Question 61: AC fundamentals

Question

The capacitor of capacitance $C$ can be charged (with the help of a resistance R ) by a voltage source V , by closing switch $\mathrm{S}_1$ while keeping switch $\mathrm{S}_2$ open. The capacitor can be connected in series with an inductor L by closing switch $\mathrm{S}_2$ and opening $\mathrm{S}_1$.

Accepted Answer

$$ \begin{aligned} &\begin{aligned} q & =\mathrm{Q}_0 \cos \omega t \ i & =\frac{d q}{d t}=\mathrm{Q}_0 \omega \sin \omega t \end{aligned}\ &	ext { From conservation of energy. }\ &\begin{aligned} \frac{1}{2} \mathrm{LI}_{\max }^2 & =\frac{1}{2} \mathrm{CV}^2 \ \mathrm{I}_{\max } & =\sqrt{\frac{\mathrm{CV}^2 	imes \frac{1}{2}}{\frac{1}{2} \mathrm{~L}}}=\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{~L}}} \end{aligned} \end{aligned} $$

Answer

at $t=0$, the energy stored in the circuit is purely in the form of magnetic energy.

Answer

at any time $t>0$, the current in the circuit is in the same direction.

Answer

at $t>0$, there is no exchange of energy between the inductor and the capacitor.

Answer

at any time $t>0$, the instantaneous current in the circuit may $\mathrm{V} \sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$.

AC fundamentals

The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V , by closing switch \mathrm{S}_1 while keeping switch \mathrm{S}_2 open. The capacitor can be connected in series with an inductor L by closing switch \mathrm{S}_2 and opening \mathrm{S}_1.

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