Sometimes the best way to find an unknown, is to attack the unknown with a black box.
This issue we continue attempting to fill your circuit analysis toolbox with methods used by electricians, electronics and electrical engineers. Of course we will be taking the least bumpy ride we can manage without making it completely Sesame Street!
While compound circuits will be the aim of the exercise, we will also look at techniques for the easiest of the complex circuits, as well as give you something to work on.
To recap from last month and other previous issues of DIYODE Magazine, common tools for circuit analysis are Ohm’s Law, Kirchhoff’s Laws (KVL and KCL) and the Power Formulae. VIRP (i.e., EIRP) tables are useful for beginners as they force you to think of all options, and develop good practices for series and parallel circuits.
Compound circuits can be solved with VIRP tables, requiring a little more thought about whether two (or more) components are actually in series or parallel. If you’re a beginner, as your skillset develops, it is recommended that you move on from using VIRP tables.
NODES AND LOOPS
A simple circuit has one loop called the “path”, which is the circuit. The loop has two nodes: positive and negative.
A series circuit still has one loop or path, and more than two nodes (i.e., one between each pair of components).
A parallel circuit has only two nodes and all components are between those two nodes.
A compound circuit has more than one path and more than two nodes. It features both series and parallel connected components, and can be broken into parts that can be solved individually, using equivalent values to replace series or parallel components.
A complex circuit features multiple voltage and/or current sources that cannot be solved with Ohm’s Law alone.
Circuit analysis is a process of mathematically evaluating all the values of all the components in a circuit, including VIRP, voltage current, resistance and power.
A circuit is typically converted to an equivalent circuit containing only pertinent components, and circuit values required for analysis. Labels, arrows and symbols are used to clarify the circuit.
To recap how a compound circuit can be solved, and sometimes a complex circuit may be simplified [1A], you should redraw it without the switch or ammeter and replace the battery with a DC source symbol (an arrow to remind you of the polarity). Also include an arrowhead to show the current direction. Re-read the article in Issue 11, if you need to remind yourself what we discussed.
Alternatively, we have also illustrated just what is required for the calculations [1B]. Note that R3 and R4 are in parallel. By definition they have the same two nodes and no other components across those nodes. Therefore, they may be replaced by a single resistor, [1C] by R3//R4 with a value calculated to be 16Ω. You could call it R5, or “Rex” if you want to, but changing the name can result in confusion, so we recommend replacement component names relate directly to their previous parts.
Now R2 is in series with R3//R4. We know that because they share one node between them and no other components share that node. Therefore they have the same current, and by definition are in series. So R2 is added to R3//R4 equivalent value; 12 + 16 = 28 Ohms.
By this time, you will hopefully see what to do next.. R1 is in parallel with (R2+R3//R4) [1D]. (Remember that the '//' symbol takes precedance, just as 'x' does over '+'.) They have the same two nodes and no other... but what about the battery?
The good news is that we replaced the battery with a voltage source, which has infinite resistance when it comes to voltage, and zero resistance when it comes to current; but we’ll get to that in a few more paragraphs.
Ignoring the voltage source for now, the circuit formula has become RT = R1//(R2+R3//R4), which I understand is scary for people who slept through algebra. The equivalent resistance, a.k.a. the Total Resistance (RT) becomes RT = 1/(1/36 + 1/28) = 15.75Ω.
We can use the equivalent resistance to calculate total current from IT = ET / RT = 1.524A.
REDEFINING A SOURCE
Just a few lines ago we alluded to some special treatment of the component we call the “source”. Remember a circuit consists of the source, the path and the sink (load). So for analysis, the source would be a pure source, with no path and no load.
In fact a perfect voltage source would have a fixed voltage value that does not change under load. A 12V car battery would remain at 12V whether the engine is cranked or the battery is charged. Of course it does not, and the reason is because all real components have imperfections. Batteries have an internal resistance, for example.
For a battery, which is intended as a voltage source, there is the internal voltage generated by the chemistry; but a real battery also has several resistances – the resistance of the terminals and the battery structure; the resistance of the electrolyte; cell interconnections; and perhaps other internal chemistry, which are all in series with the battery cells; and the actual source. There are also parallel resistances, but we can ignore them for now as they have too high a resistance to matter to our circuit.
The circuit for a battery, therefore, is a voltage source with a series resistor. The circuit [2A] is a simple 1.5V cell (battery) supplying a 15Ω load resulting in a 100mA current draw from the battery.
As the load remains applied to the battery however, we all expect that eventually the battery will go flat! So does that mean it melts into the carpet? What is meant by “going flat”? The chemistry will produce the same voltage regardless of the state of charge. Chemists out there will understand why.
As the chemical reaction is almost completed, the usually ignored internal resistance increases dramatically, until we consider the battery cell to be completely discharged, or flat.
At TAFE, when we had pre-vocational students, every electrical class was given an experiment to perform requiring them to watch a battery go flat, and record its vital signs as it died. The circuit was similar to [2A] but added a voltmeter across the load resistor, which is actually the same nodes as the battery. The voltage is called the “terminal voltage”.
You may notice in the circuit analysis version [2B] we have added a pink box, representing the battery case, with a voltage source symbol and a resistance within the box. The resistance (Ri) represents the internal resistance.
We could have placed an ammeter in series but there is no need, as any voltage below the voltage of a fully charged cell is a result of the internal resistance, and the current can be calculated from the terminal voltage divided by the load resistance. The voltmeter can actually be used as the ammeter as well.
A datalogger would be nice, but the students needed something to keep them busy.
You might like to use an Arduino to log the terminal voltage and draw a chart from the readings, using the serial link.
I suggest using cheap batteries for the experiment as the experiment will take less time to complete!
Just to give you an example of what to expect, a 1.5V AAA Energizer cell was connected to a multimeter set on voltage. The terminal voltage, unloaded, was 1.612V. It was in better condition than I imagined. Adding a 10Ω resistor across the terminals caused the terminal voltage to drop to 1.536V, a difference of 76mV, for a current of 153.6mA (1.536V/10Ω). Therefore, the internal resistance of the battery cell was approximately 0.5Ω – it was in a good condition!
Of course, if I had left it there for an hour or so, the voltage and current would both decrease, and the internal resistance would rise, giving us the load curve of the battery cell, and the changing internal resistance.
The idea of the Black Box is that a circuit can be replaced by another circuit, inside a Black Box with two terminals. The avid circuit analyser doesn’t care what is inside the box, as long as it behaves the same as the original circuit. In fact we don’t even need an original circuit, just a circuit that behaves exactly as the unknown one in the black box.
Thevenin’s Theorem states “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across a load.”
Thevenin’s Theorem recognises the internal resistance of a battery or other generation device, and uses the series source and internal resistance circuit as the Thevenin Equivalent Circuit of a battery, or any voltage source. The internal resistance is also known as the Thevenin Resistance (Rth).
The Thevenin resistance is found (in theory) by dividing the open circuit terminal voltage by the short circuit current, which is not recommended for larger batteries or generators. A better, more practical, and much safer method would use two resistor loads, one at a time, and plot or calculate the short circuit current, although the concept is the same:
Rth = Voc/Isc.
You might try this with students, but only short the AAA cell long enough to read the current, as it will get hot over time.
Some battery testers measure cell condition by the Thevenin method administered by a microprocessor, such as an ATtiny, to measure Voc, then briefly short the cell with a transistor while measuring current. Ri = Rth is then calculated.
Thevenin’s Theorem however, allows multiple sources and internal components to be replaced by a single source and Thevenin resistor.
Thevenin’s equivalent circuit is a useful tool for analysing circuits with multiple sources [3A]. Here we have a common circuit which some refer to as a simple complex circuit, but others do not consider to be complex. It is complex enough for many electrical students, and Thevenin is one means of solving it.
Solving the circuit requires a means of calculating the current supplied to the 6Ω resistor from each source. There are other means to achieve this, but one method, using Thevenin, simply converts the two sources into one Thevenin Equivalent.
A modified layout of the same circuit [3B] and components, shows how the two sources are actually in parallel to the 6Ω resistor.
The Thevenin resistance is now equal to R1//R2 = 0.8//1 = 0.444Ω.
The Thevenin voltage is a little more complex. It is related to the voltage divider and can be found without the formula by treating R1 and R2 as in series between the two battery terminals.
However, using a supplied formula:
Vth = (V1R2 + V2R1)/(R1 +R2) = (12x1 + 10x0.8)/(0.8 + 1) = 11.11V.
This transformation replaces the two sources with a single Thevenin equivalent circuit [3C].
The Load current, IL = Vth/(Rth + R3) = 11.11/(0.444 + 6) = 1.724A.
The load voltage, VL = Iload x R3 = 1.724 x 6 =10.35V.
To check these values, the current supplied by the 12V source is determined from:
I1 = (E1-VLoad)/R1 = (12 – 10.35)/0.8 = 2.063A, more than the load takes.
The 10V source supplies: I2 = (E2-VLoad)/R2 = (10 – 10.35)/1 = -0.350A, minus current.
I3 = I1 + I2 = 2.063 + (-0.350) = 1.713A
Okay, so I rounded off and missed the answer by 11mA! What this proves is that Thevenin can be used to analyse circuits with multiple sources, sometimes the current in a battery flows negative (i.e., charging the battery), and most importantly, if you want your answers to check out exactly, don’t round off your figures!
Actually, as a teacher I would have started with the answers and calculated the values to give me nice round numbers, but I’m not a teacher anymore!
Another means to analyse this circuit uses an alternative “Black Box” technique known as Norton’s Theorem.
Using the same circuit , we can analyse it using Norton’s Theorem, and Norton’s Equivalents .
Norton’s Theorem states “Any linear circuit containing several voltages and resistances can be replaced by just one single current source in parallel with a single resistance connected across a load.”
A constant current source is one that has a limited short circuit current (Isc). This can be measured safely as with the Thevenin Isc, but the true Isc has to be calculated as before.
The open circuit voltage is also calculated with the Thevenin circuit, and the resistance is calculated the same for both: Rt = Rn (Norton Resistance) = Voc/Isc.
So you might now ask, “If it is all the same, how is it different?” I’m pleased you asked!
The Norton Theorem uses a current source that supplies the same maximum current as the short circuit current in the Thevenin circuit; but the Norton Resistance is wired in parallel with the constant current source.
Most of us are not used to dealing with current sources, so here are a few concepts for you to consider.
If there is no load resistor, all of the current will flow through the Norton resistor, (i.e., the same current as would flow in a dead shorted Thevenin circuit). If you look at a shorted Thevenin circuit, and an open circuit Norton circuit, they are the same.
Of course, a 0Ω load resistor, as an external dead short, would take all of the current, and as load resistance increases, the terminal voltage increases up to the value it would be across the Norton resistance (i.e., an open circuit). In fact, the Thevenin circuit and the Norton circuit are equivalent to each other. The Thevenin resistance is in series with the voltage source, and the Norton resistance is in parallel with the current source. The resistance value is calculated the same for each, and is the same for both. They are simply two ways to look at the same black box! So yes, that means you can replace one with the other.
Now to analyse our sample circuit [3A] using Norton’s Equivalents [4A]. The batteries and their internal resistances are replaced [4B] with their Norton’s Equivalents, both of which are now shown together to the left of the diagram.
The first task is to calculate the constant current of each source under short circuit conditions. For what was E1, I1 = E1/R1 = 12/0.8 = 15A, and for E2, I2 = E2/R2 = 10/1 = 10A.
Before you start looking for a fire extinguisher, this is inside the black box for now, and we are not concerned with what seems to be a large current. Indeed, to calculate the Norton’s Equivalent Current Source, combining I1 and I2, which are obviously in parallel, results in In = I1 + I2 = 25A.
The Norton’s Equivalent resistance is also obviously Rn = R1//R2 = 0.8//1 = 0.4444Ω, making a circuit to show the new Norton Equivalent circuit [4C].
Before we calculate the voltage, or current in the circuit, we need to remember it is a circuit supplied by a constant current source. So first we need to calculate the total resistance of the circuit, from Rn//R3 = 0.4444//6 = 0.41375Ω.
Note: I’m being more careful with rounding off for this circuit!
The circuit is a parallel circuit, so the voltage is the same across every component or source, but we are particularly interested in the voltage across R3, which will be V3 = In x RT = 25 x 0.41375 = 10.3439V.
The current through R3 therefore is I3 = V/R3 = 10.3439/6 = 1.724A, which is the same answer that Thevenin gave us!
The current through R1 is still I1 = (E1-V3)/R1 = (12 – 10.3439)/0.8 = 2.07A.
The current through R2 is still I2 = (E2-V3)/R2 = (10 – 10.3439)/1 = -0.3439 Amps.
I3 = I1 + I2 = 2.07 + (-0.3439) = 1.726 Amps. OK, closer but still 2mA difference. Now I can’t vouch for other colleges, but when I marked exams, correct to three places was considered okay!
Just to recap, Thevenin's Theorem and Norton's Theorem are two points of view on the same concept. A voltage source in series with a resistor and a current source in parallel with a resistor, as a simple circuit is exactly the same circuit when Thevenin's circuit is shorted out, and Norton's circuit is open circuit.
The voltage source and the current source are both energy sources. The resistor in the Thevenin circuit limits current to a maximum value with the terminals short circuited. That same value resistance in the Norton circuit limits voltage to a maximum value when the terminals are open circuit.
To replace the Thevenin Black Box with a Norton Black Box, the internal resistance does not change. The current source in the Norton circuit is calculated from the simple Ohm's Law formula:
In = Vth/Rth
and conversely the Thevenin voltage source is calculated from the same formula, transposed and relabelled:
Vth = InRn
remembering that Rth = Rn.
Which one you use depends on the circuit you are analysing. Generally, Thevenin is preferred when the first component is in series with the load, and Norton when the first component is in parallel.
When there are multiple sources in parallel, as with figures 3 and 4, you might agree that Norton's theorem was the easiest to use. The current sources were simply added together. However, when several sources are in series, Thevenin's Theorem is easiest as the voltage sources are simply added.
Remember that you can swap Black Boxes that have the same behaviour when shorted or open circuited, whenever they are purely resistive.
Other Black Boxes may have three terminals, such as the Star/Delta transform that converts three resistors in a triangle connection into three resistors connected between the terminals and a centre point, i.e. a fourth node not being one of the terminals.
In electronics there are four terminal Black Boxes known as “2-Port” networks, and maybe we can look at them in another classroom.
Looking at the clock, or the word count, that’s enough for this lesson. Just one thing I promised that was still outstanding…
THE CUBE PROBLEM
In the Issue 10 we presented the cube problem, which is a common exercise for science and engineering students when beginning circuit analysis. The task is to calculate the total resistance of 12 resistors formed into a cube, such that each edge is represented by a 1kΩ resistor.
Each corner of the cube is a joint between three resistors, and each face has four resistors as it’s square perimeter. If an Ohm-meter is placed across diametrically opposite corners, what will the total resistance be?
If you have not tried your luck yet, don’t read on until you have! I don’t want to tell you the butler did it, just in case you didn’t guess!
To solve the puzzle, you only need a few minutes and suitable circuit analysis strategies. My method, before I studied the engineering subject, was very simple, but can only be proven via the mathematical methods: superposition, Thevenin or Norton source replacement, star/delta (star/mesh) transform, simultaneous equations, and/or matrix mathematics – these are all great tools for your circuit analysis toolbox, but a bit heavy going for most makers.
So the “Sesame Street”, or “Bush Engineer” method is to realise that to get from one corner to the diagonally opposite corner, current must pass through three resistors, dividing equally into three, then equally into six at the second set, and combining equally back into three at the far end. If three 1kΩ resistors were in parallel, then the total would be 1kΩ/3; and for six in parallel, the resistance would be 1kΩ/6; and finally another 1kΩ/3 (the same as the first end) makes 5/6kΩ, or 833Ω total, as the answer. There’s nothing like bush logic!