Saved Bookmarks
| 1. |
A 100 Omega resistance and a capacitor of 100 Omega reactance are connected in series across a 220 V source.When the capacitor is 50% charged, the peak value of the displacement current is: (a) 4.4A (b) 11sqrt(2) A (c) 2.2A (d) 11 A |
|
Answer» `4.4 A` `= 220xx sqrt(2)V` `|Z|=sqrt(R^(2)+X_(C )^(2))=sqrt((100)^(2)+(100)^(2))` `|Z|=sqrt(2+(100)^(2))` `|Z|=100 sqrt(2)Omega` `THEREFORE`Maximum value of displacement current `I_(D)=(E_(0))/(|Z|)` `therefore I_(d)=(220sqrt(2))/(100sqrt(2)) "" therefore I_(d)=2.2 A` |
|