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
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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 $X$.
Step 1: Understanding the Question:
We need to find the equivalent capacitance across terminals A and B for a network of five identical capacitors arranged in a bridge formation.
Step 2: Key Formula or Approach:
When five capacitors are arranged in a quadrilateral with one bridging the diagonal, we must check for a balanced Wheatstone bridge condition:
$$\frac{C_1}{C_2} = \frac{C_3}{C_4}$$
If balanced, no charge flows through the central diagonal capacitor, and it can be entirely removed from the circuit calculations.
Step 3: Detailed Explanation:
In the standard bridge figure, let the four outer arm capacitors be $C_1, C_2, C_3, C_4$ and the central one be $C_5$.
The problem states that every single capacitor is identical, so $C_1 = C_2 = C_3 = C_4 = C_5 = 6\ \mu\text{F}$.
Checking the balance condition:
$$\frac{6}{6} = \frac{6}{6} \implies 1 = 1$$
Since the bridge is perfectly balanced, the central capacitor $C_5$ acts as an open circuit and is ignored.
The circuit simplifies to two parallel branches, each containing two capacitors in series.
Upper branch equivalent ($C_{s1}$): $\frac{1}{C_{s1}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \implies C_{s1} = 3\ \mu\text{F}$.
Lower branch equivalent ($C_{s2}$): $\frac{1}{C_{s2}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \implies C_{s2} = 3\ \mu\text{F}$.
These two branches are in parallel, so their equivalent capacitance is their sum:
$$C_{eq} = C_{s1} + C_{s2} = 3\ \mu\text{F} + 3\ \mu\text{F} = 6\ \mu\text{F}$$
Step 4: Final Answer:
The equivalent capacity is $6\ \mu\text{F}$, matching option (B).
