It is very similar to bridge network but I cant deal with this resistor in the middle

 

Have you worked with Kirchoffs' laws?

 

Have you worked with Kirchoffs' laws?

Yes. I tried for example, to remove this middle resistor, and I found the voltage between the points that connect R5 is 3,3V. Could I just add the resistor again and appy Ohms law on it? (3.3 = 1k*I)

The biggest issue in the moment is to redraw this circuit since this middle resistor is hard to redraw for me.

 

Yes. I tried for example, to remove this middle resistor, and I found the voltage between the points that connect R5 is 3,3V. Could I just add the resistor again and appy Ohms law on it? (3.3 = 1k*I)

The biggest issue in the moment is to redraw this circuit since this middle resistor is hard to redraw for me.

Not unless you find the Thevenin (or Norton) equivalent circuit and use that. The reason is that inserting the middle resistor back in will change the voltage at the points it connects to.

What is the source? Is that 10A DC downward? I've never seen that symbology before.

Label the node at the top of the circuit Node A and label the nodes to the left and right of the middle resistor Node B and Node C, respectively. The node at the bottom of the circuit is already labeled and declared to be 0V.

In terms of the node voltages, Va, Vb, and Vc, what is an expression for the currents coming in to Node A? What about Node B? What about node C? Set each of these expressions equal to zero since KCL requires that the sum of the currents entering a node be zero.

You now have three equations and can solve for the three node voltages. With the voltages for Node B and Node C now known, can you find the current Ix?