**July 30, 2008 Educational Radio Net, PSRG 10th session**

This session is the 9th in the impedance series. Given that impedance is the combination of reactance and resistance and, further, that reactance is an alternating current phenomenon it is clear that we must have some elemental definitions under our belts to fully appreciate the subject. This multi-part narrative series is an attempt to elevate participants to an intuitive level of electrical understanding without using any serious mathematics as well as provide some review for those of us who have not spent a lot of time on fundamentals lately.

Thus far we have talked about electrical current, voltage, power, resistance, Ohm's Law, DC or direct current, and AC or alternating current. Subsequent parts of the series will introduce capacitance, inductance, then reactance, and, finally, impedance as the combination of resistance and reactance. All discussion material will be reviewed continually and be available on the blog.

Let's review what has been covered up to this point in the series.

Part 1 developed the idea of electrical current consisting of moving charge and defined the ampere as 1 coulomb of charge moving past a fixed point in 1 second. One coulomb was defined as a collection of charge numbering 6.24 x 10^18 electrons.

Part 2 developed the notion of mechanical "work" and considered objects at different "potential" levels in a gravitational field. The concept of "voltage", also known as electrical potential difference, and the relationship of voltage to current follows closely with the idea of a mechanical weight being moved between different levels. In both cases work is being done and energy is being manipulated in various ways.

Part 3 capitalized on Bob's lightning series to review electrical current in the context of a charged cloud redistributing charge in the form of lightning where modest amounts of charge make a large impression if moved rapidly.

Part 4 developed the notion of potential difference and ended with a definition of voltage. If you move 1 coulomb of charge from point A to point B in an electric field such that 1 joule of work is done then the potential difference between points A and B is defined as 1 volt. Another way to state this is that 1 joule of energy is required to push 1 coulomb through a potential difference of 1 volt.

Part 5 developed the notion of power by using a mechanical analogy. Power is the relationship between energy and time. Specifically power is the change in energy as in work done divided by the change in time to do the work. Conversely, energy is power multiplied by time.

Part 6 developed the notion of resistance by using a simple circuit to compare how well various materials conduct electrical current. We looked at a simple series circuit with fixed voltage, one D cell battery, a fuse, an ammeter, a switch, and a pair of DUT terminals as in Device Under Test. Substituting various materials across the DUT terminals yielded different measurements on the ammeter and we ranked these materials based upon their "conductance". Finally, we learned that resistance and conductance are reciprocals and that high conductance equals low resistance and vice versa.

Part 7 developed the notion of Ohm's Law by using a simple series circuit to illustrate the relationship of voltage, current, and resistance. Ohm's Law states that electrical current through a resistive device is directly proportional to the voltage across the device so, for example, doubling the voltage across the device will double the current through the device. This relationship stated in math terms is I (which is the symbol for current) equals E (the symbol for voltage) divided by R (the symbol for resistance).

Part 8 developed the notion of direct current and alternating current by using a sand filled tube with a scribed fiducial mark. By assuming that the sand particles represented electrons we could watch the action at the mark and deduce if the current, or moving electrons, was AC or DC.

Part 9, tonight's edition, will again contrast direct current and alternating current by measuring the temperature of a resistor with a suitable thermometer and then forming conclusions from observed data.

Ok, on with the heating effect of direct current and alternating current.

We will revert to our simple series circuit to study resistive heating or what is also known as joule or Johnson heating. Let me assert that electrical energy delivered to a resistive circuit by battery or power supply ultimately ends up dissipated as heat energy when all is said and done. As everyone knows heat energy can be measured with a thermometer. Let's select a 100 ohm resistor with a heat dissipation rating of 2 watts. Now attach a thermometer to this resistor in such a fashion that we can collect accurate readings of the resistor temperature. The thermometer could be liquid in glass or, perhaps, a thermocouple device. Remembering that our simple circuit consists of a variable voltage power supply, fuse, ammeter, switch, DUT terminals, and now a voltmeter in the form of a calibrated oscilloscope across the DUT terminals. Let's attach the resistor with associated thermometer device across the DUT terminals. So, we have a fixed 100 ohm resistor attached to the DUT terminals and the switch is off.

Now adjust the variable power supply voltage to +10 volts DC and move the switch to on. Note that the voltmeter in the form of an oscilloscope trace moves upscale to the +10 volt line. Ohm's Law reigns supreme here so we predict that 10 volts DC divided by 100 ohms equals 1/10 ampere or 100 milliamperes. Sure enough the simple circuit ammeter indicates 100 ma. Taking a look at the thermometer we see the resistor temperature rising from room temperature to a much higher value. After a few moments the resistor temperature comes to equilibrium and the thermometer reading is steady. At this point we log the temperature reading and the oscilloscope voltage reading and compute the actual energy dissipation in the resistor by applying Joule's Law wherein we square the current and multiply by the resistor value. So, 1/10 ampere squared times 100 ohms is 1 watt. Our resistor will not burn up since we selected a resistor rated at 2 watts dissipation initially. From this bit of circuit work we conclude that a constant DC voltage will cause the test resistor to heat up to some temperature value and stay at that value as long as the constant voltage is applied to the circuit.

Now, turn the switch off and let the resistor cool down to room temperature. We will make a simple change to our circuit by replacing the DC power supply with a variable voltage AC supply which produces a sine wave output. After the change is made we repeat the above measurements using the variable voltage AC supply. We start with the AC voltage set to zero. Turn the switch on and we notice that nothing happens and the ammeter indicates zero current, the oscilloscope voltmeter shows zero volts, and the thermometer on the resistor indicates room temperature.

Now we advance the AC power supply voltage control such that the peak of the sine wave just touches the +10 volt line on the oscilloscope and things start to happen. The thermometer moves upscale and quickly comes to equilibrium but the indicated temperature is much lower than the reading obtained in the previous DC voltage exercise and the ammeter only shows 70.7 milliamperes. Clearly something is going on here so we slowly advance the voltage control and notice that the resistor temperature continues to rise. Eventually we find the point where the resistor temperature is identical to that which we measured in the earlier DC voltage exercise. Looking at the scope we notice that the peak voltage is 14.14 volts and the ammeter indicates 100 milliamperes.

In terms of the heating capacity we must conclude that 10 volts DC and 14.14 volts peak AC will heat the resistor to the same temperature hence perform the same amount of work. So how do we process the AC voltage sine wave to account for the difference between DC and AC? There is a math process called root mean square, or more definitive, root of the sum of the means squared and, if you apply this process to the 14.14 peak voltage waveform, the answer turns out to be 10 volts rms. So, 10 volts rms in the AC world is the heating equivalent to 10 volts in the DC world. This is only true for AC sine waves. Other waveforms require special treatment to determine equivalency. Fortunately most naturally oscillating voltage sources are sine wave so it is a simple matter to determine the RMS value. Identical DC and AC rms voltages will behave exactly the same with regard to Ohm's Law. Peak AC sine wave voltage times 0.7070 will yield the rms voltage every time.

In summary, if the AC voltage waveform is a sine wave then the rms voltage of that waveform will produce the same heating effect as a DC voltage of the same magnitude. AC rms is AC peak times 0.7070. Most meters, V-O-M or volt/ohm/milliammeter for instance, are calibrated assuming that the waveform being measured is sinusoidal. There are special true rms meters which will measure any complex non-sinusoidal waveform and report correct rms values.

This concludes the set up discussion of AC rms volts vs DC volts. Are there any questions or comments?

Now I have a challenge question for those interested. This question is intended to demonstrate scale. We have been using a coulomb of sand to represent a coulomb of electrons. We know that a coulomb is 6.24 x 10^18 electrons. Given that a cubic centimeter of sand contains about 3000 sand particles figure out how many cubic centimeters would be required to contain 6.24 x 10^18 sand particles. Secondly, express the answer in the number of cubic yards to contain 6.24 x 10^18 particles, and, thirdly, express the answer in the number of cubic miles required to contain this number of particles. What does this tell you about the real size of the electron?

Given, everything you need to know:

One cc of sand contains 3000 particles (more or less depending on sand particle size but we will assume this to be the average)

one coulomb contains 6.24 x 10^18 particles

one inch = 2.54 centimeters

one yard = 36 inches

one mile = 5280 feet

You may enter your answer in the blog comments or email me at N7KC@comcast.net.I will provide the answers next week.

This is N7KC for the Wednesday night Educational Radio Net

## 3 comments:

I'll venture an answer to the challenge (thanks to Excel):

I got 2.08x10^15 cc sand per coulomb.

2,720,537,288 cubic yards of sand/coulomb.

Half a cubic mile of sand/coulomb.

KE7RJI

i amvery intrested in impedence now you are to thank!!!!!!!!!!

ke7suh

sam the ham radio man

I got the same thing as KE7RJI for the first two, but something different for mi^3. Maybe I flipped a bit somewhere. I was just using an ancient TI-85 instead of Excel so that can be my excuse.

2.08x10^15 cc / 1C of particles

2.72x10^9 yd^3 / 1C of particles

6.84x10^-4 mi^3 /1C of particles

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