Engineering Kirchoff's Tension Law, resistor circuit.

AI Thread Summary
The discussion revolves around applying Kirchhoff's Tension Law to find current intensities (I1, I2, I3, I6) in a resistor circuit with specified resistance values and voltage sources. The original poster, Wassim H., encountered issues with negative and excessively high current values while solving the circuit equations. Other participants confirmed that the loop equations appeared correct and suggested using alternative methods, such as mesh analysis and node equations, to verify results. Ultimately, Wassim resolved the discrepancies by correcting calculation errors and expressed interest in further exploring the node method for future problems. The conversation highlights the importance of careful circuit analysis and verification in electrical engineering.
Whoracle
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Homework Statement



Find I1, I2, I3, I6 (intensities at R1, R2, R3 and R6) using Kirchoff's Tension Law.

R1 = 1/3 Ohm, R2 = 1/4, R3 = 1/3, R4 = 1/2, R5 = 1, R6 = 1/4
E3 = 32V, E1 = 2V, E2 = 4V.

http://dl.dropbox.com/u/9301772/Capture.PNG


The Attempt at a Solution



The schematic has been vastly simplified, and I haven doubts about my simplifications. Anyway, I keep getting the following system, which yields me very weird negative answers.

If you notice, I1 = I2+I6 (see schematic)

Left loop: 2 = R1 I2 + R1 I6 + R2 I2 + 4
Right loop: 4 = (R3+R4) I3 + R5(I3+I6)
Upper loop: 32 = -(R3+R4) I3 - R2 I2 + R6 I6

Which yields negative and very high values for I2 and I3.

Any help would be GREATLY appreciated,
Wassim H.
 
Last edited by a moderator:
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Whoracle said:

Homework Statement



Find I1, I2, I3, I6 (intensities at R1, R2, R3 and R6) using Kirchoff's Tension Law.

R1 = 1/3 Ohm, R2 = 1/4, R3 = 1/3, R4 = 1/2, R5 = 1, R6 = 1/4
E3 = 32V, E1 = 2V, E2 = 4V.

http://dl.dropbox.com/u/9301772/Capture.PNG


The Attempt at a Solution



The schematic has been vastly simplified, and I haven doubts about my simplifications. Anyway, I keep getting the following system, which yields me very weird negative answers.

If you notice, I1 = I2+I6 (see schematic)

Left loop: 2 = R1 I2 + R1 I6 + R2 I2 + 4
Right loop: 4 = (R3+R4) I3 + R5(I3+I6)
Upper loop: 32 = -(R3+R4) I3 - R2 I2 + R6 I6

Which yields negative and very high values for I2 and I3.

Any help would be GREATLY appreciated,
Wassim H.

Hi Whoracle, welcome to Physics Forums.

I've taken a quick look at your loop equations and they appear to be okay.

What do you consider to be "very high values"?
 
Last edited by a moderator:
gneill said:
Hi Whoracle, welcome to Physics Forums.

I've taken a quick look at your loop equations and they appear to be okay.

What do you consider to be "very high values"?

I2 = -25.666
I3 = -19
I6 = 39

Pretty sure they're wrong. I tried solving with a software called Solve Elec (just gives me the values, and it gives me more reasonable ones (all positive and under 10A).
 
Whoracle said:
I2 = -25.666
I3 = -19
I6 = 39

Pretty sure they're wrong. I tried solving with a software called Solve Elec (just gives me the values, and it gives me more reasonable ones (all positive and under 10A).

The values match my own calculations for the circuit as drawn and with the component values indicated. I used mesh analysis (a form of loop analysis), and didn't depend upon your loop equations. So, two methods at least give the same result...

I'm not familiar with Solve Elec.
 
I'm starting to have doubts about my simplifications. I'll give you the original circuit in a minute.
(Iy = 6A, Iz = 12A)

I used the Norton equivalence to replace current sources with tension sources.

http://dl.dropbox.com/u/9301772/original.PNG
 
Last edited by a moderator:
Whoracle said:
I used the Norton equivalence to replace current sources with tension sources.
Ah. Once you replace a given set of components with a circuit equivalent, even if the resistor values are numerically the same they are no longer the same components. There's no guarantee that the current carried by the R in a Norton equivalent will be the same as the current through the same valued R in its Thevenin equivalent.

In such a case a way to proceed is to use your simplified circuit to find parameters that are identical in both circuits, such as node voltages or currents through components that are left unchanged. Then use those values to find the missing bits in the original circuit.

EDIT: Also, I see that your R3 should have ended up in series with the 4V supply that replaces iz, and not in series with R4.
 
Last edited:
gneill said:
EDIT: Also, I see that your R3 should have ended up in series with the 4V supply that replaces iz, and not in series with R4.

I think we're on to something here ! I'll try it and let you know. How did I not see that D: !
 
I2 = -30.08
I3 = 2.97
I6 = 43.68

... I don't get it.
 
Whoracle said:
I2 = -30.08
I3 = 2.97
I6 = 43.68

... I don't get it.

Without seeing what happened there's not much I can comment on...

Did you try a node equation approach? It seems that all of the resistances in the circuit have been chosen so that they would make nice round-figure conductances.
 
  • #10
Sorry for the late reply, but I actually found the right values :) I had just miscalculated.

The next question is finding it using the node method, as you said. Something I'm not very good at it. I'll keep you updated !
 

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