Question

The equivalent capacitance of the system shown in the figure between the points A and B is:

• A. 5 μF
• B. 10 μF
• C. 20 μF
• D. 40 μF

Hint:

In symmetric capacitor networks, check for equipotential points to eliminate central components.

Answer 1

The Correct Option is B

Step 1: Assign node potentials instead of arguing from symmetry alone.
Let A be at potential $V$ and B at $0$. Call the midpoint of the top branch $M_1$ (potential $V_1$) and the midpoint of the bottom branch $M_2$ (potential $V_2$); each branch has a 10 $\mu$F capacitor from A to its midpoint and another 10 $\mu$F from the midpoint to B, with the 5 $\mu$F bridging $M_1$ and $M_2$.

Step 2: Write charge balance at each midpoint node.
\[ 10(V - V_1) = 10(V_1 - 0) + 5(V_1 - V_2) \] \[ 10(V - V_2) = 10(V_2 - 0) + 5(V_2 - V_1) \]
Step 3: Solve the two equations together.
Adding them, the bridge terms cancel, giving $V_1 + V_2 = V$. Subtracting them gives $V_1 = V_2$. Combined, $V_1 = V_2 = \frac{V}{2}$.

Step 4: Since $V_1 = V_2$, no charge crosses the bridge capacitor, and the charge from A is simple to total.
\[ Q = 10\left(V - \frac{V}{2}\right) + 10\left(V - \frac{V}{2}\right) = 10V \]
Step 5: Get the equivalent capacitance.
\[ C_{eq} = \frac{Q}{V} = 10\ \mu\text{F} \]
Final Answer:
\[ \boxed{10\ \mu\text{F}} \]