JEE Main 2026 — Capacitance Question with Solution

Let the nodes in the circuit be labeled to identify the connections:
- Let the input terminal be Node A.
- Let the junction between $C_1$ and $C_2$ be Node X.
- The top wire connects Node A to the junction between $C_2$ and $C_3$. Thus, this junction is also at the potential of Node A.
- The bottom wire connects Node X to the output terminal Node B, with capacitor $C_4$ in this path.

Based on these nodes, the connections of each capacitor are as follows:
- $C_1$ is connected between Node A and Node X.
- $C_2$ is connected between Node X and Node A.
- $C_3$ is connected between Node A and Node B.
- $C_4$ is connected between Node X and Node B.

We can now simplify the circuit step-by-step:
Capacitors $C_1$ and $C_2$ are in parallel between nodes A and X. Their equivalent capacitance is:
$C_{12}=C_1+C_2=1\mu \text{F}+1\mu \text{F}=2\mu \text{F}$

This equivalent capacitor $C_{12}$ is in series with $C_4$ (which is connected between X and B). The equivalent capacitance of this branch is:
$C_{AXB}=\frac{C_{12}\times C_4}{C_{12}+C_4}=\frac{2\times 2}{2+2}=1\mu \text{F}$

Finally, this branch is in parallel with $C_3$ (which is connected directly between A and B). The total equivalent capacitance between A and B is:
$C_{AB}=C_{AXB}+C_3=1\mu \text{F}+1\mu \text{F}=2\mu \text{F}$

Answer: $2$