Joe,

If you consider that the 3.3V ideal voltage source is the only voltage that can be seen by the 2200 ohm resistor, how can the current in the right loop be anything but 1.5 milliamps?

hgmjr

As i mentioned above. There are two different problems in play here.

This is creating much confusion.

 

Steve & Joe & Ratch,

Oh, Cr$$$$p. I just notice that the problem I have been solving was the first one. I did not realize that the problem changed in mid thread.

Sorry guys.

hgmjr

 

Steve & Joe,

Where am I missing the boat?

hgmjr

Look carefully at the two schematics that the OP presented. They are very similar , but the 12V and 3.3V power supplies are swapped.

 

Steve1992,

You are killing me here.

hgmjr

 

No, we don't need simulators. When hgmjr said he used multisim, I used superposition and put in into tina for S&Gs. When the discrepancy came up, it was nice to find the error (casued by the two different circuits) on why my answers were so different.

 

I can see where I went wrong.

The attached schematics are small enough to quality as microfilm.

hgmjr

 

steveb,

The current I3 is the current in the 3.3V power supply. It is supposed to be found be adding the loop currents, since both loop currents flow through this common path.

Yes, but loop currents existing in a component common to two loops oppose each other. Assuming a clockwise direction of loop 1 and loop2 then the current through the 3.3 volt source is I1-I2 assuming the current direction is south.

The thing is that the math is still wrong. The mesh currents are opposing and the net current in the 3.3 V supply is -16 mA. The negative sign means that it is dissipating power and not supplying power. All power comes from the 12V supply in this example.

Yes, you are right. 17.4 ma in loop 1 and 1.5 ma in loop 2 subtract to give 16 ma of current in a south direction. The 3.3 voltage source is absorbing power from the 12 volt source.

I tried to point out the importance of getting the mesh current direction correct in post number 2. The OP original question was that he tried the left mesh current in both counter-clockwise and clockwise directions but got a positive current in both cases, which is wrong.

No, he was correct on that one. He assumed the current was clockwise in the left loop. Using KVL in either direction he got the same positive current value, which always happens with KVL. It confirmed that the assumption of clockwise current direction was correct. Read his post again.

This problem is quite trivial since the voltage on both resistors is seen by inspection. So the answer is obvious, but the purpose of the problem should be to get the correct answer by properly applying mesh current analysis.

And we do if we follow the procedure correctly. I certainly am ashamed of the simple stupid mistakes I made on this problem.

Ratch

 

OK, now I see that there are two different schemats and problems. No wonder our preceptions are skewed.

Ratch

 

No, he was correct on that one. He assumed the current was clockwise in the left loop. Using KVL in either direction he got the same positive current value, which always happens with KVL. It confirmed that the assumption of clockwise current direction was correct. Read his post again.

Yes, I saw your response after mine. We each interpreted his words differently. After rereading his original statement, I still think my interpretation is correct. His math is consistent with both interpretations, but his words seem to be in line with my interpretation. However, only the OP can verify this.

in particular he says

"but I get a positive current from either direction for the loops:"

this implies he is trying both directions for the loop

 

Sorry guys for the confusion. I feel like Ive been going around in circles - or is that loops.