E5: Electrical Principles

This group is all about what happens when you combine a resistor (R), an inductor (L), and a capacitor (C) into one circuit, an "RLC circuit", and then find the one special frequency where something remarkable happens.

First, what is reactance?

Both coils and capacitors push back against AC, but in opposite ways, and how hard they push depends on the frequency. We measure this push in ohms, just like resistance, and we call it reactance (symbol X).

• Inductive reactance (the coil's push), written X-sub-L, grows as frequency goes up. Its formula is X-sub-L = 2 × pi × f × L, where f is frequency in hertz and L is inductance in henries. A coil fights fast changes harder, so higher frequency means more push.
• Capacitive reactance (the capacitor's push), written X-sub-C, shrinks as frequency goes up. Its formula is X-sub-C = 1 ÷ (2 × pi × f × C), where C is capacitance in farads. Notice the formula is "one divided by" the rest, so as f gets bigger, X-sub-C gets smaller.

Hold onto that contrast: as frequency rises, the coil pushes harder and the capacitor pushes softer. Somewhere in the middle they must be equal. That crossing point is resonance.

Resonance: where the two pushes cancel

Resonance is the one frequency at which X-sub-L exactly equals X-sub-C. Because inductive reactance and capacitive reactance act in opposite directions, when they are equal they cancel each other out. The formula for the resonant frequency is:

f = 1 ÷ (2 × pi × the square root of (L × C))

Let's work a real exam problem. R is 22 ohms, L is 50 microhenries, and C is 40 picofarads. (A microhenry is a millionth of a henry; a picofarad is a trillionth of a farad. Notice R never enters the resonance formula at all, it is a distractor.)

1. Convert units: L = 50 microhenries = 0.000050 henries (50 × 10 to the minus 6). C = 40 picofarads = 0.000000000040 farads (40 × 10 to the minus 12).
2. Multiply L × C = (50 × 10 to the minus 6) × (40 × 10 to the minus 12) = 2000 × 10 to the minus 18, which is 2 × 10 to the minus 15.
3. Square root of (2 × 10 to the minus 15) is about 4.47 × 10 to the minus 8.
4. Multiply by 2 × pi (about 6.283): 6.283 × 4.47 × 10 to the minus 8 = about 2.81 × 10 to the minus 7.
5. Take 1 divided by that: 1 ÷ (2.81 × 10 to the minus 7) = about 3,560,000 Hz, which is 3.56 MHz.

That matches the pool answer exactly: 3.56 MHz.

One more, to lock it in. R is 33 ohms, L is 50 microhenries, C is 10 picofarads. Same recipe: L × C = (50 × 10 to the minus 6) × (10 × 10 to the minus 12) = 500 × 10 to the minus 18 = 5 × 10 to the minus 16. Its square root is about 2.236 × 10 to the minus 8. Times 2 × pi gives about 1.405 × 10 to the minus 7. One divided by that is about 7,120,000 Hz, or 7.12 MHz. That is the pool answer.

What impedance looks like at resonance

Remember, impedance (symbol Z) is the total opposition to AC, combining resistance and reactance. At resonance the two reactances cancel, so the leftover opposition is just the resistance. But the answer depends on whether the parts are wired in a line (series) or side by side (parallel):

• In a series RLC circuit at resonance, the impedance is approximately equal to the circuit resistance (a minimum value). With the reactances cancelled, only R is left, so current is at a maximum.
• In a parallel RLC circuit at resonance, the impedance is also approximately equal to the circuit resistance. (Both kinds of resonant circuit settle to "just the resistance," but for opposite physical reasons.)

Current behavior at resonance

This part trips people up, so go slow. A parallel LC circuit at resonance acts like an energy merry-go-round. Two different currents matter:

• The circulating current sloshing back and forth inside the loop, between the coil and capacitor, is at a maximum at resonance. Energy bounces between the magnetic field of the coil and the electric field of the capacitor.
• The input current coming from the source into the parallel RLC circuit is at a minimum at resonance, because the circuit looks like a very high impedance to the outside source.

Picture a playground swing: a tiny push each cycle (small input current) keeps a big swinging motion going (large circulating current).

Phase at resonance

At resonance the reactances cancel, leaving a purely resistive circuit, and in a pure resistance the voltage and current are in phase (they rise and fall together, no timing offset). That is the tell-tale sign of resonance: zero phase angle.

Q, the quality factor

Q (short for "quality factor") measures how sharp and efficient a resonant circuit is, basically how little energy it wastes per cycle. A high Q means a very narrow, selective response; a low Q means a broad, sloppy response.

For a parallel resonant RLC circuit, Q is calculated as the resistance divided by the reactance of either the inductance or the capacitance. (At resonance those two reactances are equal, so you can use either one.)

Two consequences worth memorizing:

• Increasing the Q of an impedance-matching circuit decreases its matching bandwidth. (Higher Q = narrower. A matching network that works over a wider frequency range has lower Q.)
• Increasing Q in a series resonant circuit makes the internal voltages increase. In fact, resonance is exactly what can make the voltage across the reactances in a series RLC circuit climb higher than the voltage applied to the whole circuit, a surprising but real effect, and the higher the Q, the bigger that voltage rise.

Bandwidth from Q

The half-power bandwidth is the width of frequencies, centered on resonance, over which the circuit still passes at least half its peak power. The formula is dead simple:

Bandwidth = resonant frequency ÷ Q

Example: resonant frequency 7.1 MHz, Q of 150. Bandwidth = 7,100,000 Hz ÷ 150 = about 47,300 Hz = 47.3 kHz. That is the pool answer.

Another: resonant frequency 3.7 MHz, Q of 118. Bandwidth = 3,700,000 Hz ÷ 118 = about 31,400 Hz = 31.4 kHz. Again, exactly the pool answer. Higher Q would give a narrower bandwidth, lower Q a wider one.

This group has two big themes: how fast a capacitor charges or discharges (the time constant), and how far out of step the voltage and current get (the phase angle). It also introduces two "flip-side" quantities, admittance and susceptance.

The RC time constant

When you connect a resistor (R) and capacitor (C) together, the capacitor does not charge up instantly, it fills like a glass of water through a straw, fast at first then slowing down. The natural pace of that filling is set by the time constant (symbol tau, the Greek letter). For an RC circuit:

time constant = R × C

The official definition the exam wants: one time constant is the time required for the capacitor to charge to 63.2% of the applied voltage, or to discharge down to 36.8% of its starting voltage. (Those two percentages always add to 100%, they are two views of the same curve.)

Now a worked problem, and watch the parallel-combination step carefully. Two 220-microfarad capacitors and two 1-megohm resistors, all in parallel.

1. Capacitors in parallel add: 220 + 220 = 440 microfarads total. In farads that is 440 × 10 to the minus 6.
2. Resistors in parallel: two equal 1-megohm resistors in parallel give half, so 0.5 megohms = 500,000 ohms.
3. Time constant = R × C = 500,000 × (440 × 10 to the minus 6) = 500,000 × 0.00044 = 220 seconds.

That matches the pool answer: 220 seconds. The trap here is forgetting that capacitors add when in parallel while resistors halve.

How coils and capacitors shift the timing (phase)

In a pure reactance, the voltage and current are not in step, they are a quarter-cycle (90 degrees) apart. Which one leads depends on the part:

• In a capacitor, the current leads the voltage by 90 degrees. (The current rushes in first to charge the plates, then the voltage builds.)
• In an inductor, the voltage leads the current by 90 degrees. (The coil opposes the change, so voltage appears first and current catches up.)

A classic memory aid is "ELI the ICE man." In ELI (inductor, the letter L), voltage E comes before current I. In ICE (capacitor, the letter C), current I comes before voltage E. That single mnemonic answers both questions instantly.

Calculating the phase angle of a series RLC circuit

When R, L, and C are all in series, the net reactance is X-sub-L minus X-sub-C, and the phase angle between the circuit's voltage and current is found with the inverse tangent (arctangent):

phase angle = arctangent of ((X-sub-L minus X-sub-C) ÷ R)

The sign of (X-sub-L minus X-sub-C) tells you who wins. If inductive reactance wins (the result is positive), the circuit is inductive and the voltage leads the current. If capacitive reactance wins (the result is negative), the circuit is capacitive and the voltage lags the current. Let's do all three exam problems.

Problem 1: X-sub-C is 500 ohms, R is 1 kilohm (1000 ohms), X-sub-L is 250 ohms.

1. Net reactance = X-sub-L minus X-sub-C = 250 minus 500 = minus 250 ohms (capacitive, so voltage will lag).
2. Divide by R: minus 250 ÷ 1000 = minus 0.25.
3. Arctangent of minus 0.25 = about minus 14.0 degrees.

So the answer is 14.0 degrees with the voltage lagging the current, exactly the pool answer.

Problem 2: X-sub-C is 300 ohms, R is 100 ohms, X-sub-L is 100 ohms.

1. Net reactance = 100 minus 300 = minus 200 ohms (capacitive, voltage lags).
2. Divide by R: minus 200 ÷ 100 = minus 2.0.
3. Arctangent of minus 2.0 = about minus 63 degrees.

Answer: 63 degrees with the voltage lagging the current. Matches the pool.

Problem 3: X-sub-C is 25 ohms, R is 100 ohms, X-sub-L is 75 ohms.

1. Net reactance = 75 minus 25 = plus 50 ohms (inductive this time, so voltage leads).
2. Divide by R: 50 ÷ 100 = 0.5.
3. Arctangent of 0.5 = about 27 degrees.

Answer: 27 degrees with the voltage leading the current. That is the pool answer. The recipe is always the same; only the sign and the size change.

Admittance and susceptance: the flip side

Engineers sometimes find it handier to talk about how easily AC flows rather than how hard it is opposed. These are the "reciprocal" (one-divided-by) quantities:

• Admittance is the inverse of impedance. (If impedance is the opposition, admittance is the willingness. Just as conductance is one over resistance, admittance is one over impedance.)
• Susceptance is the imaginary part of admittance, the reactive portion of that willingness. Its symbol is the letter B.

Two facts the exam likes:

• When you convert a pure reactance to susceptance, its magnitude is replaced by its reciprocal (you take one divided by it). So a small reactance becomes a large susceptance.
• To convert an impedance written in polar form into the equivalent admittance, take the reciprocal of the magnitude and change the sign of the angle. (We will explain polar form fully in the next group; for now, just remember the recipe: flip the size, flip the sign of the angle.)

An impedance is really two numbers stuck together: the resistance part and the reactance part. There are two standard ways to write those two numbers, and the exam expects you to read both fluently.

Rectangular notation: R plus or minus j-X

In rectangular notation we write an impedance as a resistance plus or minus a reactance, using the letter j to mark the reactive (imaginary) part. The form is R + jX or R minus jX:

• The number before the j is the plain resistance.
• The number with the j is the reactance. A plus j means inductive reactance; a minus j means capacitive reactance.

So 50 minus j25 ohms represents 50 ohms of resistance in series with 25 ohms of capacitive reactance. (The minus j is the giveaway: capacitive.) Likewise, a pure capacitive reactance of 100 ohms, written in rectangular notation, is 0 minus j100, no resistance, all capacitive reactance.

When you graph impedance on rectangular coordinates (the familiar x and y grid), the horizontal (X) axis is the resistive component and the vertical (Y) axis is the reactive component. So:

• A pure resistance (no reactance) plots right on the horizontal axis, because its vertical, reactive value is zero.
• Inductive reactance (plus j) plots above the horizontal axis; capacitive reactance (minus j) plots below it.

Polar notation: a length and an angle

Polar coordinates describe the same impedance as an arrow: a magnitude (how long the arrow is) and a phase angle (which direction it points). So impedances in polar form are described by magnitude and phase angle. This is the natural way to display the phase angle of a circuit that mixes resistance with inductive and/or capacitive reactance, polar coordinates are often used to display that phase angle.

• A pure inductive reactance in polar coordinates has a positive 90-degree phase angle (it points straight up).
• A pure capacitive reactance would point straight down, a negative 90-degree angle.
• A pure resistance points straight out along zero degrees.

Phasor diagrams and frequency-response graphs

• The kind of drawing used to show the phase relationship between impedances at a given frequency is a phasor diagram (little arrows showing magnitude and direction).
• For graphs of circuit frequency response, the Y-axis scale most often used is logarithmic. (A log scale lets a huge range of values, from tiny to enormous, fit neatly on one chart, the same idea behind decibels.)

Reading Figure E5-1

Figure E5-1 is a rectangular impedance grid: resistance runs along the horizontal axis, inductive reactance (plus j) is above the axis, and capacitive reactance (minus j) is below. To place a circuit, you compute its resistance (R, already given) and its reactance (X, which you calculate from frequency), then find the matching point. Remember the two reactance formulas from group E5A:

• X-sub-L = 2 × pi × f × L (a coil, plots above the axis, plus j)
• X-sub-C = 1 ÷ (2 × pi × f × C) (a capacitor, plots below the axis, minus j)

Example A: a 400-ohm resistor in series with a 38-picofarad capacitor at 14 MHz. This is a capacitor, so the reactance is negative (below the axis). Compute X-sub-C:

1. 2 × pi × f × C = 6.283 × 14,000,000 × (38 × 10 to the minus 12) = about 0.00334.
2. X-sub-C = 1 ÷ 0.00334 = about 299 ohms.

So the impedance is about 400 minus j299: resistance 400 (well to the right) and a fairly large negative reactance of about 300 (well below the axis). The point that sits far right and low is Point 4. That matches the pool answer.

Example B: a 300-ohm resistor in series with an 18-microhenry inductor at 3.505 MHz. This is a coil, so the reactance is positive (above the axis). Compute X-sub-L:

1. X-sub-L = 2 × pi × f × L = 6.283 × 3,505,000 × (18 × 10 to the minus 6) = about 396 ohms.

So the impedance is about 300 plus j396: resistance 300, and a large positive reactance of about 400 (well above the axis). The point sitting at the middle-right and high up is Point 3. That is the pool answer.

Example C: a 300-ohm resistor in series with a 19-picofarad capacitor at 21.200 MHz. A capacitor again, so negative reactance (below the axis). Compute X-sub-C:

1. 2 × pi × f × C = 6.283 × 21,200,000 × (19 × 10 to the minus 12) = about 0.00253.
2. X-sub-C = 1 ÷ 0.00253 = about 395 ohms.

So the impedance is about 300 minus j395: resistance 300, large negative reactance about 400 (well below the axis). That point is Point 1. That matches the pool answer.

The reading trick: the resistor value tells you how far right the point sits; the sign of the reactance tells you up (inductor, plus j) or down (capacitor, minus j); and the size of the reactance tells you how far up or down. Compute R and X, then just look.

This last group covers the real-world quirks that show up at high frequencies, plus the important difference between power that actually does work and power that just sloshes back and forth.

Skin effect

At DC, current uses the entire cross-section of a wire. But as frequency rises, the current crowds toward the outside surface of the conductor and abandons the center. That crowding is called skin effect. Because the current is now squeezed into a thinner shell, it has less metal to flow through, so resistance increases as frequency increases, because the RF current flows closer to the surface of the conductor. That is the exam answer, and it is why RF wire is often hollow tubing or silver-plated, only the surface matters.

A related geometry fact: as a conductor's diameter increases, its electrical length increases. (A fatter conductor behaves as if it were slightly longer at radio frequencies, which is why thick antenna elements need to be cut a bit shorter than thin-wire formulas predict.)

Why short leads matter at VHF and above

Every piece of wire, even a component's own leads, acts like a tiny inductor, and inductive reactance (2 × pi × f × L) grows with frequency. So at high frequencies even a short lead can add meaningful unwanted reactance:

• You keep lead lengths short for VHF-and-above components to minimize inductive reactance.
• At microwave frequencies, short connections are used to reduce phase shift along the connection (a long path at microwave is many degrees of phase, which throws circuits off).

Parasitic effects and self-resonance

Real components are never perfect. A real capacitor has a little unwanted inductance; a real inductor has a little unwanted capacitance. These unwanted, leftover properties are called parasitic characteristics. They cause several problems:

• Electrolytic capacitors are unsuitable at RF because of their parasitic inductance. (Their rolled-up construction acts like a coil at high frequency.)
• An inductor's self-resonance is created by its inter-turn capacitance, the tiny capacitance between neighboring turns of wire.
• More generally, self-resonance of a component is created when its nominal reactance and its parasitic reactance combine. (The intended reactance plus the accidental one form a resonant pair, and above that self-resonant frequency the part stops behaving as labeled.)
• The primary cause of loss in film capacitors at RF is skin effect (in the capacitor's own conductors).

Real power versus reactive power

This is a key distinction. Real power is the power that actually does work, heating a resistor, lighting a bulb, sending a signal. Reactive power is power that surges out into a coil's magnetic field or a capacitor's electric field and then comes right back each cycle, doing no net work. So:

• Reactive power is wattless, nonproductive power. (It is real energy in motion, but it never gets used up.)
• For reactive power, the current and voltage are 90 degrees out of phase. (That 90-degree offset is exactly why the average of the power over a cycle works out to zero.)
• In ideal inductors and capacitors, energy is stored in magnetic or electric fields, but no power is dissipated. The energy just shuttles back and forth, none is lost as heat.

Calculating real power with reactance present

Here is the punchline that makes one exam problem easy: real power is only consumed in the resistance. The pure reactance burns nothing. The formula is real power = current squared × resistance (I-squared × R), and you use only the resistive part.

Worked example: a 100-ohm resistor in series with a 100-ohm inductive reactance, drawing 1 ampere.

1. Real power = I-squared × R = (1 × 1) × 100 = 100 watts.
2. Notice we completely ignored the 100-ohm reactance, because the inductor consumes no real power. That is the whole trick.

The answer is 100 watts, exactly the pool answer. The tempting wrong answers (141.4 or 200 watts) come from mistakenly folding the reactance into the calculation. Do not. Only R counts toward real power.