Calculate the current that flows in the `1Omega` resistor in the following circuit.

Explanation

We can denote the current that flows from 9V battery as `I_(1)` and it splits into `I_(2) and I_(1)-I_(2)` in the junction according to Kirchoff.s current rule (KCR). It is shown below.
Now consider the loop EFCBE and apply KVR, we get
`1I_(2)+3I_(2)+2I_(1)=9`
`5I_(1)+I_(2)=9" "...(1)`
Applying KVR to the loop EADFE, we get
`3(I_(1)-I_(2))-1I_(2)=6`
`3I_(1)-4I_(2)=6" "...(2)`
Solving equation (1) and (2), we get
`I_(1)=1.83A and I_(2)=-0.13A`
It implies that the current in the 1 ohm resistor flows from F to E.