Engineering

Unveiling Thevenin Equivalence

We break down a complex linear circuit into its simplest form—a single voltage source and a single resistor—by applying the core principles of open-circuit voltage and deactivated source resistance.

V_{th}, R_{th}

Imagine a complex circuit as a black box with two terminals. We only care about how it behaves at those specific points.

Step-by-step solver

(a) Setup

Define the circuit loop with the 12V source, 4 ohm series resistor, and 6 ohm parallel load.

V_s=12\text{V}, R_1=4\Omega, R_2=6\Omega

(b) Voltage

Calculate the current through the series branch and find voltage across the 6 ohm load.

I = 12/(4+6) = 1.2\text{A}; V_{th} = 1.2 \times 6 = 7.2\text{V}

Short the voltage source and calculate the parallel equivalent resistance.

R_{th} = (4 \times 6) / (4 + 6) = 2.4\Omega

Original question

Find the Thevenin equivalent voltage and resistance of a circuit consisting of a 12 V source in series with a 4 ohm resistor, which is in parallel with a 6 ohm resistor. Explain the method.

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