Correct Answer - B
`(X_C)=(1)/(2 pi fC) =(10^6)/(2 pi 50 xx 500) =(20)/(11) Omega`
`(X_L) = 2 pi f L =2 pi xx 50 xx 10 xx10 ^(-3) = pi Omega`
Since `(X_L) lt (X_C)`, so inductive brach has less impeedance, and so more current. Hence `(B_2)` will be brighter.