Determine the current in each branch of the network shown in fig.

Explanation

Apply KVL in loop ABDA
`-101^(1)+5(I-2I^(1))+5(I-I^(1))=0`
`2I=5I^(')` …(1)
Apply KVL in ADCEFA loop
`-4(I-I^(1))-101^(1)+10-101=0`
`5I^(1)+15I=10`
From equiation (1) and (2)
`I=(10)/(17),I^(1)=(21)/(5)=(4)/(17)A`
Current in DB branch
`I-2I^(')=(10)/(17)-(8)/(17)=(2)/(17)A`