Determination of vTh = voc by mesh current method: Consider the
circuit shown below. Our aim in this problem is to determine the open circuit voltage voc at the terminals A and B by utilizing the mesh current method.
I drew the mesh currents in the appropriate windows but cannot get the right answer when I start writing my KVL's.
From the diagram, I said 1.2Vx = i3-i2.
I also wrote a super mesh equation that combined the windows of i2 and i3. Starting from the node directly under the current source
1ohm(i2-i1) - (i2)0.5ohm - i3(0.625ohm) = 0
working out the algebra yields
0.5i2 - 0.625i3=i1
Then I calculated the loop of i1 starting from ground
20 -(0.5i2 - 0.825i3)1 - 1(i2-(0.5i2 - 0.625i3) = 0
This gave i2 = 20A
Looking at the diagram, you can see that Vx = i2*0.5ohms. I substituted this into the equation 1.2Vx = i3-i2.
Rewriting gives 1.2(i2*0.5ohms)+i2 = i3 or 1.6i2 = i3
so i3 = 1.6(20) = 32A
I think I wrote the correct equations but the math just doesn't add up. Voc is 5V and in order for it to be 5V, i3 has to be 8A...
circuit shown below. Our aim in this problem is to determine the open circuit voltage voc at the terminals A and B by utilizing the mesh current method.
I drew the mesh currents in the appropriate windows but cannot get the right answer when I start writing my KVL's.
From the diagram, I said 1.2Vx = i3-i2.
I also wrote a super mesh equation that combined the windows of i2 and i3. Starting from the node directly under the current source
1ohm(i2-i1) - (i2)0.5ohm - i3(0.625ohm) = 0
working out the algebra yields
0.5i2 - 0.625i3=i1
Then I calculated the loop of i1 starting from ground
20 -(0.5i2 - 0.825i3)1 - 1(i2-(0.5i2 - 0.625i3) = 0
This gave i2 = 20A
Looking at the diagram, you can see that Vx = i2*0.5ohms. I substituted this into the equation 1.2Vx = i3-i2.
Rewriting gives 1.2(i2*0.5ohms)+i2 = i3 or 1.6i2 = i3
so i3 = 1.6(20) = 32A
I think I wrote the correct equations but the math just doesn't add up. Voc is 5V and in order for it to be 5V, i3 has to be 8A...