Thevenin's Theorem

Thevenin's theorem says this: if a circuit is linear and you only care about two output terminals, you can replace the whole network by one ideal voltage source $V_{th}$ in series with one resistance $R_{th}$. The replacement gives the same voltage-current behavior at those terminals, so any load connected there will behave the same way.

For basic DC resistor networks, the method is fast: remove the load, find the open-circuit voltage for $V_{th}$, find the resistance seen from the same terminals for $R_{th}$, then reconnect the load to the simpler circuit.

What Thevenin's Theorem Means

The theorem does not say the inside of the original circuit becomes physically identical to a battery and a resistor. It says the circuit looks identical from the chosen two terminals.

That condition matters. Thevenin's theorem applies to linear networks. In introductory problems, that usually means resistors with independent or dependent sources. In AC analysis, the same idea uses impedance instead of resistance.

Why Students Use Thevenin Equivalents

Without Thevenin's theorem, every time the load changes, you may need to solve the whole circuit again. With the Thevenin equivalent, the source network is reduced once, and then each new load becomes a simple series-circuit problem.

This is especially useful when the question asks for load current, load voltage, or power for several different load values.

How To Find A Thevenin Equivalent

1. Remove the load

Pick the two terminals where the load connects and disconnect the load. Everything you calculate next is with respect to those same terminals.

2. Find the open-circuit voltage

The voltage across the open terminals is the Thevenin voltage:

$$V_{th} = V_{oc}$$

3. Find the equivalent resistance

For a circuit with only independent sources, turn off the sources and look back into the terminals:

• Replace each independent voltage source with a short circuit.
• Replace each independent current source with an open circuit.

The resistance seen from the terminals is $R_{th}$.

If dependent sources are present, this shortcut is not enough by itself. In that case, keep the dependent sources active and use a test source or another valid method.

4. Reconnect the load

Now replace the original network by $V_{th}$ in series with $R_{th}$.

If a load $R_L$ is connected, the load current is

$$I_L = \frac{V_{th}}{R_{th} + R_L}$$

and the load voltage is

$$V_L = I_L R_L$$

Worked Thevenin Theorem Example

Suppose a $12$ V ideal source feeds a divider with $R_1 = 4 \, \Omega$ in series with $R_2 = 8 \, \Omega$. The output terminals are across $R_2$, and a load $R_L$ will be connected across those same terminals. Find the Thevenin equivalent seen by $R_L$.

Step 1: Remove the load

Disconnect $R_L$ if it is present. The source network that remains is still the divider formed by $R_1$ and $R_2$.

Step 2: Find $V_{th}$

The open-circuit voltage across the output terminals is the divider voltage across $R_2$:

$$V_{th} = 12 \cdot \frac{8}{4 + 8} = 8$$

So $V_{th} = 8$ V.

Step 3: Find $R_{th}$

Turn off the independent voltage source, so the $12$ V source becomes a short circuit. Looking back into the output terminals, $R_1$ and $R_2$ are both connected from the terminal to ground, so they are in parallel:

$$R_{th} = \frac{4 \cdot 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}$$

So $R_{th} = \frac{8}{3} \, \Omega \approx 2.67 \, \Omega$.

From the load's point of view, the original network is now just an $8$ V source in series with about $2.67 \, \Omega$.

Step 4: Try one load

If you now connect $R_L = 5 \, \Omega$, the load current is

$$I_L = \frac{8}{2.67 + 5} \approx 1.04$$

So $I_L \approx 1.04$ A. Then the load voltage is

$$V_L \approx (1.04)(5) \approx 5.2$$

So $V_L \approx 5.2$ V. This is the main benefit of Thevenin's theorem: once you reduce the source network, trying a new load is quick.

Common Thevenin Theorem Mistakes

• Finding $V_{th}$ with the load still attached. The standard definition uses the open-circuit terminal voltage.
• Turning off dependent sources as if they were independent sources. That gives the wrong $R_{th}$ in many circuits.
• Forgetting that Thevenin is defined at a specific pair of terminals. Change the terminals, and the equivalent can change too.
• Mixing up source transformation with simplification inside the original circuit. The equivalence is about terminal behavior, not about every internal branch value.

When Thevenin's Theorem Is Used

Thevenin equivalents appear in circuit design, measurement problems, sensor interfaces, and load-matching questions. They are also a practical way to describe how strongly a source network can drive a load.

Once the idea clicks, Norton equivalents are the natural next comparison because they describe the same terminal behavior in current-source form.

Try A Similar Circuit

Keep the same source network, but change the load to $R_L = 10 \, \Omega$. Use the same $V_{th}$ and $R_{th}$ to find the new load current and voltage. If you want to go one step further, try your own version with a different divider and check how both $V_{th}$ and $R_{th}$ change.