In the following circuit, the equivalent capacitance between terminal A and terminal B is :

The correct answer is 2μF.

Step 1: Analyze the circuit diagram
Looking at the given circuit image:
The circuit consists of five capacitors, each with a capacitance of $C=2\mathrm{\mu F}$, arranged in a bridge network between terminal A and terminal B.

Step 2: Check for Wheatstone Bridge Balance Condition
Let the top-left, top-right, bottom-left, bottom-right, and central capacitors be $C_1$, $C_2$, $C_3$, $C_4$, and $C_5$ respectively, where:
$C_1=2\mathrm{\mu F}$, $C_2=2\mathrm{\mu F}$, $C_3=2\mathrm{\mu F}$, ${\mathrm{C<!--\; <mn-->4}}_{}=2\mathrm{\mu F}$, and central capacitor ${\mathrm{C<mn>5</mn>}}_{}=2\mathrm{\mu F}$.

The condition for a balanced Wheatstone bridge is given by:

$\frac{C_1}{C_2}=\frac{C_3}{C_4}$

Substituting the given values:

$\frac{2}{2}=\frac{2}{2}=1$

Since the ratio is equal, the bridge is balanced. Therefore, the potential difference across the central capacitor ($C_5=2\mathrm{\mu F}$) is zero, and no charge flows through it. We can safely remove the central capacitor from our calculations.

Step 3: Calculate equivalent capacitance of the simplified circuit
After removing the middle capacitor:
- The top branch has two capacitors ($C_1$ and $C_2$) connected in series.
- The bottom branch has two capacitors ($C_3$ and $C_4$) connected in series.

The equivalent capacitance of the top series branch ($C_{\text{top}}$) is:

$C_{}\text{top}=\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

The equivalent capacitance of the bottom series branch ($C_{}\text{bottom}$) is:

$C_{}\text{bottom}=\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

Step 4: Find total equivalent capacitance ($C_{\text{AB}}$)
The top and bottom branches are connected in parallel, so their equivalent capacitances add directly:

$C_{}\text{AB}=C_{}\text{top}+C_{}\text{bottom}=1\mathrm{\mu F}+1\mathrm{\mu F}=2\mathrm{\mu F}$

Hence, the equivalent capacitance between terminal A and terminal B is 2μF.