The equivalent capacitance between A and B in the circuit is given below.

The equivalent capacitance between $A$ and $B$ in the circuit is given below.

$3.6 μF$
$2.4 μF$
$4.9$ $μ F$
$5.4$ $μ F$

Solution

Simplified circuit,

$\Rightarrow \frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{eq} = \frac{12}{5} = 2.4 μ F$

Asked in: JEE Main 2018 (15 Apr)

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