I first short circuited the 6V. I simplified the circuit by obtaining the equivalent resistance (2 is parallel to 4) and got 8/6 ohms. With this value, I used current division: V1 = [ 6A (2/(2+8/6)) ] 2 = 18/5 Next I created an open circuit with the 6A. I then used voltage division: V2 = [2/(4+2)]6=3/2 So V0 = V1+V2 = 51/10 ? What did I do wrong here?
I am not saying that I am right, but I think that is how it should be solved. The minus sign comes from the direction of current. Check out the first circuit. The voltage source is shorted. What is the direction of the current? It is from right to left. Now look at the second circuit. The current source is open. What is the direction of current? From left to right. What you have to do is pick which direction is positive and which is negative. So I chose one. I hope I solved it right. Let us wait and see what others say.
Look at the resistor where your making the voltage measurement. It specifies where the negative and positive leads are attached. Now look at the battery ... the negative lead is connected to battery positive. the positive meter lead is closer to the negative terminal. The answer is 6. The 2 ohm resistor has 8 volts across it when the battery source is shorted out. V1 = 6 * (1/((1/2)+(1/4))) The 2 ohm resistor has neg 2 volts across it when the current source is open. V2 = (2*-6)/(2+4) V = V1 + V2 = 8 + (-2) = 6