In the circuit shown in fig. 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 decreases
D. Only `B_(2)` will glow because the capacitor has infinite impedence

Answer

B

Explanation

Let `I_(1)` and `I_(2)` be currents through `B_(1)` and `B_(2)` then
`I_(1)xxsqrt(R^(2)xxX_(L)^(2))=220`
Let `I_(1)` and `I_(2)` be currents through `B_(1)` and `B_(2)` then
`I_(1)xxsqrt(R^(2)xxX_(L)^(2))=220`
`(I_(1))/(I_(2))=(sqrt(R^(2)xxX_(C)^(2)))/(sqrt(R^(2)xxX_(L)^(2)))=(sqrt(R^(2)+(1/(c omega))^(2)))/(sqrt(R^(2)+L^(2)omega^(2)))`
`=(sqrt(R^(2_)+(1/(500xx10^(-6)xx2pixx50))^(2)))/(sqrt(R^(2)+(10xx10^(-3)xx2pixx50)^(2)))`
`=sqrt(R^(2)+40)//sqrt(R^(2)+9.87), I_(2)gtI_(1)`
Bulb `B_(2)` will be brighter. As frequency increases `X_(C)` decreases `X_(L)` increases. `I_(2)` becomes less `I_(1)` increases. So, brightness of `B_(1)` will increases and that of `B_(2)` decreases.