Electronics Module 1B: Analysis of DC circuits
I. Circuit Analysis: Introducing Kirchhoff’s Laws
At this point we want to go back to basic physics, and ask whether there are any other physical principles that may guide us in attempting to analyze circuits.
Conservation of energy: We recall from physics that E is a “conservative” field (at least, aside from cases where there are varying magnetic fields... more on that when we talk about AC circuits!). This has three consequences:
1.
The potential can be defined as V = ∫ E ● ds, where the integral depends only on the starting and ending points, and not on the path chosen to get from one endpoint to the other. (This is known as “path independence.”
2.
From path independence it follows that if we choose a path that starts and ends at the same point (a “closed loop”), V must be equal to zero.
Therefore, if I draw any closed loop through a circuit (or anyplace at all!), the sum of the voltage drops encountered as I traverse that closed loop and return to the starting point must be exactly zero:
Σ closed loop
V = 0
This is known as Kirchhoff’s Voltage Law (KVL, or Kirchhoff’s Loop Rule), and is a simple consequence of the conservation of energy.
Conservation of charge: We know that charges cannot appear or disappear out of thin air. That means that at any point in a circuit, the sum of charges going into that point must equal the sum of charges going out of that point. Taking the time derivative, this means that the sum of currents going into that point must equal the sum of currents going out of that point.
This rule is most useful at a place where two or more wires intersect, known in circuit parlance as a
“node.” Hence
Σ node
I = 0
This is known as Kirchhoff’s Current Law (KCL, or Kirchhoff’s Node Rule), and is a simple consequence of the conservation of charge.
Pro tip 1: Note that for both of Kirchhoff’s Laws we must be careful about plus and minus signs. In fact, when working with Kirchhoff’s Laws, keeping track of minus signs will be the bane of your existence!

Pro tip 2: For all but the simplest circuits, application of Kirchhoff’s Laws will result in simultaneous equations with multiple unknowns (i.e. N equations in N unknowns). Get comfortable in dealing with these things!
II. Overview of Circuit analysis methods
In this section I give a brief outline of the circuit analysis tools in our “toolbox.”
Note that while I will develop these tools to deal with resistive networks and DC power supplies (constant voltage or constant current), the same tools will be used to analyze RC and RL circuits (in which time dependence of I and V must be considered), AC circuits (in which R is joined by its buddies X
C
, X
L
and Z, and the relevant circuit quantities must be expressed as numbers with both real and imaginary components), and in circuits including diodes, transistors and op amps (in which “ideal _____ models” will be used to describe circuit behavior, just as we used our “ideal resistor model,” “ideal voltage source model,” and “ideal conductor model,” for example).
A.
Decomposition of circuits
This method involves identifying resistors in a complicated network which have connections entirely in series or in parallel, which can be replaced by a single “equivalent” component R eq
to yield a simpler circuit. This step can be repeated until a single-loop circuit is formed, which can be analyzed using KVL and Ohm’s Law. Once that is done, you can gradually “re-compose” the circuit (replace the R eq
’s with the original R’s), using Ohm’s Law and the facts that currents in series are equal, currents in parallel add, voltages in series add, and voltages in parallel are the same, to solve for the voltages across, and the currents through, each component.
(This is usually the easiest technique to apply, so use it whenever possible!)
B.
Application of Kirchhoff’s Laws to complex circuits
When a circuit cannot be decomposed, Kirchhoff’s Laws must be used to set up a set of simultaneous equations in multiple unknowns. There are several ways to do this:
1.
The “Branch Current” method
In this method, the currents in each of the circuit’s N branches are identified (where a
“branch” is a set of components connected in series so that the currents through all are the same); these N branch currents generally serve as the “unknowns” for which you will solve.
KVL is used to write down equations for the voltage drops around each loop; at minimum you must have enough loops to include every component in the circuit once. Then, KCL is used to write relationships between the branch currents at nodes. If there are N branch currents, a total of N equations are needed to solve the system, and these will come from some combination of loop equations using KVL, and node equations using KCL. Then you use your method of choice (substitution, Gaussian elimination, or Cramer’s rule) to solve the system of simultaneous equations.
(This is the method typically taught in intro physics classes!)

2.
The “Mesh Current” method
This method involves the enumeration of “meshes,” which are the smallest loops one can create in the circuit (i.e. loops which do not have any wires or components contained within; if the circuit represented the frame of a window, the meshes would be loops encircling individual panes of glass). Each mesh has a fictitious “mesh current” circulating around it; the actual branch currents are the algebraic sums of all mesh currents passing through that branch. If there are N meshes in the circuit, there are N unknown mesh currents, and KVL is used to write N loop equations, which are then solved. Finally, you use your simultaneousequation-solving method of choice to solve for your N unknowns.
(The advantage of this method, compared to the branch current method, is that you typically have one (sometimes more) fewer equation since the use of meshes effectively eliminates the KCL equation(s) you would have needed to relate the branch currents to each other. The disadvantage is that the loop equations are a little more confusing to write down since you have to keep careful track of + and – signs, and the fact that some resistors have multiple mesh currents running through them.)
3.
The “Node Voltage” method
This method involves the identification of “nodes,” which are all places where two or more branches of a circuit meet. Generally only a subset of the total number of nodes are needed: nodes that are separated by only a length of ideal conductor are at the same voltage, and nodes separated only by a voltage source have trivially different voltages. The trick is to identify which nodes are necessary, and determine the voltages V i
at each of those
N nodes in terms of the branch currents (which are identified in the same way as in the branch current method). Of these N nodes, one is defined to be at V=0 (“grounded”), leaving only N-1 nodes which need to be solved. Then, at each of these N-1 nodes KCL is used to derive a relationship between the branch currents at that node. Solve each of your node voltage equations at a given node (i.e. V i
= …) for the branch currents, which you plug into your KCL equation at that node. There will be N-1 such equations, one from each node; these can be solved using your method of choice. With the voltages at each node determined, your node voltage equations can be used to determine the branch currents, and the circuit voltages found using Ohm’s Law.
(The advantage of this method is that the number of simultaneous equations generated is usually smaller than for either the mesh or branch current methods. In fact, even rather complex circuits will end up having only one or two nodes, and thus unknowns and simultaneous equations, to solve.)
C.
Application of Linear Network Theorems
These methods assume that the circuit is a linear network ; that is, it assumes that there are linear relationships between voltage and current which hold throughout the circuit. Ohm’s Law is an example of a linear relationship, and ideal voltage and current sources are linear (constant voltage and constant current are linear relationships with slope = 0!). These methods fail if

circuits contain non-ohmic devices (such as light bulbs or diodes), although there are some
“tricks” one can use to get around this.
1.
Superposition Theorem
This theorem holds that the actual currents in a circuit with multiple power supplies are the algebraic sums of the currents that would exist in “subcircuits” where only one of the power sources exists. To apply this method, if a circuit has N sources, all but a single power source are removed from a circuit (more strictly, they are replaced by their internal resistances – ideal voltage sources by a short circuit, and ideal current sources by an open circuit), and the
N resulting subcircuits analyzed (generally decomposition can be used!). Then the currents obtained from each of the N simplified subcircuits are algebraically added to obtain the currents that exist in the multi-source circuit. Ohm’s Law can be used to determine voltages, and your answers checked by applying KVL loop equations.
2.
Thevenin’s Theorem
This theorem holds that a linear network, no matter how complex , from the point of view of a single set of terminals attached to two points on that network , can be described by an equivalent network which contains only a voltage source V th
and a series resistance R th
. To apply Thevenin’s theorem, one first determines the “open circuit” voltage V
∞
between the terminals in order to determine V th
= V
∞
, and the “short circuit” current I sc
between the terminals to determine R th
= V th
/I sc
. (Alternately, to find R th
one may replace all voltage sources with short circuits, and all current sources with open circuits – as is done with the superposition method, and determine the equivalent resistance R eq
of the “deactivated” network, with R th
= R eq
.) Then the current and voltage between the two terminals produced by the complex network will be equal to that produced by the Thevenin equivalent network.
3.
Norton’s Theorem
Analogous to Thevenin’s Theorem, Norton’s Theorem holds that a linear network, no matter how complex, from the point of view of a single set of two terminals attached to two points on the network, can be described by an equivalent network which contains only a current source I n
and a parallel resistance R n
. I n
is the “short circuit” current I sc
as defined above, and R n
is the equivalent resistance R eq
of the circuit with all sources removed. It also follows that a Thevenin network has a Norton equivalent network, with R th
= R n
, and I n
= V th
/R th
.
A useful consequence of the Thevenin and Norton theorems is the so-called Maximum
Power Transfer Theorem . This theorem states that if a load is connected across the terminals of a linear network, maximum power is transferred to the load when R load
= R th
(or
R n
). This can be proven using calculus for either a Thevenin or Norton network, but is also true for a more complicated network that can be represented by a much simpler Thevenin or Norton equivalent.
III. Decomposition – Step by Step

Rules of the road:
A.
When components are connected in series , the currents through them are always the same!
(This is a consequence of the conservation of charge, or alternately KCL applied at any point along the wire connecting two series components.)
B.
When components are connected in series , the voltage across the string of components is equal to the sum of the voltage drops across the individual components. (This is a consequence of the conservation of energy, or alternately KVL applied on a loop going through all the components in series, then back to the beginning.)
C.
When components are connected in parallel , the currents through them add up to yield the total current delivered to the network. (This is also a consequence of conservation of charge / KCL.)
D.
When components are connected in parallel, the voltages across them are the same . (This can be shown by applying KVL with a loop going “down” through one of the parallel components and back “up” through another.)
E.
The relationship between V, I and R for a resistor is given by Ohm’s Law: V = I R. Therefore, if any two of those quantities are known, the third can be very easily determined.
F.
Ideal voltage sources maintain a constant potential difference ε between their terminals, with the drawn current determined by the load attached (through Ohm’s Law). Ideal current sources maintain a constant current ɩ through them, with an unknown potential difference ε ɩ
between their terminals given by KVL applied to a closed loop containing the current source.
1.
The first step is to inspect the circuit, and decide which, if any, components are connected in series or parallel. If so, those components can be replaced by an “equivalent component,” and a new circuit drawn with the “equivalent component” replacing the multiple components with series or parallel connections.
2.
When resistors are in series, the “equivalent resistance” is the sum of the individual resistances:
R eq
= R
1
+ R
2
+ R
3
+ …
Note that R eq
for series combinations is always greater than any of the individual resistances.
When resistors are in parallel, the conductances (G = 1/R) add, or, alternately, one may apply the “one over” rule:
G eq
= G
1
+ G
2
+ G
3
+ …. --or-- 1/R eq
= 1/R
1
+ 1/R
2
+ 1/R
3
+ …
Note that R eq
for parallel combinations is always smaller than any of the individual resistances.
If there are only two resistors in parallel, the above simplifies to:
R eq
= R
1
R
2
/ (R
1
+ R
2
) (i.e. “product over sum” rule)

which has two simple cases that are worth remembering: (1) if R
1
= R
2
= R, then R eq
= R/2 (which also generalizes: N equal resistors R in parallel have R eq
= R/N), and (2) if R
2
>> R
1
, then R eq
is approximately equal to (but slightly smaller than) R
1
.
3.
Once those series or parallel components are replaced with their equivalent resistance R eq
, and a new circuit redrawn with R eq
replacing those components, the process can be repeated (by combining R eq
with other R i
’s to yield new R eq
’s) as many times as is necessary to yield a simple circuit which can be analyzed with Ohm’s Law to find I and V
Req
.
4.
Then, once V, I and R for this overly simplified circuit have been determined, the rules above can be used to “rebuild” the circuit to its original complexity (i.e. replacing the R eq
’s with the original resistors), using the fact that currents divide among parallel components, and voltages divide among series components to solve for the individual V’s and I’s. This process can be repeated until the original circuit has been “re-composed,” and the V’s and I’s in each component have been determined.
And you’re done!
IV. Branch currents – Step by Step
1.
The first step is to use the decomposition method to simplify the circuit as much as possible, because the more components there are in the circuit, the more difficult the simultaneous equations become to solve. So if there are any voltage sources in series, add them to create a single voltage source. If any resistors are in series or parallel, replace them with a single R eq
.
2.
Once the circuit is simplified as much as possible, identify how many branches exist in the circuit. Generally, a “branch” is a section of the circuit composed of components connected entirely in series, which all have the same I flowing through them, and the number of branches
N is the number of independent currents. As soon as one branch intersects another branch, a third branch (and maybe a fourth, fifth, etc.) begins.
3.
Once you’ve identified the N branches in your circuit, assign an (unknown) current I j
to each of them. You will have to arbitrarily guess which direction to use; try to guess which way the current will actually flow, but don’t sweat over it – if you guess wrong you will simply get a negative answer. That is, the direction you guess isn’t important, but be sure to clearly indicate which way you are assuming the current to flow so that you get your plus and minus signs right, and you know what a plus or minus sign on your final answer means!
Your goal now is to set up N equations to solve for these N unknown currents I
1
, I
2
, … I
N
.
4.
Next, identify how many nodes there are in the circuit. A node is any place where two or more wires meet, or, equivalently, where any two or more branches meet. The “nodes” are all the places in the circuit where currents may branch out in different directions; nodes are the only places where the current may change as you travel along a given path. With each node

identified, use KCL to write down a relationship for the branch currents at those M nodes.
(Typically, only a small subset of the nodes will yield useful KCL equations, as multiple nodes will often have the exact same node equation since it’s the same branch currents meeting at all of them.) A useful convention is to define currents going into a node as positive, and currents going out of a node as negative.
5.
Then, you will use KVL to write down equations for the voltages around closed loops in the circuit. Since you need N total equations, and you have M equations using KCL on your nodes, you will need N-M loop equations in order to solve the circuit. You will also need enough loops so that each and every component in the circuit is counted at least once (usually, some components will be counted more than once since they are on branches shared between two or more loops). Pay very careful attention to plus and minus signs! Note that the loops don’t have to represent actual paths of current flow; they just need to be closed loops in that they start and end at the same place.
Useful things to remember:
-To avoid confusion, I prefer to traverse my loops in the clockwise direction (regardless of the assumed directions of the branch currents), and start from the lower left corner of the loop.
Sometimes I violate this rule, though.
- Values and signs for voltage sources: + ε if you traverse the source from its negative to positive terminals (mnemonics: + sign if you swim “upstream” through it, or travel from the side of the source with the little bar to the side with the big bar). ε if you traverse the source from its positive to its negative terminals (mnemonic: - sign if you go “downhill,” against the direction of the current the source is pushing, or travel from the side with the big bar to the side with the little bar.
-Values and signs for resistors: V
R
= -IR if you are swimming “downstream,” i.e. in the direction of I, from where the charges come in with high energy, to where they come out with lower energy. V
R
= +IR if you are swimming “against the current,” i.e. opposite the direction of I, from where the charges come out with low energy, to where they went in with higher energy.
-Note regarding current sources: if a branch contains a current source ɩ , then that branch current I j
is known, and is equal to ɩ . However, there is an unknown voltage ε ɩ
across the current source; your KVL equation will be used to solve for the value of ε ɩ
. This changes the mechanics of the mathematics a bit (in that ε ɩ
replaces the branch current I j
as one of your unknowns in the loop equations), but otherwise you will still solve the circuit in the same way.
6.
Once you set up your simultaneous equations, there are several methods you can use to solve them, which I have covered in class. These include:

a.
Substitution: Solve one equation for one unknown in terms of the others, and plug that into another of the equations; repeat until you have one (nasty!) equation containing only one unknown, solve it for that unknown, and then recursively solve for the other unknowns. b.
Linear combinations (aka Gaussian elimination): If you have a system of N linearly independent equations in N unknowns, there is a set of values of the unknowns which satisfies all those equations. It follows that any linear combination of those equations is also a perfectly good equation describing the system. You can take advantage of this fact, and take linear combinations of your equations which cause all but one of your unknowns to cancel out. Solve for that unknown, and go back recursively to solve for the other unknowns. c.
Cramer’s rule: You can use linear algebra techniques to more efficiently solve for the unknowns. To do this you will need to write your simultaneous equations in “matrix form”
M x = b (where M is your matrix of coefficients, x is a vector representing your unknowns, and b is a vector of known constant values; usually your ε ’s) and solve N+1 determinants (D
= det(M) and the N D i
’s where one column of M is replaced by b) to find your unknowns: I i
=
D i
/D.
7.
Now that you have solved for the unknown currents (and possibly one or more V cs
’s corresponding to current sources), you can calculate all the resistor voltages using V = I R. If any of the currents are negative, this simply means they are in the direction opposite what you assumed when you started!
8.
Finally, you should check your answers by plugging in your voltages to the loop equations and verifying that they sum up to zero, and by plugging in your currents to your node equations, and verifying that they sum up to zero.
Pro tip 1 : If numerical values are given in the circuit, and you plug in the numbers to your equations
(rather than keeping them in terms of variables such as ε and R), the equations generally are simpler to work with (i.e. equations look like 5I
2
– 3I
1
= 0 instead of (R
1
+R
2
) I
1
– 6R
1
2 /(R
3
+R
4
) I
2
= 0. However, if you keep the variables in place until the last step, you can use dimensional analysis (i.e. make sure the “units” of all terms in your loop equations are of voltage - ε , IR, IR 2 /R, etc.), which can help catch algebraic errors made in setting up your equations). Also, if you keep the variables, it’s easier for me to verify your equations along the way, which means you’re more likely to get partial credit if you make a mistake!
Pro tip 2: There is no reason to ever get the wrong answer, since you can check your work! If the sums of the voltages or currents in Step 7 do not come to zero, then your answer is wrong and you should go back through your calculations! If the sums of your voltages and currents do add up to zero, then as long as your original equations accurately represent the circuit you’re solving, your solution is correct.
V. Mesh currents – Step by Step

One thing you’ll probably notice about the branch current method is that your KCL equations are generally trivial in comparison to the KVL equations, so the first step will almost always be to use your KCL equation(s) to solve for one of the currents, and plug that into your KVL equations, to reduce the total number of equations and unknowns. Since there are N total unknowns (the branch currents) and M constraints on them (the KCL node equations), you can reduce the total number of unknowns to N-M.
It’s not a hard step, but it is one nonetheless. The mesh current method enables you to bypass it.
1.
To do this, rather than assign currents to each branch, you assign a looping current to each
“mesh.” A “mesh” is a closed loop that does not contain any smaller closed loops. That is, if the circuit were a window, with the components and conductors representing the frame, each
“mesh” would correspond to a single pane of glass. Each mesh is assigned a “mesh current” which is a fictitious current that traverses the closed loop.
General rule: A circuit with N meshes can be described by N mesh currents (where the branch currents are the algebraic sum of the mesh currents of all meshes passing through that branch).
To solve this system, you will need to write down exactly N loop equations – any less and the system is not constrained, any more and at least one of your equations will be redundant (which turns out to be very bad if you try to use Cramer’s rule, because it depends on D not being equal to zero, which is what happens if two or more of your equations are not linearly independent).
2.
Next, write down loop equations for each of your loops. Generally, the loops you use will be the same as your meshes (it almost always makes the equations simpler to solve), but this is not necessarily true (see the Pro Tips below for examples of cases where you may want to deviate).
However, your loop equations must account for each component at least once (some of them will be traversed twice or more), and there must be exactly N of them (where N equals the number of meshes in the circuit).
Here are a couple of tips: (I would advise you to use discretion with these tips, since they can easily lead you astray. But, if you apply them correctly, they can save you a lot of grief!)
Pro tip 1: While, strictly speaking, the “mesh currents” should be defined to go around meshes, this is not necessarily true. You can define your “meshes” to be larger loops, although you have to take care to go through each component at least once, and to have exactly N loops. While you may think of “creative” ways to include every component in a smaller number of loops (by having them overlap or loop around in figure-eights, for example), this will result in a pathological system since you do not have enough independent currents.
Pro tip 2: One really good reason to get creative with mesh definitions: If a resistor-containing branch is shared between two or more meshes (say, I
1
and I
2
), the voltage across that resistor will be (I
1
±I
2
)R (depending on the relative directions of I
1
and I
2
), and therefore terms containing

both I
1
and I
2
will appear in two or more of your loop equations. But if two meshes share a branch which contains only a voltage source, then neither I
1
nor I
2
will appear in the terms arising from that branch (since ideal voltage source terms are just ±
, depending on the direction of loop traversal). The more “decoupled” your equations are (i.e. the fewer times that two or more mesh currents appear in two or more equations), the easier they are to solve using the
“substitution” method.
Pro tip 3: The opposite works if you have a current source ɩ in the circuit. If a current source is shared between two meshes I
1
and I
2
, then the only constraint we have is that I
1
± I
2
= ɩ . This is actually a separate equation (the “N+1 th ”) you have to write down since there is also the “N+1 th ” unknown in the current source voltage ε ɩ
. You can use that I
1
± I
2
= ɩ to eliminate either I
1
or I
2
, and bring you back to N unknowns (with ε ɩ
replacing the eliminated current) but this is an extra step. But if you can re-arrange your meshes so that the current source is only located on a single mesh (say, I
1
), then you can immediately solve for that mesh current, I
1
= ± ɩ . That eliminates one of your mesh currents from the simultaneous equations. Depending on the circuit layout, you may also be able to make a loop containing only the current source and a voltage source to eliminate ε ɩ
in one fell swoop (transforming it into a system of N-1 equations in N-1 unknowns), but at worst it’s a wash – you can immediately solve for I
1
and remove it from all your other equations, but have a new unknown in ε ɩ
.
Pro tip 4: In textbooks, they tell you to define all mesh currents as being clockwise. If, however, you can correctly guess the direction of the actual current by looking at the circuit, you can define the mesh current to go that way. The advantage is that if you’re right there will be one less minus sign in the final answer. The disadvantage is that you have to be a little more careful about putting the plus and minus signs on your IR terms.
3.
Now that you’ve defined your N mesh currents, and written down your N loop equations, you’re ready to solve them for the unknowns using any of the three techniques we went over
(substitution, Gaussian elimination, or Cramer’s rule), or any other advanced math tricks you may know.
4.
With the values of the unknowns in hand (keeping in mind that negative values indicate propagation directions opposite what you assumed) you can determine the currents through each branch as the algebraic sum of the mesh currents through that branch.
5.
Then the resistor voltages are calculated as R times the actual branch current through the resistor.
6.
Finally, we go back into KVL to check that the sums of the voltages around our loops are zero.
VI. Node Voltages: Step by Step

While the node voltage method also uses Kirchhoff’s laws, the mechanics are somewhat different from the branch and mesh current methods. Some sources advise replacing resistances with conductances and voltage sources with current sources, but I think this is as likely to cause confusion as it is to alleviate mathematical difficulties. So, do that if you want, but wouldn’t generally bother.
1.
The first step is to identify how many nodes there are in the circuit (call this N
0
). Your goal is to determine the voltage at each of these nodes; once this is done the branch currents can be solved.
2.
You’ll note that many of the nodes are connected by ideal conductors, in which case they’re at the same voltage; these nodes are therefore not independent, and are effectively one node.
Once you account for these connected nodes, you’ll come up with some smaller number N of independent nodes. (Note that this N is almost always significantly smaller than the number of branches or meshes!)
3.
Next, since voltages are only meaningful as a difference, you will pick one node as a
“reference,” calling the voltage of this node V=0. (In electronics terms, we “connect that node to ground.”) It’s worth thinking carefully about which node to ground; in general you want to pick one that “grounds out” a good portion of the circuit’s “real estate,” since that makes your life easier in mapping out the other node voltages.
That leaves N-1 nodes. (Before I go any further I’ll state that in most circuits you’ll see in exam problem N-1 = 1, even in circuits that would have 2 or 3 branches or meshes!)
4.
Next, you want to define all the branch currents in the circuit as you did for the branch current method, and use KCL to write down relationships between the branch currents at each of the N-
1 nodes.
5.
Pick one of your nodes (call it node A). For each path from the ground node to node A, write down an equation of the form V
A
= …., where the “…” represents the sum of the voltage drops encountered along the path from the ground node to node A. If there are m branches (with m branch currents) connecting to node A, you will have m equations of this form.
6.
Solve each of these m equations for I m
= …., where the “…” represents some expression involving V
A
(and perhaps the voltages at other nodes passed along the way, if this is a really nasty circuit!).
7.
Plug these I m
= … equations into your node equation from KCL. You will now have one equation involving V
A
(and perhaps other node voltages) on the left-hand side, and 0 on the right-hand side.

8.
If there are other nodes (B, C, D, etc.) repeat Steps 5-7 until you have N-1 equations involving N-
1 node voltages.
9.
You now have N-1 equations involving N-1 node voltages. If N-1 = 1, the solution is trivial: solve for V
A
. If not, use your simultaneous equation-solving tools to solve for the V x
’s.
Pro tip: In a very complex circuit with multiple nodes, you will not be able to get directly from the ground node to an arbitrary node in the circuit. When this is the case, pick “node A” to be one that you can reach directly from the ground node. Then node B will be a node you can reach directly from the ground node and/or node A. Since the only branch currents that should appear in your node B equation are those in the branches that directly connect to node B, you should write your V
B
= … equation (step 5) in terms of only the currents in the KCL equation for node B, plugging in V
A
wherever necessary to eliminate other currents. You can repeat this step as much as necessary. In step 9 you will end up with a set of coupled simultaneous equations with the V
A
,
V
B
, etc. as unknowns, but we know how to deal with those. (Right?)
10.
Once you have solved for the node voltages, you can solve for the branch currents using the equations you wrote in step 6.
11.
With branch currents in hand, we can solve for the resistor voltages using Ohm’s Law.
12.
With resistor voltages known, we can use KVL and a few loop equations to check our answers.
Pro tip: There is something known as the “anchor effect” in which the first thing you learn about the topic is weighed more heavily than things you learn later. Since any circuit analysis you’ve seen before probably concentrated on the branch and mesh current methods (and I introduced them first in class), you’ll probably gravitate toward using those methods, even though the node voltage method is often simpler.
My advice: before writing down loop equations, think about using the node voltage method – it may save you a whole lot of trouble!
VII. Linear Network Theorems – The Superposition Theorem – Step by Step
These are theorems that can be used to simplify the analysis of circuits that are linear networks.
Therefore, to apply them, it is necessary to verify that the circuit in question is a “linear network.”
Okay, great. Well, what in the world is a “linear network?” This means that all components have a linear relationship between voltage and current. Ohm’s Law-obeying (“ohmic”) resistors, ideal voltage sources, and ideal current sources are linear. So as long as a circuit contains only those things, we’re good to go.

Superposition Theorem : In a linear network containing multiple independent (voltage or current) sources, the currents and voltages throughout the circuit are given by a linear superposition of the currents and voltages that obtain in subcircuits formed by removing all but one of the sources.
That is, a circuit with three independent sources can be represented as the sum of three subcircuits, each of which contains one of the three sources. “Independent” means that the output of each source does not depend on some other parameter in the circuit, such as a voltage or current measured elsewhere. “Dependent sources” are a bit more difficult to deal with – Schaum’s is big on them, though we will only use them a bit in later discussions on amplifier circuits. Dependent sources don’t really “exist” in the sense that you can go to Radio Shack and buy a dependent voltage source. But they’re useful in modeling certain more complex components, which behave somewhat as if they contain internal sources whose output is dependent on some other parameter in the circuit.
So we take our original circuit with three sources, and from this generate three subcircuits, each of which contains one of the original sources, and has the other two eliminated.
What happens to the eliminated sources? They are, technically speaking, replaced by their internal resistance:
For an ideal voltage source, R = 0, so ideal voltage sources are replaced by a short .
For an ideal current source, R = ∞ , so ideal current sources are replaced by an open circuit .
For a “real” source, the source is replaced by whatever internal resistance it has.
A battery, for example, typically has an internal resistance on the order of an ohm or so, so a battery would be replaced by a resistor with a small resistance on that order.
So here’s the process:
1.
Given a circuit with N independent sources, draw N “subcircuits,” each containing only one of the independent sources, with the omitted sources replaced by a short (for a voltage source), an open (for a current source), or a resistor (for a “real” source with internal resistance).
2.
Solve, using whatever method(s) you like, for the branch currents and resistor (and current source) voltages in each of the N subcircuits. (Generally, you will be able to do this using decomposition, since it’s usually the presence of multiple sources that destroys simple series/parallel relationships and requires you to use the more general Kirchhoff-based methods
(branch current, mesh current or node voltage.)
3.
Once all N subcircuits have been solved, determine the branch currents and resistor / current source voltages in the original circuit as the algebraic sum of their counterparts in the N subcircuits.

4.
Check your work by applying KVL / KCL to your original circuit and seeing that the sums are zero, and check that each resistor’s voltage and current sync up with Ohm’s Law. If there are any discrepancies, you can locate the source of the error by using KVL/KCL on each of the N subcircuits. The corollary: if the currents and voltages you have calculated do satisfy KVL, KCL and Ohm’s Law, then you know that you have the correct solution since, for a well-posed circuit, the solution is unique.
And that’s it; you’re done!
VII. Linear Network Theorems – Thevenin’s Theorem & Norton’s Theorem – Step by Step
There are theorems due to Thevenin and Norton which are extremely powerful in circuit analysis.
There are others as well, although they either follow as consequences of Thevenin and Norton, or are more specialized theorems with more limited applications so we won’t dwell on them.
In addition to the above provisos about linear networks, both theorems are predicated on the idea of a network being divided into two parts – a “load,” which is the component you’re interested in, and a “source,” which is everything else.
Equivalent network: two networks as described as “equivalent” if they will deliver the same terminal voltage and current to an arbitrarily-chosen load circuit.
One would hope that two off-the-shelf 12 V power supplies, for example, are “equivalent” in that they will have the same output at their terminals when connected to a project we’re building, regardless of the internal details of their construction.
The idea is that the “source” can be arbitrarily complicated, but we really don’t care what’s happening inside (unless you feel like reading voluminous power-supply service manuals!). So we model the source as a “black box” with two wires sticking out, or two terminal plugs on the front panel, and to facilitate our analysis of how the source acts when those two wires are connected to a load, we replace the “black box” with something simpler (the simplest possible network being a single power supply connected to a single resistor).
And that’s what the Thevenin and Norton theorems do. They replace some arbitrarily complex
“black box” network (about whose internal details we could care less, aside from it being linear or
“close enough” to linear that it doesn’t make much of a difference) with a much simpler network which is “equivalent” from the “point of view” of those two terminals.
Thevenin’s Theorem: Any linear network, which may be connected to an external load via two terminals, may be replaced by an equivalent network composed of only an ideal voltage source of voltage V th
, connected in series with a resistor of value R th
.

Norton’s Theorem: Any linear network, which may be connected to an external load via two terminals, may be replaced by an equivalent network composed of only an ideal current source of current I
N
, connected in parallel with a resistor of value R
N
.
First, one important proviso about the “equivalent networks” you get from Thevenin and Norton: these networks are only equivalent from the “point of view” of the two external terminals (or, as I like to say, the “load’s eye view.”) Internally, the original network (“black box”) and the Thevenin or
Norton equivalents are quite different.
Therefore, any internal behavior (such as the current through or the voltage across some component inside the original network) will not be reproduced in the Thevenin or Norton circuit. If I place terminals at different locations in the original “black box” circuit, Thevenin and Norton tell us nothing about how those two new terminals behave. Only the voltage and current at those two original terminals will be the same in the original circuit and its
Thevenin and Norton equivalents. (We can always re-apply Thevenin or Norton to find the equivalent circuits for the same “black box” circuit with differently-located terminals, although there is no reason to expect that our V th
and R th
(or I
N
and R
N
) will be the same!)
All that remains is to figure out the values of V th
and R th
(or I
N
and R
N
). Two little measly numbers – how hard could that be? And there are three ways you can determine them; I’ll give three ways to execute both theorems (you have the freedom to choose any two of these, so pick which of the three will be the hardest, and don’t do that one):
1.
Replace the “load” with an open circuit (i.e. R = ∞ ) and determine the “open circuit” voltage V oc between the terminals using whatever methods you like. Then, V th
= V oc
, and V
OC
can also be used with the results of either of the next two analyses to determine I
N.
2.
Replace all the independent power sources in the circuit with their internal resistances (shorts for voltage sources, opens for current sources, as we did with the superposition theorem). Then determine the equivalent resistance R eq
between the terminals of this “deactivated” circuit.
(With all sources removed, this can usually be done with series/parallel decomposition.) Then,
R th
= R
N
= R eq
, and I
N
= V technique cannot be used.
OC
/R eq
. However, if there are dependent sources in the circuit, this
3.
Short out the terminals (i.e. connect a wire between them, or a load with R = 0), and determine the short-circuit current I
SC
between the terminals. Again, you can use whatever method you want. Then I
N
= I
SC
, R
N
= R th
= V
OC
/I
SC and if you haven’t solved for it already, V th
= I
SC
R eq
.
To convert from a Thevenin circuit to its Norton equivalent: I
N
= V th
/R th
, R
N
= R th
To convert from a Norton circuit to its Thevenin equivalent: V th
= I
N
R
N
, R th
= R
N
Done!
Pro tip 1: In most “textbook” descriptions (including this one) the “source” and “load” portions of the circuit are clearly separated from each other, usually with the “load” hanging promiscuously off of

the right-hand side of the circuit. However, this need not be the case. The schematic can be drawn with the “load” anywhere, including right in the middle of the diagram, surrounded on all sides by a tangled morass of “source.” So don’t let the inconvenient position of some component in your diagram dissuade you from applying Thevenin or Norton!
Pro tip 2: There need not be a “source” and “load” in the conventional sense (such as a store-bought power supply and your electronics project). Since the “source” and “load” can be demarcated arbitrarily, Thevenin and Norton are useful on circuits such as the ones we’ve studied. To apply
Thevenin or Norton, single out one of the components (such as a resistor or capacitor whose characteristics you are trying to model) and treat it as the “load,” with the balance of the circuit consigned to “source” ignominy. Then the messy “source” can be replaced by its Thevenin or Norton equivalent using the above technique, and the voltage and current at the “load” component found straightforwardly.
Pro tip 3: This is also useful for analyzing multi-loop circuits (which heretofore have been solved by setting up simultaneous equations): by selecting the “load” to be along some branch shared between two or more loops, you can effectively replace that component with an open or short circuit, which makes the circuit much simpler to solve. This may be true even if you have to repeat the technique several times to find Thevenin equivalents with respect to several different “source”/“load” combinations!
Pro tip 4: It’s also extremely useful to effectively “delete” some nasty component, such as a nonohmic resistive device, from the circuit by treating it as the “load,” and replacing the well-behaved
(but complicated) components in the “source” with their Thevenin or Norton equivalent.
Maximum Power Transfer Theorem : Given a “source” connected to a “load,” a source with
Thevenin or Norton equivalent resistance R th
(R
N
) will deliver maximum power to a load resistance R
L if R
L
= R th
= R
N
. The maximum power that can be transferred by such a source is V th
2 / 4R th
= I
N
2 R
N
/4.
While you may be tempted to tuck this theorem away in the “novelty” category (as SNL’s Church
Lady would say, “Isn’t that special?”), it actually is quite useful. If I am trying to deliver maximum power from an amplifier to a speaker system (“But does it go up to 11?”), provide the brightest possible lighting in an area, get the maximum possible output from a power supply of a given size
(and cost), or the maximum possible transmitted power to a radio antenna (“Can you hear me now?”), this theorem allows me to match my “load” (the speaker, the lights, the radio antenna) to the “source” (an amplifier, a power supply or a transmitter) I am using, or vice versa. In more advanced circuit analysis this is known as “impedance matching.” (Impedance is the AC generalization of DC resistance.) If my source and my load are not “impedance matched,” then at best I am wasting energy by heating up the components in my source (which has to be overbuilt, at additional expense, in order to handle this wasted power), and at worst the power I would like to go to my load is finding its way back into the source and possibly causing damage!