Question

In the arrangement of the capacitors as shown in figure, each capacitor is of $6\ \mu\text{F}$, then equivalent capacity between points A and B is

• A. $12\ \mu\text{F}$
• B. $6\ \mu\text{F}$
• C. $4\ \mu\text{F}$
• D. $10\ \mu\text{F}$

Hint:

A massive time-saver for standard symmetric circuits: If a Wheatstone bridge is balanced and all four outer components (resistors or capacitors) have the exact same value $X$, the total equivalent value of the entire bridge is always simply $X$.

Answer 1

The Correct Option is B

Step 1: Understand the network.
Five equal capacitors, each $6\ \mu\text{F}$, are joined in a bridge shape between points A and B. Four sit on the outer arms and one sits across the middle.

Step 2: Recall the balanced bridge idea.
A bridge is "balanced" when the arm capacitors satisfy $\frac{C_1}{C_2}=\frac{C_3}{C_4}$. If balanced, the middle capacitor carries no charge and can be ignored.

Step 3: Check the balance.
All capacitors are equal, so $\frac{6}{6}=\frac{6}{6}$, i.e. $1=1$. The bridge is balanced.

Step 4: Remove the middle capacitor.
Because it carries no charge, the central $6\ \mu\text{F}$ acts like it is not there. The circuit becomes two simple branches.

Step 5: Combine each branch (series).
Each branch has two $6\ \mu\text{F}$ in series: \[ \frac{1}{C_s}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\Rightarrow C_s=3\ \mu\text{F}. \] So each branch gives $3\ \mu\text{F}$.

Step 6: Combine the two branches (parallel).
The two branches sit side by side between A and B, so they add: \[ C_{eq}=3+3=6\ \mu\text{F}. \]

Step 7: State the result.
The equivalent capacitance is $6\ \mu\text{F}$, which is option (2).
\[ \boxed{6\ \mu\text{F}} \]