Mesh Current method help

Thread Starter

TripleDeuce

Joined Sep 20, 2010
26
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...
 

The Electrician

Joined Oct 9, 2007
2,815
You have:

"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"

This should be:

1ohm(i1-i2) - (i2)0.5ohm - i3(0.625ohm) = 0

How does the rest of the problem work out if you make this change?
 

Thread Starter

TripleDeuce

Joined Sep 20, 2010
26
Can you explain why it is i1-i2 and not the other way around? The direction of i2 is up while i1 is down at the 1 ohm resistor
 

The Electrician

Joined Oct 9, 2007
2,815
Can you explain why it is i1-i2 and not the other way around? The direction of i2 is up while i1 is down at the 1 ohm resistor
For consistency.

You were following a clockwise path around the i2 and i3 meshes. When you pass a resistor, you can choose the voltage drop to be positive or negative for currents that are going in the same direction as your path around the loop, but you must do the same thing all the time.

Look at the subexpression "- (i2)0.5ohm". The current i2 is flowing in the same direction as your direction of travel (clockwise). You used the convention that this is a negative voltage drop.

But, look at the subexpression "1ohm(i2-i1)". You have made the voltage drop due to i2 positive even though i2 is flowing in the same direction as your direction of travel. That is inconsistent, and gives an error.
 
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