Question

In the circuit shown in figure equivalent capacitance across AB is (all capacitance in µF)

Answer 1

9 µF

Answer 2

The circuit can be simplified by identifying the common node connecting several capacitors. Let the central vertical line represent a single node, say X. Then, capacitors 17, 6, and 13 are connected between terminal A and node X. These are in parallel, so their equivalent capacitance is $17 + 6 + 13 = 36 \, \mu$F. Capacitors 10, 1 (top), and 1 (bottom) are connected between node X and terminal B. These are in parallel, so their equivalent capacitance is $10 + 1 + 1 = 12 \, \mu$F. The capacitor 5 is connected between two points on the horizontal line that are connected to the central vertical line (node X). Thus, capacitor 5 is connected between X and X, which is a short circuit and does not contribute to the equivalent capacitance. The equivalent capacitance between A and B is the series combination of the equivalent capacitance between A and X and the equivalent capacitance between X and B.

$C_{AB} = \frac{C_{AX} C_{XB}}{C_{AX} + C_{XB}} = \frac{36 \times 12}{36 + 12} = \frac{36 \times 12}{48} = \frac{36}{4} = 9 \, \mu$F.