Norton equivalent circuit

I tried to solve the problem below but I don’t know if it’s right.

Im finding I2 or the current when SW1 is open. I put a test voltage 1V between the nodes of 8 ohm and 6 ohm resistors so its in parellel with 6-ohm resistor when all independent sources are suppressed. Then I got I2=1/12 A but I'm not sure if it should be negative because of the direction of the current. So I don't know if the norton current is 2.08A or 1.83A. I thought of I2 being negative because I tried to follow the same current flow as I1.

The Rn I got is 12 ohms.

As you've found you can see the resistance is 12Ω = (12+6||8)Ω

Since you know the short-circuit current, you're done, it's 2A || 12Ω with the current source in the direction to get the correct I1.

• Shouldn't I also solve for the current with the switch open? then add that current to I1 to get the norton current? Commented Mar 15 at 13:25
• The Norton current is the short-circuit current. Commented Mar 15 at 13:43

You cannot use the test source method since there are independent sources in the circuit.

The Norton Equivalent circuit is as shown below, where RL is the load resistance:

simulate this circuit – Schematic created using CircuitLab

The norton current IN would be the short circuit current, which is 2A. RN, the norton resistance, can be calculated by replacing the voltage sources as short, and the current sources as open circuit, then calculating the resistance as seen by the load.

The other answers are correct. This is a different way of getting there.