A pure resistive circuit element X when connected to an ac supply of peak voltage 400 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same ac supply also gives the same value of peak current but the current lags behind by `90^(@)`. If the series combination of X and Y is connected to the same suply, what will be the rms value of current?A. `10/sqrt 2` AB. `5/sqrt 2` AC. 5/2 AD. 5A

A pure resistive circuit element X when connected to an ac supply of p

1.	A pure resistive circuit element X when connected to an ac supply of peak voltage 400 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same ac supply also gives the same value of peak current but the current lags behind by `90^(@)`. If the series combination of X and Y is connected to the same suply, what will be the rms value of current?A. `10/sqrt 2` AB. `5/sqrt 2` AC. 5/2 AD. 5A
Answer» Correct Answer - C R = `underset(L)(X)` = 200 / 5 = `40 Omega` `therefore` Z = `` = `40sqrt 2 Omega` `underset(rms)(I)` = `underset(rms)(V)`/Z = `200/sqrt 2` / 40`sqrt 2` = 5/2 A

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