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In the circuit shown in fig. if both the...

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

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

Text Solution

Verified by Experts

The correct Answer is:
B
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.
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