IMPEDANCE SERIES PART 12, Lee week 16

September 10, 2008 Educational Radio Net, PSRG 16th session

This session is the 12th 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, resistance, Ohm's Law, power, DC or direct current, AC or alternating current, Joule’s law, Kirchoff’s 2 circuit laws, and introductory capacitance. This 12th part of the series will expand on capacitance, introduce the idea of reactance, and define impedance for the first time. Subsequent parts of the series will introduce inductance, inductive reactance, and, finally, impedance as the combination of resistance, capacitive reactance, and inductive 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 contrasted direct current and sinusoidal alternating current by measuring the temperature of a resistor when subjected to the same maximum voltage from each waveform. The conclusion was that equal values of DC voltage and AC rms voltage, if impressed across a resistor in turn, will produce the same heating effect, or work, in that resistor hence are equivalent. Heat produced as a consequence of current through a resistance is called Joule heating. Energy losses such as this are sometimes called Johnson losses as well.

Part 10 reviewed Ohm’s law and restated the concepts from part 9 in a manner called Joule’s law wherein energy is associated with time to define power and a variable substitution from Ohm’s law produces the familiar P = (E^2)/R formulation. Additionally, the very important Kirchoff’s voltage and current laws were introduced.

Part 11 introduced the concept of capacitance. The relationship of charge denoted by symbol q, capacitance denoted by symbol C, and voltage denoted by symbol V is simply q=CV. The unit of capacitance is the farad which is defined as 1 coulomb per volt. Given that one farad is a very large unit we normally express capacitance by micro-farads, nano-farads, or pico-farads.

Part 12, tonight’s edition, will introduce the concept of reactance and combine circuit resistance with capacitive reactance to form impedance which represents the total opposition to electrical current flow and which is denoted by the symbol Z.

Ok, let’s continue with the very sophisticated idea of electrical capacitive reactance.

First, let’s set up the scenario as follows. Imagine a simple three element series circuit containing a source of alternating voltage, a single resistor, and a single capacitor. Assume the signal source to be some sort of generator connected in series with a 1000 ohm resistor which is, in turn, connected in series to a 0.1 micro-farad capacitor. Let’s further say that the variable frequency signal source can output 12 volts rms and support a 12 watt load without being strained. We need to measure circuit current so we will insert a series current metering device… one with visual output such as an oscilloscope… which is perfect in the sense that it does not change the circuit performance in any way. Additionally, we will want to measure voltage across both the resistor and capacitor so we include a perfect visual voltmeter for that purpose such as an oscilloscope. One last thing we want to measure is temperature so we will use a thermocouple meter which is not attached to our circuit in any fashion. The demonstration circuit is now complete.

Secondly, lets consider electrical current flow in a series circuit. Each connection point… generator to resistor, resistor to capacitor, and capacitor back to the generator can be thought of as a node or junction if you please. From Kirchoff’s current law we know that the current leaving a junction must equal the sum of currents entering that junction so, given that our junctions have no branches, when we apply power to the series circuit the resulting current is the same everywhere in the circuit.

Let’s energize the circuit and set the generator to produce a 1 KHz sine wave signal and set the output voltage to 14.14 volts peak. This corresponds to 10 volts rms.

The first measurement we will make is associated with the resistor. Looking at the voltage and current waveforms we note that both voltage and current reach their peak values at the same time and they move in lock step. They are said to be in "phase" as in "time". The driving waveform is a voltage sine wave so the resulting current waveform is also a sine wave. The visual current measuring device indicates that the circuit peak current is about 7.5 milliamperes or about 5.3 milliamperes rms. The thermocouple probe placed near the resistor indicates that the resistor is warmer than room temperature. Clearly the capacitor in the circuit is having some effect on the current flow since, if it were replaced by a short, the current would have been 14.14 ma peak or 10 ma rms according to Ohm’s Law and the thermocouple meter would have shown the resistor to be somewhat warmer. The main point here is that the voltage across the resistor and the current through the resistor are in phase and that the resistor heats up.

The second measurement is associated with the capacitor. Since the current through the capacitor is identical to the current through the resistor… it is a series circuit after all… we just move the voltage sensing probes to the capacitor and suffer a huge surprise. The current and voltage are not in phase but rather are 90 electrical degrees apart. Placing the thermocouple probe near the capacitor indicates no change in temperature from the room even though circuit current is flowing. Given that there are 360 electrical degrees in one cycle of alternating current or voltage then 90 degrees represents ¼ cycle displacement in time or more properly a phase shift of ¼ cycle. Now, if you take two sine waves of the same frequency and shift them along the time axis by ¼ cycle you will notice that the peak of one corresponds to the zero crossing of the other. From Joule’s Law we know that power is the product of current and voltage but if one or the other is zero then the power dissipated as a result of energy change per time is zero. The constant capacitor temperature confirms that no energy is lost in the capacitor but clearly the displacement current is moving freely back and forth through the capacitor.

So, what do we know at this point? If the capacitor were not in the circuit then the current would have been larger by almost a factor of two. The resistor heated up but the capacitor did not so the resistor is forcing an energy change per time and the capacitor is not. The voltage and current associated with the resistor are in phase but the current and voltage associated with the capacitor are 90 electrical degrees apart or in "quadrature". The peak voltages across the resistor and capacitor are never maximum at the same time. Clearly the presence of the capacitor caused the circuit current to decrease.

This artifact of decreasing current as a result of the capacitance effect is given the name reactance, in this case capacitive, denoted by the symbol X, and given the unit of ohms. In our particular circuit being driven by a voltage with frequency of 1000 Hz and a capacitor of 0.1 uF the capacitive reactance is about 1592 ohms. In fact, reactance is inversely dependent on frequency so as frequency increases then capacitive reactance decreases. In a capacitor, energy is stored during charging and returned to the circuit... then reversed charged and returned to the circuit... in contrast to the resistor where energy is converted to heat constantly. Resistance and reactance, when properly combined, yield impedance which is the measure of total opposition to electrical current flow.

Ohm’s Law is generalized by replacing R, resistance, with Z, impedance, and is applicable to both AC and DC circuits. Impedance, or Z, is easily calculated by using the Pythagorean Theorem wherein the hypotenuse squared is equal to the sum of the squares of the legs of a right triangle. Resistance is plotted along the horizontal axis and reactance is plotted along the vertical axis. These two axes represent sides of a rectangle generally and sometimes a box specifically. Complete the rectangle or box then draw a diagonal line from the origin to the point opposite in the figure. The length of this line, or hypotenuse, represents the magnitude of impedance as represented by the resistance and reactance sides of the figure. In our particular circuit the resistance R is 1000 ohms, the reactance X is about 1592 ohms and the combination of the two yields an impedance of about 1880 ohms.

There is one more important point to make. The angle of the hypotenuse, or impedance line, with respect to the horizontal, or resistance line, is the overall phase shift of the circuit due to the combination of resistance and reactance and is called theta from the Greek alphabet. Although the phase shift associated with the resistor itself is zero and the phase shift associated with the capacitor is 90 electrical degrees the combination of the two values, resistance and reactance, will produce an overall circuit phase shift someplace between zero and 90 electrical degrees. Theta in our particular circuit is 1592 divided by 1000 then find the angle whose tangent equals that value or about 58 degrees. Reactance as a result of capacitance produces a "leading" phase angle meaning that circuit current leads the voltage. If resistance is much larger than reactance then resistance predominates and the phase shift is small but always leading. Conversely, if the reactance is much larger than resistance then the reactance dominates and the phase shift can be large and leading.

You have just suffered the garden path explanation of reactance. In fact there is an elegant mathematical approach to this subject which is beyond the scope of our net. Many of you may be equipped with the tools to investigate this subject in more detail and in a more rewarding manner.

This concludes the set up for the discussion of reactance associated with capacitance. Are there any questions or comments?

Terminology
Apparent power: The multiple of voltage times current without regard to phase angle.

True power: The multiple of voltage times current times cosine of the phase angle.

Power factor: True power divided by apparent power. The degree to which current and voltage are out of phase. A power factor of 1 indicates a purely resistive circuit wherein current and voltage are in phase. If current and voltage are in phase then the phase angle is zero electrical degrees. The cosine of zero degrees is 1 hence the power factor is defined as 1. Conversely, a phase shift of 90 electrical degrees indicates a purely reactive circuit with cosine of 90 electrical degrees equal to zero.

The last challenge question was as follows: you plan to install a 910 ohm resistor in a circuit and you know that the direct current through this resistor will be 125 milliamperes. How many watts of power will the resistor dissipate? 14.22 watts How many joules per second? From the definition… 1 joule equals one watt-second hence 14.22

This weeks challenge question: if you have equal values of resistance and reactance what is the overall circuit phase angle?

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

This is N7KC for the Wednesday night Educational Radio Net