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: Formula first
$U=\dfrac{1}{2}C_{eq}V^{2}$ with the whole potential difference $V$ across the series combination.

Step 2: Substitution
$C_{eq}=\dfrac{6C}{11}$ gives $U=\dfrac{3CV^{2}}{11}$.

Step 3: Match
The option with $\dfrac{3}{11}$ is correct.

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