Saved Bookmarks
| 1. |
Using Kichhoff's law in the given circuit determine the current 'I' in the are EF. |
|
Answer» SOLUTION :Applying kirchoff's FIRST rule at E `rArr 0.5 =I_(1)=I` Where 'I' flows through 'R' Now Kirchoff's 2dn rule in closed LOOP FEABF `-2I_(2)=0.5xx2=-4+3` `-2I_(2)+1=-1` `cancel-2I_(2)=cancel-2` The CURRENT in arm `EF=12=+1A` |
|