JEE Physics 2025 Question with Solution

Given: Each capacitor has capacitance $C = 16\,\mu F$.

Find: The equivalent capacitance $C_{AB}$ between points $A$ and $B$.

Concept Used: If all capacitors are connected across the same two nodes, they are in parallel, and their equivalent capacitance is the sum of individual capacitances.

Shorted or directly connected points are at the same potential and can be treated as a single node.

Step 1: The top rail is a single conductor and finally goes down to point $B$. Hence every point on the top rail is at the potential of $B$.

Step 2: The lower rectangular wire connects the two midpoints and drops to $A$. Therefore the entire lower rectangular path is at the potential of $A$.

Step 3: With these node identifications, each of the four capacitors is connected directly between the same two nodes $A$ and $B$. Therefore, all four capacitors are in parallel.

Step 4: Add the capacitances in parallel.

Therefore, the equivalent capacitance between $A$ and $B$ is $64\,\mu F$.

Given: Four identical capacitors, each of capacitance $16\,\mu F$.

Find: Equivalent capacitance between $A$ and $B$.

Instead of reducing the circuit branch by branch, inspect the nodes. The upper conductor is one common node ending at $B$, and the lower conductor is one common node ending at $A$. Since every capacitor has one plate on the upper node and the other plate on the lower node, all capacitors are in parallel.

Therefore, the correct numerical answer is 64.