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, and then reduce 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. Aaarrggghhhh!
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 for explanation in this chapter. 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.
This is not as scary as it may seem at first. Trust me!
Later on we'll see that some clever people have found tricks to avoid having to use simultaneous equations on these types of circuits. We call these tricks network theorems, and we will explore a few later in this chapter.
- 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 okay 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 + ER2 + ER1). 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 the same 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 (not voltages!) 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. 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 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.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. Normally, we try to avoid this when establishing our mesh current directions, but in a bridge circuit it is unavoidable: two of the mesh currents will inevitably clash through a component. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit.
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:
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 will confirm the accuracy of our voltage calculations:
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
- 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).
Introduction to network theorems
Anyone who's studied geometry should be familiar with the concept of a theorem: a relatively simple rule used to solve a problem, derived from a more intensive analysis using fundamental rules of mathematics. At least hypothetically, any problem in math can be solved just by using the simple rules of arithmetic (in fact, this is how modern digital computers carry out the most complex mathematical calculations: by repeating many cycles of additions and subtractions!), but human beings aren't as consistent or as fast as a digital computer. We need "shortcut" methods in order to avoid procedural errors.In electric network analysis, the fundamental rules are Ohm's Law and Kirchhoff's Laws. While these humble laws may be applied to analyze just about any circuit configuration (even if we have to resort to complex algebra to handle multiple unknowns), there are some "shortcut" methods of analysis to make the math easier for the average human.
As with any theorem of geometry or algebra, these network theorems are derived from fundamental rules. In this chapter, I'm not going to delve into the formal proofs of any of these theorems. If you doubt their validity, you can always empirically test them by setting up example circuits and calculating values using the "old" (simultaneous equation) methods versus the "new" theorems, to see if the answers coincide. They always should!
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