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GUJCET · Physics · Alternating Current
The charge of the capacitor in LC oscillatory circuit, when the energy associated with inductor and capacitor are equal, is ______________ . ( \(Q_{0}\) is the initial charge on the capacitor.)
- A \(\frac{Q_{0}}{2}\)
- B \(Q_{0}\)
- C \(\frac{\mathrm{Q}_{0}}{\sqrt{3}}\)
- D \(\frac{\mathrm{Q}_{0}}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{Q}_{0}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
(D) \(\frac{\mathrm{Q}_{0}}{\sqrt{2}}\)
Total energy \(U=\frac{1}{2} \frac{Q_{0}^{2}}{C}\)
Assume that the charge on the capacitor is Q when the total energy is stored equally in both fields.
\(U ^{\prime}=\frac{1}{2} \frac{q^2}{ C }\)
but \(\quad U^{\prime}=\frac{U}{2}\)
\(\therefore \quad \frac{1}{2} \frac{q^2}{ C }=\frac{1}{2}\left[\frac{1}{2} \frac{ Q _0^2}{ C }\right]\)
\(\therefore \quad q^2=\frac{ Q _0^2}{2}\)
\(\therefore \quad q=\frac{ Q _0}{\sqrt{2}}\)
Total energy \(U=\frac{1}{2} \frac{Q_{0}^{2}}{C}\)
Assume that the charge on the capacitor is Q when the total energy is stored equally in both fields.
\(U ^{\prime}=\frac{1}{2} \frac{q^2}{ C }\)
but \(\quad U^{\prime}=\frac{U}{2}\)
\(\therefore \quad \frac{1}{2} \frac{q^2}{ C }=\frac{1}{2}\left[\frac{1}{2} \frac{ Q _0^2}{ C }\right]\)
\(\therefore \quad q^2=\frac{ Q _0^2}{2}\)
\(\therefore \quad q=\frac{ Q _0}{\sqrt{2}}\)
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