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Note that this assumes you have read through the Technician notes, or have a Technician or other Amateur licence. Reading ahead to Components may also be helpful.
Also, for maths, try looking at "Wu-tube" on You-Tube.
If you forget negative numbers: I have several accounts with one bank, and only transfer money to the Visa debit card linked account when I need to make a card transaction (it stops thieves stealing my money via PayWave / PayPass). If I buy, in Aussie dollars a product from a global company with local warehousing, but they process it in the Netherlands to dodge tax, then the bank will charge me a fee a few days later, thus my bank account with have a balance of, say, -$2.00, in other words they say I owe them money. If I try to buy a can of beans for $1.00, the transaction will decline, as the balance is below +$1.00. This can also be applied to a battery, where if the battery is charging, we call this +10 amps, and if it is discharging into a load call it, say, -20 amps. Historically cars had such a meter. Note in this case it is somewhat arbitrary what we call plus and minus, although charging the battery should probably be positive.
If we go to the National Zoo here (they charge in Canberra, unlike in Washington) twice, and it costs $32 each time, then your credit card could be said to have a balance of -$32 × 2 = -$64, meaning you owe the bank $64. A negative number times a positive number is negative. A bit hard to do with money, but if you multiply two negative numbers you get a positive. Perhaps we can say that a thief uses PayWave / PayPass 3 times $99 to get $297 worth of alcohol to sell for different drugs. The bank is responsible for this fraud, since they provide this fraud-enabling system, thus there are three credits or reversals of $99, which you might think of as the original transactions of -$99 times -3, returning you $297, thus: -$99 × -3 = +$297
As another example, early on transistors were expensive, with NPN far more so than PNP. Thus circuits were often designed around a positive ground, and a negative supply voltage. Base current flowed from emitter to base, and load current from emitter to collector. Electrons thus flowed into the collector, and out the emitter. Therefore many of the voltage and current parameters were / are expressed using negative numbers.
Also off the exam, in some cases we casually talk about a value which is actually an absolute value. A modern car is "negative chassis" or "negative ground", meaning the live terminal on a lamp is at +12 volts. Old vehicles used positive chassis, and heavy vehicles often use 24 volts. Thus you might find an old fire truck with -24 volts on the auxiliary power connection or lighter socket! Your radio will not like this! But an old filament lamp just cares about the absolute value of the voltage, also termed its magnitude.
This is indicated by vertical bars: x = |y| = |-5.7| = 5.7
Squaring a number means multiplying it by its self, 3² = 3 × 3 = 9. Squaring a negative number gives a positive: -3² = -3 × -3 = 9
A square root of a number is the number which can be multiplied by itself to generate that number. As we see above √9 must be 3, but it can also be -3. In some cases we are only interested in the positive result, but not always.
Useful, but for some reason ignore at school except in senior advanced maths: Normally we cannot find a square root of negative number, as both 1² and -1² equal 1. We need to use an imaginary number, i, called j in electrical engineering and electronics. Note that both j and -j = √-1 as both j² and -j² = -1.
A complex number is a real number plus or minus an imaginary number, such as Z = 37 + 19j or Z = 80 - 42j.
Complex numbers take your one dimensional number-line from primary / elementary school, and replaces it with a two dimensional one, which can define a point at any position on the page. Or perhaps they go from locating a farm gate near a mile-marker on a rural road to locating a building at 14th and 53rd in a big city.
Four important values are the square roots of 2 and 3; and their reciprocals.
√2 ≈ 1.414213562 & 1/√2 ≈ 0.707106781
√3 ≈ 1.732050808 & 1/√3 ≈ 0.577350269
You might remember the first few digits after the decimal, or obtain them from a calculator as required.
A formula can be expressed as y / x or as y × (1/x); so you may see either division by 1.4142, or multiplication by 0.7071, for example.
We use the √2 below for peak and RMS voltages.
√3 is used to convert between star (wye) and delta voltages on three phase power, such as 120 / 208 volt, and 277 / 480 volt systems in the US, 127 / 220 volts in Mexico, and 240 / 415 volt systems in Oz & India. Europe uses 230 / 400 volts in most cases, and 690 volts for large pumps and tunnel ventilation fans, etc, this being √3 × 400 volts (and 3 × 230 v).
Note that the bar above is just a convenient separator, and has no mathematical meaning. Occasionally the values are reversed.
You should remember that frequency is in Hertz (Hz). An alternative expression is rotational velocity symbol, lower case omega, ω = 2πf, expressed in radians per second.
We should be aware of resistance, expressed in ohms. This limits current in DC circuits, and also in AC circuits. All conductors, except at super-low temperatures, have some resistance, but some metals such as iron, and some alloys (such as Nichrome) have higher resistance than others. The best conductor is silver, followed by copper. Gold is a good conductor, but ranks below copper, its benefit being resistance to tarnishing (a sulphur reaction), and oxidisation, the former being a major problem with silver.
Gold is useful as a plating on small signal connectors, and on contacts switching small signals. Although, in many cases it is simply that gold is prettier than nickel.
Silver is used in tuned circuits in some transmitters and high power amplifiers, both as plating on inductors, and in silver-mica capacitors, as this can reduce resistance to the high circulating currents in a resonant circuit.
Inductance of coils causes reactance, also termed inductive reactance. This increases with frequency. Old-style fluorescent lamps, including the first CFLs, use an a "ballast" wound on a laminated steel core, to limit the current in the tube. They are selected for the supply voltage, tube power, and mains frequency. These are close to the perfect inductor, with low resistance. Inductance is in L, and is expressed in Henries.
Reactance is indicated by a capital X.
For inductance, X = ωL = 2πfL. You will note that this is positive.
The reactance adds vectorially to the resistance to create an impedance, Z. Impedance is the opposition to the flow of AC current. Just as a distance right and a distance up measured in centimetres (or metres, inches, feet, or furlongs) total distance is also in these units, ohms are used for each aspect. Pythagoras tells us that a 4 metres up, and 3 across is 5 metres displaced. Likewise the DC resistance (copper resistance) in ohms and the reactance in ohms contribute to an impedance in ohms.
Distance = √(4² + 3²) = √(16 + 9) = √25 = 5
Z = √(R² + X²)
If a load has 40 ohms of resistance and 35 ohms of inductive reactance Z = √(40² + 35²) = √(1600 + 1225) = √2825 ≈ 53.15 Ω.
The alternative way to write this is Z = 40 + j35, with j being used in the place of i to avoid confusion with I for current. Z = R + jX
Capacitors only pass AC. With DC, ones the capacitor has charged, current falls to very close to zero, only leakage current being passed. At low frequencies the reactance is high, dropping as frequency increases.
For capacitance X = 1/(ωC) = 1/(2πfC). Some express this as X = -1/(ωC) = -1/(2πfC).
The negative is used, as the alternative expression is Z = R - jX, so an example for a capacitive circuit might be Z = 40 - j35.
Inductors and capacitors can be termed reactive components, and they can be further grouped with resistors as passive components. These can be used for filters, and also tasks such as blocking the flow of DC, or passing it while blocking RF. Likewise, 50 or 60 Hertz low voltage AC can be passed while RF is blocked.
This is used in power feeds for masthead amplifiers, DC or AC from a plug-pack (wall-wart) sent up the coaxial cable to the amplifier, passed into the cable via a coil, but blocked from entering the TV or radio receiver using a capacitor. Likewise, it is separated using an inductor to feed the power to the amplifier, but with a capacitor preventing it reaching the transistor or MMIC output.
Apparently ceramic filters should also be isolated from DC voltages using a small series capacitor in the signal path.
A trend you may notice is that for low frequencies inductors and capacitor are of high value, and may be physically large. This is because a large inductor is required to have significant impedance, and a large capacitor is required to have a reactance which is not excessively high. For high frequencies, often the components are small, in both value and size. Here a low value inductor has usefully low impedance, and a small capacitance is needed to have any significant impedance.
When comparing power due to the flow of the new-fangled AC current, it is helpful to be able to compare it to that delivered by good, old-fashioned DC. This equivalent voltage is called the RMS voltage, standing for "root mean squared". If we know it is a sine wave, as the voltage and current from a rotating generator is, we can use a simple meter with a rectifier which gets an average voltage, and apply a factor to get a useful estimate of the RMS voltage, which is what is displayed on the AC scale, or the digital display. Spend more on a meter, and it is "True RMS", useful for unusual waveforms. Another way to assess an AC signal or supply is to observe the waveform on an oscilloscope (CRO). This allows us to assess the peak and peak-to-peak voltages with ease, and the more expensive (digital) ones will even put these values on the screen for you.
The relationships, for a sine wave are:
VP-P = 2 x VP
VP = VRMS × √2 ≈ VRMS × 1.4142
VP-P = VRMS × 2 x √2 ≈ VRMS × 2.8284
VRMS = VP / √2 ≈ VP / 1.4142 ≈ VP x 0.7071
VRMS = VP-P / (2 × √2) ≈ VP / 2.8284 ≈ VP x 0.35355
If you want, you can look up the relationships to average voltage.
One useful function of the peak to / from RMS calculation is to determine the transformer needed to supply a certain filtered, but unregulated DC voltage, or the voltage generated by rectifying and filtering the output of a transformer of which we know the RMS voltage. It is important to note the the voltage drop of the diode(s) and that mains voltage variation, plus the tendency of the output of an unloaded supply to float upwards.
Off the exam, this is handy to determine the voltage on the capacitors on the high-voltage side of a switch-mode power supply. For 120 volts this is 170 volts, but can float higher. For 240 volts it is 340 volts. Plug a 240 volts supply into a 277 volt outlet (a 2 pin Aussie / Kiwi one will probably fit a NEMA 7-15R or 7-20R) this reaches a potentially "exciting" 391.7 volts, bumping against the 400 volt rating of the electrolytic capacitors used in most of these supplies, especially if the voltage is a little above nominal. If 277 volts floats 10% high, it reaches 305 volts AC, meaning 431 volts DC on the capacitors. Some supplies are sold rated from 85 to 305 volts, covering sagging 100 volts supplies to unloaded 277 volt ones.
Measuring the average or RMS power of complex waveforms, such as those resulting from the SSB modulation of speech is difficult, and would vary for each speaker. Instead a two-tone test is conducted, with the modulated signal observed on an oscilloscope.
If we look at the waveform, we will see a peak in the envelope. If we can zoom in (by using a faster timebase), we see that there is an RF sinewave at this point, and it is the power that we want to know the power of. We thus need to perform the same calculations as we would on DC power or AC mains power. Despite the term "peak", we never-the-less need to find the RMS voltage to insert into the power formula, thus, noting we first have to halve VP-P:
P = (VP x 0.7071)² / R
R is most often 50 ohms, but this does depend on the system. Some TV broadcast systems are 75 ohms, for example.
For some waveforms, notably CW (Morse), and FM, the PEP and average power is the same. In the US Amateur all power limits for 1.8 MHz (160 m) and up are in PEP, noting that for 60 metres the limit is the ERP of that power into a dipole. In many cases the limit is 1500 watts, but for 30 metres, the limit is 200 watts PEP. This limit also applies to Technician CW operations on HF. In this case, that is also 200 watts average power. This exceeds the power Australian Advanced operators are allowed to use when using CW, 120 watts average.
Suppose the output of our amplifier has a constant 780 volt peak-to-peak voltage on CW, into 50 ohms.
VP = VP-P / 2 = 780 / 2 = 390. Thus VRMS = 390 * 0.707106781 = 275.77164459. Plug this into V²/R = 275.77164459² / 50 = 76049.999959873 / 50 = 1520.999999197 ≈ 1521 watts, so just a tiny bit "hot". Using the 1 / √2 x 390 gets 275.77164463, and this gives us 76050 over 50, and 1521 exactly.
The above would also be valid if a two-tone test gave this voltage as the maximum peak-to-peak value.
Equipment has a certain output or input impedance. For a Ham transceiver this is typically 50 ohms, and so the cable and if possible the antenna should be this impedance, something which applies to dipoles and resonant ¼ or ⅝ λ verticals. However, some antennas have higher impedance. Delta and similar loops are 102 ohms. Some folded dipoles are 300 ohms. End-fed halfwave antennas have very high impedance, but work quite well without a ground-plane, handy for use on fibreglass or timber boats, and for some field operations. An air-cored auto-transformer (tapped coil) in the base the half-wave can be used to provide a 50 ohm match. An antenna used on the "wrong" band, or a random wire will have a all manner of "interesting" impedances.
Various means can be used to match the radio to the antenna, and from the antenna to the radio.
For the 102 ohm loop, the simplest option is a quarter-wave section of 75 Ω coaxial cable (transmission line), taking the velocity factor into account.
Pi-networks, named for the similarity to the Greek latter π, typically consist of a capacitor to ground, a series inductor, and another capacitor to ground. To allow use at various frequencies, and to match random impedance antennas, the capacitors can be adjustable, or be selected from a bank by relays; and the inductors can be tapped, these taps selected by a switch, combinations of coils can be selected by relays, or be "roller inductors" with a continuously adjustable tap. When relays selection is used,
Transformers are either air-cored, or wound on ferrite or iron-powder cores. Baluns (balanced to unbalanced) or ununs (unbalanced to unbalanced) are examples. These may or may not provide DC isolation. A transformer can be a single coil, with a tap. If the signal from the transmitter is fed into the tap, the transmitter will see a lower impedance, say 50 ohms, while the antenna has high impedance. Thus the voltage at the top of the coil is significantly higher than at the tap.
An LED, or light emitting diode, makes a convenient indicator in DC circuits. These need to operate from a current source, and the simplest is a resistor in series with the supply. In most cases a quarter-watt (250 mW) or similar small resistor, costing a few cents, is fine, but we should calculate the power dissipated for safety, to prevent overheating and/or failure.
If we needed an indicator for a 24 volt truck or boat, or a repeater's battery charger, the maximum voltage might be 28.8 volts, while a red LED is rated at 1.8 volts. Thus we need to drop 27 volts at 20 milliamps. R = E/I = 27 / 0.02 = 1350 ohms. P = I²R = 0.02² × 1350 = 0.0004 × 1350 = 0.54 watts or 540 mW. Therefore we need a resistor rated at around 1 watt.
I skipped looking for a commercially available value for simplicity, and while 1.35 kΩ is available at specialist suppliers, this is only in 0.25 watt versions, so it is probably easier to parallel two 2.7 kΩ half-watters from a High Steet hobbyist shop. What parallel these, and not use two 680 ohms (1.36k) in serises? Either is fine, but the parallel arrangement may be more reliable.
These calculations also apply when making a zener based supply, as shown in the Technician exam. On the Circuits page you will see resistor feeding a zener be used to provide a clean, fixed voltage, low current supply to a varactor controlled oscillator. Likewise, they can generate a clean supply of say 8 volts, within a 12 volt radio, to power a small signal audio amplifier stage.
As always, these are actual questions from the General exam pool.
What is impedance?
A. The electric charge stored by a capacitor
B. The inverse of resistance
C. The opposition to the flow of current in an AC circuit
D. The force of repulsion between two similar electric fields
This is the total opposition to the flow of AC current, answer C.
What is reactance?
A. Opposition to the flow of direct current caused by resistance
B. Opposition to the flow of alternating current caused by capacitance or inductance
C. A property of ideal resistors in AC circuits
D. A large spark produced at switch contacts when an inductor is de-energized
This is the reactance, answer B.
These components are called reactive components, because that cause reactance when AC is applied.
Which of the following causes opposition to the flow of alternating current in an inductor?
This is the reactance, answer D.
Which of the following causes opposition to the flow of alternating current in a capacitor?
Once again, it is reactance, answer C.
How does an inductor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases
B. As the amplitude of the applied AC increases, the reactance increases
C. As the amplitude of the applied AC increases, the reactance decreases
D. As the frequency of the applied AC increases, the reactance increases
Reactance is a function of frequency, not amplitude, so we have a 50/50. However increasing frequency means increasing reactance, answer D.
How does a capacitor react to AC?
A. As the frequency of the applied AC increases, the reactance decreases
B. As the frequency of the applied AC increases, the reactance increases
C. As the amplitude of the applied AC increases, the reactance increases
D. As the amplitude of the applied AC increases, the reactance decreases
Again a 50-50, as we can discount the amplitude. Specifically, as frequency increases, the reactance decreases, answer A.
What happens when the impedance of an electrical load is equal to the output impedance of a power source, assuming both impedances are resistive?
A. The source delivers minimum power to the load
B. The electrical load is shorted
C. No current can flow through the circuit
D. The source can deliver maximum power to the load
This transfers the maximum power to the load, answer D.
What is one reason to use an impedance matching transformer?
A. To minimize transmitter power output
B. To maximize the transfer of power
C. To reduce power supply ripple
D. To minimize radiation resistance
This is to transfer the maximum power to the load, answer B.
These are often used to transfer power from a 50 ohm radio and coax to a high impedance halfwave vertical or end-fed wire antenna.
What unit is used to measure reactance?
This is the Ohm, answer B.
Ohm, Ohm on the range - resistors on the stove / cook-top / hob.
What unit is used to measure impedance?
Once again, it is the Ohm, answer B.
Which of the following describes one method of impedance matching between two AC circuits?
A. Insert an LC network between the two circuits
B. Reduce the power output of the first circuit
C. Increase the power output of the first circuit
D. Insert a circulator between the two circuits
You may remember that an antenna tuner, one form of impedance matching device, consists of inductors (L) and Capacitors (C), so it is the LC network, answer A.
For small signal parts of a circuit, these would be physically small, low power components.
What dB change represents a two-times increase or decrease in power?
A. Approximately 2 dB
B. Approximately 3 dB
C. Approximately 6 dB
D. Approximately 12 dB
A doubling or halving of the power is very close to 3dB, answer B.
How does the total current relate to the individual currents in each branch of a purely resistive parallel circuit?
A. It equals the average of each branch current
B. It decreases as more parallel branches are added to the circuit
C. It equals the sum of the currents through each branch
D. It is the sum of the reciprocal of each individual voltage drop
Our friend Kirchhoff tells us about current flowing through circuit nodes, and from this we can deduce that the total current in a circuit is equal to the sum of the current in the branches, answer C.
This applies whether the circuit branches are powered by a reasonably constant voltage one, like a car battery or a shack power supply, or is powered by a current source.
How many watts of electrical power are used if 400 VDC is supplied to an 800 ohm load?
A. 0.5 watts
B. 200 watts
C. 400 watts
D. 3200 watts
The less mentally stressing method, for someone writing this at a rather late hour, is to have Herr Ohm tell us that 400 V / 800 Ω is half an amp, then multiply thus by 400, and get 200, answer B.
You will notice that there is a 400 on the top of the division, and later we multiply our answer by 400, so we could do 400 x 400 = 160,000 (16 x 10,000), then divide by the 800 to get 200 watts. Not a usual path, but you could apply I²R to the ½ amp (and resistance) to get ¼ x 800 = 200, if that got you there. When you square a number smaller (in absolute value) than 1, you get an even smaller, always positive, number.
How many watts of electrical power are used by a 12 VDC light bulb that draws 0.2 amperes?
A. 2.4 watts
B. 24 watts
C. 6 watts
D. 60 watts
Power is volts multiplied by amps, so 12 x 0.2 amps is 2.4 watts, answer A.
How many watts are dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms resistance?
A. Approximately 61 milliwatts
B. Approximately 61 watts
C. Approximately 11 milliwatts
D. Approximately 11 watts
You can apply I²R, so P = 7² x 1.25 since everything is milli and kilo, so they cancel. However, the proper method is 0.007² x 1250 = 0.06125 watts, or 61.25 mW, answer A.
A standard ¼ watt resistor would be fine in this case, or even a ⅛ W one. You can calculate the voltage across the resistor if you wish.
What is the output PEP from a transmitter if an oscilloscope measures 200 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
A. 1.4 watts
B. 100 watts
C. 353.5 watts
D. 400 watts
200 VP-P is 100 VP. P = (VP x 0.7071)² / R = 70.71² / 50 = 4999.9 / 50 = 99.998 watts. If you use a 1/&radic2; you get to 5000, and thus 100 watts. It is answer B.
What value of an AC signal produces the same power dissipation in a resistor as a DC voltage of the same value?
A. The peak-to-peak value
B. The peak value
C. The RMS value
D. The reciprocal of the RMS value
The reason we so strongly want to determine the RMS voltage of a waveform is that this is the voltage which causes the same power dissipation as that DC voltage, answer C.
Applying 12 volts RMS to a resistor has the same result as 12 volts DC.
What is the peak-to-peak voltage of a sine wave with an RMS voltage of 120.0 volts?
A. 84.8 volts
B. 169.7 volts
C. 240.0 volts
D. 339.4 volts
I could work out the maths, or, living in a 240 volt country, I can remember that the peak voltage, or rectified and smoothed DC voltage for 240 volts (RMS) is 340 volts, or thereabouts, and this figure would apply to the peak-to-peak voltage figure for half the RMS voltage.
But since you are paying me for a maths answer:
VP-P = VRMS × 2 × √2 = 120 × 2 × √2 ≈ 120 × 2.8284 ≈ 339.4 volts, answer D.
The parameters here can be displayed on an oscilloscope, and the voltage is not a real issue, but great care needs to be taken when taking measurements directly from mains power. Some CROs ground the shield of the probe lead. Alternatively, if they don't you can have potentially dangerous voltages on the BNC nut connecting the probe. It is possible for large fault currents to flow in the probe shield, if a connection error is made. A far safer option is to find an old 12 volts AC output "wall-wart" or similar plug-in power supply at a flea market sale, and view the output of this on a CRO. You should thus see around 17 volts as the peak, and 34 volts as the peak-to-peak. Some loading may be necessary to prevent it floating high.
Off on a tangent, you may notice that the top of the mains waveform is a little flat on the top. This is due to all the switch-mode supplies in computers, TVs, iToy chargers, LED lamps, most compact fluoro lamps, even some appliances, only drawing current at the peak of the waveform, unlike traditional lamps which draw current proportional to the voltage through the whole cycle.
What is the RMS voltage of a sine wave with a value of 17 volts peak?
A. 8.5 volts
B. 12 volts
C. 24 volts
D. 34 volts
The ratio is 1.4142, so 17 divided by this is 12 volts, answer B.
You could also get there using 17 × 0.7071.
What percentage of power loss would result from a transmission line loss of 1 dB?
A. 10.9 percent
B. 12.2 percent
C. 20.5 percent
D. 25.9 percent
It is very close to 20.5% power loss, answer C.
This means 79.5% of power is left. You can try to remember this 20.5%, but there is a way to check: if three such section were put in series, the remaining power is 0.795 x 0.795 × 0.795 = 0.795³ = 0.502, near enough to a half, the power remaining after a 3dB loss. I suppose the cube root of 0.5 is close to 0.795, in fact it is 0.7937. Some other fun with this is that 0.79510 is around 0.1, a 10 dB loss, and 0.79520 is 0.01, a 20 dB loss.
What is the ratio of peak envelope power to average power for an unmodulated carrier?
For an unmodulated carrier (or the elements of a Morse signal), the average and PEP power are the same. Thus specifying 200 watts PEP for CW is the same as specifying 200 watts average. Thus the ratio is 1:1, or 1.00, answer B.
Some administrations separately specify the PEP power for SSB, and average for CW, at different levels.
What would be the RMS voltage across a 50 ohm dummy load dissipating 1200 watts?
A. 173 volts
B. 245 volts
C. 346 volts
D. 692 volts
If P = V² / R, then V² = P × R = 1200 × 50 = 60000. √60000 = 244.95 ≈ 245 volts, answer B.
What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts?
A. 530 watts
B. 1060 watts
C. 1500 watts
D. 2120 watts
An AM, FM, or key-down CW signal giving a 1060 watt reading on a standard wattmeter is the same as 1060 watts PEP, answer B.
On most bands this is a legal power level.
What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50 ohm resistive load connected to the transmitter output?
A. 8.75 watts
B. 625 watts
C. 2500 watts
D. 5000 watts
Halving the 500 volts gets 250 volts, and multiplying this by 0.7071 gives us 176.775 V RMS. Squaring this, then dividing by 50 gives 624.9880125, a fraction under 625 watts , answer B.
Using a closer approximation of 1/√2, or simply dividing by your calculator's value for √2 gives 625 exactly. 176.776695296636881² = 31250. 31250 / 50 = 625 watts.
Forgetting to halve the voltage gives one of the other answers.
The answer to the voltage across the resistor in G5B05 is E = IR = 0.007 × 1250 = 8.75 volts. Pure supposition, but subtract this from 12 volts, and get 3.25 volts, the typical forward voltage of an InGaN blue or pure green LED.
On to: Electrical Principles 2 - Rs, Cs, & Ls in series & parallel; Transformers
You can find links to lots more on the Learning Material page.
Written by Julian Sortland, VK2YJS & AG6LE, April 2022.
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