Question

When three capacitors of capacitances ' C ', ' 2 C '? and ' 3 C ' are connected in series. The potential difference applied across the combination is ' \(V\) '. The energy stored in the circuit is

• A. \(\frac{1}{11}\mathrm{CV}^{2}\)
• B. \(\frac{3}{11}\mathrm{CV}^{2}\)
• C. \(\frac{5}{11}\mathrm{CV}^{2}\)
• D. \(\frac{7}{11}\mathrm{CV}^{2}\)

Hint:

Find the equivalent capacitance and use U = half C V squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For series capacitors, \(\dfrac{1}{C_{eq}}=\dfrac{1}{C}+\dfrac{1}{2C}+\dfrac{1}{3C}\).

Step 2: Equivalent capacitance
\(\dfrac{1}{C_{eq}}=\dfrac{6+3+2}{6C}=\dfrac{11}{6C}\Rightarrow C_{eq}=\dfrac{6C}{11}\).

Step 3: Energy
\(U=\dfrac{1}{2}C_{eq}V^{2}=\dfrac{1}{2}\times\dfrac{6C}{11}V^{2}\).

Step 4: Result
\(U=\dfrac{3}{11}CV^{2}\).

Final Answer:
The energy stored is \(\dfrac{3}{11}CV^{2}\). \[ \boxed{\text{(B) }\dfrac{3}{11}\mathrm{CV}^{2}} \]