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. \(8\,\mu C\)
• B. \(10\,\mu C\)
• C. \(9\,mC\)
• D. \(10\,mC\)

Hint:

For a balanced capacitor bridge: \[ \frac{C_1}{C_2} = \frac{C_3}{C_4} \] the bridge capacitor has zero potential difference across it and can be removed from the analysis. After removing it, reduce the circuit using simple series and parallel combinations.

Answer 1

The Correct Option is B

Step 1: Check whether the bridge is balanced. The capacitor bridge has: \[ \frac{4}{2} = \frac{6}{3} = 2 \] Since \[ \frac{C_1}{C_2} = \frac{C_3}{C_4}, \] the bridge is balanced. Therefore, no charge flows through the central capacitor. Hence, the \[ 5\,\mu F \] capacitor can be ignored.

Step 2: Find equivalent capacitance of the upper branch. The \(4\,\mu F\) and \(2\,\mu F\) capacitors are in series. \[ C_{\text{upper}} = \frac{4\times2}{4+2} = \frac{8}{6} = \frac{4}{3}\,\mu F \]

Step 3: Find equivalent capacitance of the lower branch. The \(6\,\mu F\) and \(3\,\mu F\) capacitors are in series. \[ C_{\text{lower}} = \frac{6\times3}{6+3} = \frac{18}{9} = 2\,\mu F \]

Step 4: Combine the two branches. The upper and lower branches are in parallel. \[ C_{\text{eq}} = \frac{4}{3} + 2 = \frac{10}{3}\,\mu F \]

Step 5: Calculate total charge stored. The network is connected across \[ V=3\,V \] Therefore, \[ Q=C_{\text{eq}}V \] \[ Q= \left(\frac{10}{3}\right)(3) \] \[ Q=10\,\mu C \] Therefore, the total charge stored in the network is \[ \boxed{10\,\mu C} \]