MHT CET · Physics · Alternating Current
A capacitor of capacity ' \(\mathrm{C}\) ' is charged to a potential ' \(\mathrm{V}\) '. It is connected in parallel to an inductor of inductance ' \(L\) '. The maximum current that will flow in the circuit is
- A \(V \sqrt{\frac{L}{C}}\)
- B \(\mathrm{V} \sqrt{\mathrm{LC}}\)
- C \(\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}\)
- D \(\frac{\mathrm{VC}^2}{\mathrm{~L}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}\)
Step-by-step Solution
Detailed explanation
The charge oscillates according to the equations
\(
\begin{aligned}
& \mathrm{q}=\mathrm{q}_0 \cos \omega \mathrm{t} \\
& \mathrm{i}=-\frac{\mathrm{dq}}{\mathrm{dt}}=\omega \mathrm{q}_0 \sin \omega \mathrm{t}=\mathrm{i}_0 \sin \omega \mathrm{t} \\
& \text { where } \mathrm{i}_0=\omega \mathrm{q}_0, \omega=\frac{1}{\sqrt{\mathrm{LC}}}, \mathrm{q}_0=C V \\
& \therefore \mathrm{i}_0=\frac{1}{\sqrt{\mathrm{LC}}} \cdot \mathrm{CV}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{q}=\mathrm{q}_0 \cos \omega \mathrm{t} \\
& \mathrm{i}=-\frac{\mathrm{dq}}{\mathrm{dt}}=\omega \mathrm{q}_0 \sin \omega \mathrm{t}=\mathrm{i}_0 \sin \omega \mathrm{t} \\
& \text { where } \mathrm{i}_0=\omega \mathrm{q}_0, \omega=\frac{1}{\sqrt{\mathrm{LC}}}, \mathrm{q}_0=C V \\
& \therefore \mathrm{i}_0=\frac{1}{\sqrt{\mathrm{LC}}} \cdot \mathrm{CV}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}
\end{aligned}
\)
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