Alternating Current - Result Question 23

23. A condenser of capacity $C$ is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is

[2010]

(a) $(\frac{C(V_1^{2}-V_2^{2})}{L})^{1 / 2}$

(b) $(\frac{C(V_1-V_2)^{2}}{L})^{1 / 2}$

(c) $\frac{C(V_1^{2}-V_2^{2})}{L}$

(d) $\frac{C(V_1-V_2)}{L}$

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Answer:

Correct Answer: 23. (a)

Solution:

  1. (a) $q=C V_1 \cos \omega$

$\Rightarrow i=\frac{d q}{d t}=-\omega C v_1 \sin \omega t$

Also, $\omega^{2}=\frac{1}{L C}$ and $V=V_1 \cos \omega t$

At $t=t_1, V=V_2$ and $i=-\omega C V_1 \sin \omega t_1$

$\therefore \cos \omega t_1=\frac{V_2}{V_1}$ (-ve sign gives direction)

Hence, $i=V_1 \sqrt{\frac{C}{L}}(1-\frac{V_2^{2}}{V_1^{2}})^{1 / 2}$

$=(\frac{C(V_1^{2}-V_2^{2})}{L})^{1 / 2}$