Electro Magnetic Induction And Alternating Currents Question 176
Question: 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
Options:
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: A
Solution:
-
$ q=CV_{1}\cos \omega t\Rightarrow i=\frac{dq}{dt}=-\omega Cv_{1}\sin \omega t $
Also, $ {{\omega }^{2}}=\frac{1}{LC} $
and $ V=V_{1}\cos \omega t $ At $ t=t_{1} $ ,
$ V=V_{2} $
and $ i=-\omega CV_{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}} $
BETA
