## The KCRC TechNet Puzzle for September 27, 2017

Which statement below best describes the function of an antenna tuning unit (ATU)?

A An ATU removes the reactive portion of the feed impedance.

B An ATU adjusts the length of the antenna to bring it to resonance.

C An ATU adjusts the input impedance so that Z=50Ω.

D An ATU brings the input impedance to 50Ω resistive and non-reactive.

## The KCRC TechNet Puzzle for September 13, 2017

And now for the answer to our September 13th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

First off, RMS stands for “Root Mean Square”. It has many different meanings in different disciplines, but for cyclical alternating electrical current, like those seen in RF (radio frequency) generation, RMS is equal to the value of the direct current that would produce an equivalent amount of power. For Sine waves (sinusoidal waves), that are usually used in RF generation, this amounts to “peak amplitude” (labeled arrow “2” in the above illustration) divided by the square root of two, or 1.4142135623730950488016887242097… (you get the idea, it’s an irrational number). The only amplitude that is lower than the peak amplitude in this illustration is arrow “1”.

Arrow “3” is the Peak-to-Peak amplitude, and arrow “4” is the period of the wave = 1/f(Hz).

So, the correct answer is “A” Arrow “1”!

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## The KCRC TechNet Puzzle for August 23, 2017

And now for the answer to our August 23rd, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Probably the first thing to consider is what the heck a “*Q Factor*” is? Yeah, sure, go look it up and you will see that it is a “*Quality Factor*” of an oscillator of some kind, but that isn’t how it started. An engineer by the name of K.S. Johnson over at Western Electric Co. just used “*Q*” because it was one of the few letters not overused in engineering terms. The “Quality” explanation came later on…

“*Q Factor*” was first related to how damped a given oscillator was, but it has found many definitions that explore different aspects of the same inherent characteristic.

One of the many definitions for “*Q Factor*”, which has a unitless value, is that as it increases, an oscillator has greater amplitude oscillations over a narrower frequency range. Some circuits can have a “*Q*” in the single digits, but some crystal based oscillator circuits can have a “*Q*” as high as in the millions!

One definition for a circuit’s “*Q*” is the resonant frequency divided by the frequency range before the signal drops by 3 dB (in half). If you look at the above illustration you can “guesstimate” that at -3dB (a third of the way down in the first “grid line”, the “offset” intercept looks like more than 1,000 Hz in each direction. So, let’s say that the offset is 1.4kHz on one side and 1.4kHz on the other, and the frequency range is something like 2.8kHz and the resonant frequency is 10.7MHz or 10,700kHz, so the “*Q*” would be 10,700/2.8 which is 3,821. The closest number that is available from your “eyeballing” of the 3dB point is answer “**C**” 3200, which in this case is the correct answer!

Now go out there and calculate some *Q Factors**!*

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## The KCRC TechNet Puzzle for August 9, 2017

Which schematic shows a way to make a non-polar capacitor from polar electrolytic capacitors?

Will the total capacitance be higher or lower than the individual capacitor’s capacitance?

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And now for the answer to our August 9th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

An extremely common polarized capacitor is the ubiquitous electrolytic capacitor – it has an anode (+) made from a metal that can form an oxidized insulating layer by the process of *anodization* (think of the process that gives many Apple products their colored aluminum finish). Any reverse DC current might destroy this anodized insulating layer and “poof” goes your capacitor!

“A” is the worst of all worlds – the diode will only conduct when your polarized capacitor is getting current in the opposite direction that it was designed for, when connected with the correct polarity for the capacitor, the diode will block current!

“B” is just two polarized capacitors in series – their total capacitance is reduced to 5 μF, and if you reverse the current you still are going to kill at least one of the capacitors and the other one would soon follow it and fail as well.

“D” is almost as bad as “A” – one of those two capacitors will be ruined the moment that the current travels through them in the wrong direction, and if you were lucky enough to have the capacitor fail as an open circuit, the remaining capacitor is only safe as long as the current does not change direction!

Well, by the process of elimination, it would seem that “C” was the correct answer, and it IS! This is called a ** bi-polar **electrolytic capacitor, with their anodes connected in series. The total capacitance is lessened to 5μF, but at least you don’t have to worry about how to install it!

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## The KCRC TechNet Puzzle for July 26, 2017

The primary winding of a transformer has 80 turns and the secondary has 40 turns. If the input impedance is 250Ω, what is the impedance across the secondary terminals?

…….. A) 62.5Ω

…….. B) 125Ω

…….. C) 176Ω

…….. D) 500Ω

And now for the answer to our July 26th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Transformers are AC devices. Effectively, they are magnetically coupled electromagnets. Alternating currents on the first, “*primary winding*” induce an always changing magnetic field, and that induced alternating magnetic field, in turn, induces another electric field onto the second “*secondary winding*”. How the output voltage and current on the secondary winding, compares to the voltage and current input on the primary winding, is determined by the ratio of the number of wire turns used in a given transformer.

E_{S}/E_{P} = N_{S}/N_{P}

E_{P} = Input AC Voltage

E_{S} = Output AC Voltage

N_{P} = Number of turns on the primary winding

N_{S} = Number of turns of the secondary winding

If your primary and secondary windings have the same number of turns, you have a 1:1 ratio, the voltage and current induced in the secondary winding is identical to the voltage and current applied to the primary winding – you would have something used all the time – an isolation transformer, that isolates the electrical power from its source, usually for safety measures, or ground loop protection. What about the winding ratio of the question? 80 turns on the primary and 40 turns on the secondary give a 2:1 ratio – Two volts of AC in the input will get you 1 volt on the output, 20 volts on the input will get you 10 volts on the output…

But the question is not about a transformed “voltage” – they are asking for the transformed “impedance”?

We mentioned that your transformer can induce half the voltage it had in the primary winding – does that mean your transformer can make your energy disappear? Sorry! The Law of Conservation of Mass-Energy is not a suggestion, it’s THE LAW!

So, what happens to the current when you are halving the voltage?

I_{P}/I_{S} = N_{S}/N_{P}

I_{P} = Input AC Current

I_{S} = Output AC Current

N_{P} = Number of turns on the primary winding

N_{S} = Number of turns of the primary winding

So, the current is being doubled! No energy *disappeared* today!

Of course, this is quite theoretical, there are other factors *afoot* that waste energy, turning it into heat that escapes into the system during the transfer of energy. The devil is always in the details…

Everyone remembers Ohm’s Law:

E = IR or R=E/I

Well, Ohms Law for us big boys is actually

Z=E/I

** Impedance** is equal to the

**divided by the**

*Voltage***.**

*Current*So, the transformer cited in the question halves the input voltage while doubling the input’s current on its output winding:

Z_{S}/Z_{P} =N_{S}/N_{P} x N_{S}/N_{P}= (N_{S}/N_{P})^{2}

Z_{P} = Input Impedance

Z_{S} = Output Impedance

So, halve the turn ration and you halve the output voltage, you double the output current and you transform the impedance from its value at the input to one fourth its original value on the output!

250Ω/4 = 62.5Ω

Answer “A” is the right answer!

(Or you can just remember that a transformer “*transforms”* the input impedance by the *square* of its winding ratio.)

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## The KCRC TechNet Puzzle for July 12, 2017

The antenna shown in the diagram is a

…….. A) Quad

…….. B) Delta loop

…….. C) Trap dipole

…….. D) Folded dipole

And now for the answer to our June 28th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, even if you are not well versed with each type of antenna, look at their names as an indicator of their likely shapes. A quad is a four-sided loop – it can simply be a single rectangular radiator, or much like a Yagi, it can be made of additional rectangular loops, of a slightly larger dimension behind the radiating element, working as a reflector and a series of smaller dimension rectangular loops in front of the radiator acting as directors, improving the antenna directionality!

A “delta loop” is named for the Greek letter delta “Δ”, so it’s a loop shaped like a triangle. The picture above is clearly not triangular!

A *trap* dipole is antenna with two equal lengths of wire, going in opposite directions, each interrupted by a tuned circuit “*traps*”. The *trap* operates as a low pass filter, by letting the RF pass through it at a lower frequency, for which the entire length of the dipole is resonant, but does not conduct the RF at a high enough frequency, effectively shortening the antenna to a resonant length at the higher frequency. In this way, a single length dipole can be made resonant for two or more different frequencies. The same principle can be used with numerous “*traps*” on the same dipole for even more resonant frequencies! The price that you pay is the power loss inefficiency introduced by the numerous “*tra*p” filters. The picture above is not a length of two wires going off in opposite directions.

A folded dipole is just a simple pair of wires extending in the opposite direction from a central feed point, that have been folded back upon itself either once or more than once. The picture above does not appear to be any form of a dipole, neither a trap dipole, nor a folded dipole…

So, the answer would seem to be “A” a quad!

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## The KCRC TechNet Puzzle for June 28, 2017

If a 9-turn inductor has a measured value of 20μH and the turns are squeezed closer together the inductance might change to approximately

………A) 20mH

………B) 22μH

………C) 18μH

………D) 10μH

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And now for the answer to our June 28th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

There is an “easy” way to answer this question and a slightly harder one – let’s try the easy one first!

This method employs the amazing power of sheer logic, without any fundamental understanding of inductance! If you change the physical characteristic of an object there are only three possible outcomes – it will “increase” some property, it will “decrease” some property or it will leave that random property unchanged. Inductors are infamous for being affected by its surrounding environment – planning to adjust a variable slug inductor shielded in an IF “can”? Better use a nonmetallic screwdriver! Winding some wire with only air around it can make a modest inductor – wrap it around a ferrite doughnut and you have a MUCH more powerful inductor!

Your inductor is unlikely to be unchanged, so answer “A” is equally unlikely to be true. So, let’s guess that it changes its inductance for the worse – the trouble is that both answers “C” and “D” could be true, and there really isn’t that much information to tell you which value would be more likely! So, logically alone, you would be left with answer “B” 22μH, which is the correct answer!

Now for the less easy way:

Capacitors and inductors – two very important types of devices in electronics and especially so in the field of RF electronics. Capacitors store and release the energy of a stored electrical charge into a voltage potential and the flow of current, and inductors store and release the energy of a stored magnetic field into a voltage potential and the flow of current! Inductance causes current flow to “lag” voltage potential, and capacitance causes voltage potential to “lag” current flow! One cancels the other.

Capacitance values are relatively simply determined by the surface area and geometry of the opposing conductive plates, the distance between them and the “dielectric constant” of the material separating them. The equation describing the inductance of an inductor is far less clear cut:

L = (1/ℓ)(μ_{0}KN^{2}A)

Where:

*L = inductance in Henries (H)
μ0 = permeability of free space = 4 × 10−7 H/m
K = Nagaoka coefficient
N = number of turns
A = area of cross-section of the coil in square meters (m2)
*ℓ

*= length of coil in meters (m)*

(ℓ is the important variable in your question)

Sufficiently “clear as mud”?

Suffice it to say, there are four factors that principally determine the inductance of a given inductor:

The number of turns of the coils – the more coils, the more inductance.

The coil area – the greater the coil area’s cross-section (the bigger the coils), the more the inductance.

The core material – the greater the magnetic permeability of the core that the coil is wrapped around, the greater the resultant inductance.

The coil length – the greater the coil length, the lesser the resulting inductance (with a longer path for the magnetic field flux to travel there is more opposition to the formation of magnetic flux, for any given amount of field force). Stretch a given coil out until it is a straight piece of wire and the inductance shrinks to its smallest amount (never actually zero – all wire has some inherent inductance AND capacitance!).

So, bunch up a given coil, and its resultant inductance increases…

Giving you answer “B” 22μH!

[Phew!]

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## The KCRC TechNet Puzzle for June 14, 2017

And now for something completely different – a little amateur radio trivia about a revolutionary technological invention (try not to “Google” it immediately) – the superheterodyne radio receiver circuit!

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1) When was the superheterodyne (superhet) radio first invented?

………. A) 1898

………. B) 1918

………. C) 1929

………. D) 1942

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2) What does the “super” in superheterodyne stand for?

………. A) Nothing, it’s just a marketing gimmick.

………. B) It stands for “super-duper”!

………. C) It stands for “supersonic”.

………. D) It stands for “super accurate”.

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3) Who invented the superheterodyne radio?

………. A) Edwin Armstrong

………. B) Lee DeForest

………. C) Guglielmo Marconi

………. D) Hedy Lamar

BONUS QUESTION:

What event, framed the end of life of the superheterodyne inventor?

………. A) A Nobel Prize

………. B) An Oscar Lifetime Achievement Award

………. C) An episode of “This Is Your Life”

………. D) Took a “gainer” out of an open window in their 13th floor apartment, plummeting to their death on a nearby roof.

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And now for the answer to our June 14th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

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So, here are the answers:

1) “B” – Yup, that brand new shiny rig is, at its heart, operating using technology patented almost a *century ago*! Not exactly bleeding edge technology there…

2) “C” – They tagged a “supersonic” on heterodyne technology, supposedly because the intermediate frequency used in this technology, was above the range of human hearing? Personally, it sounds more like marketing BS, but that was their reason and they’re sticking with it.

3) “A” – Yes, Edwin Armstrong, one of the most brilliant engineers of the early 20^{th} Century. He developed all kinds of things that you may have heard of, like the super-regenerative circuit, and FM radio. He was involved in a legal patent case, during which he testified on how DeForest’s Audion tube actually worked (Le DeForest didn’t seem to have a clue at the time!). Marconi’s research took place for the most part before the turn of the 20^{th} Century and was more involved in promoting the technology rather than developing it further when Armstrong was active. Hedy Lamar, in addition to being a successful actress, was a smart lady and was very interested in military weapon technology and came up with the principle on which spread spectrum radio technology is based.

Extra Credit Question: “D” – Yup, life ain’t fair. Edwin Armstrong was a brilliant engineer, but an awful patent lawyer. A parade of larcenous individuals cheated him out of most of the royalties from his designs. In the throes of financial ruin and despair, he sought solace his 13^{th} Floor window and the nature of gravity and ended his own problems, the hard way. His estate was able to arrange for a more equitable sharing of royalties that he deserved, but that didn’t really help him much after he killed himself.

Remember George Santayana “Those who do not learn from history are doomed to repeat it”! So, if you find yourself in a similar situation, spend your time looking for a good patent lawyer, not a window on the top floor!

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## The KCRC TechNet Puzzle for May 24, 2017

.And now for the answer to our May 24th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, this one is kind of easy to guess – it is a voltage divider. The voltage drops across each resistor, and adds up to the total voltage of the source – 20 volts. It would be great if it multiplied voltages, but that would defy a couple of laws of physics, so answer “D”, 82 volts is definitely out!

If it divides voltage it couldn’t be answer “C” 20 volts – that’s the entire source’s voltage potential!

Most people with an idea of how a voltage divider works should realize that the only way to evenly divide the voltage drop equally would be to have both resistors be of the same value, which is NOT the case here, so answer “B” 10 volts can’t be right.

So, by the simple process of elimination the answer is “A” 2 volts!

But wouldn’t it be nice to know why the answer is “A”?

Consider good olde “Ohm’s Law”:

E = IR

Or

I = E/R

Simplify the circuit by adding both resistors, as you can quite easily do when they are in series, and the total current flowing through the circuit is:

20 volts/ 92,000 ohms = 217 microamps

Now, by using ohms law, E= IR you can figure out the voltage potential drop occurring across each resistor – the 82K ohm resistor would be:

E = 217 microamp x 82,000 ohms = 17.8 volts

And the 10K ohm resistor would be:

E = 217microamps x 10,000 ohms = 2.2 volts

If you want a formula, the voltage drop across R1 with two resistors in series, R1 and R2 could be written as:

E_{voltage drop} = (R1/(R1+R2)) x E_{source}

But isn’t it nicer knowing why?

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## The KCRC TechNet Puzzle for May 10, 2017

And now for the answer to our May 10th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Sorry, but you just have to figure out what the darn device is. It looks a lot like a *strip line* connected to coaxial cables externally with a pair of *strip lines* internally inductively coupled to the main *strip line*. Why two couplers? Well, if you take a look at the way the diodes are connected, they appear to be *directional * power couplers, one detecting *forward *power, the other, reverse power. So, what would switch “S” do? It switches the forward power coupler to to the power meter in one position, and the reverse power coupler in the other position!

Just like choice “D” Forward-Reverse.!

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## The KCRC TechNet Puzzle for April 26, 2017

And now for the answer to our April 26th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, the illustrated schematic is a simple RC circuit. The instant that the switch closes the circuit, current flows into the charging capacitor slowed down by the resistor in series with it – over time as the capacitor develops a greater and greater charge the current flow declines until at the point at which the capacitor is fully charged when the current ceases to flow at all.

At “zero time” the capacitor acts as if it were a short circuit and the voltage potential between both sides of it are zero – it is only after a charge begins to develop in the capacitor, and current flow is diminished by the charge repulsion of the partially charged capacitor, that a voltage differential, voltage drop, rises and finally approaches a constant value, giving you a curve that looks like this:

Does that look anything like the curve in the question?

Nope, so it ain’t “A”.

Talk about coincidence – that’s pretty much the same curve for the charge developing over time on the capacitor…

So, it ain’t “B” either!

Well, what about it being a curve showing the time constant, like answer “C”? The problem is that an RC circuit’s time constant isn’t a curve it’s a…

Constant!

τ=RC

Or, in the wise words of Wikipedia:

“Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1/e ≈ 36.8 % in this time (say from a step decrease). In an increasing system, the time constant is the time for the system’s step response to reach 1-1/e ≈ 63.2% of its final (asymptotic) value (say from a step increase).”

(Try to say that three times fast!)

So, if the graph looked like this:

It might be choice “C”…

But, since that isn’t the same illustration in the question…

It ain’t “C” either!

So, by the process of elimination (or knowing what the voltage potential drop between a resistor in an RC circuit actually looks like) you have the answer, “D”! Initially when the circuit is closed the current flows as if the capacitor was a simple piece of conducting wire – under that circumstance the voltage drop between the two leads of a single resistor in a simple resistor circuit would be the entire voltage potential applied to the circuit, BUT as the capacitor develops its charge, the current slows down with the capacitor acting as a component whose resistance to the flow of current, increases over time. Eventually the capacitor acts as an open circuit not allowing any current to flow – at that point the voltage potential difference between both sides of a resistor, its voltage drop, when NOT connected to a circuit with an actual current, is zero.

So, “D” is the answer, and if ya think about it for just a little while, it makes a lot of sense!

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## The KCRC TechNet Puzzle for April 12, 2017

And now for the answer to our April 12th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, there are four basic types of filters:

Curve 1 isn’t really a filter, it is a straight wire connection that allows the entire frequency spectrum to be passed unaffected. Curve 2 is a notch filter – it allows DC to low frequencies as well as high frequencies be passed and only blocks a small selection of frequencies in the middle range. Curve 4 is a low pass filter, blocking higher frequency signals. If you look at the circuit shown and just think of how it would deal with the ultimate low frequency 0 Hz – DC, you can see that a DC current would short itself out by the shorting inductor – this would rule out curves 1, 2, and 4 and only leave a band pass filter curve 3! Indeed, the schematic is for a band pass filter, but even if you did not recognize that, you could still work out how it would behave logically!

So the answer is “C” curve 3!

## The KCRC TechNet Puzzle for March 22, 2017

And now for the answer to our March 22nd, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

At its resonant frequency a quarter wave vertical acts just like any other serial resonant LC circuit – this means that it’s voltage peaks at its tip (this is why it is more risky to get an RF burn at the tips of a resonant antenna), and its current peaks at the site of the antenna feed (this is why the position of the antenna feed is more important than any other portion of the antenna and why you see A LOT more inverted “V” dipoles than you do noninverted “V” dipoles (the radiating centers are higher and less likely to suffer from ground wave absorption).

So, the answer is B, drawing 2!

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## The KCRC TechNet Puzzle for March 8, 2017

And now for the answer to our March 8th, 2017 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Ultimately, this question is asking for the length of a quarter-wave coaxial stub for 145 MHz. You can calculate wavelength as:

Wavelength = The speed of light in a vacuum / the frequency of the wave

Wavelength = 299,792,458 meters per second / 145,000,000 cycles per second

Wavelength = 2.06 Meters

1/4 wavelength = 1/4 x 2.06 Meters = 0.50 meters = 50 cm

(Of course the simplest way to approximate is:)

Wavelength = 300/Frequency in MHz = 300/145 = 2.06 Meters

1/4 Wavelength = 1/4 x 2.06 Meters = 50 cm

So…. answer “B” is, right??????

WRONG! Notice that you have been using the speed of light IN A VACUUM. Coaxial cable is nothing like a vacuum, it has a high dielectric material that causes the transmission of electromagnetic waves to slow down…

That’s what ** velocity factor** means – the percentage of the speed in a medium compared to its speed in a vacuum.

In the case of a solid polyethylene dielectric coaxial cable its * velocity factor* is 66 percent (in the case of something like LMR400, which uses a closed cell foam it is 85 percent).

As you slow down a wave’s velocity at the same frequency, you shorten its wavelength. In this case you shorten it to 66% of your previous calculation:

50 cm x 66% = 33 cm

So, the answer is “A”, not “B” and you are now more aware of how *velocity** factor *can screw up a calculation!

## The KCRC TechNet Puzzle for February 22, 2017

And now for the answer to our February 22, 2017 TechNet Puzzle.

(It is in “Invisotext” and will be visible if you highlight the area below!)

Simple diodes may supply a voltage drop when forward biased, but they do not regulate the voltage output. Zener diodes are designed for a specific reverse bias voltage breakdown threshold – if you reverse bias the Zener diode across the output power leads, when the voltage supply is a higher than the reverse bias breakdown voltage, the Zener diode conducts, until the voltage output is just below the breakdown voltage. It’s not efficient, but it works. Zener diodes are somewhat affected by temperature variations and current drain. Regulator IC are probably the best method to regulate power, and relatively inexpensive. The only schematic showing a Zener diode reverse biassed is “circuit 2”, which is answer “B”!

## The KCRC TechNet Puzzle for February 8, 2017

The circuit shown is fed from an AC source and two of the three voltmeters show the readings obtained. What should the third meter read?

A 4V

B 5V

C 6V

D 7V

E 8V

And now for the answer to our February 8, 2017 TechNet Puzzle.

(It is in “Invisotext” and will be visible if you highlight the area below!)

Everyone here knows Ohm’s Law:

E = I R

Well, you can expand that a bit for alternating currents:

E = I Z

Where “Z” is, of course, the complex impedance, rather than the pure resistance of a circuit.

In a series circuit the current going through all components is identical, so you can consider it to be a constant “c”:

E = cZ

In other words, voltage is PROPORTIONAL to impedance.

Therefore, if you sum the AC voltage drops as you would their impedances, you can discover the AC voltage of the power source powering this circuit.

Whether you wish to solve it as a rudimentary bit of vector math or a simple Pythagorean Equation, is up to you. The equation for the sum of the voltage drop across a resistor, because of resistance (3V) and an inductor, because of inductive reactance (4V) is:

sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5

So, the answer IS 5 Volts (choice “B”). If both components were resistors (or inductors for that matter) the answer would have been 7 Volts, or choice “D”

THE END (and the resistor and the inductor lived happily ever after.)

## Comments

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