Saved Bookmarks
| 1. |
Determine the current in each branch of the network shown in fig. |
Answer» Solution : Apply KVL in loop ABDA `-10I^(1)=5(I-2I^(1))+5(I-I^(1))=0` `2I=5I."………(i)"` Apply kVL in ADCEFA loop `-5(I-I^(1))-10I^(1)+10-10I=0` `2I=5I."…..(i)"` `"from equation (1) and (2) "I=(10)/(17),I^(1)=(21)/(5)=(4)/(17)A` CURRENT in AB BRANCH `=(4)/(17)=I-I^(1)=(10)/(17)-(4)/(17)=(6)/(17)A` Current in DB branch `I-2I.=(10)/(17)-(8)/(17)=(2)/(17)A` |
|