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According to my textbook, abca with 2 ohm resistor is independent. A second loop with 3 ohm resistor and the current source is independent. The third loop, with 2 ohm resistor in parallel with 3 ohm resistor is also independent.

Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop.

Let's take the first loop, abca with 2 ohm resistor. Say the unique branch is 2 ohm resistor. now bc with parallel resistors 3 and 2 ohm is also said to be independant. but the latter contains the 2 ohm resistor, meaning 2 ohm resistor is not unique to a loop after all. Same goes for the current source and 3 ohm resistor loop, 3 ohm is also not unique.

So according to the definition, why are these three loops independant?

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  • \$\begingroup\$ I think, if your book is in English (or maybe also if it's not) a verbatim quote of the text might be helpful in determining what they mean. In principle your personal analysis is right in that the schematic has many interactions, but maybe the book means something else in context to do with the analysis of voltages and currents. \$\endgroup\$ – Asmyldof Jun 7 '15 at 16:15
  • \$\begingroup\$ Probably when they say "independent", they mean loops that when the KVL equations are written, produces a set of linearly independent equations. It should be the same as what Wikpedia calls essential meshes. But Wiki's definition is "loops that don't contain any other loops", which would make your three proposed loops "essential". \$\endgroup\$ – The Photon Jun 7 '15 at 17:44
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I teach circuits and use the same textbook that the figure is from. My students have asked me about this and it took a while to come up with a sound definition. The statement from the text "an independent loop is a loop that contains a branch that is not part of any other independent loop" is ambiguous and as was previously stated is necessary but not sufficient. If you go strictly by this definition you are correct that you can form the left loop IL (abca) which leaves the 3 ohm resistor and the 2A source. You can then form the right loop IR which starts at b goes through the 2A source to c and then back to b through the 3 ohm resistor. At this point there are no more unique branches but you have not found all the independent loops.

What needs to be added to this definition is "Can any nodes be expanded creating a unique branch without breaking any independent loops? If a unique branch is created without breaking an independent loop then a new loop must be formed which contains this unique branch."

If you expand node a into two points (a,d) then you will break the left loop, IL. Therefore the unique branch created between points a and d must become part of loop IL so that IL can be closed. No independent loop is added. If you expand node b into two points (b,e) then neither the left loop, IL, nor the right loop, IR, are broken. This creates a new branch between points b and e that must be part of a new independent loop, IM. If you then expand node c into (c,f) you create a new branch but break loop IM so the new branch must become part of IM to close that loop. If you continue to expand any nodes no new unique branches are formed and you are left with the 3 independent loops, IL, IM, and IR.

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Now the definition of an independent loop is a loop that contains a branch that is not part of any other independent loop.

If a loop has a branch that's not part of any other loop, that does guarantee independence, but I don't think it's required. (Mathematically, it's sufficient but not necessary.)

In mesh analysis, you're trying to solve a system of equations. For that, you need one equation per variable. But the equations must be linearly independent -- if you can make one equation by adding, subtracting, and/or multiplying the other equations, it doesn't count. For example:

$$x + y = 5$$ $$2x + 2y = 10$$

The second equation can be produced by doubling every value in the first equation. This doesn't give you any new information, so you can't solve for x and y. But in this example:

$$x + y = 5$$ $$x + 2y = 7$$

you can't get the second equation by manipulating the first. So you can find the solution: x = 3 and y = 2.

Back to circuits. Your system has three variables -- the mesh currents \$I_L\$ (on the left), \$I_M\$ (in the middle), and \$I_R\$ (on the right). Here are the equations, assuming the mesh currents flow clockwise:

$$10\mathrm V - I_L \cdot 5 \Omega - (I_L - I_M) \cdot 2 \Omega = 0$$ $$-(I_M - I_L) \cdot 2 \Omega - (I_M - I_R) \cdot 3 \Omega = 0$$ $$I_R = -2\mathrm A$$

Grouping the variables gives:

$$10 \mathrm V - I_L \cdot 7 \Omega + I_M \cdot 2 \Omega = 0$$ $$I_L \cdot 2 \Omega -I_M \cdot 5 \Omega + I_R \cdot 3 \Omega= 0$$ $$I_R = -2 \mathrm A$$

There's no way we can make one of these equations out of the other. The first has a constant term, the second doesn't, and the third just gives us the value of one variable. If we substitute \$-2\mathrm A\$ for \$I_R\$ and try to make the signs match, it's even more obvious:

$$I_L \cdot 7 \Omega - I_M \cdot 2 \Omega - 10 \mathrm V = 0$$ $$I_L \cdot 2 \Omega - I_M \cdot 5 \Omega - 6 \mathrm V = 0$$

The ratios of the coefficients and constants are totally different. These equations are linearly independent.

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An independent loop contains at least one branch that doesn't belong to another loop. So using your two loops, ABC and BC as examples: ABC contains branch 10V/5 Ohm that isn't in BC. BC contains the 3 Ohm which isn't in ABC, therefore they are independent. Additionally ABC through the 2A/5ohm/10V contains the 2A source which isn't in either of the first two loops, so that too is independent. This not the only possible combination of three independent loops necessary to solve the problem.

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