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

Answer

D

Explanation

Current will be maximum at the condition of resonance so `i_(max)=V/R=V/10A`
Energy stored in the coil `W_(L)=1/2 Li_(max)^(2)=1/2L(E/10)^(2)`
`=1/2xx10^(-3)((E^(2))/100)=1/2xx10^(-5)E^(2)` joule
`:.` Energy stored in the capacitor
`W_(C)=1/2CE^(2)=1/2xx2xx10^(-6)E^(2)=10^(-6)E^(2)` joule
`:. (W_(C))/(W_(L))=1/5`