KCET · Physics · Electromagnetic Induction
A fully charged capacitor \(C\) with initial charge \(q_0\) is connected to a coil of selfinductance \(L\) at \(t=0\). The time at which the energy is stored equally between the electric and the magnetic field is
- A \(\sqrt{L C}\)
- B \(\pi \sqrt{L C}\)
- C \(\frac{\pi}{4} \sqrt{L C}\)
- D \(2 \pi \sqrt{L C}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4} \sqrt{L C}\)
Step-by-step Solution
Detailed explanation
In \(L C\)-oscillation, maximum stored energy in inductor coil \(L\) is \(\frac{1}{2} L I_0^2\) and maximum stored energy in capacitor \((C)\) is \(\frac{q_0^2}{2 C}\).
Since, energy is equally divided in \(L\) and \(C\). Hence, total stored energy
\( E=\frac{1}{2} \times \text { maximum stored energy in either } L \text { o } \)
\( \Rightarrow \frac{1}{2} L I^2=\frac{1}{2} \times \frac{1}{2} L I_0^2 \Rightarrow I^2=\left(\frac{I_0}{\sqrt{2}}\right)^2 \Rightarrow I=\frac{I_0}{\sqrt{2}}\)
We know that, \(I=I_0 \sin \omega t\)
\( \Rightarrow \frac{I_0}{\sqrt{2}}=I_0 \sin \omega t \Rightarrow \sin \frac{\pi}{4}=\sin \frac{2 \pi}{T} \cdot t \)
\( \Rightarrow \frac{\pi}{4}=\frac{2 \pi}{T} \cdot t \Rightarrow t=\frac{T}{8}=\frac{2 \pi \sqrt{L C}}{8}=\frac{\pi \sqrt{L C}}{4}\)
Since, energy is equally divided in \(L\) and \(C\). Hence, total stored energy
\( E=\frac{1}{2} \times \text { maximum stored energy in either } L \text { o } \)
\( \Rightarrow \frac{1}{2} L I^2=\frac{1}{2} \times \frac{1}{2} L I_0^2 \Rightarrow I^2=\left(\frac{I_0}{\sqrt{2}}\right)^2 \Rightarrow I=\frac{I_0}{\sqrt{2}}\)
We know that, \(I=I_0 \sin \omega t\)
\( \Rightarrow \frac{I_0}{\sqrt{2}}=I_0 \sin \omega t \Rightarrow \sin \frac{\pi}{4}=\sin \frac{2 \pi}{T} \cdot t \)
\( \Rightarrow \frac{\pi}{4}=\frac{2 \pi}{T} \cdot t \Rightarrow t=\frac{T}{8}=\frac{2 \pi \sqrt{L C}}{8}=\frac{\pi \sqrt{L C}}{4}\)
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