Calculate Equivalent Capacitance Between Terminals in Circuit

To find the equivalent capacitance between terminals $A$ and $B$, we analyze the node connections in the given circuit:

1. Identify the Nodes and Potentials:

   • Let terminal $A$ be at potential $V_A$. Terminal $A$ is directly connected to:
     • The left plate of capacitor $C_1$.
     • The junction between capacitors $C_2$ and $C_3$ (via the top connecting wire).
   • Let terminal $B$ be at potential $V_B$. Terminal $B$ is directly connected to:
     • The right plate of capacitor $C_3$.
     • The right terminal of capacitor $C_4$.
   • Let $X$ be the intermediate node between $C_1$ and $C_2$ at potential $V_X$. Node $X$ is also connected to the left terminal of $C_4$.

2. Analyze the Connection of Each Capacitor:

   • Capacitor $C_1$: Connected between Node $A$ and Node $X$.
   • Capacitor $C_2$: Connected between Node $X$ and Node $A$.
   • Capacitor $C_3$: Connected directly between Node $A$ and Node $B$.
   • Capacitor $C_4$: Connected between Node $X$ and Node $B$.

3. Calculate Equivalent Capacitances Step-by-Step:

   • Step 1: Parallel combination of $C_1$ and $C_2$ Since $C_1$ and $C_2$ are both connected between Node $A$ and Node $X$, they are in parallel: $C_{12} = C_1 + C_2 = 1\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 2\text{ }\mu\text{F}$

   • Step 2: Series combination of $C_{12}$ and $C_4$ The branch from terminal $A$ to terminal $B$ through node $X$ consists of $C_{12}$ in series with $C_4$: $C_{AXB} = \frac{C_{12} \cdot C_4}{C_{12} + C_4} = \frac{2 \times 2}{2 + 2} = 1\text{ }\mu\text{F}$

   • Step 3: Parallel combination of $C_{AXB}$ and $C_3$ The branch $AXB$ and capacitor $C_3$ are connected in parallel between terminals $A$ and $B$: $C_{AB} = C_{AXB} + C_3 = 1\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 2\text{ }\mu\text{F}$

Thus, the equivalent capacitance between terminals $A$ and $B$ is $2\text{ }\mu\text{F}$.

Correct Option: A