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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.

Linear CircuitAB
Vth,RthV_{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
1

(a) Setup

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

Vs=12V,R1=4Ω,R2=6ΩV_s=12\text{V}, R_1=4\Omega, R_2=6\Omega
2

(b) Voltage

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

I=12/(4+6)=1.2A;Vth=1.2×6=7.2VI = 12/(4+6) = 1.2\text{A}; V_{th} = 1.2 \times 6 = 7.2\text{V}
3

(c) Resistance

Short the voltage source and calculate the parallel equivalent resistance.

Rth=(4×6)/(4+6)=2.4ΩR_{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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