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I've been practising loop analysis questions, but don't really understand why the method is valid.

Say you have a circuit that only consists of resistors and some power source. Assume the circuit is valid (contains no contradictions)

How do we know that the currents in each node can be expressed as a sum of loop currents? Just because current exists in the circuit, doesn't necessarily imply that the current can be expressed as loop currents.

So could someone give me a proof that if a current exists through the circuit, its necessarily expressible as a sum of "loop currents"?

Example

For instance let C1, C2 be actual currents flowing through some nodes. They aren't abstractions - these are measurable currents that actually exist.

Now essentially loop analysis expresses these real, measurable currents as the sum of abstract loop currents. You represent the current in nodes that are on the "sides" of two loops as sums of the loop currents in those loops. Without loss of generality, for some loop currents L1, L2 say you express the physical currents as:

C1 = L1 + L2

C2 = L1 - L2

What essentially confuses is that loop analysis assumes that some values L1 and L2 exist that meet the above criteria. In other words, it assumes that the above system of equations has a unique solution.

So why is it okay to assume that the actual currents, the physical currents, flowing through the nodes can always be expressed as loop currents?

For an even more concrete example, consider this circuit:

enter image description here

Note how the real, physical currents are being expressed as sums of loop currents. I don't understand why this step is always valid. How do we know there exists some Ia, Ib, Ic and Id that fulfill the above system of equations?

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    \$\begingroup\$ Kirchoff's Current Law must always be satisfied. \$\endgroup\$ – Matt Young Mar 29 '14 at 19:52
  • \$\begingroup\$ Are you asking for a proof of the validity of mesh current analysis? \$\endgroup\$ – Alfred Centauri Mar 29 '14 at 20:10
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    \$\begingroup\$ Doesn't it all pretty much break down to conservation of charge? \$\endgroup\$ – dext0rb Apr 3 '14 at 20:56
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    \$\begingroup\$ I don't have anything else to add that hasn't been said. At the end of the day, this is stuff you will no longer care about in a few more years. It is purely academic to weed out the non-hackers. \$\endgroup\$ – Matt Young Apr 4 '14 at 12:26
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    \$\begingroup\$ I have never even thought about mesh or nodal analysis at work. \$\endgroup\$ – Matt Young Apr 4 '14 at 20:37
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Loop currents exist because we define them mathematically to exist.

There are certain things that we take as axiomatic in circuit analysis: That components (resistors, capacitors and inductors) are linear, that the total current at a circuit node must be zero, and that the voltage around a complete loop must add up to zero.

The fact that components are linear gives us the principle of superposition: The response of a component can be divided up into parts; these parts can analyzed independently and then combined at the end to give the overall response.

Loop analysis is just one way of dividing the current through a component into parts that are easier to analyze. The current in any particular loop can't in general be measured directly — it's just a mathematical abstraction. But if we use that abstraction in the right way, it tells us useful things about the overall circuit.

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  • \$\begingroup\$ I don't understand how they can be considered axiomatic. You're assuming that the current in each branch can be represented as the sum of loop currents. How can you know this for sure? How do you know that there exists some current value for each "loop current", so that the right combinations of them can add up to the current in each branch? \$\endgroup\$ – dfg Apr 1 '14 at 5:46
  • \$\begingroup\$ Perhaps an analogy would better convey what I'm having trouble with: Say you have 3 different shapes. Creating loop currents and using their sums to represent the current in each branch is analogous to creating n number of smaller shapes and arranging them to create the 3 bigger shapes. This isn't necessarily possible. You don't know for sure that their exists n smaller shapes that can be arranged to form the 3 bigger shapes, so how do you know they're exists n loop currents that can be "arranged" in equations to add up to the current in each node? \$\endgroup\$ – dfg Apr 1 '14 at 5:47
  • \$\begingroup\$ A little help please? \$\endgroup\$ – dfg Apr 2 '14 at 16:20
  • \$\begingroup\$ @dfg: Since you haven't filled out your profile with details such as age, level of education and/or professional experience, I'm at a loss to formulate a response that would be suitable for your level of understanding. Do you understand what an axiom is? Do you understand that any real number can be expressed as a sum of two or more other real numbers? \$\endgroup\$ – Dave Tweed Apr 2 '14 at 18:47
  • \$\begingroup\$ I'm a freshman in college doing ECE, so I have a good understanding of linear algebra, and calculus. I know what an axiom is, and I know that a real number can be expressed as a sum of two or more other real numbers. \$\endgroup\$ – dfg Apr 3 '14 at 16:22
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No, the loop currents do not exist.

A component between two nodes such as a resistor carries a single, concrete current, say 1.0A.

We can can pretend that this 1.0A is is actually the combination of two currents, say 1.5A - 0.5A. This is only an algebraic trick to help us simplify and balance some equations.

How do we know that these currents do not exist?

If they existed, they would create extra power dissipation. If 1.5A is flowing through a purely resistive component, independently of another -0.5A is flowing through the same resistor, each of these two currents have to dissipate \$I^2R\$ W of heat, which is greater than the dissipation of a single 1.0A current.

But this is not the case: the power dissipation is consistent with 1.0A flowing through that component. In other words, the component currents which add up to the total are not "real" in the energy-based view of the situation, which we cannot ignore if we want a complete picture.

Never mind the energy view; we also cannot measure these component currents. An ammeter interposed into that branch of the circuit will produce a single value: 1.0A. The -0.5A and 1.5A cannot be observed, whereas the 1.0A total can be. What is observable is "more real" in some sense than what isn't, especially if the quantity was just postulated for algebraic convenience, and is not hypothesized to be a real entity.

What justifies the algebraic trick is that the math works out: no algebraic rule is broken, and there is no unwarranted use of these currents, like squaring their values inside wattage calculations.

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  • \$\begingroup\$ As I answered elsewhere, arguing whether loop currents do or don't exist is moot -- each loop current is simply a relabeling of a sum of one or more currents. But to the heart of Kaz's answer: there's a flaw in the argument. If you were to replace 2-currents-through-1-resistor with 2-currents-through-2-resistors, then the 2 resistors chosen must be such as to produce the same voltage drop as the original, given the separated currents. The two separate resistors will generally not have the same value as the original... \$\endgroup\$ – gwideman Apr 6 '14 at 5:02
  • \$\begingroup\$ Example: I1 = 1.5A and I2 = -0.5A, Icombo = 1A. Suppose a single 3 ohm resistor. Voltage drop = 3V. Power = VI, = 1 * 3 = 3W. \$\endgroup\$ – gwideman Apr 6 '14 at 5:04
  • \$\begingroup\$ Separating into two paths: R1 = V/I1 = 3/1.5 = 1 ohm, P1 = I2R = 2.25 * 2 = 4.5W; R2 = V/I2 = 3/-0.5 = -6 ohm, P2 = I^2R = 0.25 * -6 = -1.5W; Total power = 3W \$\endgroup\$ – gwideman Apr 6 '14 at 5:05
  • \$\begingroup\$ So, the power in the separated case is the same as in the combined case. This analysis requires negative resistors through which a positive current produces negative energy (absorbs energy from the environment). We don't have a real-life device like this (Peltier devices are somewhat similar, though very inefficient), but that is what what allows you to perform the decomposition to two resistors, if one of the decomposed currents is negative. \$\endgroup\$ – gwideman Apr 6 '14 at 5:07
  • \$\begingroup\$ @gwideman We can also add this to the answer: between two nodes, there is only one potential difference. Multiple simultaneous currents require different potentials. \$\endgroup\$ – Kaz Apr 6 '14 at 6:06
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I'll try to give you an answer based on how I visualize it - something that I try to do whenever an algebraic answer feels unsatisfying. It is as informal as it can possibly get, but I think it helps understand how come it is possible (and correct) to reduce so many branch currents to the much fewer mesh currents.

Imagine a node with 4 branches coming out of it. Each branch carries a "real" current \$i_{k}\$. Now visualize it in 3D by putting each branch at a height equal to its current:

enter image description here

For convenience, positive currents (heights) come out of the node. Since the sum of the currents is zero, then the sum of the heights is also zero.

On the other hand, mesh currents would be like edifices/blocks, because it is the same current component all around each innermost loop/pane:

enter image description here

Since each "pane" shares any give side with its neighbor, then the vertical distance between edifices must be equal to the "real" branch current. Therefore they edifices are "locked" vertically to one another, and the height difference is the branch current.

enter image description here

If the group of edifices around a node are looked from the top:

enter image description here

And each one is "locked" vertically to one another, you could traverse the locking conditions: 1>>2>>3>>4, and the resulting height of pane 4 should magically coincide with the alternate path 1>>4.

The magic comes from the pre-established fact that the sum of the currents (or "heights") of the branches coming out of the center node is zero. Whatever altitude was gained in the path, must be given back at the end. The loop edifices equivalent to the original currents would be as follows:

enter image description here

So the height locks are self-consistent, but the overall height can have, at this point, any value, as long as the differences are equal to the branch currents (you can push up or down the group of edifices as a whole and still be consistent).

The single solution is achieved thanks to the outermost edges of the circuit, where there is no neighbor, and the mesh current is equal to the branch current. You could also consider that the outermost neighbor lying outside of the circuit boundary is of known zero height. This boundary condition makes the solution unique.

So I'm basically saying that BECAUSE the currents from a node add up to zero, it is possible to assign equivalent mesh currents. If you violate this and assign any current to any branch, then it would NOT be possible to use mesh currents to represent the same system.


You could also reach the same conclusion by imagining a large planar circuit and assuming that it is possible to express it as mesh currents. You start with the outer edge, since the mesh currents are equal to the branch currents. All neighbors are consistent because of their simple relation of currents at the edge. Then you work your way inward, each mesh current being the difference between the branch current and the already assigned outer neighbor mesh current. It all works out to the last loop.

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  • \$\begingroup\$ apalopohapa: There's probably something clever going on here, but I'm having difficulty understanding how the red and blue lines correspond to loops. I assume the grey lines are a measurement grid, but what's the significance of the red versus the blue? \$\endgroup\$ – gwideman Apr 4 '14 at 11:27
  • \$\begingroup\$ tl;dr. +1 for the style of your drawings. \$\endgroup\$ – realtime Apr 4 '14 at 22:31
  • \$\begingroup\$ @apalopohapa: I see you've changed the drawings so there are mainly blue lines. I understand the first one (with -3 0 1 2), basically KCL at one node. But in subsequent drawings, what is the meaning of the blue lines which are vertical? \$\endgroup\$ – gwideman Apr 4 '14 at 22:47
  • \$\begingroup\$ @gwideman The vertical lines are just part of the block, it is what gives the blocks height, and I am equating height with current. The blocks are mesh currents, and they are blocks (with a flat top) because the current is the same on all its sides. All branches only have two common blocks (or meshes), and because of the clockwise convention of mesh currents, the branch current is always equal to the difference between mesh currents, or in this geometrical interpretation, the difference in height of adjacent blocks. \$\endgroup\$ – apalopohapa Apr 5 '14 at 0:09
  • \$\begingroup\$ @gwideman The main point is that any branch current can be represented by 2 adjacent blocks (or values) whose height difference is equal to the branch current. But the trick is that that height is the same on all sides, in other words the mesh current has the same current contribution on all branches that make it, so it is reasonable to ask if it is possible to have this city of blocks all balance out and produce the correct current branches (hence representing the same system). It is (we all knew that), and it seems obvious because we're so used to it... \$\endgroup\$ – apalopohapa Apr 5 '14 at 0:19
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I don't understand why this step is always valid. How do we know there exists some Ia, Ib, Ic and Id that fulfill the above system of equations?

A proof of the validity of mesh current analysis can be found in section 12.2, "The topology of one-dimensional complexes", of chapter 12, "The theory of electrical networks", in the 2nd book of "a course in mathematics for students of physics". Some of this is available on Google Books

This proof may be inaccessible to some. Here's a sample:

"Our immediate goal is to give some geometric interpretation to the spaces \$H_0\$ and \$Z_1\$; in the process we shall get some understanding of the mesh-current method. We wish to prove the following two facts (whose precise statement we shall give in the course of the discussion):

(i) dim \$H_0\$ is the number of connected components of the complex;

(ii) We can find a basis of \$Z_1\$ consisting of meshes "

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  • \$\begingroup\$ Thank you, from the excerpt your proof seems to be what I'm looking for. Others, however, are saying that the proof directly follows from the fact that all real numbers can be expressed as sums. You're proof seems a lot more sophisticated. How is this possible? \$\endgroup\$ – dfg Apr 4 '14 at 1:28
  • \$\begingroup\$ @dfg, writing the equations in terms of mesh currents instead of branch currents amounts to a change in basis. To prove that mesh current is valid, one must prove that the mesh basis exists (the 2nd fact to prove). \$\endgroup\$ – Alfred Centauri Apr 4 '14 at 2:02
  • \$\begingroup\$ The other answers are reasoning by analogy, not formal proofs - this is the most authoritative answer with a formal proof. Although apalopohapa is approaching a sketch of a proof with his planar decomposition. \$\endgroup\$ – pjc50 Apr 4 '14 at 10:01
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Most of the answers so far are correct, but are perhaps not getting at your particular obstacle. So here's yet another try :-).

First, I think you are probably already comfortable with two points:

  1. The sum of currents in and out of a particular node (junction point) is zero. All current into a node has to be matched by current out of a it, since a node itself can neither absorb nor produce any other current. This is "KCL", but hopefully this is intuitively obvious.

  2. The "loop currents" that are drawn for these calculations do not represent distinguishable separate currents that actually flow in a circle in the direction of the arrow. This is simply an accounting device for calculating actual currents in various branches attributable to each closed path in the net. That said, for a branch which is a member of two loops, the claim is that the actual current is the sum of the two loop currents. But why?

Perhaps you can make some headway by proceeding from the outside in, as follows:

Your four-loop example figure already makes each loop current synonymous with one or another of the outer edge currents (or its negative).

iA = i1
iB = i3
iC = -i6
iD = -i8

Now you can look at each of the nodes in the outer border, such as node (1). Here, by KCL,

i1 = i2 + i3. But i1 = iA, and i3 = iB, so we get
iA = i2 + iB, or 
i2 = iA - iB.

And the same for the other three nodes on the outer edges. And one more KCL at node (3) involving the current in all 4 legs. This simple system of equations can then solved, given some set of values for the components and sources.

These equations are all as written in your list of equations. Note that we didn't get there by assuming that the "loop current procedure" is true. We got there by simply writing down KCL for each node.

Really, the main utility of "loop currents" is to give variable names to particular sums and/or directions (clockwise) of currents, to make it a little easier to avoid mistakes when assembling the many individual node current-sum equations. (Especially in more complicated meshes than your example.) The loop currents idea doesn't draw on any phenomenon beyond KCL.

We could solve the system just fine without relying on the loop currents mental helper (that is, using only the numbered currents and KCL). But, absent the clockwise loop convention, with a large mesh you quickly get confused about current direction and sign at each node.

Hope that helps some!

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Part A: Theory

Currents, numbers, units, adjectives, mathematical operations like "+", equations, direction and time (to name a few) do not exist in the same sense that beer and donkeys exist. They are, however, mental tools invented by mankind to help us understand the universe.

Real numbers are a small subset of mathematical ideas, which we use to express quantities (the how-muchness of something). Units are to numbers like adjectives are to nouns (only more rigidly defined). A current is just a real number with a unit (named after André-Marie Ampère), which expresses how much charge (held by electrons) is moving through a specific volume (or area or line) during a certain time. Current still does not exist like your multimeter exists.

Loop currents are no different. A loop current is simply an imaginary part of a sum. Any real number (like a current) can be expressed as a finite (and sometimes even infinite) sum of other real numbers. Therefore the fact that currents can be written as the sum of loop currents follows directly from the closure of real numbers under addition (http://www.icoachmath.com/math_dictionary/closure_property_of_real_numbers_addition.html).

Part B: Practice

Look at the circuit in your question. Start at i1, which is a known current (real number). Follow the circuit in the direction of i1. The current splits at node 1. A part of it (a real number called i2) goes through R1, and another (a real number called i3) goes through R2. So i1 = i2 + i3 (Eq1). If you look at the top two branches of the circuit (the left and right 'arms' of node 1), you will see that the only loop current flowing there is iA on the left and iB on the right. Therefore i1 = iA and i3 = iB. Substituting this into Eq1, and solving for i2, we get: i2 = i1 - i3 = iA - iB . You can test this for every branch in the circuit. Loop analysis holds as long as KCL and KVL holds (like you said, when there are no contradictions in the circuit).

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  • \$\begingroup\$ Oops... by "imaginary part" I do NOT mean imaginary as in complex numbers. I mean imaginary as in "made-up". \$\endgroup\$ – rudolfbyker Apr 4 '14 at 8:41
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Dave Tweed's answer is correct, but maybe it would help to think about it differently.

Current is a flow of charge (usually electrons). An analogy can be made to the flow of water through a pipe. Strange effects aside, the amount of water that flows into the pipe will always be equal to the amount flowing out of it. If we had a closed loop of pipes, with a pump forcing the water to flow in the loop (your power source), the same would also be true -- the flow of water into each part of the loop will be the same. If it were not true, then we would know that there was a leak somewhere in the system (which is where the analogy breaks down).

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  • \$\begingroup\$ Isn't this an explanation of KCL and not loop analysis? \$\endgroup\$ – dfg Apr 3 '14 at 21:03
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    \$\begingroup\$ @dfg, the equivalent for KVL is, if you trace a closed path through the pipe network, and add up the pressure drops through each segement of pipe, it will add up to zero. Another way to say this is that if you start at location "A" in the pipes, and the pressure there is \$P_A\$, once you trace a path back to "A", the pressure there will still be \$P_A\$. \$\endgroup\$ – The Photon Apr 3 '14 at 21:20

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