Electro Magnetic Induction And Alternating Currents Question 152
Question: In a series circuit $ C=2\mu F,,L=1mH $ and $ R=10,\Omega , $ when the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be
Options:
A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 5
Show Answer
Answer:
Correct Answer: D
Solution:
Current will be maximum in the condition of resonance so $
{i_{\max }}=\frac{V}{R}=\frac{V}{10}A $ Energy stored in the coil $ W_{L}=\frac{1}{2}Li_{\max }^{2} $
$ =\frac{1}{2}L{{( \frac{E}{10} )}^{2}} $
$ =\frac{1}{2}\times {{10}^{-3}}( \frac{E^{2}}{100} ) $
$ =\frac{1}{2}\times {{10}^{-5}}E^{2}\ joule $
$ \therefore $ Energy stored in the capacitor $ W_{C}=\frac{1}{2}CE^{2}=\frac{1}{2}\times 2\times {{10}^{-6}}E^{2}={{10}^{-6}}E^{2}\ joule $
$ \therefore \frac{W_{C}}{W_{L}}=\frac{1}{5} $
BETA
