 Three resistors of $2\,\Omega$, $4\,\Omega$ and $6\,\Omega$ are connected in **series** as shown. Find the equivalent resistance.
A$4\,\Omega$
B$\dfrac{12}{11}\,\Omega$
C$8\,\Omega$
D$12\,\Omega$
Answer & Solution
Correct answer: D. $12\,\Omega$
For resistors in series, equivalent resistance is the sum: $R_{eq} = R_1 + R_2 + R_3 = 2 + 4 + 6 = 12\,\Omega$.
Option D ($\tfrac{12}{11}$) is what you'd get for the same three resistors in *parallel* — a common trap.
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