Need help with Mesh example
- Thread starter ENG99
- Start date
The Electrician
- Joined Oct 9, 2007
- 2,970
Look at the very first equation under the schematic, where you have:
-6 + 5*i1 - 1.5(i2-i1) = 0
You have incompatible units. The -6 is in volts; 5*i1 is 5 ohms*i1 amps which gives volts; 1.5(i2-i1) has units of amps. That won't work, will it?
You have a supermesh in this circuit; have you studied supermeshes?
Also, it's not true that ix = (i2-i1); rather 1.5*ix = (i2-i1) and ix = i2.
-6 + 5*i1 - 1.5(i2-i1) = 0
You have incompatible units. The -6 is in volts; 5*i1 is 5 ohms*i1 amps which gives volts; 1.5(i2-i1) has units of amps. That won't work, will it?
You have a supermesh in this circuit; have you studied supermeshes?
Also, it's not true that ix = (i2-i1); rather 1.5*ix = (i2-i1) and ix = i2.
Thanks for showing your work. In the future, try reducing the size and resolution of images so that they don't take so long to load (many of us have slow connections and bandwidth limitations).
As The Electrician already pointed out, your first problem is in your first equation. As people around here well know, I am big on units and this is a big part of why. Most mistakes you make will screw up the units and, if you religiously track your units throughout your work, you will put yourself in a position to catch them without even trying too hard right at the point you make them. But if you just tack units onto a number at the end because that's what you think the answer should have, then you rob yourself of perhaps the single most powerful error detection tool available to the engineer. Just consider all of the effort you put into this and it was all guaranteed to be wasted right from the start.
Your first equation, as you have it, should have been written as:
-6V + 5Ω(i1) - 1.5(i2 - i1) = 0
Now it is obvious that the units don't work out and that this equation must therefore be wrong. Once this is recognized, then you can tend to figuring out what you did wrong at this point, fix it, and move on. The key to recognizing what you did wrong is also in the units. You know that your mesh equations are summing up voltages and that your last term is not a voltage, so the first place to focus is that last term. Hopefully you would quickly recognize that you have no way, at this point, to write a term that directly expresses the voltage across a current source and so you need to consider using a super mesh or rearranging the circuit to put the current source on an exterior branch.
So with that, and the other point that The Electrician made, in mind, try again and if you are still having problems post your revised work.
As The Electrician already pointed out, your first problem is in your first equation. As people around here well know, I am big on units and this is a big part of why. Most mistakes you make will screw up the units and, if you religiously track your units throughout your work, you will put yourself in a position to catch them without even trying too hard right at the point you make them. But if you just tack units onto a number at the end because that's what you think the answer should have, then you rob yourself of perhaps the single most powerful error detection tool available to the engineer. Just consider all of the effort you put into this and it was all guaranteed to be wasted right from the start.
Your first equation, as you have it, should have been written as:
-6V + 5Ω(i1) - 1.5(i2 - i1) = 0
Now it is obvious that the units don't work out and that this equation must therefore be wrong. Once this is recognized, then you can tend to figuring out what you did wrong at this point, fix it, and move on. The key to recognizing what you did wrong is also in the units. You know that your mesh equations are summing up voltages and that your last term is not a voltage, so the first place to focus is that last term. Hopefully you would quickly recognize that you have no way, at this point, to write a term that directly expresses the voltage across a current source and so you need to consider using a super mesh or rearranging the circuit to put the current source on an exterior branch.
So with that, and the other point that The Electrician made, in mind, try again and if you are still having problems post your revised work.
