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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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4th Mar 2018, 12:16 pm | #21 | |
Dekatron
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Re: Correct units for Reactance/Impedance Calculations
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4th Mar 2018, 12:23 pm | #22 | |
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Re: Correct units for Reactance/Impedance Calculations
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Surely there aren't any "mind-bogglingly weird concepts behind the -j bit" ? Start with the definition of j [=√(-1)] and then simply show that j² = -1 as a consequence: the concept of -j then simply arises from that: it's trivial to show that (-j)² = -1. The "mind-boggling" stuff begins when the ramifications of j are taken to ever-increasing depths, not at its introductory stage. As for its use as a handy tool, I learnt that j can be explored in two related areas: within the realm of pure mathematics (academic) or within physics and engineering (practical). The greater the depth of the latter, the greater the need for comprehension in the former. But such a requirement is not limited to j, (or -j), is it? For example, the deeper one delves into physics, the need for enhanced mathematical ability increases in all areas on mathematics, not only j (and -j). Al. |
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5th Mar 2018, 1:40 pm | #23 |
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Re: Correct units for Reactance/Impedance Calculations
'j' just comes at the end(?) of a sequence of new numbers, forced on us by discovering holes in the numbers we previously knew.
We know about positive integers - we use them to count objects. We can add two of them to get another one. Then someone invented subtraction. 5-3 is OK, we know about 2. What about 3-5 or 5-5? So we had to invent negative numbers and zero. What does -2 bananas look like? Along came multiplication, which was basically just repeated addition. 5x3 is 15; we already know about 15. We can even do -5x3 = -15. No new numbers needed. Then someone wondered about undoing multiplication, so we then had division. 15/5 is 3; we know about 3. What about 16/5? A bit tricky. We started by saying that it is 3, with remainder 1. Then someone said why not just write it as 3 1/5? We then had rational fractions. We discovered that 3 1/5 is a number (you can do addition, subtraction, multiplication, division with it) but it isn't an integer. Just as multiplication is repeated addition, someone clever said what about repeated multiplication? So we had powers 5^3 = 5x5x5 = 125. No new numbers needed. Now the fun starts. What about the inverse of powers? Roots, they were called. We know that the cube root of 125 is 5. We know that the square root of 4 is 2. What about the square root of 2? There surely must be such a number, but it was found that no rational fraction would do. 1 2/5 is a bit too small, while 1 1/2 is too big. No matter how we tried we could not find the right number but there had to be one. Eventually we accepted that the square root of 2 was sqrt(2). We had invented irrational numbers - numbers we could precisely describe but not write down as a rational fraction. We later discovered that irrational numbers could be simple roots of integers or rational numbers, or could be the roots (i.e. the solution) of equations. The story somewhat splits into two paths now. First there are numbers which we can define but are not the roots of any equation using rational coefficients. Pi is an example; the ratio of diameter to circumference of a circle. It exists, yet not only is it not a rational number, not a square/cube root etc., but you cannot even find any combination of powers of pi which are equal to any other combination or any rational number. Pi is a transcendental number; a special type of irrational number. The second path at this point arises when someone asked about sqrt(-2). None of our existing numbers will do the job, yet this sort of number crops up when solving some quadratic equations (for example, damped harmonic motion such as tuned circuits or turntable suspensions). It turns out that if we say that sqrt(-2)= j sqrt(2) and jxj=-1 than everything is fine. We can still do algebra and arithmetic, and solve all quadratic equations. So we have a long line of types of number, so we don't have to give up on arithmetic: addition -> positive integers subtraction -> negative integers and zero multiplication -> no new numbers needed division -> rational numbers powers -> no new numbers needed roots (+) -> irrational numbers non roots -> transcendental numbers roots (-) -> complex numbers The next step in this journey is quarternions, which are bit like 4-dimensional complex numbers (complex itself being 2-D). Every step has been useful for doing physics. |
5th Mar 2018, 3:46 pm | #24 |
Nonode
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Re: Correct units for Reactance/Impedance Calculations
Can I say from the start that I truly admire and respect any one who can master the Mathematics as just described.
But I would like to ask a question. How much of the maths we engineering students studied doing our C&G/HNC/ etc was later used in our careers? From my own experience very little. Many engineers who seek promotion have to go down the management route. This requires additional studying including Health and Safety, Employment Law etc. When I debate this topic (math) with my now elderly colleagues almost 95% say they have never used or had a requirement to use the maths studied during their apprenticeships. So I wonder what other members think? Cheers John |
5th Mar 2018, 4:46 pm | #25 |
Dekatron
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Re: Correct units for Reactance/Impedance Calculations
I suspect it depends on which area of your speciality you end up. And whether in your work you need to understand the physics behind the engineering or simply to use it in order to get the job done.
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5th Mar 2018, 4:53 pm | #26 |
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Re: Correct units for Reactance/Impedance Calculations
I have used (in my hobby of electronics) a significant amount of the maths I learnt. In order to understand the physics I learnt (part of which underpins my hobby) I needed virtually all the maths. If you want to understand electronics, then you need maths. Fortunately, it is possible to get quite a long way in doing electronics without properly understanding it - but you will trip up from time to time, or even deny the truth.
A good example of this is the inductive loading of telephone lines, invented by Oliver Heaviside using his mathematical theory of transmission lines. He could not convince the head of the GPO that this counter-intuitive (to someone who did not understand the maths) idea was right, so the UK lagged behind the US in long distance telephony for a while. Lack of maths, and hence lack of physics, lies behind some of the audiofoolery and antenna myths which some people sincerely believe, including some of those who make their living from electronics. |
5th Mar 2018, 4:55 pm | #27 |
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Re: Correct units for Reactance/Impedance Calculations
Indeed it should! Which silly idiot typed West when he meant East! Oh, well.... David
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5th Mar 2018, 5:44 pm | #28 |
Dekatron
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Re: Correct units for Reactance/Impedance Calculations
If
ax squared plus bx plus c equals nought Then x equals minus b plus or minus the square root of b squared minus 4ac ALL over 2a That is only etched om my brain because Mr Mcdonnell promised to cane all the boys who couldn't recite it the following Monday. B a s t a r d. At college, where there thankfully there were no public floggings, I has difficulty with the B-H curve/hysteresis-loop concept, If only the lecturer had used the analogy of mechanical backlash, a concept I was familiar with, I think I would have got it.
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5th Mar 2018, 7:20 pm | #29 | |
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Re: Correct units for Reactance/Impedance Calculations
There are also negative capacitors and negative inductors to consider although these aren't passive devices.
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Regards, Jeremy G0HZU Last edited by G0HZU_JMR; 5th Mar 2018 at 7:35 pm. |
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5th Mar 2018, 9:26 pm | #30 |
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Re: Correct units for Reactance/Impedance Calculations
It is interesting to look at the actual units in the MKS system.
For example voltage is J.Q^-1 or Joules per Coulomb. ( Voltage is in an energy density, just like pressure, though people think of pressure in terms of Newtons per square meter, if you multiply both by meters m (meters) it's obvious that pressure is Joules per cubic meters of volume. A coulomb of charge is merely a volume or amount of charge. So I will list the electrical units below (and for comparison in brackets the fluid equivalent) J joules, s seconds, Q coulomb, m meters: (notice how it turns out that Q=Coulombs and cubic meters look the same) Voltage. JQ^-1. (Jm^-3). (Called pressure) Current. Qs^-1. (m^3s^-1). (Called flow rate) Inductance . Js^2Q^-2 ( Js^2m^-6) (called fluidic inertia) Capacitance. Q^2J^-1. (m^6J^-1). (Called compliance) Resistance. JsQ^-2. (Jsm^-6). (Called resistance both systems) So a quick check if voltage is Joules per Coulomb and you divide that by current or Coulombs per second, the units of resistance must be Joule.Seconds per square Coulomb. Notice how the units of resistance are Joule.Seconds per square Coulomb. And for inductance the units are similar except the units are now Joule.Seconds(squared) per square coulomb. This gives insight as to why an inductor has a time domain property not seen with a resistor. So you can have an AC circuit where some inductor and some resistor would result in the same circuit current, but this is impossible for a DC circuit where the inductance then ceases to exist for practical purposes. In most electronics textbooks the units for L and C are seldom documented as we are more interested in the reactances in AC circuits and complex quantities, but still it is useful to know them at times depending on the type of problem being solved and even for a quick check that some equation solved has ended up with the expected units. |
5th Mar 2018, 11:27 pm | #31 |
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Re: Correct units for Reactance/Impedance Calculations
For resistance, we have the ohm. For reactance, we also have the ohm. But they're different types of 'ohms'! You can add two resistive ohms together using simple arithmetic, but you can't combine resistive ohms and reactive ohms together in the same way. But you can, if they are both 'reactive ohms'. Personally, all of that has never troubled me, but using the same unit name has always struck me as a serious inconsistency. After all, we do have V-A and V-AR (Or should that be VA-R?) as a precedent. Something like ohms and 'rohms' (reactive ohms, +ve and -ve, or even johms) would eliminate that inconsistency.
Just a thought. Al. |
5th Mar 2018, 11:36 pm | #32 |
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Re: Correct units for Reactance/Impedance Calculations
You use miles for travelling East-West, You use miles for travelling North-South and you also use miles for the crow-flight distance. Multidimensional bookkeeping keeps it all in check.
David
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6th Mar 2018, 1:09 am | #33 |
Nonode
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Re: Correct units for Reactance/Impedance Calculations
Nicely put! (Well, except that probably >90% of the world’s population use kilometres rather than miles for travel distances…)
As far as I know, it is a basic tenet of SI to keep its units “clean”, not modified by methods of test or vectorial direction issues, etc. So “var” for example has not become an SI unit. Electrical impedance, measured in ohms, is a complex quantity, in general with both real and imaginary components. So, it is subject to complex arithmetic rules - multidimensional bookkeeping as it were. Only in the special case where the imaginary component is zero – i.e. pure resistance with zero reactance – does complex arithmetic reduce to simple arithmetic. The ohms themselves are the same. Starting with the complex plane as the general case and then treating the real axis as a special case is I think an easier way to look at these things than the other way around. Cheers, |
6th Mar 2018, 8:48 am | #34 |
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Re: Correct units for Reactance/Impedance Calculations
After a hard day's mulitidimensional bookkeeping, you may feel the need for a pan-galactic gargleblaster in order to unwind.
David
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6th Mar 2018, 11:50 am | #35 |
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Re: Correct units for Reactance/Impedance Calculations
One thing it pays to remember about the imaginary number plane is that components of currents on this plane can do no real work. I once saw a power quartz crystal driver where the designers wanted to know the power delivered to the crystal so they multiplied the rms current by the rms voltage with no consideration for the phase angle which was also swinging around with mechanical loading of the crystal. It was the usual power factor problem, but it got ignored.
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6th Mar 2018, 12:33 pm | #36 |
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Re: Correct units for Reactance/Impedance Calculations
Yes, resistive ohms and reactive ohms are the same sort of ohms; it is the number which is complex, not the unit.
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