Question

Two capacitors, \(C_1\) and \(C_2\) have their capacitances in the ratio 1:2. \(V_s\) and \(V_p\) are the potential differences applied across the series and parallel combination of \(C_1\) and \(C_2\) respectively, so that the energy stored in the two cases becomes same. The ratio \(V_s\) to \(V_p\) is

• A. \(\sqrt{2}:3\)
• B. \(3:\sqrt{2}\)
• C. \(2:\sqrt{3}\)
• D. \(\sqrt{3}:2\)

Hint:

Energy is \(\frac12C_{eq}V^2\); equate for the two combinations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Let \(C_1=C\) and \(C_2=2C\). Series: \(C_s=\dfrac{C\cdot2C}{3C}=\dfrac{2C}{3}\). Parallel: \(C_p=3C\).

Step 2: Key Formula or Approach
Energy stored is \(U=\tfrac12C_{eq}V^2\).

Step 3: Detailed Explanation
Equal energies: \(\tfrac12C_sV_s^2=\tfrac12C_pV_p^2\).
\[ \frac{V_s^2}{V_p^2}=\frac{C_p}{C_s}=\frac{3C}{2C/3}=\frac92 \]
\[ \frac{V_s}{V_p}=\frac{3}{\sqrt2} \]

Final Answer:
The ratio \(V_s:V_p\) is \(3:\sqrt2\), option (B). \[ \boxed{3:\sqrt2\ \text{(B)}} \]