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?

Options

A. `(10)/(sqrt2) A`
B. `(5)/(sqrt2)A`
C. `(5/2)A`
D. `5A`

Answer

C

Explanation

As current is in phase with the applied voltage, X must be R.
` R = (V_0)/(I_0) = (200V)/(5A) = 40 Omega `
As current lags behind the voltage by `90^@` , Y must be an inductor .
`X_L = (V_0)/(I_0) = (200V)/(5A) = 40 Omega`
In the series combination of X and Y
`Z = sqrt(R^2 + X_L^2) = sqrt(40^2 + 40^2) = 40sqrt2 Omega `
`I_(rms) = (V_(rms))/(Z) = (V_0)/(sqrt2 Z) = (200)/(sqrt2(40sqrt2) ) = 5/2 A `