A circuit shown in the figure in which $k_{1}$ is closed and $k_{2}$ is open. Inductor L can be connected to capacitor $C_{1}$ by closing switch $k_{2}$ and opening $k_{1}$ then pick out the correct option(s)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Let the switch $k_{1}$ be closed and $k_{2}$ is opened for long time, then the charge on the capacitor $C_{2}$ will be $24 μ C$

At t=0 when the capacitors are fully charged, switch $k_{1}$ is opened and the switch $k_{2}$ is closed so that inductor is connected in series with capacitor $C_{1}$ . The maximum charge will appear on capacitor $C_{1}$ at time $t = \frac{π}{500} \sec$

The maximum energy stored in the inductor is $0.144 m J$

The maximum energy stored in the inductor is $0.072 m J$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$O p t i o n 1 : B o t h c a p a c i t o r s a r e i n s e r i e s, s o h a v e s a m e c h a r g e \frac{q}{C_{1}} + \frac{q}{C_{2}} = 20 \Rightarrow q = (\frac{2 \times 3}{2 + 3}) 10^{- 6} \times 20 = \frac{6}{5} \times 20 μ C = 24 μ C O p t i o n 2 : C i r c u i t b e c o m e s L C o s c i l l a t o r w i t h t i m e p e r i o d T = 2 π \sqrt{L C} = \frac{π}{250} s S i n c e, g i v e n t i m e t = \frac{π}{500} = \frac{T}{2}, m e a n s c a p a c i t o r i s a g a i n f u l l y c h a r g e d w i t h o p p o s i t e p o l a r i t y . O p t i o n 3 : M a x e n e r g y s t o r e d i n i n d u c t o r i s s a m e a s m a x e n e r g y s t o r e d i n c a p a c i t o r . U = \frac{q^{2}}{2 C} = \frac{24 \times 24 \times 10^{- 12}}{2 \times 2 \times 10^{- 6}} = 0.144 m J$