Question

Four capacitors each of capacitance $16\,\mu F$ are connected as shown in the figure. The capacitance between points A and B is __ (in $\mu F$) 

Hint:

Use the parallel connection of capacitors to find the equivalent capacitance.

Answer 1

Correct Answer: 64

Each capacitor has \(C=16\,\mu F\). We need \(C_{AB}\).

Concept Used:

If all capacitors share the same two nodes, they are in parallel and the equivalent is the sum: \[ C_{\text{eq}}=\sum C_i. \] Shorted (equipotential) points can be merged into a single node.

Step-by-Step Solution:

Step 1: Identify the top rail as a single conductor that 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 mid points (either side of the middle capacitor) and drops to \(A\). Therefore, the entire lower rectangular path is at the potential of \(A\).

Step 3: With these node identifications:

• Left vertical capacitor connects the top rail (\(B\)) to the lower rail (\(A\)).
• Middle horizontal capacitor has both its plates tied to the same two rails (\(B\) and \(A\)).
• Right horizontal capacitor also connects the same two rails (\(B\) and \(A\)).
• Right vertical capacitor connects the same two rails (\(B\) and \(A\)).

Thus all four capacitors are connected directly between the same pair of nodes \(A\) and \(B\); they are in parallel.

 

Step 4: Add the capacitances in parallel.

\[ C_{AB}=C+C+C+C=4C=4\times 16\,\mu F=64\,\mu F. \]

Final Computation & Result:

Equivalent capacitance between A and B: \(C_{AB}=64\,\mu F\).

Answer 2

1. Redraw the circuit: - The capacitors are connected in parallel.
2. Equivalent capacitance: \[ C_{\text{eq}} = 4C = 4 \times 16 \mu \mathrm{F} = 64 \mu \mathrm{F} \] Therefore, the correct answer is (64) $64 \mu \mathrm{F}$.