Most Important Selected Qs for JEE AdvancedPhysicsAlternating Current
Paragraph: 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 . Question: The maximum energy across inductor will be
Options
- A0.144 mJ
- B0.288 mJ
- C0.072 mJ
- DNone of these
Correct answer
A. 0.144 mJ
Step-by-step solution
For long time capacitor gets full charged and charge on each capacitor must be same. aligned & Q 2 + Q 3 =20 & Q =24 C & [ = 1 LC = 1 2 2 10⁻⁶ = 10 2 =50 rad / s ] & Q ( t )=24 t C & Q ( t )=(24 t ) C & 500 t = 2 (-1)^ n + n & t = n 500 +(-1)^ n 1000 & E _ = Q ₀^2 2 C = 24 10⁻³ 24 10⁻⁶ 2 2 10⁻⁶ =144 10⁻⁶ Joule =0.144 ~mJ aligned