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In the circuit shown in fig. if both the bulbs `(B_1) and (B_2)` are identical A. their brightness will be the sameB. `B_(2)` will be brighter than `B_(1)`C. as frequency and that of `B_(2)` will becreaseD. Only `B_(2)` will glow because the capacitor has infinite impedance |
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Answer» Correct Answer - 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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