In a series circuit `C=2muF,L=1mH` and `R=10 Omega`, when the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Options

A. `1:2`
B. `5:1`
C. `1:5`
D. `1:1`

Answer

B

Explanation

Maximum current flows in a series L-C-R circuit, when the condition of resonance is satisfied. In that case,
`X_(L) = X_(C) and Z = R and I = I_(max) = V/R = V/10 A`
Energy stored in the inductor
`E_(L) = 1/2 LI_(max)^(2) = 1/2 L ((V)/(10))^(2)` ........(1)
and the energy stored in the capacitor
`E_(C) = 1/2 CV^(2)` .......(2)
`:. (E_L)/(E_C) = (1/2L((V)/(10))^(2))/(1/2CV^(2)) = L/100C`
`=(10^(-3))/(100 xx 2 xx 10^(-6)) = 1000/200 = 5/1`.