Three resistors of values 2 Ω, 3 Ω and 6 Ω are connected in parallel between two nodes. The equivalent resistance of the combination is:
A0.5 Ω
B11 Ω
C2 Ω
D1 Ω
Answer & Solution
Correct answer: D. 1 Ω
$1/R_{eq} = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1$ ⇒ $R_{eq} = 1\ \Omega$. The parallel combination is always smaller than the smallest individual resistor.
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