Question

A pure resistive circuit element X when connected to an ac supply of peak voltage 200 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?
A. `(10)/(sqrt(2))A`
B. `(5)/(sqrt(2))A`
C. `(5)/(2)A`
D. 5A

Answer 1

Correct Answer - C
As current is in phase with the applied voltage, X must be R.
`R=(V_(0))/(I_(0))=(400V)/(5A)=80Omega`
As current lags behind the voltage by `90^(@)`, Y must be an inductor.
`X_(L)=(V_(0))/(I_(0))=(400V)/(5A)=80Omega`
In the series combination of X and Y,
`Z=sqrt(R^(2)+X_(L)^(2))=sqrt(80^(2)+80^(2))=80sqrt(2)Omega`
`I_("rms")=(V_("rms"))/(Z)=(V_(0))/(sqrt(2)Z)=(400)/(sqrt(2)(80sqrt(2)))=(5)/(4)A`