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Thévenin Equivalent of a Divider Network

A $12\ \text{V}$ ideal source drives $R_1 = 4\ \Omega$ in series with $R_2 = 6\ \Omega$ to ground.

The load terminals are taken across $R_2$ .

Find the Thévenin equivalent voltage $V_{th}$ seen at those terminals.

Given

Vs	12 V	Source voltage
R1	4 ohm	Series resistance
R2	6 ohm	Shunt resistance

Hint 1

Thévenin voltage is measured with the terminals open — nothing draws current.

Hint 2

That reduces the circuit to a plain voltage divider.

Hint 3

V_th = V_s·R₂/(R₁+R₂), with R₂ the element you measure across.

Worked solution — try the problem first

$V_{th}$ is the open-circuit voltage at the terminals. With no load connected, $R_1$ and $R_2$ form a simple divider:

$V_{th} = V_s\frac{R_2}{R_1 + R_2} = 12 \times \frac{6}{4 + 6} = 12 \times 0.6 = 7.2\ \text{V}$

$\boxed{V_{th} = 7.2\ \text{V}}$

The companion result. $R_{th}$ is found by zeroing independent sources — shorting the ideal voltage source — which puts $R_1$ in parallel with $R_2$ :

$R_{th} = \frac{4 \times 6}{4 + 6} = 2.4\ \Omega$

So any load can be analysed against a simple 7.2 V source behind 2.4 Ω, no matter how complex the original network was.

Note on divider direction. The numerator is the resistance across which you measure, here $R_2$ . Using $R_1$ would give 4.8 V — the voltage across the other element.

Concepts:Thévenin equivalentVoltage dividerCircuit reduction

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