Kirchoff's Tension Law, resistor circuit.

In summary, the conversation revolved around finding intensities at specific points in a circuit using Kirchoff's Tension Law. The attempt at a solution involved simplifying the circuit and using loop equations, which initially yielded negative and high values. It was suggested to use the original circuit and find parameters that are identical in both circuits to solve for the missing values. Eventually, the correct values were found and the next step was to use the node method to solve the problem.
  • #1
Whoracle
6
0

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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  • #2
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:
  • #3
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).
 
  • #4
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.
 
  • #5
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:
  • #6
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:
  • #7
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: !
 
  • #8
I2 = -30.08
I3 = 2.97
I6 = 43.68

... I don't get it.
 
  • #9
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 !
 

What is Kirchoff's Tension Law?

Kirchoff's Tension Law, also known as Kirchoff's Voltage Law, states that the algebraic sum of the voltage drops in a closed loop of a circuit is equal to the voltage supplied to the loop.

How is Kirchoff's Tension Law applied in a resistor circuit?

In a resistor circuit, Kirchoff's Tension Law is applied by using it to calculate the voltage drop across each individual resistor. The sum of these voltage drops should equal the total voltage supplied to the circuit.

What happens if Kirchoff's Tension Law is violated in a circuit?

If Kirchoff's Tension Law is violated in a circuit, it means that the voltage drops do not add up to the total voltage supplied. This could be due to a malfunctioning component or an error in the circuit design.

Can Kirchoff's Tension Law be applied to circuits with non-ideal components?

Yes, Kirchoff's Tension Law can be applied to circuits with non-ideal components. However, it may not be as accurate as it is with ideal components, as non-ideal components have additional factors that may affect the voltage drop.

How does Kirchoff's Tension Law relate to Ohm's Law?

Kirchoff's Tension Law is closely related to Ohm's Law, as it can be used to calculate the voltage drop across each resistor in a circuit, while Ohm's Law can be used to calculate the current in a circuit. Together, these laws can be used to analyze and design complex circuits.

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