Correct Answer - C
At resonance, `i = e/R = 0.1/2 = 5 xx 10^(-2) A`
At resonance, `omega = 1/(sqrt(LC))`
`:. V_(L) = iX_(L) = i omega L = (iL)/(sqrt(LC))`
`=(5 xx 10^(-2)xx10^(-1))/(sqrt(10^(-1)xx 40 xx 10^(-6))) = (5 xx 10^(-3))/(2 xx 10^(-3))` ltbr gt`:. V_(L) = 2.5V`.