JEE Main 2026 — Physics Electromagnetism

Let the nodes in the circuit be labeled to identify the connections:

• Let the input terminal be Node A.

• Let the junction between $C_1$C1​ and $C_2$C2​ be Node X.

• The top wire connects Node A to the junction between $C_2$C2​ and $C_3$C3​. 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 $C_4$C4​ in this path.

Based on these nodes, the connections of each capacitor are as follows:

• $C_1$C1​ is connected between Node A and Node X.

• $C_2$C2​ is connected between Node X and Node A.

• $C_3$C3​ is connected between Node A and Node B.

• $C_4$C4​ is connected between Node X and Node B.

We can now simplify the circuit step-by-step:

Capacitors $C_1$C1​ and $C_2$C2​ are in parallel between nodes A and X. Their equivalent capacitance is:

$$C_{12} = C_1 + C_2 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}$$C12​=C1​+C2​=1μF+1μF=2μF

This equivalent capacitor $C_{12}$C12​ is in series with $C_4$C4​ (which is connected between X and B). The equivalent capacitance of this branch is:

$$C_{AXB} = \dfrac{C_{12} \times C_4}{C_{12} + C_4} = \dfrac{2 \times 2}{2 + 2} = 1\,\mu\text{F}$$CAXB​=C12​+C4​C12​×C4​​=2+22×2​=1μF

Finally, this branch is in parallel with $C_3$C3​ (which is connected directly between A and B). The total equivalent capacitance between A and B is:

$$C_{AB} = C_{AXB} + C_3 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}$$CAB​=CAXB​+C3​=1μF+1μF=2μF

Answer: $2$2