Question

A network of capacitors is as shown below. If the voltage supply is \(100\,V\), find the energy stored in the \(6\,\mu F\) capacitor.

• A. \(1.2\,\text{mJ}\)
• B. \(12\,\text{mJ}\)
• C. \(2.2\,\text{mJ}\)
• D. \(4.2\,\text{mJ}\)

Hint:

In a balanced capacitor bridge, the middle capacitor has no charge. Always check symmetry first to simplify circuits.

Answer 1

The Correct Option is A

Step 1: Identify the configuration.
The circuit forms a bridge network. Capacitors \(C_1\) and \(C_3\) are on vertical branches, \(C_2\) is on top, and \(C_4\) is in the middle.
Given:
\[ C_1 = 3\mu F,\; C_2 = 6\mu F,\; C_3 = 3\mu F,\; C_4 = 4\mu F \]

Step 2: Check symmetry of the circuit.
Since \(C_1 = C_3 = 3\mu F\), the bridge is balanced.
Hence, no potential difference appears across \(C_4\), so it does not affect the circuit.

Step 3: Reduce the circuit.
Now the circuit reduces to a series combination:
\[ C_3 - C_2 - C_1 \]
i.e., three capacitors in series: \(3\mu F, 6\mu F, 3\mu F\).

Step 4: Find equivalent capacitance.
\[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{3} \]
\[ \frac{1}{C_{eq}} = \frac{2 + 1 + 2}{6} = \frac{5}{6} \]
\[ C_{eq} = \frac{6}{5} = 1.2\mu F \]

Step 5: Find charge in the series circuit.
\[ Q = C_{eq}V = 1.2\mu F \times 100 \]
\[ Q = 120\mu C \]

Step 6: Voltage across \(6\mu F\) capacitor.
In series, charge is same:
\[ V_2 = \frac{Q}{C_2} = \frac{120}{6} = 20V \]

Step 7: Energy stored in \(6\mu F\).
\[ U = \frac{1}{2}CV^2 \]
\[ U = \frac{1}{2} \times 6\mu F \times (20)^2 \]
\[ U = 3 \times 400 \mu J \]
\[ U = 1200 \mu J = 1.2\,\text{mJ} \]
\[ \boxed{1.2\,\text{mJ}} \]