A pure resistive circuit element `'x'` when connected to an `A.C` supply of peak voltage `100 V` gives a peak current of `4 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 to of `'x'` and `'y'` is connected to the same supply. `R.M.S` value of current is

Options

A. `(5)/(sqrt(2)) A`
B. `2 A`
C. `1//2 A`
D. `(sqrt(2))/(5) A`

Answer

2

Explanation

`X_(L) = (epsilon_(0))/(I_(0)) = 25 Omega , R = (epsilon_(0))/(I_(0)) = 25 Omega , Z = sqrt(R^(2) + X_(C )^(2))`,
`I_(0)^(1) = epsilon_(0)//Z = 4 // sqrt(2) A, I_("r.m.s") = I_(0)^(1) // sqrt(2) = (4//sqrt(2))/(sqrt(2)) = 2A`