How does Thevenin's theorem replace a complex network by a single source and resistance?

In one line

Thevenin's theorem says any linear two-terminal network can be replaced by one voltage source \( V_{th} \) in series with one resistance \( R_{th} \) as seen from the load terminals.

Explained simply

When you plug a lamp into a wall socket, you do not care about the whole power grid behind it; to the lamp, the socket looks like "some voltage with some internal resistance". Thevenin's theorem makes that idea exact: hide the whole circuit behind two terminals and replace it with one source and one resistor.

Technically, \( V_{th} \) is the open-circuit voltage across the load terminals, and \( R_{th} \) is the resistance seen from those terminals with all independent sources replaced by their internal resistances. The load current is then \( I_L = V_{th}/(R_{th} + R_L) \).

How it works

1. Remove the load resistor \( R_L \) and mark terminals A and B.
2. Find the open-circuit voltage across A-B using KVL, voltage divider, mesh or nodal methods. This is \( V_{th} \).
3. Short voltage sources and open current sources. Find the resistance between A and B. This is \( R_{th} \).
4. Draw the Thevenin equivalent: \( V_{th} \) in series with \( R_{th} \), and reconnect \( R_L \).
5. Find \( I_L = V_{th}/(R_{th}+R_L) \).

Big advantage: if the load changes many times, you solve the network only once.

Key formula

\( I_L = \dfrac{V_{th}}{R_{th} + R_L} \)

\( V_{th} \) = open-circuit voltage at the load terminals (V), \( R_{th} \) = equivalent resistance seen from those terminals (\( \Omega \)), \( R_L \) = load resistance (\( \Omega \)), \( I_L \) = load current (A).

Quick example

12 V source, 6 \( \Omega \) in series, then 3 \( \Omega \) from the node to ground; load \( R_L = 2\,\Omega \) across the 3 \( \Omega \).

\( V_{th} = 12 \times 3/(6+3) = 4 \) V. \( R_{th} = 6 \parallel 3 = 2\,\Omega \). \( I_L = 4/(2+2) = 1 \) A.

Common mistakes

• Finding \( V_{th} \) with the load still connected (it must be removed).
• Including the load resistor when calculating \( R_{th} \).
• Forgetting that the drop across a resistor carrying no current (open terminals) is zero when finding \( V_{th} \).

How MU asks it

One of the most common questions: "Find the current through \( R_L \) using Thevenin's theorem" (7–10 marks), or "State and explain Thevenin's theorem" (5 marks). For full marks draw three or four circuits: the original, the circuit for \( V_{th} \), the circuit for \( R_{th} \), and the final Thevenin equivalent with the load. Show every step and give units.