DC Network Analysis

What is network analysis?

Generally speaking, network analysis is any structured technique used to mathematically analyze a circuit (a “network” of interconnected components). Quite often the technician or engineer will encounter circuits containing multiple sources of power or component configurations which defy simplification by series/parallel analysis techniques. In those cases, he or she will be forced to use other means. This chapter presents a few techniques useful in analyzing such complex circuits.
To illustrate how even a simple circuit can defy analysis by breakdown into series and parallel portions, take start with this series-parallel circuit:

To analyze the above circuit, one would first find the equivalent of R₂ and R₃ in parallel, then add R₁ in series to arrive at a total resistance. Then, taking the voltage of battery B₁ with that total circuit resistance, the total current could be calculated through the use of Ohm's Law (I=E/R), then that current figure used to calculate voltage drops in the circuit. All in all, a fairly simple procedure.
However, the addition of just one more battery could change all of that:

Resistors R₂ and R₃ are no longer in parallel with each other, because B₂ has been inserted into R₃'s branch of the circuit. Upon closer inspection, it appears there are no two resistors in this circuit directly in series or parallel with each other. This is the crux of our problem: in series-parallel analysis, we started off by identifying sets of resistors that were directly in series or parallel with each other, reducing them to single equivalent resistances. If there are no resistors in a simple series or parallel configuration with each other, then what can we do?
It should be clear that this seemingly simple circuit, with only three resistors, is impossible to reduce as a combination of simple series and simple parallel sections: it is something different altogether. However, this is not the only type of circuit defying series/parallel analysis:

Here we have a bridge circuit, and for the sake of example we will suppose that it is not balanced (ratio R₁/R₄ not equal to ratio R₂/R₅). If it were balanced, there would be zero current through R₃, and it could be approached as a series/parallel combination circuit (R₁--R₄ // R₂--R₅). However, any current through R₃ makes a series/parallel analysis impossible. R₁ is not in series with R₄ because there's another path for electrons to flow through R₃. Neither is R₂ in series with R₅ for the same reason. Likewise, R₁ is not in parallel with R₂ because R₃ is separating their bottom leads. Neither is R₄ in parallel with R₅.
Although it might not be apparent at this point, the heart of the problem is the existence of multiple unknown quantities. At least in a series/parallel combination circuit, there was a way to find total resistance and total voltage, leaving total current as a single unknown value to calculate (and then that current was used to satisfy previously unknown variables in the reduction process until the entire circuit could be analyzed). With these problems, more than one parameter (variable) is unknown at the most basic level of circuit simplification.
With the two-battery circuit, there is no way to arrive at a value for “total resistance,” because there are two sources of power to provide voltage and current (we would need two “total” resistances in order to proceed with any Ohm's Law calculations). With the unbalanced bridge circuit, there is such a thing as total resistance across the one battery (paving the way for a calculation of total current), but that total current immediately splits up into unknown proportions at each end of the bridge, so no further Ohm's Law calculations for voltage (E=IR) can be carried out.
So what can we do when we're faced with multiple unknowns in a circuit? The answer is initially found in a mathematical process known as simultaneous equations or systems of equations, whereby multiple unknown variables are solved by relating them to each other in multiple equations. In a scenario with only one unknown (such as every Ohm's Law equation we've dealt with thus far), there only needs to be a single equation to solve for the single unknown:

However, when we're solving for multiple unknown values, we need to have the same number of equations as we have unknowns in order to reach a solution. There are several methods of solving simultaneous equations, all rather intimidating and all too complex . However, many scientific and programmable calculators are able to solve for simultaneous unknowns, so it is recommended to use such a calculator when first learning how to analyze these circuits.

• REVIEW:
• Some circuit configurations (“networks”) cannot be solved by reduction according to series/parallel circuit rules, due to multiple unknown values.
• Mathematical techniques to solve for multiple unknowns (called “simultaneous equations” or “systems”) can be applied to basic Laws of circuits to solve networks.

Branch current method

The first and most straightforward network analysis technique is called the Branch Current Method. In this method, we assume directions of currents in a network, then write equations describing their relationships to each other through Kirchhoff's and Ohm's Laws. Once we have one equation for every unknown current, we can solve the simultaneous equations and determine all currents, and therefore all voltage drops in the network.
Let's use this circuit to illustrate the method:

The first step is to choose a node (junction of wires) in the circuit to use as a point of reference for our unknown currents. I'll choose the node joining the right of R₁, the top of R₂, and the left of R₃.

At this node, guess which directions the three wires' currents take, labeling the three currents as I₁, I₂, and I₃, respectively. Bear in mind that these directions of current are speculative at this point. Fortunately, if it turns out that any of our guesses were wrong, we will know when we mathematically solve for the currents (any “wrong” current directions will show up as negative numbers in our solution).

Kirchhoff's Current Law (KCL) tells us that the algebraic sum of currents entering and exiting a node must equal zero, so we can relate these three currents (I₁, I₂, and I₃) to each other in a single equation. For the sake of convention, I'll denote any current entering the node as positive in sign, and any current exiting the node as negative in sign:

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the currents. Remember that the “upstream” end of a resistor will always be negative, and the “downstream” end of a resistor positive with respect to each other, since electrons are negatively charged:

The battery polarities, of course, remain as they were according to their symbology (short end negative, long end positive). It is OK if the polarity of a resistor's voltage drop doesn't match with the polarity of the nearest battery, so long as the resistor voltage polarity is correctly based on the assumed direction of current through it. In some cases we may discover that current will be forced backwards through a battery, causing this very effect. The important thing to remember here is to base all your resistor polarities and subsequent calculations on the directions of current(s) initially assumed. As stated earlier, if your assumption happens to be incorrect, it will be apparent once the equations have been solved (by means of a negative solution). The magnitude of the solution, however, will still be correct.
Kirchhoff's Voltage Law (KVL) tells us that the algebraic sum of all voltages in a loop must equal zero, so we can create more equations with current terms (I₁, I₂, and I₃) for our simultaneous equations. To obtain a KVL equation, we must tally voltage drops in a loop of the circuit, as though we were measuring with a real voltmeter. I'll choose to trace the left loop of this circuit first, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is arbitrary). The result will look like this:













Having completed our trace of the left loop, we add these voltage indications together for a sum of zero:

Of course, we don't yet know what the voltage is across R₁ or R₂, so we can't insert those values into the equation as numerical figures at this point. However, we do know that all three voltages must algebraically add to zero, so the equation is true. We can go a step further and express the unknown voltages as the product of the corresponding unknown currents (I₁ and I₂) and their respective resistors, following Ohm's Law (E=IR), as well as eliminate the 0 term:

Since we know what the values of all the resistors are in ohms, we can just substitute those figures into the equation to simplify things a bit:

You might be wondering why we went through all the trouble of manipulating this equation from its initial form (-28 + E_R2 + E_R1). After all, the last two terms are still unknown, so what advantage is there to expressing them in terms of unknown voltages or as unknown currents (multiplied by resistances)? The purpose in doing this is to get the KVL equation expressed using thesame unknown variables as the KCL equation, for this is a necessary requirement for any simultaneous equation solution method. To solve for three unknown currents (I₁, I₂, and I₃), we must have three equations relating these three currents (notvoltages!) together.
Applying the same steps to the right loop of the circuit (starting at the chosen node and moving counter-clockwise), we get another KVL equation:

















Knowing now that the voltage across each resistor can be and should be expressed as the product of the corresponding current and the (known) resistance of each resistor, we can re-write the equation as such:

Now we have a mathematical system of three equations (one KCL equation and two KVL equations) and three unknowns:

For some methods of solution (especially any method involving a calculator), it is helpful to express each unknown term in each equation, with any constant value to the right of the equal sign, and with any “unity” terms expressed with an explicit coefficient of 1. Re-writing the equations again, we have:

Using whatever solution techniques are available to us, we should arrive at a solution for the three unknown current values:

So, I₁ is 5 amps, I₂ is 4 amps, and I₃ is a negative 1 amp. But what does “negative” current mean? In this case, it means that our assumed direction for I₃ was opposite of its real direction. Going back to our original circuit, we can re-draw the current arrow for I₃ (and re-draw the polarity of R₃'s voltage drop to match):

Notice how current is being pushed backwards through battery 2 (electrons flowing “up”) due to the higher voltage of battery 1 (whose current is pointed “down” as it normally would)! Despite the fact that battery B₂'s polarity is trying to push electrons down in that branch of the circuit, electrons are being forced backwards through it due to the superior voltage of battery B₁. Does this mean that the stronger battery will always “win” and the weaker battery always get current forced through it backwards? No! It actually depends on both the batteries' relative voltages and the resistor values in the circuit. The only sure way to determine what's going on is to take the time to mathematically analyze the network.
Now that we know the magnitude of all currents in this circuit, we can calculate voltage drops across all resistors with Ohm's Law (E=IR):

Let us now analyze this network using SPICE to verify our voltage figures.[spi] We could analyze current as well with SPICE, but since that requires the insertion of extra components into the circuit, and because we know that if the voltages are all the same and all the resistances are the same, the currents must all be the same, I'll opt for the less complex analysis. Here's a re-drawing of our circuit, complete with node numbers for SPICE to reference:

network analysis example   
v1 1 0  
v2 3 0 dc 7    
r1 1 2 4      
r2 2 0 2
r3 2 3 1
.dc v1 28 28 1  
.print dc v(1,2) v(2,0) v(2,3)  
.end    

v1            v(1,2)      v(2)        v(2,3)          
2.800E+01     2.000E+01   8.000E+00   1.000E+00

Sure enough, the voltage figures all turn out to be the same: 20 volts across R₁ (nodes 1 and 2), 8 volts across R₂ (nodes 2 and 0), and 1 volt across R₃ (nodes 2 and 3). Take note of the signs of all these voltage figures: they're all positive values! SPICE bases its polarities on the order in which nodes are listed, the first node being positive and the second node negative. For example, a figure of positive (+) 20 volts between nodes 1 and 2 means that node 1 is positive with respect to node 2. If the figure had come out negative in the SPICE analysis, we would have known that our actual polarity was “backwards” (node 1 negative with respect to node 2). Checking the node orders in the SPICE listing, we can see that the polarities all match what we determined through the Branch Current method of analysis.

• REVIEW:
• Steps to follow for the “Branch Current” method of analysis:
• (1) Choose a node and assume directions of currents.
• (2) Write a KCL equation relating currents at the node.
• (3) Label resistor voltage drop polarities based on assumed currents.
• (4) Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equations.
• (5) Solve for unknown branch currents (simultaneous equations).
• (6) If any solution is negative, then the assumed direction of current for that solution is wrong!
• (7) Solve for voltage drops across all resistors (E=IR).

Mesh current method

The Mesh Current Method, also known as the Loop Current Method, is quite similar to the Branch Current method in that it uses simultaneous equations, Kirchhoff's Voltage Law, and Ohm's Law to determine unknown currents in a network. It differs from the Branch Current method in that it does not use Kirchhoff's Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equations, which is especially nice if you're forced to solve without a calculator.

Mesh Current, conventional method

Let's see how this method works on the same example problem:

The first step in the Mesh Current method is to identify “loops” within the circuit encompassing all components. In our example circuit, the loop formed by B₁, R₁, and R₂ will be the first while the loop formed by B₂, R₂, and R₃ will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops. In fact, this method gets its name from the idea of these currents meshing together between loops like sets of spinning gears:

The choice of each current's direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I₁ and I₂ are both going “up” through resistor R₂, where they “mesh,” or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value.
The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the “upstream” end of a resistor will always be negative, and the “downstream” end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not “agree” with the resistor polarities (assumed current directions):

Using Kirchhoff's Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor's voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents.
Tracing the left loop of the circuit, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is ultimately irrelevant), counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead on the point behind, we get this equation:

Notice that the middle term of the equation uses the sum of mesh currents I₁ and I₂ as the current through resistor R₂. This is because mesh currents I₁ and I₂ are going the same direction through R₂, and thus complement each other. Distributing the coefficient of 2 to the I₁ and I₂ terms, and then combining I₁ terms in the equation, we can simplify as such:

At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counter-clockwise:

Simplifying the equation as before, we end up with:

Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I₁ and I₂:

Knowing that these solutions are values for mesh currents, not branch currents, we must go back to our diagram to see how they fit together to give currents through all components:

The solution of -1 amp for I₂ means that our initially assumed direction of current was incorrect. In actuality, I₂ is flowing in a counter-clockwise direction at a value of (positive) 1 amp:

This change of current direction from what was first assumed will alter the polarity of the voltage drops across R₂ and R₃ due to current I₂. From here, we can say that the current through R₁ is 5 amps, with the voltage drop across R₁ being the product of current and resistance (E=IR), 20 volts (positive on the left and negative on the right). Also, we can safely say that the current through R₃ is 1 amp, with a voltage drop of 1 volt (E=IR), positive on the left and negative on the right. But what is happening at R₂?
Mesh current I₁ is going “up” through R₂, while mesh current I₂ is going “down” through R₂. To determine the actual current through R₂, we must see how mesh currents I₁ and I₂ interact (in this case they're in opposition), and algebraically add them to arrive at a final value. Since I₁ is going “up” at 5 amps, and I₂ is going “down” at 1 amp, the real current through R₂ must be a value of 4 amps, going “up:”

A current of 4 amps through R₂'s resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR), positive on the top and negative on the bottom.
The primary advantage of Mesh Current analysis is that it generally allows for the solution of a large network with fewer unknown values and fewer simultaneous equations. Our example problem took three equations to solve the Branch Current method and only two equations using the Mesh Current method. This advantage is much greater as networks increase in complexity:

To solve this network using Branch Currents, we'd have to establish five variables to account for each and every unique current in the circuit (I₁ through I₅). This would require five equations for solution, in the form of two KCL equations and three KVL equations (two equations for KCL at the nodes, and three equations for KVL in each loop):





I suppose if you have nothing better to do with your time than to solve for five unknown variables with five equations, you might not mind using the Branch Current method of analysis for this circuit. For those of us who have better things to do with our time, the Mesh Current method is a whole lot easier, requiring only three unknowns and three equations to solve:





Less equations to work with is a decided advantage, especially when performing simultaneous equation solution by hand (without a calculator).
Another type of circuit that lends itself well to Mesh Current is the unbalanced Wheatstone Bridge. Take this circuit, for example:

Since the ratios of R₁/R₄ and R₂/R₅ are unequal, we know that there will be voltage across resistor R₃, and some amount of current through it. As discussed at the beginning of this chapter, this type of circuit is irreducible by normal series-parallel analysis, and may only be analyzed by some other method.
We could apply the Branch Current method to this circuit, but it would require six currents (I₁ through I₆), leading to a very large set of simultaneous equations to solve. Using the Mesh Current method, though, we may solve for all currents and voltages with much fewer variables.
The first step in the Mesh Current method is to draw just enough mesh currents to account for all components in the circuit. Looking at our bridge circuit, it should be obvious where to place two of these currents:

The directions of these mesh currents, of course, is arbitrary. However, two mesh currents is not enough in this circuit, because neither I₁ nor I₂ goes through the battery. So, we must add a third mesh current, I₃:

Here, I have chosen I₃ to loop from the bottom side of the battery, through R₄, through R₁, and back to the top side of the battery. This is not the only path I could have chosen for I₃, but it seems the simplest.
Now, we must label the resistor voltage drop polarities, following each of the assumed currents' directions:

Notice something very important here: at resistor R₄, the polarities for the respective mesh currents do not agree. This is because those mesh currents (I₂ and I₃) are going through R₄ in different directions. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit. Though later, we will show how to avoid the R₄ current clash. (See Example below)
Generating a KVL equation for the top loop of the bridge, starting from the top node and tracing in a clockwise direction:

In this equation, we represent the common directions of currents by their sums through common resistors. For example, resistor R₃, with a value of 100 Ω, has its voltage drop represented in the above KVL equation by the expression 100(I₁ + I₂), since both currents I₁ and I₂ go through R₃ from right to left. The same may be said for resistor R₁, with its voltage drop expression shown as 150(I₁ + I₃), since both I₁ and I₃ go from bottom to top through that resistor, and thus work together to generate its voltage drop.
Generating a KVL equation for the bottom loop of the bridge will not be so easy, since we have two currents going against each other through resistor R₄. Here is how I do it (starting at the right-hand node, and tracing counter-clockwise):

Note how the second term in the equation's original form has resistor R₄'s value of 300 Ω multiplied by the difference between I₂ and I₃ (I₂ - I₃). This is how we represent the combined effect of two mesh currents going in opposite directions through the same component. Choosing the appropriate mathematical signs is very important here: 300(I₂ - I₃) does not mean the same thing as 300(I₃ - I₂). I chose to write 300(I₂ - I₃) because I was thinking first of I₂'s effect (creating a positive voltage drop, measuring with an imaginary voltmeter across R₄, red lead on the bottom and black lead on the top), and secondarily of I₃'s effect (creating a negative voltage drop, red lead on the bottom and black lead on the top). If I had thought in terms of I₃'s effect first and I₂'s effect secondarily, holding my imaginary voltmeter leads in the same positions (red on bottom and black on top), the expression would have been -300(I₃ - I₂). Note that this expression is mathematically equivalent to the first one: +300(I₂ - I₃).
Well, that takes care of two equations, but I still need a third equation to complete my simultaneous equation set of three variables, three equations. This third equation must also include the battery's voltage, which up to this point does not appear in either two of the previous KVL equations. To generate this equation, I will trace a loop again with my imaginary voltmeter starting from the battery's bottom (negative) terminal, stepping clockwise (again, the direction in which I step is arbitrary, and does not need to be the same as the direction of the mesh current in that loop):

Solving for I₁, I₂, and I₃ using whatever simultaneous equation method we prefer:

Example:
Use Octave to find the solution for I₁, I₂, and I₃ from the above simplified form of equations. [octav]
Solution:
In Octave, an open source Matlab® clone, enter the coefficients into the A matrix between square brackets with column elements comma separated, and rows semicolon separated.[octav] Enter the voltages into the column vector: b. The unknown currents: I₁, I₂, and I₃ are calculated by the command: x=A\b. These are contained within the x column vector.

 
        octave:1>A = [300,100,150;100,650,-300;-150,300,-450]
        A =
          300  100  150
          100  650  -300
          -150 300  -450
 
        octave:2> b = [0;0;-24]
        b =
          0
          0
          -24
               
        octave:3> x = A\b
        x =
          -0.093793
           0.077241
           0.136092

The negative value arrived at for I₁ tells us that the assumed direction for that mesh current was incorrect. Thus, the actual current values through each resistor is as such:

Calculating voltage drops across each resistor:

A SPICE simulation confirms the accuracy of our voltage calculations:[spi]

unbalanced wheatstone bridge   
v1 1 0  
r1 1 2 150      
r2 1 3 50       
r3 2 3 100      
r4 2 0 300      
r5 3 0 250      
.dc v1 24 24 1  
.print dc v(1,2) v(1,3) v(3,2) v(2,0) v(3,0)    
.end    

v1            v(1,2)      v(1,3)      v(3,2)      v(2)        v(3)            
2.400E+01     6.345E+00   4.690E+00   1.655E+00   1.766E+01   1.931E+01

Example:
(a) Find a new path for current I₃ that does not produce a conflicting polarity on any resistor compared to I₁ or I₂. R₄ was the offending component. (b) Find values for I₁, I₂, and I₃. (c) Find the five resistor currents and compare to the previous values.
Solution: [dvn]
(a) Route I₃ through R₅, R₃ and R₁ as shown:

Note that the conflicting polarity on R₄ has been removed. Moreover, none of the other resistors have conflicting polarities.
(b) Octave, an open source (free) matlab clone, yields a mesh current vector at “x”:[octav]

        octave:1> A = [300,100,250;100,650,350;-250,-350,-500]
        A =
          300  100  250
          100  650  350
          -250  -350  -500
      
        octave:2> b = [0;0;-24]
        b =
          0
          0
        -24
              
        octave:3> x = A\b
        x =
          -0.093793
          -0.058851
           0.136092

Not all currents I₁, I₂, and I₃ are the same (I₂) as the previous bridge because of different loop paths However, the resistor currents compare to the previous values:

        IR1 = I1 + I3 = -93.793 ma + 136.092 ma = 42.299 ma
        IR2 = I1 = -93.793 ma
        IR3 = I1 + I2 + I3 = -93.793 ma -58.851 ma  + 136.092 ma =  -16.552 ma
        IR4 = I2 = -58.851 ma
        IR5 = I2 + I3 = -58.851 ma  + 136.092 ma = 77.241 ma

Since the resistor currents are the same as the previous values, the resistor voltages will be identical and need not be calculated again.

• REVIEW:
• Steps to follow for the “Mesh Current” method of analysis:
• (1) Draw mesh currents in loops of circuit, enough to account for all components.
• (2) Label resistor voltage drop polarities based on assumed directions of mesh currents.
• (3) Write KVL equations for each loop of the circuit, substituting the product IR for E in each resistor term of the equation. Where two mesh currents intersect through a component, express the current as the algebraic sum of those two mesh currents (i.e. I₁ + I₂) if the currents go in the same direction through that component. If not, express the current as the difference (i.e. I₁ - I₂).
• (4) Solve for unknown mesh currents (simultaneous equations).
• (5) If any solution is negative, then the assumed current direction is wrong!
• (6) Algebraically add mesh currents to find current in components sharing multiple mesh currents.
• (7) Solve for voltage drops across all resistors (E=IR).

Mesh current by inspection

We take a second look at the “mesh current method” with all the currents running counterclockwise (ccw). The motivation is to simplify the writing of mesh equations by ignoring the resistor voltage drop polarity. Though, we must pay attention to the polarity of voltage sources with respect to assumed current direction. The sign of the resistor voltage drops will follow a fixed pattern.
If we write a set of conventional mesh current equations for the circuit below, where we do pay attention to the signs of the voltage drop across the resistors, we may rearrange the coefficients into a fixed pattern:

Once rearranged, we may write equations by inspection. The signs of the coefficients follow a fixed pattern in the pair above, or the set of three in the rules below.

• Mesh current rules:
• This method assumes electron flow (not conventional current flow) voltage sources. Replace any current source in parallel with a resistor with an equivalent voltage source in series with an equivalent resistance.
• Ignoring current direction or voltage polarity on resistors, draw counterclockwise current loops traversing all components. Avoid nested loops.
• Write voltage-law equations in terms of unknown currents currents: I₁, I₂, and I₃. Equation 1 coefficient 1, equation 2, coefficient 2, and equation 3 coefficient 3 are the positive sums of resistors around the respective loops.
• All other coefficients are negative, representative of the resistance common to a pair of loops. Equation 1 coefficient 2 is the resistor common to loops 1 and 2, coefficient 3 the resistor common to loops 1 an 3. Repeat for other equations and coefficients.

  +(sum of R's loop 1)I1 - (common R loop 1-2)I2 - (common R loop 1-3)I3   = E1
  -(common R loop 1-2)I1 + (sum of R's loop 2)I2 - (common R loop 2-3)I3   = E2
  -(common R loop 1-3)I1 - (common R loop 2-3)I2 + (sum of R's loop 3)I3   = E3

• The right hand side of the equations is equal to any electron current flow voltage source. A voltage rise with respect to the counterclockwise assumed current is positive, and 0 for no voltage source.
• Solve equations for mesh currents:I₁, I₂, and I₃ . Solve for currents through individual resistors with KCL. Solve for voltages with Ohms Law and KVL.

While the above rules are specific for a three mesh circuit, the rules may be extended to smaller or larger meshes. The figure below illustrates the application of the rules. The three currents are all drawn in the same direction, counterclockwise. One KVL equation is written for each of the three loops. Note that there is no polarity drawn on the resistors. We do not need it to determine the signs of the coefficients. Though we do need to pay attention to the polarity of the voltage source with respect to current direction. The I₃counterclockwise current traverses the 24V source from (+) to (-). This is a voltage rise for electron current flow. Therefore, the third equation right hand side is +24V.

In Octave, enter the coefficients into the A matrix with column elements comma separated, and rows semicolon separated. Enter the voltages into the column vector b. Solve for the unknown currents: I₁, I₂, and I₃ with the command: x=A\b. These currents are contained within the x column vector. The positive values indicate that the three mesh currents all flow in the assumed counterclockwise direction.

           octave:2> A=[300,-100,-150;-100,650,-300;-150,-300,450]
           A =
             300  -100  -150
             -100  650  -300
             -150  -300  450

           octave:3> b=[0;0;24]
           b =
              0
              0
             24

           octave:4> x=A\b
           x =
             0.093793
             0.077241
             0.136092

The mesh currents match the previous solution by a different mesh current method.. The calculation of resistor voltages and currents will be identical to the previous solution. No need to repeat here.
Note that electrical engineering texts are based on conventional current flow. The loop-current, mesh-current method in those text will run the assumed mesh currents clockwise.[aef] The conventional current flows out the (+) terminal of the battery through the circuit, returning to the (-) terminal. A conventional current voltage rise corresponds to tracing the assumed current from (-) to (+) through any voltage sources.
One more example of a previous circuit follows. The resistance around loop 1 is 6 Ω, around loop 2: 3 Ω. The resistance common to both loops is 2 Ω. Note the coefficients of I₁ and I₂ in the pair of equations. Tracing the assumed counterclockwise loop 1 current through B₁ from (+) to (-) corresponds to an electron current flow voltage rise. Thus, the sign of the 28 V is positive. The loop 2 counter clockwise assumed current traces (-) to (+) through B₂, a voltage drop. Thus, the sign of B₂ is negative, -7 in the 2nd mesh equation. Once again, there are no polarity markings on the resistors. Nor do they figure into the equations.

The currents I₁ = 5 A, and I₂ = 1 A are both positive. They both flow in the direction of the counterclockwise loops. This compares with previous results.

• Summary:
• The modified mesh-current method avoids having to determine the signs of the equation coefficients by drawing all mesh currents counterclockwise for electron current flow.
• However, we do need to determine the sign of any voltage sources in the loop. The voltage source is positive if the assumed ccw current flows with the battery (source). The sign is negative if the assumed ccw current flows against the battery.
• See rules above for details.

Node voltage method

The node voltage method of analysis solves for unknown voltages at circuit nodes in terms of a system of KCL equations. This analysis looks strange because it involves replacing voltage sources with equivalent current sources. Also, resistor values in ohms are replaced by equivalent conductances in siemens, G = 1/R. The siemens (S) is the unit of conductance, having replaced the mho unit. In any event S = Ω^-1. And S = mho (obsolete).
We start with a circuit having conventional voltage sources. A common node E₀ is chosen as a reference point. The node voltages E₁ and E₂ are calculated with respect to this point.

A voltage source in series with a resistance must be replaced by an equivalent current source in parallel with the resistance. We will write KCL equations for each node. The right hand side of the equation is the value of the current source feeding the node.

Replacing voltage sources and associated series resistors with equivalent current sources and parallel resistors yields the modified circuit. Substitute resistor conductances in siemens for resistance in ohms.

           I1 = E1/R1 = 10/2 = 5 A
           I2 = E2/R5 = 4/1  = 4 A
           G1 = 1/R1 = 1/2 Ω   = 0.5 S
           G2 = 1/R2 = 1/4 Ω   = 0.25 S
           G3 = 1/R3 = 1/2.5 Ω = 0.4 S
           G4 = 1/R4 = 1/5 Ω   = 0.2 S
           G5 = 1/R5 = 1/1 Ω   = 1.0 S

The Parallel conductances (resistors) may be combined by addition of the conductances. Though, we will not redraw the circuit. The circuit is ready for application of the node voltage method.

           GA = G1 + G2 = 0.5 S + 0.25 S = 0.75 S
           GB = G4 + G5 = 0.2 S + 1 S = 1.2 S 

Deriving a general node voltage method, we write a pair of KCL equations in terms of unknown node voltages V₁ and V₂ this one time. We do this to illustrate a pattern for writing equations by inspection.

           GAE1 + G3(E1 - E2) = I1             (1)
           GBE2 - G3(E1 - E2) = I2             (2)


           (GA + G3 )E1         -G3E2 = I1     (1)
                  -G3E1 + (GB + G3)E2 = I2     (2)

The coefficients of the last pair of equations above have been rearranged to show a pattern. The sum of conductances connected to the first node is the positive coefficient of the first voltage in equation (1). The sum of conductances connected to the second node is the positive coefficient of the second voltage in equation (2). The other coefficients are negative, representing conductances between nodes. For both equations, the right hand side is equal to the respective current source connected to the node. This pattern allows us to quickly write the equations by inspection. This leads to a set of rules for the node voltage method of analysis.

• Node voltage rules:
• Convert voltage sources in series with a resistor to an equivalent current source with the resistor in parallel.
• Change resistor values to conductances.
• Select a reference node(E₀)
• Assign unknown voltages (E₁)(E₂) ... (E_N)to remaining nodes.
• Write a KCL equation for each node 1,2, ... N. The positive coefficient of the first voltage in the first equation is the sum of conductances connected to the node. The coefficient for the second voltage in the second equation is the sum of conductances connected to that node. Repeat for coefficient of third voltage, third equation, and other equations. These coefficients fall on a diagonal.
• All other coefficients for all equations are negative, representing conductances between nodes. The first equation, second coefficient is the conductance from node 1 to node 2, the third coefficient is the conductance from node 1 to node 3. Fill in negative coefficients for other equations.
• The right hand side of the equations is the current source connected to the respective nodes.
• Solve system of equations for unknown node voltages.

Example: Set up the equations and solve for the node voltages using the numerical values in the above figure.
Solution:

           (0.5+0.25+0.4)E1 -(0.4)E2=  5 
          -(0.4)E1 +(0.4+0.2+1.0)E2 = -4
           (1.15)E1 -(0.4)E2=  5 
          -(0.4)E1 +(1.6)E2   = -4
           E1 =  3.8095
           E2 = -1.5476

The solution of two equations can be performed with a calculator, or with octave (not shown).[octav] The solution is verified with SPICE based on the original schematic diagram with voltage sources. [spi] Though, the circuit with the current sources could have been simulated.

           V1 11 0 DC 10
           V2 22 0 DC -4
           r1 11 1 2
           r2 1 0 4
           r3 1 2 2.5
           r4 2 0 5
           r5 2 22 1
           .DC V1 10 10 1 V2 -4 -4 1 
           .print DC V(1) V(2)
           .end

                v(1)            v(2)
            3.809524e+00    -1.547619e+00

One more example. This one has three nodes. We do not list the conductances on the schematic diagram. However, G₁ = 1/R₁, etc.

There are three nodes to write equations for by inspection. Note that the coefficients are positive for equation (1) E₁, equation (2) E₂, and equation (3) E₃. These are the sums of all conductances connected to the nodes. All other coefficients are negative, representing a conductance between nodes. The right hand side of the equations is the associated current source, 0.136092 A for the only current source at node 1. The other equations are zero on the right hand side for lack of current sources. We are too lazy to calculate the conductances for the resistors on the diagram. Thus, the subscripted G's are the coefficients.

           (G1 + G2)E1              -G1E2             -G2E3      = 0.136092
                 -G1E1  +(G1 + G3 + G4)E2             -G3E3      = 0
                 -G2E1              -G3E2 +(G2 + G3 + G5)E3      = 0

We are so lazy that we enter reciprocal resistances and sums of reciprocal resistances into the octave “A” matrix, letting octave compute the matrix of conductances after “A=”.[octav] The initial entry line was so long that it was split into three rows. This is different than previous examples. The entered “A” matrix is delineated by starting and ending square brackets. Column elements are space separated. Rows are “new line” separated. Commas and semicolons are not need as separators. Though, the current vector at “b” is semicolon separated to yield a column vector of currents.

           octave:12> A = [1/150+1/50 -1/150 -1/50
           > -1/150 1/150+1/100+1/300 -1/100
           > -1/50 -1/100 1/50+1/100+1/250]
           A =
              0.0266667  -0.0066667  -0.0200000
             -0.0066667   0.0200000  -0.0100000
             -0.0200000  -0.0100000   0.0340000

           octave:13> b = [0.136092;0;0]
           b =
              0.13609
              0.00000
              0.00000

           octave:14> x=A\b
           x =
              24.000
              17.655
              19.310

Note that the “A” matrix diagonal coefficients are positive, That all other coefficients are negative.
The solution as a voltage vector is at “x”. E₁ = 24.000 V, E₂ = 17.655 V, E₃ = 19.310 V. These three voltages compare to the previous mesh current and SPICE solutions to the unbalanced bridge problem. This is no coincidence, for the 0.13609 A current source was purposely chosen to yield the 24 V used as a voltage source in that problem.

• Summary
• Given a network of conductances and current sources, the node voltage method of circuit analysis solves for unknown node voltages from KCL equations.
• See rules above for details in writing the equations by inspection.
• The unit of conductance G is the siemens S. Conductance is the reciprocal of resistance: G = 1/R