In the circuit shown in fig. if both the bulbs `(B_1) and (B_2)` are identical

Options

A. (A) their brightness will be the same
B. (B) `(B_2)` will be brighter than `(B_1)`
C. (C) `(B_1)` will be brighter than `(B_2)`
D. (D) only `(B_2)` will glow because the capacitor has infinite impedance

Answer

B

Explanation

`(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.