In the given circuit, initially switch S 1 is closed and S 2 and S 3 are open. After charging of 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 = 4CR 2 then find the time (in sec) after which current passing through capacitor and inductor will be same.
(given R = ln 2 m Ω , L = 2 mH)
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(1)
Sol. After charging, charge on capacitor = C ε
Now at t = 0 two circuits formed
Discharging of capacitor
∴ q = C ε 
= C ε 
∴ i 1 = 
Growth of current in L-R circuit
i 2 = 
now i 1 = i 2
= 
…(1)
given L = 4CR 2 ∴ 
= 2RC = 
⇒ ln from equation (1) 2 
= 1
⇒ t ln2 = ln2
⇒ t = 1sec.
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