Branch current method Step by Step Procedure/Explanation

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.
The first and most straightforward network analysis technique is called the
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 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.[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).

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What is network analysis?

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.
Generally speaking,
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. 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.

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-Y and Y- conversions Step By Step Procedure/Explanation

In many circuit applications, we encounter components connected together in one of two ways to form a three-terminal network: the “Delta,” or Δ (also known as the “Pi,” or π) configuration, and the “Y” (also known as the “T”) configuration.

It is possible to calculate the proper values of resistors necessary to form one kind of network (Δ or Y) that behaves identically to the other kind, as analyzed from the terminal connections alone. That is, if we had two separate resistor networks, one Δ and one Y, each with its resistors hidden from view, with nothing but the three terminals (A, B, and C) exposed for testing, the resistors could be sized for the two networks so that there would be no way to electrically determine one network apart from the other. In other words, equivalent Δ and Y networks behave identically.
There are several equations used to convert one network to the other:

Δ and Y networks are seen frequently in 3-phase AC power systems (a topic covered in volume II of this book series), but even then they're usually balanced networks (all resistors equal in value) and conversion from one to the other need not involve such complex calculations. When would the average technician ever need to use these equations?
A prime application for Δ-Y conversion is in the solution of unbalanced bridge circuits, such as the one below:

Solution of this circuit with Branch Current or Mesh Current analysis is fairly involved, and neither the Millman nor Superposition Theorems are of any help, since there's only one source of power. We could use Thevenin's or Norton's Theorem, treating R₃ as our load, but what fun would that be?
If we were to treat resistors R₁, R₂, and R₃ as being connected in a Δ configuration (R_ab, R_ac, and R_bc, respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit:

After the Δ-Y conversion . . .

If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration.




Resistors R₄ and R₅, of course, remain the same at 18 Ω and 12 Ω, respectively. Analyzing the circuit now as a series/parallel combination, we arrive at the following figures:

We must use the voltage drops figures from the table above to determine the voltages between points A, B, and C, seeing how the add up (or subtract, as is the case with voltage between points B and C):




Now that we know these voltages, we can transfer them to the same points A, B, and C in the original bridge circuit:

Voltage drops across R₄ and R₅, of course, are exactly the same as they were in the converted circuit.
At this point, we could take these voltages and determine resistor currents through the repeated use of Ohm's Law (I=E/R):

A quick simulation with SPICE will serve to verify our work

unbalanced bridge circuit   
v1 1 0  
r1 1 2 12       
r2 1 3 18       
r3 2 3 6
r4 2 0 18       
r5 3 0 12       
.dc v1 10 10 1  
.print dc v(1,2) v(1,3) v(2,3) v(2,0) v(3,0)    
.end    

v1            v(1,2)      v(1,3)      v(2,3)      v(2)        v(3)            
1.000E+01     4.706E+00   5.294E+00   5.882E-01   5.294E+00   4.706E+00

The voltage figures, as read from left to right, represent voltage drops across the five respective resistors, R₁ through R₅. I could have shown currents as well, but since that would have required insertion of “dummy” voltage sources in the SPICE netlist, and since we're primarily interested in validating the Δ-Y conversion equations and not Ohm's Law, this will suffice.

• REVIEW:
• “Delta” (Δ) networks are also known as “Pi” (π) networks.
• “Y” networks are also known as “T” networks.
• Δ and Y networks can be converted to their equivalent counterparts with the proper resistance equations. By “equivalent,” I mean that the two networks will be electrically identical as measured from the three terminals (A, B, and C).
• A bridge circuit can be simplified to a series/parallel circuit by converting half of it from a Δ to a Y network. After voltage drops between the original three connection points (A, B, and C) have been solved for, those voltages can be transferred back to the original bridge circuit, across those same equivalent points.

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Maximum Power Transfer Theorem Step By Step Procedure/Explanation

The Maximum Power Transfer Theorem is not so much a means of analysis as it is an aid to system design. Simply stated, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power. If the load resistance is lower or higher than the Thevenin/Norton resistance of the source network, its dissipated power will be less than maximum.
This is essentially what is aimed for in radio transmitter design , where the antenna or transmission line “impedance” is matched to final power amplifier “impedance” for maximum radio frequency power output. Impedance, the overall opposition to AC and DC current, is very similar to resistance, and must be equal between source and load for the greatest amount of power to be transferred to the load. A load impedance that is too high will result in low power output. A load impedance that is too low will not only result in low power output, but possibly overheating of the amplifier due to the power dissipated in its internal (Thevenin or Norton) impedance.
Taking our Thevenin equivalent example circuit, the Maximum Power Transfer Theorem tells us that the load resistance resulting in greatest power dissipation is equal in value to the Thevenin resistance (in this case, 0.8 Ω):

With this value of load resistance, the dissipated power will be 39.2 watts:

If we were to try a lower value for the load resistance (0.5 Ω instead of 0.8 Ω, for example), our power dissipated by the load resistance would decrease:

Power dissipation increased for both the Thevenin resistance and the total circuit, but it decreased for the load resistor. Likewise, if we increase the load resistance (1.1 Ω instead of 0.8 Ω, for example), power dissipation will also be less than it was at 0.8 Ω exactly:

If you were designing a circuit for maximum power dissipation at the load resistance, this theorem would be very useful. Having reduced a network down to a Thevenin voltage and resistance (or Norton current and resistance), you simply set the load resistance equal to that Thevenin or Norton equivalent (or vice versa) to ensure maximum power dissipation at the load. Practical applications of this might include radio transmitter final amplifier stage design (seeking to maximize power delivered to the antenna or transmission line), a grid tied inverter loading a solar array, or electric vehicle design (seeking to maximize power delivered to drive motor).
The Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.
Similar to AC power distribution, high fidelity audio amplifiers are designed for a relatively low output impedance and a relatively high speaker load impedance. As a ratio, "output impdance" : "load impedance" is known as damping factor, typically in the range of 100 to 1000.
Maximum power transfer does not coincide with the goal of lowest noise. For example, the low-level radio frequency amplifier between the antenna and a radio receiver is often designed for lowest possible noise. This often requires a mismatch of the amplifier input impedance to the antenna as compared with that dictated by the maximum power transfer theorem.

• REVIEW:
• The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power.
• The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency.

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Thevenin-Norton equivalencies Step By Step Procedure/Explanation

Since Thevenin's and Norton's Theorems are two equally valid methods of reducing a complex network down to something simpler to analyze, there must be some way to convert a Thevenin equivalent circuit to a Norton equivalent circuit, and vice versa (just what you were dying to know, right?). Well, the procedure is very simple. .......
You may have noticed that the procedure for calculating Thevenin resistance is identical to the procedure for calculating Norton resistance: remove all power sources and determine resistance between the open load connection points. As such, Thevenin and Norton resistances for the same original network must be equal. Using the example circuits from the last two sections, we can see that the two resistances are indeed equal:..........







Considering the fact that both Thevenin and Norton equivalent circuits are intended to behave the same as the original network in suppling voltage and current to the load resistor (as seen from the perspective of the load connection points), these two equivalent circuits, having been derived from the same original network should behave identically. .......
This means that both Thevenin and Norton equivalent circuits should produce the same voltage across the load terminals with no load resistor attached. With the Thevenin equivalent, the open-circuited voltage would be equal to the Thevenin source voltage (no circuit current present to drop voltage across the series resistor), which is 11.2 volts in this case. With the Norton equivalent circuit, all 14 amps from the Norton current source would have to flow through the 0.8 Ω Norton resistance, producing the exact same voltage, 11.2 volts (E=IR). Thus, we can say that the Thevenin voltage is equal to the Norton current times the Norton resistance: .......

So, if we wanted to convert a Norton equivalent circuit to a Thevenin equivalent circuit, we could use the same resistance and calculate the Thevenin voltage with Ohm's Law.
Conversely, both Thevenin and Norton equivalent circuits should generate the same amount of current through a short circuit across the load terminals. With the Norton equivalent, the short-circuit current would be exactly equal to the Norton source current, which is 14 amps in this case. With the Thevenin equivalent, all 11.2 volts would be applied across the 0.8 Ω Thevenin resistance, producing the exact same current through the short, 14 amps (I=E/R). Thus, we can say that the Norton current is equal to the Thevenin voltage divided by the Thevenin resistance:
.......
This equivalence between Thevenin and Norton circuits can be a useful tool in itself, as we shall see in the next section. .......

• REVIEW:.......
• Thevenin and Norton resistances are equal.
• Thevenin voltage is equal to Norton current times Norton resistance.
• Norton current is equal to Thevenin voltage divided by Thevenin resistance.

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