MHT CET · Physics · Electromagnetic Induction
In the given circuit, when \(S_1\) is closed, the capacitor \(C\) gets full charged. Then \(S_1\) is kept open and \(S_2\) is closed. Hence
- A The current in the circuit is in the same direction.
- B The instantaneous current in the circuit may be \(V \sqrt{\frac{C}{L}}\).
- C The energy stored in the circuit is purely in the form of magnetic energy.
- D There is no exchange of energy between inductor \(L\) and capacitor \(C\).
Answer & Solution
Correct Answer
(B) The instantaneous current in the circuit may be \(V \sqrt{\frac{C}{L}}\).
Step-by-step Solution
Detailed explanation
When \(S_1\) is closed, potential difference across capacitor is \(V\).
Direction of current gets reversed when \(S_2\) is closed.
When \(S_1\) is opened and \(S_2\) is closed,
\(\begin{aligned}
& \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\
& \Rightarrow I=V \sqrt{\frac{C}{L}}
\end{aligned}\)
Total energy oscillates between capacitor and inductor.
Direction of current gets reversed when \(S_2\) is closed.
When \(S_1\) is opened and \(S_2\) is closed,
\(\begin{aligned}
& \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\
& \Rightarrow I=V \sqrt{\frac{C}{L}}
\end{aligned}\)
Total energy oscillates between capacitor and inductor.
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