NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
In the given circuit, initially switch S 1 is closed and S 2 and S 3 are open. After charging of the capacitor, at t = 0 , S 1 is open and S 2 and S 3 are closed. If the relation between inductance, capacitance and resistance is L = 4 C R 2 , then find the time (in s ) after which the current passing through the capacitor and inductor will be the same. [given R = ln 2 m Ω , L = 2 mH ]
Correct answer
1
Step-by-step solution
Charge on capacitor after complete charging = C ε Now at t   =   0 two circuits are formed, (A) Discharging of capacitor ∴   q = Cεe - t / τ L =   Cεe - t / 2 RC ∴   i 1 =   ε 2 R e - t / 2 RC (B) Growth of current in L - R circuit i 2 =   ε 2 R 1 - e - t / τ L now i 1 =   i 2 ε 2 R e - t / τ C   =   ε 2 R 1 - e - t / τ L       … 1 given L   =   4 C R 2 ∴