Impedance & Reactance

01 / 12

Why resistance is no longer enough

Three loads all have an impedance magnitude of 50 Ω at 2.45 GHz. If the same 0.7071 Vrms sinusoidal voltage is imposed across each, do they draw the same current, absorb the same average power, and return the same phase relationship?

Predict before opening the answer. The loads are an ideal 50 Ω resistor, an ideal 3.248060 nH inductor, and an ideal 1.299224 pF capacitor. Those component values are chosen so the two reactances have magnitude 50 Ω at exactly 2.45 GHz.

Think about itIf |Z| and imposed Vrms are equal, which results must be equal—and which can differ?

Answer

Current magnitude is equal: 0.7071 V / 50 Ω = 14.142 mA. Current phase and average real power differ. The resistor absorbs 10 mW. The ideal L and C absorb zero average real power in periodic sinusoidal steady state, although energy moves in and out during every cycle.

R / L / C energy-flow explorer

Same voltage magnitude. Three different energy stories.

Compare synchronized terminal voltage, current, instantaneous power, and stored energy for the three ideal 50 Ω-magnitude loads.

Ideal load

Phase scrubber

Voltage: +1.000 V
Current: +20.000 mA
Instantaneous power: +20.000 mW
Stored energy now: 0.0000 pJ
RMS / phase / average: 0.7071 V · 14.142 mA0° · in phase; 10.000 mW

Assumptions: ideal imposed terminal voltage, linear lumped component, sinusoidal steady state, and the e^jωt convention. This is not a claim that a real “50 Ω generator” holds 0.7071 Vrms across every load. The GHz waveform is normalized and slowed. Ideal L and C return energy, so their instantaneous power changes sign while average real power is zero.

Without JavaScript: all three loads draw 14.142 mArms in magnitude. Only the resistor absorbs 10 mW average; the ideal L and C exchange energy with the source.

Same ideal imposed terminal voltage at 2.45 GHz
Load	Impedance	Current	Average real power	Relationship
Ideal resistor	50 + j0 Ω	14.142 mArms · 0°	10.000 mW	Voltage and current in phase
Ideal inductor	0 + j50 Ω	14.142 mArms · −90°	0 W	Current lags; magnetic energy returns
Ideal capacitor	0 − j50 Ω	14.142 mArms · +90°	0 W	Current leads; electric energy returns

This is an ideal imposed terminal voltage. It isolates the loads so their behavior can be compared. It does not mean a practical “50 Ω generator” will maintain 0.7071 Vrms across every reactive termination.

Common misconceptionImpedance is just AC resistance, and an ohm always describes dissipation.

Resistance is the real part of impedance. Reactance is measured in ohms too, but it records quadrature behavior associated with stored and returned energy. The unit alone does not tell you whether average real power is absorbed.

Common misconceptionA 50 Ω impedance must be a 50 Ω resistor.

At this one frequency, +j50 Ω, −j50 Ω, and 50 + j0 Ω all have magnitude 50 Ω. Magnitude has discarded the angle that distinguishes them.

02 / 12

Resistance: where electrical energy leaves

What makes the resistor different from the two energy-storage elements?

Under the passive sign convention, current enters the terminal marked positive for voltage. A positive value of $p(t)=v(t)i(t)$ means electrical energy enters the modeled element. For an ideal positive resistor, $v(t)=Ri(t)$ , so voltage and current remain in phase and instantaneous power never becomes negative.

$P_{\mathrm{avg}} = V_{\mathrm{rms}}I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}^{2}}{R} = I_{\mathrm{rms}}^{2}R$ These equalities use an ideal resistance and RMS voltage and current at the same port.

The modeled resistance tells us that real energy leaves the electrical port. It may become conductor heat, dielectric loss, useful load power, or radiation. Antenna radiation resistance, for example, accounts for radiated power in a port model; it is not necessarily one hot, visible resistor.

Think about itCould a measured positive resistance represent both radiation and heat?

Answer

Yes. A terminal model can contain radiation resistance and loss resistance. Both absorb real power from the port, but only one represents desired radiation.

Common misconceptionMatched means lossless, and maximum power transfer automatically means maximum efficiency.

A matched input can still contain substantial conductor, dielectric, or radiation loss. In the restricted resistive Thevenin maximum-power case, equal source and load resistance also means half the generated power is lost in the source resistance. Matching, loss, and efficiency are separate claims.

03 / 12

Capacitance: energy in an electric field

Why must current arrive before a capacitor voltage reaches its crest?

A capacitor stores separated charge. The charge is proportional to terminal voltage: $q=Cv$ . Current is the rate at which that charge changes, so current is largest while voltage crosses zero most steeply and becomes zero at a voltage crest.

$q=Cv\qquad i=C\frac{\mathrm dv}{\mathrm dt}$ $W_{\mathrm C}=\frac12Cv^2$ $Z_{\mathrm{C}} = \frac{1}{j\omega C} = -\frac{j}{\omega C}$ C is capacitance [F], ω = 2πf [rad/s], and W_C is stored electric-field energy [J].

With the e^jωt convention, capacitor current leads capacitor voltage by 90°. The signed reactance is $X_{\mathrm C}=-\frac1{\omega C}$ ; its positive magnitude is 1/(ωC). At 2.45 GHz, the recurring capacitor stores at most about 0.6496 pJ under the imposed 0.7071 Vrms sine.

Think about itIf voltage is at its positive maximum, what are ideal capacitor current and instantaneous power?

Answer

Both are zero at that instant. A quarter-cycle later the capacitor returns stored energy and instantaneous power is negative under the passive sign convention.

Common misconceptionNegative reactance means negative resistance, negative energy, or negative average power.

The minus sign encodes capacitive phase under the stated convention. Stored energy remains nonnegative, and an ideal capacitor has zero average real power over a complete steady-state cycle.

Common misconceptionAn ideal capacitor has zero instantaneous power, or simply ‘passes AC.’

Its instantaneous power alternates in sign as the electric field charges and discharges. Its opposition is frequency-dependent, never a universal pass/fail rule, and a DC-changing transient can drive current too.

04 / 12

Inductance: energy in a magnetic field

Why does inductor current reach its crest after the voltage does?

Inductor voltage is proportional to how quickly current changes. A positive voltage drives current upward; it does not set the current value directly. Current therefore reaches a maximum only after the voltage has fallen through zero.

$v=L\frac{\mathrm di}{\mathrm dt}$ $W_{\mathrm L}=\frac12Li^2$ $Z_{\mathrm{L}} = j\omega L$ L is inductance [H], and W_L is stored magnetic-field energy [J].

Under e^jωt, inductor current lags voltage by 90° and $X_{\mathrm L}=\omega L$ . In the ideal lumped model, increasing frequency increases the voltage needed for the same current amplitude. The recurring inductor also reaches about 0.6496 pJ maximum stored energy.

Think about itIf frequency doubles while L stays fixed, what happens to ideal inductive reactance?

Answer

It doubles. The current magnitude under the same imposed voltage magnitude therefore halves.

Common misconceptionAn inductor simply blocks AC.

Inductive reactance rises continuously with frequency in this ideal model. At any finite reactance, finite voltage produces finite current. Real inductors add loss and parasitic capacitance, so they can eventually stop looking inductive at all.

05 / 12

Reactance: frequency changes the relationship

What single signed quantity records whether ideal L or C dominates at one frequency?

Reactance is the imaginary part of impedance, measured in ohms. Positive reactance is inductive and negative reactance is capacitive under our convention. It records the frequency-dependent quadrature relation that came from magnetic or electric energy storage.

$X = X_{\mathrm{L}} + X_{\mathrm{C}} = \omega L - \frac{1}{\omega C}$ $P_{\mathrm{avg}} = V_{\mathrm{rms}}I_{\mathrm{rms}} \cos \phi$ φ is the voltage phase minus current phase. Magnitudes alone cannot determine average real power.

Think about itBelow the L–C cancellation frequency, which reactance dominates in the recurring pair?

Answer

The capacitor. The negative magnitude 1/(ωC) grows as frequency falls, while positive ωL shrinks.

Reactance-versus-frequency explorer

Frequency pulls L and C in opposite directions.

Move frequency on a logarithmic scale or change L and C. The plot uses sign and line pattern: inductive reactance is positive; capacitive reactance is negative.

Frequency 2.4500 GHz

Inductance 3.248 nH

Capacitance 1.299 pF

Inductor · positive: X_L = +50.00 ΩωL
Capacitor · negative: X_C = −50.00 Ω−1/(ωC)
Series net reactance: −0.00 Ωapproximately zero
Ideal cancellation frequency: 2.4500 GHzf₀ = 1/(2π√LC)

Model: ideal lumped L and C in sinusoidal steady state. The plot is centered on the calculated f₀ and clips beyond ±125 Ω for readability; the textual values are not clipped. Real components add loss and parasitics.

Common misconceptionImpedance is one fixed value independent of frequency.

Even ideal L and C prove otherwise. A broadband signal generally encounters a function Z(f); each frequency component can see a different magnitude and phase.

06 / 12

Complex impedance: magnitude and phase together

How can one number preserve both current magnitude and phase?

Before using phasors, make the model contract explicit:

Network: a linear lumped model.
Signal: sinusoidal steady state at one stated frequency.
Phasors: RMS values using e^jωt.
Location: voltage and current defined at the same port and reference plane.

A complex number keeps two perpendicular components together. Resistance lies on the real axis. Reactance lies on the imaginary axis. Their vector length is impedance magnitude and their direction is impedance angle.

$Z = \frac{V}{I} = R + j X$ $|Z| = \sqrt{R^{2} + X^{2}} \qquad \angle Z = \operatorname{atan2}(X, R)$ Because I = V/Z, current phase relative to voltage is −∠Z.

Think about itCan 30 + j40 Ω have the same magnitude as a 50 Ω resistor?

Answer

Yes. √(30² + 40²) = 50 Ω, but its angle is +53.13°. Under the stated convention, current lags voltage by 53.13°.

Complex-impedance vector laboratory

Keep magnitude and phase instead of throwing one away.

Adjust resistance and signed reactance. Rectangular form names the components; polar form names the resulting length and angle.

Reading the vector

30.0 + j40.0 Ω

The horizontal projection is resistance. The vertical projection is signed reactance. The vector length is |Z|, not another resistor value.

Current lags voltage by 53.13°. Because I = V/Z, current phase relative to voltage is −∠Z.

Useful presets

Resistance R 30 Ω

Reactance X +40 Ω

Rectangular form: 30.00 + j40.00 Ω
Magnitude: |Z| = 50.00 Ω
Impedance angle: ∠Z = +53.13°
Current phase: −53.13°relative to voltage

Assumptions: RMS phasors at one frequency, e^jωt, voltage and current at the same reference plane. The symbol j rotates a phasor by 90°; it does not describe physically imaginary electricity.

Common misconceptionj represents a physically imaginary voltage or current.

Real instruments measure real waveforms. Complex notation is mathematical bookkeeping that carries magnitude and quadrature phase compactly. Multiplication by j represents a 90° rotation in the phasor plane.

07 / 12

Combining impedances without losing phase

What exactly should be added when components share current—or share voltage?

Series elements carry the same current, so their complex voltage drops add and their impedances add. Parallel elements share voltage, so their complex branch currents add and their admittances add.

$Z_{\mathrm{series}} = \sum Z_{k}$ $Y = \frac{1}{Z} = G + j B \qquad Y_{\mathrm{parallel}} = \sum Y_{k}$ G is conductance [S] and B is susceptance [S]. Invert the final parallel admittance to recover impedance.

Think about itAt 2.45 GHz, what is the series sum of 50 Ω, +j50 Ω, and −j50 Ω?

Answer

It is 50 + j0 Ω. Add the signed complex terms first. Adding their three magnitudes would incorrectly give 150 Ω.

Series / parallel impedance builder

Combine complex terms before taking the magnitude.

This deliberately small builder contains one R, one L, and one C. It shows every term so the arithmetic stays visible.

One current path

Series R–L–C

Every term at 2.450 GHz

Z_R50.000 + j0.000 Ω
Z_L0.000 + j50.000 Ω
Z_C0.000 − j50.000 Ω

Connection

Frequency 2.450 GHz

Resistance 50 Ω

Inductance 3.248 nH

Capacitance 1.299 pF

Actual calculationZ = ZR + ZL + ZC = 50.00 + j0.00 Ω + 0.00 + j50.00 Ω + 0.00 − j50.00 Ω = 50.00 + j0.00 Ω

Equivalent impedance: 50.00 + j0.00 Ω
Magnitude: 50.00 Ω
Angle: −0.00°
Behavior: approximately resistive

Assumptions: ideal lumped elements, RMS phasors, one stated frequency, and one common reference plane. Series impedances add. Parallel admittances add. This is a constrained teaching model, not a general circuit simulator.

A compact complex voltage divider

With $V_s=1\angle0^\circ\,\mathrm{V_{rms}}$ , $Z_1=30+j40\,\Omega$ , and $Z_2=50+j0\,\Omega$ , the voltage across Z₂ is:

$V_{2} = \frac{V_{s}Z_{2}}{Z_{1} + Z_{2}} = \frac{50}{80 + j 40} = 0.5 - j 0.25 V = 0.559\angle -26.57^{\circ} V$ The divider ratio is complex; it changes both magnitude and phase.

Common misconceptionSeries impedance magnitudes add, and parallel impedances add directly.

Series complex impedances add; opposing reactances can cancel. Parallelcomplex admittances add. Magnitude is taken only after the signed calculation.

08 / 12

Resonance: cancellation without disappearance

If terminal reactance becomes zero, where did the inductor and capacitor go?

They did not go anywhere. At series resonance, their voltage contributions are equal and opposite at the input port. Current can still drive large electric and magnetic stored-energy oscillations inside the network.

$Z_{\mathrm{series}} = R + j(\omega L - \frac{1}{\omega C})$ $f_{0} = \frac{1}{2\pi \sqrt{L C}}$ $Y_{\mathrm{parallel}} = \frac{1}{R} + j(\omega C - \frac{1}{\omega L})$ The parallel expression uses a shunt-loss resistor. Topology is part of the definition.

Series resonance commonly gives minimum input impedance.
Parallel resonance commonly gives maximum input impedance.
Real loss keeps both results finite.
Internal L/C voltages or branch currents can be much larger than port values.
Zero terminal reactance means cancellation at that port—not zero stored energy.

Think about itIs a resonant 25 + j0 Ω load automatically matched to a 50 Ω reference?

Answer

No. It is purely resistive at that frequency, but its 25 Ω resistance still differs from the 50 Ω reference.

Common misconceptionAt resonance the inductor and capacitor disappear, and resonance always means maximum current.

In a series voltage-driven RLC, input current peaks because impedance is smallest. In a parallel current-driven RLC, input impedance and voltage commonly peak while source current is fixed. The internal reactive elements continue exchanging energy in both cases.

Common misconceptionZero reactance automatically means a 50 Ω match.

Zero reactance says the input is purely real. A 50 Ω reference is matched only if that real value is also 50 Ω under the relevant line and port definition.

09 / 12

Q, bandwidth, and ring-down

How does the same loss appear in frequency response, stored energy, and settling time?

Quality factor compares the resonator's maximum stored energy with the real energy lost during one cycle. Less loss allows energy to circulate longer, sharpens the simple resonant response, and slows the decay after excitation stops.

$Q = 2\pi \times \frac{\text{maximum stored energy}}{\text{energy dissipated per cycle}}$ $Q_s=\frac{\omega_0L}{R}=\frac1{\omega_0CR}$ $Q_p=\frac R{\omega_0L}=\omega_0CR$ $\mathrm{BW} = \frac{f_{0}}{Q}$ Q_s uses series loss; Q_p uses the stated shunt-loss topology. BW uses the simple resonator and half-power definition shown here.

Think about itIf the recurring series resistance falls from 50 Ω to 5 Ω, what happens at resonance?

Answer

Q rises to 10. With 0.7071 Vrms ideally imposed, current becomes about 141.42 mArms. Each ideal reactive component has about 7.071 Vrms even though the L and C terminal contributions cancel in the input sum.

Resonance and Q explorer

Cancellation at the port can hide large internal exchange.

Compare a series voltage-driven resonator with a parallel current-driven resonator. Adjust loss, sweep frequency, and watch the ring-down envelope.

Purposeful ring-down view

Amplitude envelope after excitation stops

Ring-down position

Loss topology

Loss setting Q = 10.0

Sweep frequency 2.4500 GHz

Loss resistance: 5.000 Ωseries R
Impedance at cursor: 5.000 Ω∠Z -0.00°
Half-power points: 2.33056 / 2.57556 GHz
Bandwidth: 245.000 MHzf₀/Q for this topology
Amplitude ring-down τ: 1.299 ns2Q/ω₀

Model: simple lumped RLC with frequency-independent loss. The upper series trace is |I| normalized to resonant current under an ideal imposed voltage; the parallel trace is |Z|/R for current drive. “Half power” refers to the appropriate resistor power under those stated drives. BW = f₀/Q is not asserted for arbitrary networks. The envelope omits literal carrier cycles.

At Q = 10, the simple series half-power bandwidth is 245 MHz, with f₁ ≈ 2.33056 GHz and f₂ ≈ 2.57556 GHz. The amplitude ring-down time constant is about 1.299 ns. Higher Q also raises sensitivity to tolerance, drift, and internal stress.

Which Q?

Component Q: stored energy versus loss in one component under stated conditions.
Unloaded resonator Q: internal resonator loss without the intended external loading.
External Q: loading introduced by a coupling path.
Loaded Q: the combined internal and external loss seen in operation.

Common misconceptionHigher Q is always better, and Q = f₀/BW is universal.

Higher Q trades bandwidth and settling time for selectivity and stronger internal energy circulation. The bandwidth shortcut depends on a stated response, topology, loading, and half-power definition; it is not a universal way to assign Q to every network.

10 / 12

Real RF components are not ideal symbols

Can a capacitor become inductive—or an inductor capacitive?

Yes. A useful first-order capacitor model adds equivalent series resistance (ESR) and equivalent series inductance (ESL). Its impedance falls toward self-resonance, reaches a loss-limited minimum, then rises with positive phase as ESL dominates.

$Z_{\mathrm{capacitor}} \approx R_{\mathrm{ESR}} + j(\omega L_{\mathrm{ESL}} - \frac{1}{\omega C})$ This first-order model is useful, not complete; ESR itself can vary with frequency.

A real inductor adds DC resistance, frequency-dependent conductor and core loss, and parasitic capacitance. Near self-resonance, its apparent inductance and Q can change rapidly; above resonance, the parallel capacitance can make the part appear capacitive.

Think about itWould a nominal 1.3 pF capacitor necessarily still look capacitive above its self-resonance?

Answer

No. In a first-order ESR–ESL–C model, ESL dominates above self-resonance and the terminal phase becomes inductive.

Ideal / real component and reference-plane explorer

A part number is not a context-free impedance.

Switch between ideal and first-order parasitic models, then move the reporting plane outward to include an illustrative pad-and-lead fixture.

Illustrative model · not manufacturer data

Equivalent circuit

series ESR, ESL, and capacitance C

Component

Electrical model

Reference plane

Frequency 2.4500 GHz

Reported impedance: 0.300 − j37.655 Ω
Magnitude: 37.656 Ω
Phase: -89.54°appears capacitive
Device-model SRF: 4.935 GHzfixture can shift the reported response

Illustrative values: capacitor ESR 0.3 Ω and ESL 0.8 nH; inductor DCR 0.25 Ω and parasitic C 0.5 pF; fixture series L 0.35 nH and shunt C 0.12 pF. Real values vary with package, pads, vias, traces, fixture, bias, level, temperature, and frequency. Use manufacturer models or calibrated measurement for design.

Without JavaScript: a real capacitor reaches a minimum |Z| near self-resonance and can look inductive above it; a real inductor can look capacitive above its self-resonance.

Package size, electrode geometry, pads, vias, traces, fixture, bias, signal level, temperature, and nearby structures can all alter the result. A nominal value measured under one low-frequency condition does not completely describe GHz behavior.

Common misconceptionA nominal capacitor or inductor remains ideal at every frequency, and one low-frequency LCR value predicts GHz behavior.

Real components are networks of intended and parasitic effects. Use-frequency data, manufacturer models, and an appropriate fixture or calibrated RF measurement are needed when those effects matter.

11 / 12

Measuring impedance

When an instrument reports 47.3 − j12.8 Ω, what conditions belong to that number?

At minimum: frequency, signal level, DC bias when relevant, port, calibrated reference plane, fixture treatment, temperature or environment when relevant, and the model used to convert raw observations into impedance.

Choose an instrument and method for the frequency and device
Instrument	What it is good at	Watch carefully
LCR meter	Convenient R, L, C, dissipation, or Q at supported spot frequencies	Test frequency, level, bias, chosen equivalent model, and fixture compensation
Impedance analyzer	Broad swept impedance magnitude and phase with component-oriented fixtures	Instrument method, range, residual fixture errors, calibration, and compensation
Vector network analyzer	Calibrated complex reflection or transmission across RF and microwave frequency	Calibration standards, connector repeatability, cables, fixture, reference plane, and conversion model

A VNA separates incident and reflected travelling-wave information at a calibrated port. It can convert calibrated complex reflection information into impedance relative to a stated reference impedance. Reflection coefficient, S-parameters, and Smith-chart construction come later.

VNA portcalibration starts hereFixture inputlaunch, pad, and lead includedDevice terminalsdesired reporting plane

Calibration establishes known behavior at a plane. Fixture compensation or de-embedding attempts to move that plane by removing a model or measurement of intervening structures. Move the plane and the reported impedance changes because a different set of parasitics is included.

Think about itIf the same capacitor is reported at its pads and at the end of a leaded fixture, should the two impedances be identical?

Answer

No. Unless the fixture is removed accurately, the outer plane includes extra series inductance, shunt capacitance, loss, and propagation behavior.

Common misconceptionA lumped 50 Ω resistor and a 50 Ω transmission line are the same object.

A lumped resistor relates terminal voltage and current while dissipating power. A line's characteristic impedance relates travelling voltage and current waves along a distributed structure. A terminated line can present 50 Ω at its input, but the concepts are not interchangeable.

Common misconceptionThere is a universal MHz or GHz threshold where lumped analysis stops working.

Electrical size, propagation delay, rise time, bandwidth, geometry, discontinuities, and acceptable error decide whether a lumped approximation is adequate. Frequency alone cannot set one universal boundary.

Go deeperWhat this lesson deliberately defers

Transmission-line transformation, reflection coefficient, return loss, VSWR, S-parameters, full Smith-chart construction, matching-network synthesis, de-embedding, nonlinear impedance, load pull, and Bode–Fano limits need travelling-wave and network concepts not yet developed here.

12 / 12

Putting it together — one 2.45 GHz impedance journey

Can one table connect energy, phase, resonance, Q, real parts, and measurement?

Start with an ideal 0.7071 Vrms sine imposed at one terminal plane. At 2.45 GHz, ω ≈ 15.3938 × 10⁹ rad/s. The exact teaching values $L=3.248060063\ldots\,\mathrm{nH}$ and $C=1.299224025\ldots\,\mathrm{pF}$ create reactances of ±50 Ω. Displayed component values elsewhere are rounded approximations; calculations use the exact relationships.

01Impose V at one plane0.7071 Vrms · sinusoidal steady state
02Keep phaseR absorbs; L and C store and return
03Combine complex termssigned reactance cancels at f₀
04Add realityloss, parasitics, fixture, calibration

The recurring R, L, C values under each stated condition
Network	Frequency	Input impedance	Current magnitude	Interpretation
Standalone R	2.45 GHz	50 + j0 Ω	14.142 mArms	10.000 mW average
Standalone L	2.45 GHz	+j50 Ω	14.142 mArms	0 W ideal average; current lags 90°
Standalone C	2.45 GHz	−j50 Ω	14.142 mArms	0 W ideal average; current leads 90°
Series RLC	1.225 GHz	50 − j75 Ω	7.845 mArms	\|Z\| ≈ 90.14 Ω; ∠Z ≈ −56.31°
Series RLC	2.45 GHz	50 + j0 Ω	14.142 mArms	Series resonance; terminal reactance cancels
Series RLC	4.90 GHz	50 + j75 Ω	7.845 mArms	\|Z\| ≈ 90.14 Ω; ∠Z ≈ +56.31°
Low-loss series RLC	2.45 GHz	5 + j0 Ω	141.42 mArms	Q = 10; each reactive component ≈ 7.071 Vrms

At half resonance, the series capacitor contributes −j100 Ω while the inductor contributes +j25 Ω, leaving −j75 Ω. At twice resonance, the inductor contributes +j100 Ω while the capacitor contributes −j25 Ω, leaving +j75 Ω. The signs reverse the current lead/lag relation.

The connected mental model

Resistance accounts for real energy leaving an electrical port. Capacitance and inductance store energy in electric and magnetic fields, then return it. Reactance records the resulting frequency-dependent phase behavior. Complex impedance keeps resistance and reactance together. Put these quantities into a network and resonance, Q, bandwidth, and settling follow from how energy storage and loss interact.

Think about itIf the frequency moves from 2.45 GHz to 4.90 GHz while R, L, and C remain unchanged, what changes?

Answer

Resistance stays 50 Ω in the ideal model. Inductive reactance doubles from +50 to +100 Ω. Capacitive reactance halves in magnitude from −50 to −25 Ω. The series network becomes 50 + j75 Ω, |Z| ≈ 90.14 Ω, ∠Z ≈ +56.31°, and current now lags the imposed voltage.

The next module, Fields & Waves, steps outside the lumped port model. It will connect these voltage, current, and stored-energy ideas to fields that vary across physical structures and space.

Ungraded review

Check your understanding

Answer each question in your own words, then reveal the model answer.

01A 0.7071 Vrms sine is applied separately to 50 Ω, +j50 Ω, and −j50 Ω impedances. Which result is shared, and which is not?
Model answer
Each draws 14.142 mArms in magnitude. The resistor absorbs 10.000 mW average; the ideal inductor and capacitor absorb zero average power because their currents are 90° out of phase with voltage.
02At 2.45 GHz, a series network is 50 + j50 − j50 Ω. Is resonance the same as the reactive components disappearing?
Model answer
No. The terminal reactances cancel, leaving 50 + j0 Ω, but the inductor and capacitor still store and return energy. It is a 50 Ω match only because the remaining resistance is 50 Ω.
03The same series RLC network moves from 2.45 GHz to 4.90 GHz. What is its new impedance?
Model answer
Inductive reactance doubles to +j100 Ω and capacitive reactance falls to −j25 Ω. The total is 50 + j75 Ω, with magnitude about 90.14 Ω and angle +56.31°, so current lags voltage.

Sources and further study

This lesson synthesizes the sources below. MIT and Keysight provide the main theory and measurement backbone; Murata and Coilcraft ground the real-component models; the remaining references preview Smith charts and matching without reproducing their prose or figures.

Core theory

Components, measurement, and next steps

Continue in RF FundamentalsFields & Waves

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