Question

Six capacitors each of capacitance $2\mu F$ are connected as shown in the figure. The effective capacitance between $A$ and $B$ is

(A) $12\mu F$
(B) $\dfrac{8}{3}\mu F$
(C) $3\mu F$
(D) $6\mu F$

Answer 1

Hint In the given figure, we have a circuit with a combination of six capacitors. The capacitance of each capacitor is given. The type of connection is not specified in the question. We have to find whether the connection is series or parallel from the figure and find the effective capacitance between the two points $A$ and $B$.

Complete Step by step solution
In the given connection of capacitors, one plate of each capacitor is connected to the common point $A$ and the other plate is connected to the common point $B$. Hence we can say that the given connection is a parallel combination of capacitors.
In a parallel combination of capacitors, the capacitance is equal to the sum of individual capacitance. If there are $n$ capacitors connected in parallel, then effective capacitance $C$ is given by
$C = {C_1} + {C_2} + ...............{C_n}$
Here we have $6$ capacitors of capacitance $2\mu F$, hence the total effective capacitance can be written as,
$C = 6 \times 2\mu F = 12\mu F$
The answer is:
Option (A): $12\mu F$

Additional information
In a series combination, the capacitors are connected using an end to end connection. Each capacitor carries the same amount of charge in a series combination of capacitors. A series combination is used to reduce the effective capacitance. The effective capacitance will be less than the lowest capacitance in the combination.

Note
In a parallel combination, the charge on each capacitor will be different. A parallel combination of capacitors is used to increase the effective capacitance of the circuit. The effective capacitance in a parallel combination of capacitors will be higher than the highest value of capacitance in the combination.