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. 2.5A
B. 2A
C. 1A
D. 0A

Answer

A

Explanation

For X, current is in phase
`therefore R=(e)/(I)=40Omega`
For Y curret lags by is `90^(@)`
`therefore X_(L)=(e)/(I)=40Omega`
`therefore Z=sqrt(R^(2)+X_(L)^(2))=40sqrt(2)Omega`
`I_(rms)=(e_(rms))/(Z)=2.5A`