Question

An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mc. The resistance of the circuit is negligible. At the instant the circuit is closed be t = 0. At what time is the energy stored completely magnetic. 

(a) t = 0ms 

(b) t = 1.54ms 

(c) t = 3.14ms 

(d) t = 6.28ms

Answer 1

The correct option  (b) t = 1.54ms

Explanation:

L = 20 mH

C = 50 μF

Q = 10 mc

∴ Frequency  = f = [1/{2π√(LC)}]

= [1/{2π√(20 × 10^–3 × 50 × 10^–6)}]

= [1/{2π√(10^–6)}]

= [(10³)/(2π)] Hz

as Time period = T = (1/f) hence T = [(2π) / (1000)] sec  (1)

given: energy stored in completely magnetic i.e. electric energy is zero.

also change on capacitor = Q = Q_o cos ωt = Q_o cos (2π/T)t.

Electric energy = 0 hence Q = 0 i.e. Q_o cos (2πt/T) = 0

means cos (2πt/T) = 0   i.e. (2πt/T) = (nπ/2)

∴ t = (nπ/4). hence t = (T/4), (3T/4), (5T/4)…..

∴ from (1),

t = (T/4)

= [{2π}/{1000}] × (1/4)

= 1.54 ms.