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In a series circuit C=2muF,L=1mH and R=1...

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

A

`1 : 1`

B

`1 : 2`

C

`2:1`

D

`1 : 5`

Text Solution

Verified by Experts

The correct Answer is:
D
At resonance
Current is maximum `(i_(max)) = (V)/(R)`
Energy stored in inductor `(U_(B)) = (1)/(2) Li^(2)`
`rArr U_(B) = (1)/(2) xx 1 xx 10^(-3) xx ((V)/(10))^(2)= 5 xx 10 ^(-6) V^(2)`
Energy stored in capacitor
`U_(E) = (1)/(2) CV^(2)`
`rArr U_(E) = (1)/(2) xx 2 xx 10^(-6) xx V^(2) = 1 xx 10^(-6) V^(2)`
`(U_(E))/(U_(B)) = (1)/(5)`
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