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

Engineering

Physics

Parallel Plate Capacitor

Question

3.6 µF

2.4 µF

5.4 µF

4.9 µF

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$\frac{1}{c_{eq}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12}$ ⇒ $C_{eq} = \frac{12}{5} = 2 .4 µ F$