Given that the total charge stored in the LC circuit is , for , the charge on the capacitor is [2006]

Question

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch $S_{1}$ . Also when $S_{1}$ is opened and $S_{2}$ is closed the capacitor is connected in series with inductor (L).

Q. Given that the total charge stored in the LC circuit is $Q_{0}$ , for $t \geq 0$ , the charge on the capacitor is [2006]

Smart Achievers · Accepted Answer

(3)

$Applying Kirchhoff's law$

$\frac{Q}{C} - L \frac{d I}{d t} = 0 \Rightarrow \frac{Q}{C} = L \frac{d I}{d t}$

$\Rightarrow Q = L C \frac{d}{d t} (- \frac{d Q}{d t}) = - L C \frac{d^{2} Q}{d t^{2}}$

Solution

1 $Q = Q_{0} \cos (\frac{π}{2} + \frac{t}{\sqrt{L C}})$

2 $Q = Q_{0} \cos (\frac{π}{2} - \frac{t}{\sqrt{L C}})$

3 $Q = - L C \frac{d^{2} Q}{d t^{2}}$

4 $Q = - \frac{1}{\sqrt{L C}} \frac{d^{2} Q}{d t^{2}}$

Ans.
(3)

$Applying Kirchhoff's law$

$\frac{Q}{C} - L \frac{d I}{d t} = 0 \Rightarrow \frac{Q}{C} = L \frac{d I}{d t}$

$\Rightarrow Q = L C \frac{d}{d t} (- \frac{d Q}{d t}) = - L C \frac{d^{2} Q}{d t^{2}}$

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Solution

1 Q=Q0cos(π2+tLC)

2 Q=Q0cos(π2-tLC)

3 Q=-LCd2Qdt2

4 Q=-1LCd2Qdt2