Question

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?
A. `10/sqrt2` amp
B. `5/sqrt2` amp
C. `5/2` amp
D. `5` amp

Answer 1

Correct Answer - C
`R=V_(0)/I_(0)=200/5=40 Omega`(For circuit `x`)
`X=V_(0)/I_(0)=40 Omega`(For circuit `y`)
if `x` & `y` are in series
`I=200/(40xxsqrt2)=5/sqrt2A`Amp. `rArr=I_(rms)=I_(0)/sqrt2=5/2` amp.