Question

Find the equivalent capacitance across points $\text{A}$ and $\text{B}$. All the capacitors have capacitance $C$.

• A. $\frac{7C}{8}$
• B. $\frac{7C}{4}$
• C. $\frac{9C}{2}$
• D. $\frac{9C}{4}$

Answer 1

The correct option is A $\frac{7C}{8}$

The uppermost capacitor of capacity $C$ can be considered as the equivalent of two capacitors joined in series, each having a capacity of $2C$.

Now, the circuit diagram can be redrawn as shown below.


This circuit is symmetrical about the perpendicular bisector $\text{XY}$ of $\text{AB}$. Hence, point $\text{X}$ and point $\text{Y}$ are at the same potential.


To the left side of $\left(\text{X,Y}\right),2C$ and $C$ are parallel.
So, their equivalent $=2C+C=3C$

Similar case in the right part of the circuit.

The circuit further reduces to a setup shown below.


Now, $C$ and $3C$ are in series.

So their equivalent $=\frac{C\times 3C}{C+3C}=\frac{3C}{4}$

Now reduced form of circuit will be as shown below.


Now, $\frac{3C}{4}$ and $C$ are in parallel, so their equivalent capacitance will be

$C^{\prime }=\frac{3C}{4}+C=\frac{7C}{4}$


So equivalent capacitance of the given circuit will be

$C_{eq}=\frac{\frac{7C}{4}\times \frac{7C}{4}}{\frac{7C}{4}+\frac{7C}{4}}=\frac{7C}{8}$

Therefore, option $\left(A\right)$ is the correct choice.

Why this question? Tip: For this type of question, check for symmetry pattern and use equipotential method to simplify circuit.