Question

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` A
B. `5/sqrt 2` A
C. 5/2 A
D. 5A

Answer 1

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