G5: Electrical Principles

This group is about opposition: the various ways electrical parts push back against alternating current. In a plain DC circuit there is only one kind of push-back, called resistance. But once the current starts reversing direction (AC), two brand-new kinds of opposition show up, and learning their names and behavior is what G5A is all about.

Resistance versus reactance

You already know resistance: it is the steady opposition to current that a resistor provides, measured in ohms. ("Ohm" is simply the unit we count electrical opposition in, the way "pounds" is the unit we count weight in. Its symbol is the Greek letter omega.) Resistance does not care whether the current is DC or AC; it pushes back the same either way.

Now meet the new idea. Reactance is opposition to the flow of alternating current that is caused by capacitance or inductance, not by an ordinary resistor. (We will define capacitance and inductance in a moment.) The crucial thing is that reactance only appears with AC, and unlike plain resistance, the amount of reactance changes with frequency. Reactance is measured in the same unit as resistance, the ohm, and its symbol on the exam is the capital letter X.

So already you can answer several questions just from definitions:

• Reactance is opposition to the flow of alternating current caused by capacitance or inductance.
• The unit used to measure reactance is the ohm.
• The letter used to represent reactance is X.

The two parts that create reactance

An inductor is a coil of wire. ("Inductance" is the property a coil has of resisting changes in current; an inductor is the part that provides it.) A capacitor is two metal surfaces with a gap between them that can store up electric charge. ("Capacitance" is the property of storing charge; a capacitor is the part that provides it.) Both of these oppose AC, but in opposite ways, and that opposite behavior is the key to this whole group.

Here is the behavior to memorize. It is just two sentences, but they unlock four exam questions:

• An inductor (coil): as the frequency of the AC goes up, its reactance goes up too. Higher frequency, more push-back. Think "a coil hates fast wiggling."
• A capacitor: as the frequency of the AC goes up, its reactance goes down. Higher frequency, less push-back. Think "a capacitor likes fast wiggling."

So the opposition to AC in an inductor is called reactance, and the opposition to AC in a capacitor is also called reactance. They are opposites in direction but they share the same name. A handy memory hook: "ELI the ICE man" is the old saying hams use, but for G5A you really only need the up-and-down rule above. Coils block highs more, capacitors block lows more.

Impedance: the grand total of opposition

When a circuit has both resistance and reactance in it (which most real circuits do), we need one word for the combined total of all the opposition. That word is impedance. Impedance is the overall opposition to current in an AC circuit, and the cleanest way the exam defines it is this: impedance is the ratio of voltage to current. In other words, take the voltage pushing on a part and divide by the current flowing through it, and the answer is the impedance. Impedance is measured in ohms too, and its symbol is the capital letter Z.

One more vocabulary item the test asks directly. The inverse of impedance (its mathematical flip-side, one divided by impedance) has its own name: admittance. Just as impedance measures how much a circuit opposes current, admittance measures how easily it admits current. So the term for the inverse of impedance is admittance.

Resonance: when the two reactances cancel

Now for the beautiful part. Remember that a coil's reactance goes up with frequency while a capacitor's reactance goes down with frequency. If you put a coil (L) and a capacitor (C) together in a circuit, then there is one special frequency where the coil's reactance and the capacitor's reactance become exactly equal. Because they push back in opposite directions, at that one frequency they cancel each other out. This magic frequency, and the condition itself, is called resonance.

The exam states it two ways:

• At resonance in an LC circuit, inductive reactance and capacitive reactance cancel. (That is literally what resonance means.)
• When the inductive and capacitive reactance are equal in a series LC circuit, the result is that resonance causes the impedance to be very low.

Why does a series circuit go to very low impedance at resonance? Picture it: the two reactances are equal and opposite, so they erase each other, leaving almost nothing to oppose the current. With the reactance gone, current flows very easily, which is exactly what "very low impedance" means. (For now, just memorize the result for the series case: equal reactances in a series LC circuit means resonance and very low impedance.)

Impedance matching

One last practical idea. Different parts of a radio station "like" to see different impedances, and to make them work together smoothly you sometimes need to convert one impedance into another. This is called impedance matching (or impedance transformation), and it is like using the right adapter so two mismatched plugs can connect. At radio frequencies, several devices can do this job: a transformer, a Pi-network (a particular arrangement of coils and capacitors), and a length of transmission line (a specially-sized piece of feed-line cable) can all transform impedance. Because more than one of those works, when the test lists them together the answer is "all these choices are correct."

This group has the most arithmetic on the whole General exam, but do not let that worry you. We will break every type into a tiny recipe you can follow. There are three families here: decibels, electrical power, and the different ways to measure an AC wave. Grab a calculator and let's take them one at a time.

The decibel

A decibel (written dB) is a compact way of describing how many times bigger or smaller one power is compared to another. Instead of saying "a thousand times more powerful," which is a big clumsy number, we can say "30 dB," which is small and tidy. You do not need the logarithm formula for the General exam; you just need to memorize a couple of landmark values.

• A factor of 2 in power (doubling or halving) is about 3 dB. So if you double your power, that is a +3 dB change; if you cut it in half, that is a -3 dB change. This is the single most-asked decibel fact: a factor-of-two increase or decrease in power is approximately 3 dB.
• A factor of 10 in power is 10 dB. (Not asked as often, but worth knowing.)

There is one more decibel question that looks tricky but has a memorized answer. How much power do you lose if you lose 1 dB? A 1 dB loss means you keep about 79.4 percent of your power, so you lose the rest. The answer the exam wants is 20.6 percent. Just remember the pair: 1 dB lost equals about 20.6 percent of the power lost. (No calculation needed on test day, simply recognize the number.)

Electrical power: the three faces of one formula

Power is how fast electrical energy is used, and it is measured in watts. The base formula is wonderfully simple: power equals voltage times current (in symbols, P = E × I, where E is voltage and I is current). From that one idea, plus Ohm's law, we get three handy forms. You pick whichever form matches the two numbers a question gives you.

• If you know voltage and current: Power = Voltage × Current.
• If you know voltage and resistance: Power = (Voltage × Voltage) ÷ Resistance. (In words: square the voltage, then divide by the resistance.)
• If you know current and resistance: Power = (Current × Current) × Resistance. (In words: square the current, then multiply by the resistance.)

Worked example 1 (voltage and current). A 12-volt light bulb draws 0.2 amperes. How many watts does it use? Use Power = Voltage × Current = 12 × 0.2 = 2.4 watts. One multiplication, done.

Worked example 2 (voltage and resistance). You supply 400 volts DC to an 800-ohm load. How much power is consumed? We have voltage and resistance, so use Power = (Voltage × Voltage) ÷ Resistance. First square the voltage: 400 × 400 = 160,000. Then divide by the resistance: 160,000 ÷ 800 = 200 watts.

Worked example 3 (current and resistance). A current of 7.0 milliamperes flows through a 1,250-ohm resistor. How many watts? First convert the current to plain amperes: 7.0 milliamperes is 0.007 amperes. ("Milli" means one-thousandth, so divide by 1,000.) Now use Power = (Current × Current) × Resistance. Square the current: 0.007 × 0.007 = 0.000049. Multiply by resistance: 0.000049 × 1,250 = 0.06125 watts. That is about 61 thousandths of a watt, which we write as approximately 61 milliwatts.

Measuring an AC wave: peak, peak-to-peak, and RMS

A DC voltage just sits at one flat value, easy to measure. But an AC voltage is a wave that swings up to a high point, back down through zero, to a low point, and back again, over and over. So which number do we call "the voltage"? There are three useful answers, and the exam wants you to convert between them.

• Peak voltage is the highest point the wave reaches above zero (the top of the hill).
• Peak-to-peak voltage is the full distance from the lowest valley to the highest peak. For a normal sine wave that is exactly twice the peak (top of the hill to bottom of the valley).
• RMS voltage stands for "root-mean-square," and it is the most useful of the three. The exam's definition: the RMS value of an AC signal is the value that produces the same power (heat) in a resistor as a DC voltage of the same number. In plain words, a 120-volt-RMS AC source heats a resistor exactly as much as a steady 120-volt battery would. That is why your wall outlet is rated in RMS volts.

You only need two conversion factors, and a calculator does the rest:

• To go from peak to RMS, multiply the peak by 0.707 (that is one divided by the square root of two). To go the other way, RMS to peak, multiply by 1.414 (the square root of two).
• Peak-to-peak is always twice the peak. So peak-to-peak to peak means divide by 2; peak to peak-to-peak means multiply by 2.

Worked example 4 (RMS to peak-to-peak). A sine wave has an RMS voltage of 120 volts. What is its peak-to-peak voltage? First go from RMS up to peak by multiplying by 1.414: 120 × 1.414 = 169.7 volts peak. Then double it to get peak-to-peak: 169.7 × 2 = 339.4 volts peak-to-peak.

Worked example 5 (peak to RMS). A sine wave has a peak value of 17 volts. What is its RMS voltage? Multiply the peak by 0.707: 17 × 0.707 = 12.02, which rounds to 12 volts RMS.

PEP: the power of a radio signal

PEP stands for Peak Envelope Power, and it is the way we rate transmitter power. To find PEP across a known load, you turn the voltage into RMS, then use the power-from-voltage-and-resistance recipe from above. A "dummy load" mentioned in these questions is just a resistor that safely absorbs your signal for testing, so you treat it exactly like a resistance.

Worked example 6 (peak-to-peak to PEP). You measure 200 volts peak-to-peak across a 50-ohm dummy load. What is the PEP? Step 1, peak-to-peak to peak: 200 ÷ 2 = 100 volts peak. Step 2, peak to RMS: 100 × 0.707 = 70.7 volts RMS. Step 3, power from voltage and resistance: (70.7 × 70.7) ÷ 50 = 5,000 ÷ 50 = 100 watts.

Worked example 7 (a bigger one). You measure 500 volts peak-to-peak across a 50-ohm load. Step 1: 500 ÷ 2 = 250 volts peak. Step 2: 250 × 0.707 = 176.8 volts RMS. Step 3: (176.8 × 176.8) ÷ 50 = 31,250 ÷ 50 = 625 watts.

Worked example 8 (power back to RMS voltage). A 50-ohm dummy load is dissipating 1,200 watts. What is the RMS voltage across it? Here we run the recipe backwards: multiply power by resistance, then take the square root. 1,200 × 50 = 60,000, and the square root of 60,000 is about 245 volts RMS.

PEP versus average power for a plain carrier

One more PEP idea, and it is easy. An unmodulated carrier means a steady, unchanging signal with no voice or data riding on it, just a constant tone. Because it never changes, its peak power and its average power are the same. So:

• The ratio of PEP to average power for an unmodulated carrier is 1.00 (they are equal).
• If an unmodulated carrier has an average power of 1,060 watts, then its PEP is also 1,060 watts. (Same number, because the ratio is 1.)

Splitting up current: parallel branches

One small non-decibel fact lives in this group. When resistors are wired in parallel (side by side across the same two points), the current splits up among them. The total current equals the sum of the currents through each branch. Picture a river splitting into several channels: add up the flow in every channel and you get the total flow. So in parallel, currents add.

This last group is a set of combine-the-parts recipes, plus transformers. The good news is that every problem here is just plugging numbers into a short formula. The only thing you must do first is notice which arrangement you are looking at, series or parallel, because the rule flips.

Series versus parallel, in plain words

• Series means the parts are connected end-to-end in a single line, so the same current must pass through every one of them, like railcars coupled in a train.
• Parallel means the parts are connected side-by-side across the same two points, giving the current several paths to choose from, like several doorways into the same room.

Resistors

Resistors behave the way your common sense expects:

• Resistors in series add up. Two resistors in a row give more total opposition than either alone.
• Resistors in parallel give a total that is smaller than the smallest one. More paths means current flows more easily, so the total resistance drops. The recipe for parallel is: add up one-divided-by-each resistor, then flip the answer over (take one divided by that sum).

Worked example 1 (two resistors in parallel). A 100-ohm and a 200-ohm resistor in parallel. Add the reciprocals: (1 ÷ 100) + (1 ÷ 200) = 0.01 + 0.005 = 0.015. Now flip it: 1 ÷ 0.015 = 66.7, which rounds to about 67 ohms. Notice it came out smaller than 100, the smallest resistor, exactly as predicted.

Worked example 2 (three resistors in parallel). A 10-ohm, a 20-ohm, and a 50-ohm resistor in parallel. Add the reciprocals: (1 ÷ 10) + (1 ÷ 20) + (1 ÷ 50) = 0.1 + 0.05 + 0.02 = 0.17. Flip it: 1 ÷ 0.17 = 5.88, which rounds to 5.9 ohms. Again, smaller than the smallest resistor (10 ohms). Good sanity check.

Inductors: same rules as resistors

Here is a gift: inductors (coils) combine by the exact same rules as resistors.

• Inductors in series add up.
• Inductors in parallel use the reciprocal recipe (smaller than the smallest).

Worked example 3 (inductors in series). A 20-millihenry inductor in series with a 50-millihenry inductor. Just add: 20 + 50 = 70 millihenries. ("Henry" is the unit of inductance; "milli" means one-thousandth.)

Worked example 4 (inductors in parallel). Three 10-millihenry inductors in parallel. Add reciprocals: (1 ÷ 10) + (1 ÷ 10) + (1 ÷ 10) = 0.3. Flip: 1 ÷ 0.3 = 3.33, or 3.3 millihenries. (Shortcut: equal parts in parallel just divide by how many there are, 10 ÷ 3 = 3.3.)

Two quick "which part to add" questions follow from this. To increase the inductance of a circuit, you add an inductor in series (because series inductors add up). To increase a capacitor's capacitance, you add a capacitor in parallel (explained next).

Capacitors: the rules are flipped

Capacitors are the oddball. Their combining rules are the opposite of resistors and inductors. This trips people up, so highlight it:

• Capacitors in parallel add up. (Parallel capacitors are like widening a bucket, more total storage. That is why adding a capacitor in parallel increases the capacitance.)
• Capacitors in series use the reciprocal recipe (the total is smaller than the smallest).

Worked example 5 (capacitors in parallel). Two 5.0-nanofarad capacitors and one 750-picofarad capacitor, all in parallel. In parallel they simply add, but first put them in the same unit. A "nanofarad" (nF) is a thousand "picofarads" (pF), so 750 pF is 0.75 nF. Now add: 5.0 + 5.0 + 0.75 = 10.75 nanofarads.

Worked example 6 (capacitors in series, all equal). Three 100-microfarad capacitors in series. Add reciprocals: (1 ÷ 100) × 3 = 0.03. Flip: 1 ÷ 0.03 = 33.3, or 33.3 microfarads. (Shortcut for equal parts in series: divide by how many, 100 ÷ 3 = 33.3.) Notice it is smaller than the smallest, just like resistors in parallel.

Worked example 7 (two unequal capacitors in series). A 20-microfarad capacitor in series with a 50-microfarad capacitor. Add reciprocals: (1 ÷ 20) + (1 ÷ 50) = 0.05 + 0.02 = 0.07. Flip: 1 ÷ 0.07 = 14.28, or 14.3 microfarads.

Memory hook for the whole group: resistors and inductors add in series; capacitors add in parallel. Whenever the wiring matches the "adding" case, just add. Otherwise, use the reciprocal recipe and the answer comes out smaller than the smallest part.

Transformers: trading volts for amps

A transformer is two coils of wire wound near each other. You feed AC into the first coil (the primary winding) and a voltage appears on the second coil (the secondary winding) even though no wires connect them. What carries the energy across the gap? Mutual inductance, which is the magnetic linkage between the two coils. (The changing current in the primary makes a changing magnetic field, and that field induces a voltage in the secondary.) So the cause of the secondary voltage is mutual inductance.

The voltage you get out depends on the ratio of turns (loops of wire) on each coil. The rule: the secondary voltage equals the primary voltage times the secondary turns divided by the primary turns. More turns on the secondary means more voltage out (a "step-up" transformer); fewer means less voltage out (a "step-down" transformer).

Worked example 8 (find the output voltage). A transformer has a 500-turn primary and a 1,500-turn secondary, with 120 volts AC applied to the primary. Output = 120 × (1,500 ÷ 500) = 120 × 3 = 360 volts. (The secondary has 3 times the turns, so it gives 3 times the voltage.)

Worked example 9 (feeding the wrong end). Suppose you have a "4-to-1 voltage step-down" transformer, meaning when used normally it cuts voltage to one-fourth. What if you instead apply your signal to the secondary winding? Then it works in reverse and becomes a step-up: the input voltage is multiplied by 4. (Running a transformer backwards flips step-down into step-up.)

Why the primary wire is sometimes thicker. In a step-up transformer, the primary side carries the higher current (a transformer gives you more voltage only by drawing more current on the other side, energy in equals energy out). Thicker wire is needed to carry more current without overheating. So the primary winding of a step-up transformer usually uses larger wire to accommodate the higher current of the primary.

Transformers for impedance matching

Transformers can also convert one impedance into another (the impedance-matching idea from G5A). The trick: the turns ratio is the square root of the impedance ratio. In words, divide the two impedances, take the square root, and that is the turns ratio.

Worked example 10 (matching an antenna). You want to match a 600-ohm antenna feed point to a 50-ohm coaxial cable. First the impedance ratio: 600 ÷ 50 = 12. Then the square root of 12 is about 3.46, which we read as a turns ratio of 3.5 to 1.