Question

The effective capacitance between points \(X\) and \(Y\) in the figure shown below is. Assume all the capacitors are \(4\,\mu\text{F}\) each.

• A. \(3\,\mu\text{F}\)
• B. \(1\,\mu\text{F}\)
• C. \(4\,\mu\text{F}\)
• D. \(2\,\mu\text{F}\)

Hint:

For capacitor networks that are not simple series-parallel combinations, assign potentials to internal junctions and apply charge balance at floating nodes. \[ \sum C(V_{\text{node}}-V_{\text{connected node}})=0 \] This method quickly gives the equivalent capacitance.

Answer 1

The Correct Option is C

Step 1: Apply nodal analysis at internal junctions.
Let potentials be V_X = V, V_Y = 0, with internal nodes A and B. Write KCL: (V_A–V)+(V_A–V_B)+V_A = 0 and (V_B–V_A)+V_B+(V_B–V) = 0.

Step 2: Solve the symmetric system.
3V_A – V_B = V, 3V_B – V_A = V → by symmetry V_A = V_B = V/2.

Step 3: Compute charge drawn from terminal X.
Q = C(V–V_A) + C(V–V_B) = C(V/2) + C(V/2) = CV.

Step 4: Find equivalent capacitance.
C_eq = Q/V = C = 4 μF.

Step 5: Final Answer:
4 μF.