A pure resistive circuit element `X` when connected to an sinusoidal `AC` supply peak voltage `200 V` gives a peak current of `5A` 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 currrent but the current lags behind by `90^(0)`. If the series combination of `X` and `Y` is connected to the same supply. the rms value of current is

Options

A. `(10)/(sqrt(2)) A`
B. `(5)/(sqrt(2)) A`
C. `(5)/(2) A`
D. `5 A`

Answer

C

Explanation

Initially `5 = (E_(0))/(R )` (for `X`) and `E_(0) = 200 V`
`5 = (E_(0))/(X_(L))` (for `Y`)
In the series combination of `X` and `Y`
`i_(0) = (E_(0))/(sqrt(R^(2) + X_(L)^(2)))` , `:. i_("rms") = (i_(0))/(sqrt(2))`