Question

The capacitance between the points \(A\) and \(B\) in the following figure is

• A. \(\frac{3}{8}\,\mu F\)
• B. \(\frac{9}{4}\,\mu F\)
• C. \(\frac{4}{5}\,\mu F\)
• D. \(2\,\mu F\)

Hint:

In a balanced capacitor bridge, the capacitor in the middle branch has zero potential difference across it, so it can be ignored while finding equivalent capacitance.

Answer 1

The Correct Option is B

Step 1: Identify the circuit topology.
The circuit between points A and B is a Wheatstone bridge-type network with capacitors. Let the capacitors be $C_1 = 1\,\mu F$, $C_2 = 3\,\mu F$ (upper branch), $C_3 = 2\,\mu F$, $C_4 = 6\,\mu F$ (lower branch), and $C_5 = 5\,\mu F$ (bridge arm between the two junctions).
Step 2: Check the Wheatstone bridge balance condition.
For a capacitor bridge, the balance condition is: \[ \frac{C_1}{C_2} = \frac{C_3}{C_4} \] Substituting: $\dfrac{1}{3}$ and $\dfrac{2}{6} = \dfrac{1}{3}$. Since both ratios are equal, the bridge is balanced.
Step 3: Apply the consequence of balance.
When the bridge is balanced, there is no potential difference across the bridge arm ($C_5 = 5\,\mu F$). Therefore, no charge accumulates on $C_5$ and it plays no role in the equivalent capacitance. We can remove $C_5$ from the circuit.
Step 4: Find the equivalent capacitance of the upper branch.
$C_1 = 1\,\mu F$ and $C_2 = 3\,\mu F$ are in series: \[ C_{\text{upper}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 \times 3}{1 + 3} = \frac{3}{4}\,\mu F \]
Step 5: Find the equivalent capacitance of the lower branch.
$C_3 = 2\,\mu F$ and $C_4 = 6\,\mu F$ are in series: \[ C_{\text{lower}} = \frac{C_3 C_4}{C_3 + C_4} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = \frac{3}{2}\,\mu F \]
Step 6: Find the total capacitance between A and B.
The two branches are now in parallel: \[ C_{AB} = C_{\text{upper}} + C_{\text{lower}} = \frac{3}{4} + \frac{3}{2} = \frac{3}{4} + \frac{6}{4} = \frac{9}{4}\,\mu F \] \[ \boxed{C_{AB} = \frac{9}{4}\,\mu F} \]