.4 V.


Note: This node would not be at 17.4 V had the ground been connected elsewhere.

The next step is to find the two node voltages, Va and Vb.  Because there is a resistor between node b and ground, Vb must be found in addition to Va.  The current source combined with the arrangement of resistors make voltage division all but impossible.  Instead, try the node voltage method for node a.

I_{1} + I_{2}+ I_{3} = 0 \rightarrow \frac{V_{a} - 17.4}{26} - 0.1 + \frac{V_{a}}{14} = 0 \rightarrow V_{a} = 7V


Notice that the last term used 14 instead of 10.  This is allowable because the 4 Ω and 10 Ω resistors are in series.  Find Vb by using either node voltage or voltage division.  In this case, it’s faster to use voltage division.  If this isn’t obvious, redraw the branch containing node b as shown in Figure 9.

V_{b} = 7(\frac{4}{4+10}) = 2V

Now solve for Vth using the equation from Figure 7 and draw the Thevenin equivalent.

V_{th} = V_{a} - V_{b} = 7 - 2 = 5V

See, Thevenin’s Equivalent is a lot of fun!  Get good at this and you have mastered quite a few important engineering concepts!  Thanks to Ryan Eatinger ( for contribution of this lesson. :)

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Sep 9, 2010 at 8:33 am

how did you get 2v as vth shouldnt it be 3v? thanks

Sep 14, 2010 at 12:56 am

Hi Rayman2o,

Thanks for the question. I just reworked the problem and it came out to Vth = 2 V and Rth = 5.98 k. Be aware of the source polarity when doing your initial combination of the voltage source and the 3k resistor. Remember to first short your current source and open your voltage source, then combine all resistors to calculate your Rth. Then, use a series of Norton and Thevenin voltage and current source combinations to calculate your Vth.

Good luck!

Oct 4, 2010 at 1:36 pm


According to Thevenin, a current source in parallel with a short should be equivalent to a voltage source. If I connect a very small valued resistor across a real current source, would I be able to produce a practical voltage source? I am trying to convert a solar cell( modeled as a current source and some other components) into a working voltage source as I need a voltage source for a boost converter.

Oct 4, 2010 at 1:53 pm

If you have a small resistor (approx. 0) in parallel with a current source and then do a source transformation on that current source to model it as a voltage source in series with this resistor (approx. 0), you will have the small supply resistance you’re looking for. HOWEVER, when performing source transformations, you multiply this resistance (approx. 0) by the current from the current source to get I(source) x 0 = V(source) = approx. 0. I don’t see this working too well. Hope this helped.


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