Alternating Current Ques 18
18. 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: 18.(a)
Solution:
- (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, $i=V_1 \sqrt{\frac{C}{L}}(1-\frac{V_2^{2}}{V_1^{2}})^{1 / 2}$
$=(\frac{C(V_1^{2}-V_2^{2})}{L})^{1 / 2}$
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