A pure resistive circuit element `X` when connected to an a.c. supply of peak voltage `200 V` gives a peak current of `5 A` . A second current element `Y` when connected to same a.c. supply gives the same value of peak current but the current lags behind by `90^(@)`. If series combination of `X` and `Y` is connected to the same supply, what is the impedance of the circuit ?

Options

A. `40 sqrt2 ohm`
B. `40 Omega`
C. `80 Omega`
D. `2 sqrt40 ohm`

Answer

A

Explanation

`R = (E_(0))/(I_(0)) = (200)/(5) = 40 Omega`
As current lags behind tha applied voltage by `90^(@)` therefore, element `Y` must be a pure inductor.
`X_(L) = (E_(0))/(I_(0)) = (200)/(5) = 40 Omega`
`Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(40^(2) + 40^(2)) = 40 sqrt2 ohm`