Question

A network of five capacitors is shown in the figure. The total charge stored in the network connected between A and B is:

• A. $8\ \mu\text{C}$
• B. $10\ \mu\text{C}$
• C. $9\ \text{mC}$
• D. $10\ \text{mC}$

Hint:

Always look for symmetry first! In most exam questions featuring five capacitors, the bridge is balanced, which lets you ignore the middle capacitor and solve the circuit much faster.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When five capacitors are arranged in a specific "diamond" shape, we first check if it is a Balanced Wheatstone Bridge. If the ratio of the capacitors in the arms is equal, no charge flows through the central capacitor.

Step 2: Calculating Equivalent Capacitance:
1. If $\frac{C_1}{C_2} = \frac{C_3}{C_4}$, the bridge is balanced. The central capacitor is removed. 2. The network simplifies to two parallel branches, each having capacitors in series. 3. Calculate $C_{\text{eq}}$ using series and parallel combination rules.

Step 3: Finding the Total Charge:
Use the formula: \[ Q = C_{\text{eq}} \times V \] Where $Q$ is the total charge, $C_{\text{eq}}$ is the equivalent capacitance, and $V$ is the applied potential difference.

Step 4: Final Answer:
By substituting the specific circuit values, the total charge stored is $10\ \mu\text{C}$.