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6.3 Resistance, Capacitance & Inductance

Key Takeaways

  • Resistance is measured in ohms; capacitance in farads; inductance in henrys—three passive properties that dominate RF network behavior

  • Inductive reactance is XL = 2πfL; capacitive reactance is XC = 1/(2πfC); both are frequency-dependent oppositions measured in ohms but are not resistance

  • Series resistances and series inductances add; parallel capacitances add; parallel resistances and series capacitances use reciprocal (product-over-sum) style combinations

  • Inductors store energy in a magnetic field; capacitors store energy in an electric (electrostatic) field

  • Bypass capacitors provide a low-impedance path to ground for AC; reducing coil turns reduces inductance

Last updated: August 2026

6.3 Resistance, Capacitance & Inductance

Quick Answer: R in ohms, C in farads, L in henrys. XL=2πfLX_L = 2\pi f L, XC=1/(2πfC)X_C = 1/(2\pi f C). Series L values add; parallel C values add. Inductors store magnetic energy; capacitors store electric energy. A bypass capacitor shunts AC to ground through a low impedance; fewer turns means less inductance.

If §6.1 named power and §6.2 named materials, this section quantifies the three passive properties you adjust every time you tune a tank, set a coupling network, or troubleshoot an RF deck.

Units and physical meaning

PropertyUnitSymbolStores energy in…DC steady-state ideal behavior
Resistanceohm (Ω)RNone (dissipates as heat)Passes current per Ohm’s law; drops voltage
Capacitancefarad (F)CElectric / electrostatic fieldOpen circuit (after charged); blocks DC
Inductancehenry (H)LMagnetic fieldShort circuit (after current established); passes DC

Practical values: RF caps are often pF–µF; inductors µH–mH; resistors from fractions of an ohm to megaohms. The farad and henry are large base units—prefixes dominate lab talk—but the exam still wants the base unit names.

Resistance opposes current and dissipates true power as heat: P=I2RP = I^2 R. It does not return energy to the circuit each cycle the way pure L and C do.

Reactance: frequency-dependent opposition

Reactance is measured in ohms but is not resistance. It causes phase shift and contributes to impedance (Z = R + jX) without the continuous heat of a pure resistor (ideal case).

Inductive reactance

XL=2πfLX_L = 2\pi f L
  • (f) in hertz, (L) in henrys → XLX_L in ohms.
  • Double frequency or double L → double XLX_L.
  • Coils look more open at higher RF frequencies (until self-resonance and parasitics intervene).

This is the exact formula tested in Topic 3-A: if frequency and coil inductance are known, XL=2πfLX_L = 2\pi f L (not πfL\pi f L, not the capacitive formula).

Capacitive reactance

XC=12πfCX_C = \frac{1}{2\pi f C}
  • Higher frequency or larger C → smaller XCX_C.
  • Capacitors look more like shorts to high-frequency AC and opens to DC—foundation of coupling and bypass practice.
DeviceReactance vs frequencyPhase (ideal)
InductorXLX_L rises with (f)Current lags voltage by 90°
CapacitorXCX_C falls with (f)Current leads voltage by 90°
ResistorR independent of (f) (ideal)V and I in phase

Worked example — coil reactance

An antenna loading coil is 50 µH at 2.0 MHz (near the 160 m / marine MF neighborhood for scale).

XL=2π(2.0×106)(50×10−6)≈628 ΩX_L = 2\pi (2.0\times 10^6)(50\times 10^{-6}) \approx 628\,\Omega

At 4.0 MHz, the same coil presents about 1.26 kΩ of XLX_L. Frequency doubled → reactance doubled. That is why a coil that “looks modest” at low HF can dominate impedance higher up.

Worked example — capacitor reactance

A 0.01 µF bypass at 1 MHz:

XC=12π(1×106)(0.01×10−6)≈16 ΩX_C = \frac{1}{2\pi (1\times 10^6)(0.01\times 10^{-6})} \approx 16\,\Omega

At audio (say 1 kHz) the same capacitor is roughly 16 kΩ—poor as a bypass, fine as a coupling thought experiment. Frequency agility of XCX_C is the whole point of choosing bypass values.

Series and parallel combination formulas

Resistance

  • Series: RT=R1+R2+R3+⋯R_T = R_1 + R_2 + R_3 + \cdots (always larger than any single resistor).
  • Parallel: 1RT=1R1+1R2+⋯\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots, or for two resistors RT=R1R2R1+R2R_T = \frac{R_1 R_2}{R_1 + R_2} (always smaller than the smallest branch).

Inductance

  • Series inductors (no mutual coupling): LT=L1+L2+⋯L_T = L_1 + L_2 + \cdots — same pattern as series R. Element 3 explicitly asks for LT=L1+L2L_T = L_1 + L_2 as the series formula.
  • Parallel inductors (ideal, no mutual coupling): reciprocal sum, analogous to parallel R.

Capacitance

  • Parallel capacitors: CT=C1+C2+C3C_T = C_1 + C_2 + C_3 — add, like series R. Element 3 tests CT=C1+C2+C3C_T = C_1 + C_2 + C_3 for three capacitors in parallel.
  • Series capacitors: reciprocal sum; total C is less than the smallest capacitor.
ConfigurationResistancesInductances (ideal)Capacitances
SeriesAddAddReciprocal sum
ParallelReciprocal sumReciprocal sumAdd

Memory hook: Series L like series R; parallel C like series R (they add).

Energy storage and time-domain behavior (overview)

Inductor: energy EL=12LI2E_L = \frac{1}{2} L I^2 in the magnetic field. Current through an inductor cannot change instantaneously; the coil “resists” current change with induced voltage. At switch-on, an inductor initially behaves like an open; after many time constants in a resistive circuit, DC current settles as if the inductor were a short (wire resistance only).

Capacitor: energy EC=12CV2E_C = \frac{1}{2} C V^2 in the electric field. Voltage across a capacitor cannot change instantaneously. At switch-on to a DC source through resistance, the capacitor initially looks like a short (charging surge), then approaches an open once charged.

Time-constant sketch (RC / RL): τ=RC\tau = RC or τ=L/R\tau = L/R. After about 5τ, transients are essentially settled for exam-level work. Detailed RC time-constant math returns in Topic 3-B; here you only need the qualitative “L magnetic / C electric” storage picture and that reactance—not resistance—sets sinusoidal AC phase shift.

Practical RF component behaviors from the pool

Bypass capacitor purpose: it removes alternating current by providing a low-impedance path to ground (for the AC component), keeping DC bias undisturbed. It does not primarily “remove DC by shunting DC to ground”—that would destroy bias supplies. Think: AC short, DC open.

Reducing the inductance of an antenna coil: reduce the number of turns. Adding turns, raising core permeability, or compressing turns generally increases inductance (or coupling), not decreases it.

Conductors again (RLC topic overlap): good conductors → many free electrons; best low-resistance metals → gold, silver, copper. Same material literacy as §6.2, now tied to minimizing R losses in coils and tank circuits.

Putting R, L, and C together on one mental schematic

Imagine a series R–L–C filter feeding an RF stage:

  1. R sets damping and true-power loss.
  2. L contributes XL=2πfLX_L = 2\pi f L and magnetic energy storage.
  3. C contributes XC=1/(2πfC)X_C = 1/(2\pi f C) and electric energy storage.
  4. Net reactance X=XL−XCX = X_L - X_C (series convention) goes to zero at resonance—preview of later RLC circuit topics—but even before resonance math, you must compute each reactance correctly.
  5. Parallel bypass caps on supply pins use large enough C (low enough XCX_C at the frequencies of interest) so AC noise sees ground.

When Element 3 shows a formula stem, match symbols carefully: 2πfL2\pi f L is inductive; 1/(2πfC)1/(2\pi f C) is capacitive; series L adds; parallel C adds. Those four facts clear most of key topic 004 and the reactance item from 001.

Test Your Knowledge

What formula gives the inductive reactance of a coil when frequency and inductance are known?

A

XL = 1 / (2πfC)

B

XL = πfL

C

XL = 2πfL

D

XL = 1 / (R² + X²)

Test Your Knowledge

How do you calculate the total inductance of two series inductors (no mutual coupling) and the total capacitance of three capacitors in parallel?

A

LT = L1 / L2; CT = 1/C1 + 1/C2 + 1/C3

B

LT = L1 + L2; CT = C1 + C2 + C3

C

LT = 1/L1 + 1/L2; CT = C1 × C2 × C3

D

LT = L1 × L2; CT = C1 / C2 + C3

Test Your Knowledge

Where do ideal inductors and capacitors store energy, respectively?

A

Both store energy only as heat in resistance

B

Electric field; magnetic field

C

Magnetic field; electric (electrostatic) field

D

Chemical electrolyte; nuclear field

Test Your Knowledge

What is the purpose of a bypass capacitor, and how can you reduce the inductance of an antenna coil?

A

It removes AC by providing a low-impedance path to ground; reduce the number of turns

B

It removes DC by shunting DC to ground; add more turns

C

It increases the resonant frequency only; add higher-permeability core material

D

It blocks all AC from ground; compress the coil turns to raise inductance intentionally

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