A) 2 ohm
B) 18 ohm
C) 6 ohm
D) 3.6 ohm
Correct Answer: D
Solution :
The circuit reduces to
\[{{R}_{AB}}=\frac{9\times 6}{9+6}=\frac{9\times 6}{15}=\frac{18}{5}=3.6\Omega \]
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