16) Two capacitors \(C_1\) and \(C_2\) have their capacitances in the ratio \(1:2\). Let \(V_s\) and \(V_p\) be the potential differences applied across the series and parallel combinations of \(C_1\) and \(C_2\) respectively, so that the energy stored in the two cases is the same. The ratio \(V_s:V_p\) is:
Solution:
Let the capacitances be: \[ C_1 = C \quad \text{and} \quad C_2 = 2C. \]
Parallel Combination: \[ C_{\text{parallel}} = C_1 + C_2 = C + 2C = 3C. \] The energy stored when a voltage \(V_p\) is applied is: \[ U_p = \frac{1}{2} C_{\text{parallel}} V_p^2 = \frac{1}{2} (3C) V_p^2 = \frac{3C V_p^2}{2}. \]
Series Combination: \[ \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C}. \] Hence, \[ C_{\text{series}} = \frac{2C}{3}. \] The energy stored when a voltage \(V_s\) is applied is: \[ U_s = \frac{1}{2} C_{\text{series}} V_s^2 = \frac{1}{2} \left(\frac{2C}{3}\right) V_s^2 = \frac{C V_s^2}{3}. \]
Since the energies are equal (\(U_p = U_s\)): \[ \frac{3C V_p^2}{2} = \frac{C V_s^2}{3}. \]
Cancel \(C\) (assuming \(C \neq 0\)) and rearrange: \[ \frac{3 V_p^2}{2} = \frac{V_s^2}{3} \quad \Longrightarrow \quad V_s^2 = \frac{9 V_p^2}{2}. \]
Taking the square root of both sides: \[ V_s = \frac{3}{\sqrt{2}}\, V_p. \]
Thus, the ratio is: \[ \frac{V_s}{V_p} = \frac{3}{\sqrt{2}}, \] which can be written as \(3:\sqrt{2}\). Hence, the correct answer is option B.
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