The Thevenin's equivalent across AB is
A Thevenin equivalent needs two numbers — the open-circuit voltage and the resistance looking back in — and here each takes one line.
Step 1 — the open-circuit voltage. With terminals A and B left open, no current leaves node A, so the current arriving through one 2 Ω resistor must leave through the other. Applying KCL at A with B as reference:
\(\dfrac{V_{A}-6}{2}+\dfrac{V_{A}-12}{2}=0\)
\(2V_{A}-18=0\quad\Rightarrow\quad V_{TH}=V_{A}=9\ \text{V}\)
The answer is simply the average of the two source voltages, which it must be because the two resistances are equal — a useful check.
Step 2 — the Thevenin resistance. Deactivate the independent sources: an ideal voltage source becomes a short circuit, so both 2 Ω resistors now run from A straight to B, in parallel:
\(R_{TH}=2\parallel2=\dfrac{2\times2}{2+2}=1\ \Omega\)
So the equivalent is 9 V in series with 1 Ω — option 2.
| Option | V | R | Verdict |
|---|---|---|---|
| 1 | 6 V | 2 Ω | ✗ One source only, sources not combined |
| 2 | 9 V | 1 Ω | ✓ |
| 3 | 12 V | 4 Ω | ✗ Resistors added instead of paralleled |
| 4 | 18 V | 2 Ω | ✗ Sources added as if in series |
The distractors are each built on one specific error : option 4 adds the two sources as though they were in series aiding, option 3 adds the resistances in series, and option 1 ignores the right-hand branch entirely. Working the two steps separately makes all three impossible.
A cross-check by source transformation. Convert each branch to its Norton form: 6 V with 2 Ω becomes 3 A with 2 Ω, and 12 V with 2 Ω becomes 6 A with 2 Ω. In parallel these give 9 A driving 1 Ω, and converting back gives 9 V in series with 1 Ω — the same result by a different route.
Why the equivalent is worth having : whatever load is now connected across AB, its current is \(9/(1+R_{L})\), and maximum power reaches it when \(R_{L}=1\ \Omega\) — conclusions the original two-source circuit does not display at a glance.
Hence, the Thevenin equivalent is 9 V in series with 1 Ω.
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(b) The Thevenin’s voltage is calculated at the open circuit terminals.
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