Most Important Selected Qs for JEE AdvancedPhysicsAlternating Current
The circuit shown in figure is in the steady state with switch S ₁ closed and S₂ to open. At t=0, S₁ is opened and S₂ is closed. The first instant t , when the energy in the inductor becomes one-third of that in the capacitor C ₂ is millisecond. Find the value of .
Correct answer
300
Step-by-step solution
= 1 L C = 1 2 10⁻⁵ Equation of oscillation of charge on capacitor is q = q ₀ t Energy in capacitor = q^2 2 C and energy in inductor = 1 2 L i^2 Given : 1 2 L i^2= 1 3 q^2 2 C ; solving t= 300 ~ms =300