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An ideal choke coil takes a current fo 8 ampere when connected to an `AC` supply of `100` volt and `50 Hz`. A pure resistor under the same conditions takes a current of 10 ampere. If the two are connected to an `AC` supply of `150` volts and `40 Hz` then the current in a series combination of the above resistor and inductor is |
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Answer» For pure inductor, `X_(L) = (E_(0))/(I_(v)) = (100)/(8) = (25)/(2) Omega` `omega L = (25)/(2), L = (25)/(2 omega) = (25)/(2 xx 2 pi xx 50) = (1)/(8pi) H` `R = (V)/(1) = (100)/(10) = 10 Omega` For the combination, the supply is `150 v, 40 Hz`. `:. X_(L) = omega L = 2 pi xx 40 xx (1)/(8 pi) = 10 Omega` `Z = sqrt(X_(L)^(2) + R^(2)) = sqrt(10^(2) + 10^(2)) = 10 sqrt(2)` ohm `I_(v) = (E_(v))/(Z) = (150)/(10 sqrt(2)) A = (15)/(sqrt(2)) A` |
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