Question
In the $LC$ circuit shown below, the current is in the direction shown. At this time :-
✓
Answer
positive charge flow outside
so, charge decreases
thus voltage $=\frac{\mathrm{Q}}{\mathrm{C}}$ also decreases and $\mathrm{v}=\frac{-\mathrm{L} \mathrm{di}}{\mathrm{dt}}$
so $\frac{\mathrm{di}}{\mathrm{dt}}$ increases
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