Question

What will be the equivalent capacitance across X and Y?

Answer 1

As we can see that the capacitors are connected both in parallel way and in series way. As we all know, the effective value of capacitance when they are connected parallel will be equivalent to the sum of the magnitude of each of the capacitances. When the capacitors are connected in series, the reciprocal of the effective resistance will be equivalent to the sum of the reciprocal of each value of the capacitances.

Complete step-by-step answer:
as we can see that the capacitors are connected both in parallel way and in series way. As we all know, the effective value of capacitance when they are connected parallel will be equivalent to the sum of the magnitude of each of the capacitances. When the capacitors are connected in series, the reciprocal of the effective resistance will be equivalent to the sum of the reciprocal of each value of the capacitances. These all are exactly opposite to that of the resistor networks.
Here first of all let us look at the $3\mu F$ capacitor network. Here two capacitors are in series and this combination is parallel to the other with two capacitors in series. The series combination here will give the value of the capacitance as,
$\begin{aligned} & \dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\mu F \\\ & \Rightarrow C=\dfrac{3}{2}\mu F \\\ \end{aligned}$
Now this capacitance in both the branches will be parallel to each other. Therefore,
$C=\dfrac{3}{2}+\dfrac{3}{2}=\dfrac{6}{2}=6\mu F$
Now let us look at the branch of $6\mu F$. Here also first of all we can find the capacitance in the series manner.
$\begin{aligned} & \dfrac{1}{C}=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6} \\\ & \Rightarrow C=2\mu F \\\ \end{aligned}$
This value of capacitance in three branches will be parallel to each other. That is,
$C=2+2+2=6\mu F$
In the branch containing $4\mu F$, we can write that,
$\begin{aligned} & \dfrac{1}{C}=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=1 \\\ & \Rightarrow C=1\mu F \\\ \end{aligned}$
Now these $1\mu F$capacitors are parallel to each other, that is we can write that,
$C=1+1+1+1=4\mu F$
These four effective capacitances will be in series. Therefore the final equivalent capacitance can be found as,
$\begin{aligned} & \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{13}{12} \\\ & \therefore {{C}_{eq}}=\dfrac{12}{13}\mu F \\\ \end{aligned}$
Therefore the final value of equivalent capacitance has been calculated.

So, the correct answer is “Option A”.

Note: As we all know, the effective value of resistors when they are connected series will be equivalent to the sum of the magnitude of each of the resistances. When the resistors are connected in series, the reciprocal of the effective resistance will be equivalent to the sum of the reciprocal of each value of the resistances.