I would convert the current source and 2 ohm resistor to the Thevenin equivalent of 1*30 volts in series with 2 ohms. Then you can write 40=I(10+4+2+30) and I=.87

If the current in the 2 ohm resistor is I’, I’ = I +15 where 15 is current from the current source.

The voltage across the 10 and 4 ohm resistors is 14I. The voltage across the 2 ohm resistor is 2 X (I + 15) or 2I + 30. So the voltage around the loop with the voltage source is 40 – 14I -2I – 30. So 16I = 10 and I = 10/16 or 0.625.

current = 40V / ( 10 + 4 + 2 ) = 2.5A

yees

I've tried. I'm pretty new to this. Please explain and solve this problem. Thank you.