Saved Bookmarks
| 1. |
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. 2.5AB. 2AC. 1AD. 0A |
|
Answer» Correct Answer - A 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` |
|