In the circuit shown in the figure, if both the bulbs `B_(1) and B_(2)` are identical

Options

A. their brightness will be the same
B. `B_(2)` will be brighter than `B_(1)`
C. as frequency and that of `B_(2)` will becrease
D. Only `B_(2)` will glow because the capacitor has infinite impedance

Answer

B

Explanation

Let `I_(1) and i_(2)` be current through `(B_1) and (B_2)` then
`I_(1)xx sqrt(R^(2)xxX_(L)^(2))=220`
`(I_2)/(I_1)=(sqrt(R^(2)xx X_(C)^(2)))/(sqrt(R^(2)xx X_(L)^(2)))=(sqrt(R^(2)+((1)/(c omega))^(2)))/(sqrt(R^(2)+L^(2)omega^(2)))`
`(sqrt(R^(2)+((1)/(500xx10^(-6)xx2 pi xx 50))^(2)))/(sqrt(R^(2)+(10xx10^(-3)xx2 pi xx50)^(2)))`
`=sqrt(R^(2)+40)//sqrt(R^(2)+9.87),I_(2)gtI_(1)`
Bulb `(B_2)` will be brighter. As frequency and increases, `X_(C)` decrease `(X_L)` increases. `(I_2)` becomes less and `(I_1)` increases.
`:.` brighter of `(B_1)` will increase and that of `(B_2)` decreases.