Use thevenin theorem to find the current through 2 ohm resistor also find thevenin voltage and thevenin resistence.
My approach:First I open circuited the resistance through 2 ohm resistor then used voltage division to find the thevenin voltage but got the wrong answer.
My calculations:
Vth = 9V * ((6+4)/(4+4+6))= 6.43V.
But the Thevenin voltage in answer is 2.571 volts.
Best Answer
When you remove the load, you end up with:
simulate this circuit – Schematic created using CircuitLab
It follows that \$I_y=0\$ since it is an open circuit.
$$I_x=\dfrac{9\text{V}}{4\Omega+4\Omega+6\Omega}=0.643\text{A}$$
The potential difference between the terminals A and B is the \$V_{\text{TH}}\$. This is just the voltage across the farthest to the right \$4\Omega\$ resistor.
Then
$$V_{oc}=V_{\text{TH}}=(0.643\text{A})(4\Omega)=2.57\text{V} $$
You need to find \$R_{\text{TH}}\$. Short circuit the 9V voltage source and find the equivalent resistance at the A and B terminals. That would be:
$$ R_{\text{TH}}=((4\Omega+6\Omega)||4\Omega)+5\Omega=7.86\Omega$$
With that, you can find \$I_2\$.