Impedance & Reactance

01 / 13

Why resistance is no longer enough

Is knowing how much current flows enough to describe an AC load?

For a resistor, the relationship between voltage and current is simple:

$i=\frac vR$

At every instant, the voltage determines the current. If the voltage is a sine wave, the current follows the same shape: it rises, falls, and changes direction at exactly the same time.

This means that resistance tells us two things at once: how much current flows and that the current is in phase with the voltage.

Capacitors and inductors behave differently. They can temporarily store energy and later return it to the circuit. Because of this, their current does not necessarily rise and fall at the same time as the voltage.

So for an AC load, knowing only the amount of current is no longer enough. We also need to know its phase relative to the voltage.

Impedance gives us both pieces of information for a sine wave at a given frequency:

Its magnitude tells us how strongly the circuit opposes the current.
Its angle tells us how far the current and voltage are shifted in phase.

This phase difference also affects power, because instantaneous power depends on the voltage and current that exist at the same moment.

To understand where these differences come from, we will now look at the basic relationship between voltage and current for a resistor, capacitor, and inductor.

02 / 13

Resistance and real power transfer

What happens to energy in a resistor?

Recall from RF Power: an ideal resistor's voltage and current are in phase. They change sign together, so current can reverse while electrical energy still enters the resistor and becomes heat. The waveforms below are our reference for comparing capacitors and inductors.

Both halves deliver energy

One sine-wave cycle · 1 V peak across 50 Ω

Voltage and current
both positiveVoltage and current
both negative

Voltage (V)

Current (mA)

Power into the resistor (mW)

0½ cycle1 cycle → time

Voltage and current reverse together. Positive × positive and negative × negative both give positive power: energy enters the resistor in either half-cycle.

Reading the direction of energy flow

Capacitors and inductors can also return energy. To distinguish the two directions of energy transfer, we will use the same voltage and current references for all three components.

Label the component's terminals A and B. We define voltage as $v(t)=V_A(t)-V_B(t)$ . Positive current enters terminal A; negative current leaves A. Here, current means conventional current: the direction positive charge would move.

The current's sign tells us its direction. To determine which way energy flows, we use both voltage and current. With these references, instantaneous power—the rate of energy transfer at that moment—is:

$p(t)=v(t)i(t)$

$p(t)>0$ : the component receives energy.
$p(t)<0$ : the component returns energy to the circuit.

This choice of references is called the passive sign convention. For an ideal resistor with positive resistance, substituting $v(t)=Ri(t)$ gives $p(t)=Ri^2(t)\geq0$ : it absorbs energy whenever current flows and never returns it.

What resistance can represent in RF

Power averaged over a complete cycle is called real power, measured in watts. It tells us the net energy transferred per second. For a resistor, this is the average heating power you calculated using RMS values in RF Power.

The resistive part of a load represents real power taken from the electrical signal. In a physical resistor, that power becomes heat. An antenna's radiation resistance represents power carried away as radio waves: a resistance in the circuit model, rather than a physical resistor heating up inside the antenna.

Go deeperWhat is physically inside a resistor?

Outside

Inside · enlarged section

A surface-mount device (SMD) is soldered directly to metal contacts on a circuit board. In this film resistor, the resistive film connects the metal ends and converts electrical energy into heat.One common construction, simplified and not to scale. ROHM construction reference.

An ideal resistor only absorbs energy. Capacitors and inductors behave differently: during part of each AC cycle, they return stored energy to the circuit.

03 / 13

Capacitance: energy in an electric field

What does a capacitor store, and what makes current flow?

A capacitor stores energy in an electric field. In its simplest form, two conducting plates are separated by an insulating material called a dielectric.

Connecting a voltage source moves electrons through the external wires. One plate gains electrons and becomes negative; the other loses electrons and becomes positive. The opposite charges create an electric field between the plates, storing energy.

In an ideal capacitor, electrons do not cross the insulating gap. Current in the external wires changes the charge on the plates. When that charge stops changing, current stops, even though a voltage and an electric field can remain.

Change the voltage. Watch the field.

Slow-motion visualization. A sine-wave source changes the voltage; the plates stay fixed.

Voltage across the capacitor (V)

Plate A relative to plate B · V means volts

Paused at this instant

+3 V: A is 3 volts higher than B.
0 V: A and B have the same voltage.
−3 V: A is 3 volts lower than B.

The point at Now and the capacitor show the same instant. At a voltage zero crossing, current still flows. Copper dots show conventional current, opposite to electron motion. No charge crosses the gap.

Static snapshot: the sine-wave voltage is positive and rising, so current enters A and the field grows. The curve shows the preceding cycle, ending at the same instant as the capacitor.

About this illustration

An ideal source drives two air-separated plates. Their area and spacing stay fixed. Field lines are schematic, and the gap is enlarged for visibility. Fainter rear lines suggest depth; the pattern darkens as the field strengthens. The + and − signs show plate polarity and relative charge. Neither the markers nor their speed represent individual electrons.

This shows a repeating sine wave, after switch-on. At a voltage peak, current is momentarily zero; at a zero crossing, current reaches a positive or negative peak. Pause freezes the same instant in both views, including its current direction. The animation is slowed for viewing; its playback speed is not the signal frequency.

Capacitance links voltage and charge

For the ideal, linear capacitor used here, doubling the voltage doubles the charge on each plate. Capacitance, $C$ , tells us how much charge is needed per volt. A larger capacitance needs more charge to reach the same voltage.

$q=Cv$

q

is the signed charge on the plate connected to A, in coulombs; the plate connected to B has the opposite charge.

C

is capacitance, in farads [F]. One farad means one coulomb per volt.

Plate area, spacing and the insulating material determine the capacitance. In Fields & Waves, we will vary the spacing and see how the electric field changes.

Current depends on how quickly voltage changes

Current is the rate at which charge moves: one ampere means one coulomb per second. To change the capacitor voltage, charge must move onto or off its plates. A faster voltage change requires a faster transfer of charge, and therefore more current. With capacitance held constant:

$i=\frac{\mathrm dq}{\mathrm dt}=C\frac{\mathrm dv}{\mathrm dt}$ Lowercase

i

and

v

are values at an instant. The term

\frac{\mathrm dv}{\mathrm dt}

means the rate of voltage change at that instant, measured in volts per second.

The sign follows the direction of change. Rising voltage means positive current into A; falling voltage means negative current, out of A. For a fixed capacitance, current is proportional to the slope of the voltage waveform.

Example — a 1 μF capacitor

Raise its voltage steadily from 0 V to 1 V in 1 ms (0.001 s). This adds 1 μC of charge to plate A. The current during that rise is:

i=(1\,\mathrm{\mu F})\frac{1\,\mathrm V}{0.001\,\mathrm s}=1\,\mathrm{mA}

Make the same voltage change in half the time, and the current doubles to 2 mA.

Now look more closely at a repeating sine-wave voltage, centered on zero. Unlike the steady rise above, a sine wave's slope changes throughout the cycle. Follow the two marked moments below; this shows the repeating response after switch-on.

Current follows the voltage slope

One sine-wave cycle. Each curve is scaled to its own peak; compare timing and sign.

Voltage

Current · dashed

Power into the capacitor

+ In− Back+ In− Back

Time → one complete cycle

1 Rising zero crossing

Voltage is zero but rising fastest. Charge changes fastest, so current is at its positive peak.

2 Positive voltage peak

Voltage momentarily stops changing. Plate charge momentarily stops changing too, so current is zero.

Current reaches its positive peak ¼ cycle before voltage: current leads voltage by 90°. At the falling voltage zero crossing, current reaches its negative peak. Power is zero exactly at every voltage peak and zero crossing.

Stored energy depends on voltage

For this ideal capacitor, the energy stored in its electric field is:

$W_{\mathrm C}=\frac12Cv^2$

W_{\mathrm C}

is stored energy in joules [J],

C

is capacitance in farads [F], and

v

is voltage at that instant, in volts. Use the instantaneous voltage, not its RMS value.

Why one-half? Charge the same 1 μF capacitor from 0 V to 1 V, adding 1 μC of plate charge. Voltage tells us the energy needed per coulomb: 1 V means 1 J per coulomb. The first charge is added at low voltage, so it takes less energy than the same amount added later.

Voltage rises in a straight line as charge is added. Its average over those charge additions is 0.5 V, half the final voltage. The stored energy is therefore:

W_{\mathrm C}=(1\,\mathrm{\mu C})(0.5\,\mathrm V)=0.5\,\mathrm{\mu J}

This averages voltage over added charge, not time. The shaded triangle is half of the 1 V × 1 μC rectangle.

As voltage moves away from zero in either direction, the capacitor stores more energy. As voltage moves back toward zero, it returns that energy to the circuit. The squared voltage in the formula keeps stored energy nonnegative: the 1 μF capacitor stores 0.5 μJ at either +1 V or −1 V, and none at 0 V.

Think about itAn ideal capacitor holds a constant, nonzero voltage. Does current need to keep flowing for that energy to remain stored?

Answer

No. Its plate charges and electric field are constant, so current is zero while stored energy remains. Current is needed to change the voltage; it is not needed to maintain it in this ideal capacitor.

Keep the capacitance and voltage amplitude fixed, and increase the sine-wave frequency. Voltage now changes faster, so more current is needed. This is the first clue that a capacitor's opposition to AC depends on frequency, which we will develop in Reactance.

Go deeperWhat is physically inside a capacitor?

Outside

Inside · enlarged section

In a multilayer ceramic capacitor (MLCC), many metal plates form two connected groups. Alternate plates connect to A or B; the ceramic dielectric keeps the groups apart. Each gap stores energy in its electric field.One common construction, simplified and not to scale. Murata construction reference.

04 / 13

Inductance: energy in a magnetic field

What does an inductor store, and why does current take time to change?

An inductor stores energy in the magnetic field created by current. It is often a wire wound into a coil: the turns reinforce the field through the coil. More current makes a stronger field. Our ideal coil has no resistance, so it does not turn electrical energy into heat.

Current creates the magnetic field

Slow-motion visualization. The current rises, falls, and reverses through the same coil.

Current through the coil (mA)

mA means milliamps. Positive current enters A; negative current leaves A.

Paused at this instant

Darker field lines show a stronger field: it is strongest at either current peak and disappears at zero current. When current reverses, the field reverses too. The dot at Now and the coil show the same instant.

Static snapshot: current is positive and rising at 50 mA. The magnetic field is strengthening, with north at the top of the coil. The curve shows the preceding cycle, ending at the same instant.

About this illustration

An ideal current source drives a repeating sine-wave current through the same linear coil. It supplies the voltage needed to change that current. The arrow inside the source marks the positive current direction. The coil's field is proportional to its current; switch-on behavior is omitted.

Field lines show direction and relative strength, not a measured field in teslas. N and S label the coil's north and south magnetic poles. Copper markers show conventional current, opposite to electron motion; their speed is schematic. Pause freezes the same instant in both views. The animation is slowed for viewing; its playback speed is not the signal frequency.

Voltage sets how quickly current changes

When current changes, its magnetic field changes too. This changing field induces a voltage in the coil that opposes the change in current, whether it is growing or shrinking. This effect is called self-induction. It does not prevent change: a finite applied voltage makes current change gradually, rather than jump instantly to a new value.

Inductance, $L$ , tells us how much voltage is needed to change current at a given rate. For the ideal, linear inductor used here, inductance stays constant:

$v=L\frac{\mathrm di}{\mathrm dt}$

L

is inductance, in henries [H]. The term

\frac{\mathrm di}{\mathrm dt}

means how quickly current changes, in amperes per second. One henry needs one volt to change current by one ampere per second.

A larger voltage changes current faster. A larger inductance needs more voltage for the same rate of change. Voltage sets the slope of the current waveform, not how much current is already flowing.

Example — a 10 mH inductor

Start with zero current. Apply a steady 1 V across a 10 mH inductor (0.010 H) for 1 ms (0.001 s). Current rises steadily. Its increase, $\Delta i$ , is:

\Delta i=\frac{1\,\mathrm V}{0.010\,\mathrm H}(0.001\,\mathrm s)=0.1\,\mathrm A

Current reaches 100 mA. Double the inductance with the same voltage and time, and it reaches only 50 mA.

Now compare sine-wave voltage and current, both centered on zero. Positive voltage keeps current increasing, even while the voltage itself is falling. Follow the two marked moments in this repeating cycle.

Current peaks after voltage

One sine-wave cycle. Each curve is scaled to its own peak; compare timing and sign.

Voltage

Current · dashed

Power into the inductor

− Back+ In− Back+ In

Time → one complete cycle

1 Positive voltage peak

Voltage is greatest, so current increases fastest. Current is crossing zero here: the magnetic field is momentarily zero.

2 Following voltage zero crossing

Current has kept increasing while voltage was positive. It now reaches its peak and the field is strongest. Negative voltage then makes current decrease.

Current reaches its positive peak ¼ cycle after voltage: current lags voltage by 90°, or voltage leads current by 90°. Positive power adds energy to the field; negative power returns energy to the circuit.

Stored energy depends on current

For this ideal inductor, the energy stored in its magnetic field is:

$W_{\mathrm L}=\frac12Li^2$

W_{\mathrm L}

is stored energy in joules [J],

L

is inductance in henries [H], and

i

is current at that instant, in amperes. Use the instantaneous current, not its RMS value.

Why one-half? Return to the 10 mH example. With a constant 1 V, current rises steadily from 0 to 100 mA in 1 ms. Power, $p=vi$ , therefore rises steadily from 0 to 100 mW. Its average during that time is 50 mW, half the final value. All the energy supplied goes into the field:

W_{\mathrm L}=(50\,\mathrm{mW})(1\,\mathrm{ms})=50\,\mathrm{\mu J}

As current moves away from zero in either direction, the field stores more energy. As current moves back toward zero, the field returns energy to the circuit. Squaring current keeps stored energy nonnegative: this inductor stores 50 μJ at either +100 mA or −100 mA, and none at zero current. In the sine wave above, energy is greatest at both current peaks.

Think about itThe positive voltage across an ideal inductor falls to half its previous value. Does its positive current immediately fall to half?

Answer

No. The current keeps increasing while voltage is positive, but at half the previous rate. Current starts decreasing only when the voltage becomes negative. The current value depends on what happened before this instant.

Keep inductance and voltage amplitude fixed, and increase the sine-wave frequency. Each half-cycle is shorter, so voltage has less time to change the current before reversing. The current amplitude becomes smaller. We will connect this to an inductor's increasing opposition to AC in Reactance.

Go deeperWhat is physically inside an inductor?

Two ways to build the coil

Wirewound

Wire around a ceramic core

One insulated wire wraps around a non-magnetic ceramic core. Its two ends attach to thin metal terminals on the ceramic, labelled A and B.

Multilayer

Finished chip →

Layers separated for clarity. Dashed lines show where the vias meet the tracks below. The finished chip is solid ceramic.

Each sheet carries a printed metal track. Short metal-filled holes, called vias, join the tracks when the layers are stacked, forming one coil between the chip's metal terminals A and B.

Ceramic supports and insulates the conductor; it is not part of the conducting path.

The stripe on the small multilayer package marks mounting orientation, not +/− polarity. Its asymmetrical internal layout can make RF behavior depend on mounting direction. Why Murata uses the mark.

Representative constructions, simplified and not to scale. Wire insulation and any protective cover are omitted. Construction references: Coilcraft wirewound and Murata multilayer.

05 / 13

Same RMS values. Different average power.

Does the same RMS current mean the same average power?

We have seen how each component responds to voltage and current. Now compare their average power over a complete cycle.

In three separate tests, an ideal source applies the same sine-wave voltage across each load: 0.7071 V RMS at 2.45 GHz. The resistor is 50 Ω; we chose the inductor and capacitor values so each also draws about 14.1 mA RMS. The voltage and current are already repeating each cycle. This is called sinusoidal steady state; switch-on is not shown.

Where does the energy go?

Slow-motion visualizationEach load: 0.707 V RMS · 14.1 mA RMS

Compare ideal loads

Voltage and current are in phase.

Voltage: A relative to B. Positive current enters A; negative current leaves A.

Power never goes below zero. Energy enters the resistor and becomes heat.

About the example values

This example: A 50 Ω resistor, a 3.248 nH inductor and a 1.299 pF capacitor draw the same RMS current at 2.45 GHz. Each tab shows a separate ideal load with the same applied voltage. The repeating response is already established; switch-on is not shown. The drawings are enlarged and field strength is schematic. Moving wire markers show conventional current. Plot scales stay fixed.

Zero average power does not mean zero current. The ideal capacitor and inductor both carry current, but each returns as much energy as it receives over a cycle. The resistor keeps receiving energy and turns it into heat.

All three loads have the same ratio of RMS voltage to RMS current. This is their impedance magnitude, written $|Z|$ :

$|Z| = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} \approx \frac{0.707\,\mathrm{V}}{0.0141\,\mathrm{A}} \approx 50\,\Omega$ The vertical bars mean magnitude. This ratio tells us how much current flows for a given sine-wave voltage, at this frequency.

That 50 Ω value alone does not tell us how much average power the load absorbs. We also need the phase difference. For sine-wave voltage and current at the same frequency, average power is:

$P_{\mathrm{avg}} = V_{\mathrm{rms}}I_{\mathrm{rms}} \cos \phi$

\phi

(phi) is voltage phase minus current phase, in this repeating sine-wave response. Voltage is A relative to B; positive current enters A.

For the resistor, $\phi=0^\circ$ and $\cos\phi=1$ : multiply the RMS voltage and current to get about 10 mW. For the ideal inductor, $\phi=+90^\circ$ ; for the ideal capacitor, $\phi=-90^\circ$ . Both have $\cos\phi=0$ , giving zero average power.

These three loads have equal impedance magnitudes at our chosen frequency. Next, we will see why the impedance magnitude of a capacitor or inductor changes with frequency.

06 / 13

Reactance: how inductors and capacitors respond to frequency

If we keep the voltage amplitude the same, why does frequency change the current?

Use the same capacitor and inductor as before, each connected separately to a sine-wave voltage source. Keep the component values and voltage amplitude fixed. Change only the frequency: how often the voltage completes a cycle.

Capacitor: faster voltage changes need more current. The same voltage swing requires the same amount of charge to move onto and off the plates. Double the frequency, and that charge must move in half the time. The current amplitude doubles.

Inductor: current has less time to change. Voltage sets how quickly current rises or falls. With the same voltage amplitude, its maximum rate of change stays the same. At double the frequency, voltage reverses in half the time, so current changes over a smaller range. The current amplitude halves.

Reactance puts a number on this opposition to current

For an ideal capacitor or inductor, its impedance magnitude is its reactance magnitude, written $|X|$ . It is voltage amplitude divided by current amplitude, in ohms. For example, at the original 2.45 GHz, each load draws 20 mA peak from 1 V peak:

$|X|=\frac{V_{\mathrm{peak}}}{I_{\mathrm{peak}}}=\frac{1\,\mathrm V}{0.020\,\mathrm A}=50\,\Omega$ Using RMS voltage divided by RMS current gives the same result. With voltage fixed, more current means a smaller reactance magnitude.

Compare the three traces below. Higher frequency fits more cycles into the same time window. The voltage peaks stay at ±1 V; the two current amplitudes change in opposite ways.

Same voltage. Change the frequency.

Slow-motion visualizationSeparate tests · same applied voltage

Frequency 2.45 GHz

Dotted guides: the original ±20 mA current peaks. Values in Ω are reactance magnitudes.

Same 1.299 pF capacitor and 3.248 nH inductor throughout. Each setting shows the repeating response at that frequency.

Calculate reactance at any frequency

The component laws from Sections 03 and 04 give these relationships for a sine wave:

$X_{\mathrm C}=-\frac1{2\pi fC}\qquad X_{\mathrm L}=2\pi fL$ Use frequency

f

in hertz, capacitance

C

in farads and inductance

L

in henries. The result is in ohms.

Frequency is in the denominator for the capacitor, so increasing it makes the reactance magnitude smaller. For the inductor, frequency multiplies inductance, so increasing it makes the reactance larger.

The sign records phase: negative reactance means capacitor current leads voltage; positive reactance means inductor current lags voltage. Both currents still reverse direction every cycle. Changing frequency changes their amplitudes; the ideal components keep their quarter-cycle phase shifts.

At double the frequency, 4.9 GHz, keep the source at 1 V peak:

Capacitor: reactance magnitude falls from 50 Ω to 25 Ω. Peak current is 1 V ÷ 25 Ω = 40 mA.
Inductor: reactance magnitude rises from 50 Ω to 100 Ω. Peak current is 1 V ÷ 100 Ω = 10 mA.

Think about itSuppose frequency becomes three times as high, with voltage amplitude unchanged. How do the two currents and reactance magnitudes change?

Answer

The capacitor current triples, while its reactance magnitude falls to one-third. The inductor current falls to one-third, while its reactance magnitude triples.

You will also see $\omega=2\pi f$ , pronounced “omega,” as shorthand. It is angular frequency: one cycle is $2\pi$ radians, so it measures radians per second. The same formulas then read $X_{\mathrm C}=-1/(\omega C)$ and $X_{\mathrm L}=\omega L$ .

Think about itWhat capacitor and inductor values give each load a reactance magnitude of 50 Ω at 2.45 GHz?

Answer

Set each reactance magnitude to 50 Ω and rearrange the formulas to find the component values:

C=\frac{1}{2\pi(2.45\times10^9\,\mathrm{Hz})(50\,\Omega)}\approx1.299\,\mathrm{pF}

L=\frac{50\,\Omega}{2\pi(2.45\times10^9\,\mathrm{Hz})}\approx3.248\,\mathrm{nH}

The explorers use the unrounded values; displayed component values are rounded.

Think about itA 1.299 pF capacitor has 1 V peak across it. A 3.248 nH inductor carries 20 mA peak. What is the maximum energy stored by each?

Answer

Each reaches a maximum stored energy of about 0.65 pJ. Use peak voltage for the capacitor and peak current for the inductor:

W_{\mathrm{C,max}}=\frac12(1.299\,\mathrm{pF})(1\,\mathrm V)^2\approx0.65\,\mathrm{pJ}

W_{\mathrm{L,max}}=\frac12(3.248\,\mathrm{nH})(20\,\mathrm{mA})^2\approx0.65\,\mathrm{pJ}

Maximum energy uses peak values. Substituting RMS voltage or current would give the cycle-average stored energy, half the maximum for these sine waves.

Common misconceptionImpedance is one fixed value independent of frequency.

Even ideal L and C have impedance that changes with frequency. We write this as $Z(f)$ . For a signal containing several frequencies, calculate the voltage-to-current ratio at each frequency separately.

07 / 13

Complex impedance: magnitude and phase together

How can one number preserve both current magnitude and phase?

We now have two quantities in ohms: resistance, $R$ , and signed reactance, $X$ . Complex impedance keeps them together without adding their magnitudes. Start with a concrete example:

$Z=(30+j40)\,\Omega$ The input has 30 Ω of resistance and +40 Ω of inductive reactance at the stated frequency.

The symbol $j$ marks the perpendicular direction used for reactance. Draw resistance horizontally to the right and positive reactance upward: this impedance reaches a point 30 units right and 40 units up. The arrow from the origin to that point has a length and an angle. Those are the impedance magnitude and impedance phase.

1. Find the magnitude: how much current?

Resistance and reactance form the two sides of a right triangle. Use Pythagoras to find its diagonal, rather than adding 30 and 40:

$|Z|=\sqrt{R^2+X^2}=\sqrt{30^2+40^2}\,\Omega=50\,\Omega$

With 1 V RMS across this input, the current magnitude is $1\,\mathrm V/50\,\Omega=20\,\mathrm{mA}$ RMS.

2. Find the angle: does current lead or lag?

For positive resistance, use inverse tangent with your calculator in degree mode. The angle is measured from the horizontal resistance axis:

$\phi=\tan^{-1}\!\left(\frac XR\right)=\tan^{-1}\!\left(\frac{40}{30}\right)=+53.13^\circ$

Impedance angle is voltage phase minus current phase. Choose voltage as the 0° reference: current must then be at −53.13°, so it lags voltage. The same impedance written as magnitude and angle is $50\angle53.13^\circ\,\Omega$ . This is polar form; $(30+j40)\,\Omega$ is rectangular form.

3. Keep magnitude and phase in the calculation

As in RF Signals, a phasor records a sine wave's magnitude and phase. Here its magnitude is the RMS value. Capital $V$ and $I$ denote these voltage and current phasors, measured across and into the same terminal pair. Ohm's law becomes:

$Z=\frac VI=R+jX\qquad I=\frac VZ$

To divide in polar form, divide the magnitudes and subtract the angles:

$I=\frac{1\angle0^\circ\,\mathrm V}{50\angle53.13^\circ\,\Omega}=0.020\angle-53.13^\circ\,\mathrm A$ The result says both things we need: 20 mA RMS, lagging voltage by 53.13°.

Average power is $P=I_{\mathrm{rms}}^2R=(0.020)^2(30)=0.012\,\mathrm W$ , or 12 mW. Here $R$ is the input resistance of the whole load. It need not be the value of one physical resistor inside it.

Using voltage instead, substitute $I_{\mathrm{rms}}=V_{\mathrm{rms}}/|Z|$ into the same power formula:

$P=\frac{V_{\mathrm{rms}}^2R}{|Z|^2}=\frac{(1\,\mathrm V)^2(30\,\Omega)}{(50\,\Omega)^2}=12\,\mathrm{mW}$

The 1 V RMS is across the whole load, so both resistance and reactance determine how much current flows. The shortcut $P=V_{\mathrm{rms}}^2/R$ applies when that voltage is across a pure resistance: then $|Z|=R$ .

These calculations describe sine waves at one frequency after the switch-on response has settled. The circuit is linear: doubling the applied voltage doubles current. We also treat its components as lumped elements, neglecting travel time between them.

Go deeperWhat j means, and how to handle angle boundaries

The imaginary unit satisfies $j^2=-1$ . Multiplying a phasor by $j$ rotates it 90° counterclockwise; $1/j=-j$ rotates it clockwise. “Imaginary” names a mathematical axis; the current and voltage are measurable. The rotating pointers below use the counterclockwise convention from RF Signals.

A purely inductive impedance has angle +90°; a purely capacitive one has angle −90°. When both resistance and reactance are zero, the angle is undefined. In software, $\operatorname{atan2}(X,R)$ handles the signed coordinates and their quadrant; the inverse-tangent calculation above assumes $R>0$ .

Write the three component impedances

A resistor has no reactive part. An ideal inductor points along the positive reactance axis; a capacitor points along the negative axis. Combining this with the frequency formulas gives:

$Z_{\mathrm R}=R\qquad Z_{\mathrm L}=j\omega L\qquad Z_{\mathrm C}=-\frac j{\omega C}$ The capacitor expression is also written

1/(j\omega C)

The earlier three loads at 2.45 GHz, each with 0.7071 V RMS across it
Load	Impedance	Current	Average real power	Relationship
Ideal resistor	$(50+j0)\,\Omega$	≈ 14.1 mA RMS · 0°	≈ 10 mW	Voltage and current in phase
Ideal inductor	$j50\,\Omega$	≈ 14.1 mA RMS · −90°	0 W	Current lags; magnetic energy returns
Ideal capacitor	$-j50\,\Omega$	≈ 14.1 mA RMS · +90°	0 W	Current leads; electric energy returns

Think about itReplace $j40\,\Omega$ with $-j40\,\Omega$ while keeping $R=30\,\Omega$ and the same terminal voltage. What changes in current and average power?

Answer

The magnitude remains 50 Ω, so current magnitude is unchanged. Current now leads by 53.13° instead of lagging. Average real power is unchanged because resistance and current magnitude are unchanged: $P=I_{\mathrm{rms}}^2R$ . Reversing the reactive sign reverses the phase relation, not the sign of the average absorbed power.

From impedance to current

Slow-motion visualization1 V RMS · one frequency

Impedance

These three examples all have $|Z|=50\,\Omega$ . Choose one to compare timing, or change R and X below.

$Z$ = 30 + j40 Ω
$|Z|$ = 50 Ω · angle 53.13°

Resistance R 30 Ω

Reactance X +40 Ω

Impedance angle +53.13° → current angle -53.13°

Voltage · solid
Current · dashed

Current lags by 53.13°. An impedance magnitude of 50 Ω gives 20 mA RMS. Watch the fixed gap between the pointers become a time offset between the waves.

Both signals are divided by their own peak: compare timing, not amplitude. The rotating pointers are a mathematical picture.

Inspect one cycle 45° of voltage phase

Waveform equations and phase convention

$v(t)=\sqrt2 V_{\mathrm{RMS}}\sin(\omega t),\qquad i(t)=\sqrt2 I_{\mathrm{RMS}}\sin(\omega t-\angle Z)$

The notation $e^{j\omega t}$ describes a unit-length reference arrow rotating counterclockwise as time advances. The pointers here use its vertical projection as their sine reference, at the same port. At zero reactance the solid and dashed traces overlap.

Return to practice questions

08 / 13

Combining impedances without losing phase

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

We can now calculate current when we know the input impedance. To find that impedance from individual components, first ask how they are connected.

Series: the same current flows through every element

Components connected end-to-end share one current. Their voltages add, so their impedances add too. Add the resistance terms and the signed reactance terms separately:

$Z_{\mathrm{series}}=Z_1+Z_2+\cdots$

For example, a 30 Ω resistor in series with a $j60\,\Omega$ inductor and a $-j20\,\Omega$ capacitor gives:

$Z=30+j60-j20=(30+j40)\,\Omega$ This is the 50 Ω-magnitude load from the previous section. Its three component magnitudes do not add to 50 Ω.

Think about itAt 2.45 GHz, what is the series sum of 50 Ω, $j50\,\Omega$ , and $-j50\,\Omega$ ?

Answer

It is $(50+j0)\,\Omega$ . The opposite reactances cancel in the sum. Adding the three magnitudes would incorrectly give 150 Ω.

Parallel: every branch has the same voltage

Now put an 80 Ω resistor and a $-j80\,\Omega$ capacitor in parallel across 1 V RMS, with voltage phase 0°. Calculate each branch current using that same voltage:

I_{\mathrm R}=\frac{1\angle0^\circ\,\mathrm V}{80\,\Omega}=12.5\,\mathrm{mA}

I_{\mathrm C}=\frac{1\angle0^\circ\,\mathrm V}{80\angle-90^\circ\,\Omega}=j12.5\,\mathrm{mA}

The resistor current is in phase with voltage; the capacitor current leads by 90°.

The source supplies both branch currents. Add their phasors—not just their magnitudes:

$I_{\mathrm{in}}=(12.5+j12.5)\,\mathrm{mA}\approx17.68\angle45^\circ\,\mathrm{mA}$

Input impedance is the common voltage divided by that total current. As before, divide magnitudes and subtract angles:

$Z_{\mathrm{in}}=\frac{1\angle0^\circ\,\mathrm V}{0.01768\angle45^\circ\,\mathrm A}\approx56.57\angle-45^\circ\,\Omega$

To recover resistance and reactance from this arrow, take its horizontal and vertical parts: $R=|Z|\cos\phi$ and $X=|Z|\sin\phi$ . They give $Z_{\mathrm{in}}=(40-j40)\,\Omega$ . The 40 Ω is the whole circuit's input resistance; the physical resistor is still 80 Ω. Use total input current with 40 Ω to calculate total power, or the resistor's own branch current with 80 Ω. Both give 12.5 mW.

Admittance makes the parallel calculation shorter

Each branch current above was voltage multiplied by $1/Z$ . This reciprocal is called admittance, $Y$ : current per volt, measured in siemens [S]. One millisiemens (mS) means one milliampere per volt. In parallel, add these current-per-volt values first, then invert the total:

$Y_{\mathrm{parallel}}=\frac1{Z_1}+\frac1{Z_2}+\cdots\qquad Z_{\mathrm{parallel}}=\frac1{Y_{\mathrm{parallel}}}$

Y=\frac1{80}+\frac1{-j80}=(0.0125+j0.0125)\,\mathrm S

The same branch-current calculation, expressed per volt. Inverting gives the same

(40-j40)\,\Omega

Admittance is sometimes written $Y=G+jB$ . Its real part $G$ is conductance; its imaginary part $B$ is susceptance. Both are in siemens.

In the explorer, switch between series and parallel. Follow the circuit connections and the arrows together: series adds impedances in ohms; parallel adds admittances in millisiemens.

Follow the connection, then add

Series and parallel explorerIdeal lumped elements · one frequency

Connection

One current through R, L and C

$Z_{\mathrm R}$ 50 Ω
$Z_{\mathrm L}$ +j30.61 Ω
$Z_{\mathrm C}$ −j81.67 Ω

R = 50 Ω · L = 3.248 nH · C = 1.299 pF

Add $Z$ directly

$Z_{\mathrm R}$
$Z_{\mathrm L}$
$Z_{\mathrm C}$
Total Z

Frequency 1.5 GHz

Reactive terms cancel at 2.45 GHz. In each connection mode, the scale stays fixed as you adjust any slider.

Z = 50.00 − j51.05 ΩMagnitude 71.46 Ω · angle −45.6°

The reactive voltages oppose each other. Add their signed impedances before finding the total magnitude.

Set frequency to 2.45 GHz, then adjust L or C. Which values make the reactive terms cancel?

Resistance 50 Ω

Inductance 3.248 nH

Capacitance 1.299 pF

Inspect the numerical sum

Z=Z_{\mathrm R}+Z_{\mathrm L}+Z_{\mathrm C}

Z=\left(50+j30.61-j81.67\right)\,\Omega

Z=\left(50-j51.05\right)\,\Omega

Ideal lumped R, L and C at one frequency and one reference plane. This example starts below their 2.45 GHz cancellation frequency.

A compact complex voltage divider

With $V_s=1\angle0^\circ\,\mathrm{V_{rms}}$ , $Z_1=30+j40\,\Omega$ , and $Z_2=50+j0\,\Omega$ , first add the two series impedances to find the common current. Then multiply that current by $Z_2$ to find the voltage across it:

The same current

I

flows through both impedances. The source voltage

V_s

is across both; the output voltage

V_2

is across

Z_2

only. Measuring the output draws no extra current in this ideal circuit.

$Z_1+Z_2=80+j40\,\Omega\approx89.44\angle26.57^\circ\,\Omega$ $V_2=\frac{V_sZ_2}{Z_1+Z_2}\approx\frac{1\times50}{89.44}\angle(0^\circ-26.57^\circ)\,\mathrm V$ $V_2\approx0.559\angle-26.57^\circ\,\mathrm V=(0.5-j0.25)\,\mathrm V$ The divider ratio is complex; the output is about 0.559 V RMS and lags the source by about 26.57°.

Using impedance magnitudes alone would lose that phase shift and give the wrong voltage magnitude. The $j40\,\Omega$ term belongs in the series sum before division, even though the voltage is measured across the purely resistive second element.

Go deeperDividing directly in rectangular form

Multiply numerator and denominator by the denominator's complex conjugate: the same real part with the opposite imaginary part. The denominator then becomes real because the cross terms cancel and $j^2=-1$ .

\frac{50}{80+j40}=\frac{50(80-j40)}{80^2+40^2}=\frac{4000-j2000}{8000}=0.5-j0.25

The same operation works for $\frac1Z$ when calculating each parallel branch's admittance.

Common misconceptionSeries impedance magnitudes add.

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

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09 / 13

Resonance: when L and C cancel at the input

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

If the inductor and capacitor are connected in series, they carry the same current at every frequency. At resonance, their voltages have equal amplitudes and opposite phases, so they cancel when added at the input terminals. Each component still stores and returns energy during the cycle.

$Z_{\mathrm{series}} = R + j(\omega L - \frac{1}{\omega C})$ $f_{0} = \frac{1}{2\pi \sqrt{L C}}$ At

f_0

, the series reactances cancel and input impedance is just R.

To find series resonance, set the net reactance to zero: $\omega L=\frac1{\omega C}$ . Multiplying by $\omega C$ gives $\omega^2LC=1$ , which produces the stated $f_0$ after using $\omega=2\pi f$ .

For this series circuit, impedance magnitude is smallest at resonance. A source holding voltage fixed therefore produces the largest current there. If each reactance magnitude at resonance is greater than R, each component voltage exceeds the source voltage. Their opposite phases still make them cancel in the input sum.

In the animation below, a sine-wave source holds its voltage at 1 V peak across R, L and C in series. The 50 Ω resistor limits the current. Change frequency and compare the L and C voltages: at $f_0$ , their sum stays zero throughout the cycle. The source voltage then appears across R, and current reaches its largest amplitude: 20 mA peak.

Frequency changes the balance

Slow-motion visualizationSine source · 1 V peak

Dots: voltage now. Outlines: full voltage swing. Dashed ticks: the source's ±1 V limits.

Frequency 1.225 GHz

Current peak 11.1 mA

The capacitor voltage has the larger amplitude. The L and C voltages only partly cancel.

Reactance: L +25 Ω · C −100 Ω · L + C −75 Ω.

L = 3.248 nH · C = 1.299 pF · f₀ = 2.45 GHz. The source amplitude and component values stay fixed. Each frequency shows the settled sine-wave response.

At half the resonance frequency, current peaks at 11.1 mA. The inductor and capacitor voltages peak at 0.277 V and 1.109 V; their sum peaks at 0.832 V. At resonance, their voltages cancel and current peaks at 20 mA.

Parallel resonance cancels branch currents instead

Put R, L and C in parallel. All three now share voltage, so add their admittances:

$Y_{\mathrm{parallel}}=\frac1R+j\left(\omega C-\frac1{\omega L}\right)$

At the same $f_0$ formula above, the capacitor and inductor branch currents have equal magnitudes and opposite phases. The source supplies only the resistor current, even though current still flows in L and C. Input impedance magnitude is largest here, equal to the parallel resistance. Under a fixed current drive, voltage therefore peaks at resonance.

Think about itIs a resonant $(25+j0)\,\Omega$ 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.

In both connections, cancellation at the input leaves energy moving between the fields. Resistance determines how much of that energy is lost, and how sharply the circuit responds around resonance.

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10 / 13

Q, bandwidth, and ring-down

How does resistance affect the width of a resonance and the time its oscillations take to fade?

Charge a capacitor, then connect it to an inductor and resistor in a closed loop, with no source attached. The capacitor drives current through the inductor. When capacitor voltage reaches zero, current is still flowing; it charges the capacitor with the opposite polarity. Energy moves between the electric and magnetic fields. The resistor turns some into heat, so successive voltage and current peaks become smaller. This fading oscillation is called ring-down.

In the animation below, follow the first polarity reversal, then change resistance to compare how quickly the oscillations fade.

A charged capacitor sets the loop in motion

Slow-motion visualizationStarts at 1 V · no source connected

Resistance 5 Ω

Initially, all the energy is in the charged capacitor; current is zero.

Elapsed time 0 ns

Pale curves: the full 2 ns view. Dots: now. Voltage is top plate relative to bottom; positive current is clockwise.

Share of starting energyElectric field 100%Magnetic field 0%Converted to heat 0%

Changing resistance resets the capacitor to 1 V. L and C, the starting energy and both plot scales stay fixed.

Reading the circuit and energy values

Voltage is the top capacitor plate relative to the bottom. Positive current is clockwise, leaving the top plate. Copper dots show conventional current in the wires, not individual electrons. Charge does not cross the insulating gap. Field-line visibility follows voltage or current magnitude; resistor marks show heating rate, not temperature.

L = 3.248 nH and C = 1.299 pF. The initial 0.650 pJ is shared between the electric field, magnetic field and heat. The percentages use that same starting energy; rounded values may not add to exactly 100%.

$W_C=\tfrac12Cv_C^2,\quad W_L=\tfrac12Li^2,\quad \frac{d(W_C+W_L)}{dt}=-Ri^2$

Ideal L and C with constant series resistance. This is a freely decaying oscillation, not a source-maintained sine wave. The view stops at 2 ns; replay explicitly restores the initial charge.

The full curves show the decay. Initially the capacitor holds all the energy. At about 0.105 ns its voltage is zero but current is still 18.4 mA. At 0.204 ns its voltage reaches −0.854 V and current reaches zero. Resistance makes each reversal smaller.

Q compares stored energy with loss

The quality factor, or $Q$ , describes how much energy a resonator stores relative to how much it loses per cycle. Higher Q means less loss relative to storage. To define it precisely, evaluate the energies at resonance with a source maintaining steady sine-wave oscillations:

$Q=2\pi\times\frac{\text{maximum stored energy}}{\text{energy dissipated per cycle}}$ Q has no units. For these simple RLC circuits, a higher Q also means a slower decay when the source stops supplying energy.

Add the energy in the capacitor's electric field and the inductor's magnetic field at the same instant. “Maximum stored energy” means the largest value of that combined total. The components reach their individual energy peaks at different times, so adding those separate peaks would overcount the stored energy.

In the animation above, 5 Ω gives Q = 10, while 25 Ω gives Q = 2. The graph below compares these two cases: follow the solid curve for Q = 10 and the dashed curve for Q = 2. These curves show their amplitude envelopes: how the size of the oscillations decreases over time, without drawing the individual cycles. Each starts at 1, meaning 100% of its initial amplitude.

Q = 10 · solid
Q = 2 · dashed

Higher Q means a slower decay. The circled point on the solid curve shows where the Q = 10 amplitude has fallen to about 37% of its starting value. The time to reach that level is called the amplitude time constant: about 1.30 ns for Q = 10, compared with 0.26 ns for Q = 2.

Bandwidth measures the frequency range around resonance

Ring-down shows what happens without a source. Now keep a sine-wave source connected and vary its frequency, letting the circuit settle at each setting. We will compare two setups: fixed voltage across series RLC, and fixed current into parallel RLC. In both, the average power absorbed by the circuit peaks at resonance. The resistor converts that power into heat; the ideal inductor and capacitor return the energy they take in over each cycle.

The two half-power frequencies, $f_1$ and $f_2$ , are where this average power falls to half its resonant value. One is below resonance and the other is above it. The interval between them is the bandwidth:

$\mathrm{BW}=f_2-f_1=\frac{f_0}{Q}$ The endpoints are approximately 3 dB below peak power. The relationship to Q applies to the two simple RLC setups used here.

At the same resonance frequency, doubling Q halves the bandwidth. Less loss therefore makes oscillations take longer to fade, and makes the range of frequencies with more than half the resonant power smaller.

Series RLC: fixed voltage amplitude

Hold the source's RMS voltage fixed while changing frequency. Current is greatest at resonance, where the inductor and capacitor voltages cancel. Away from resonance, their net reactance reduces current.

Average power is $P=I_{\mathrm{rms}}^2R$ . It falls to half its resonant value when current falls to $1/\sqrt2\approx0.707$ of its resonant magnitude. These two frequencies mark the bandwidth.

At a given current, a smaller series resistance converts less energy into heat, so it raises the series quality factor, $Q_s$ :

$Q_s=\frac{\omega_0L}{R}=\frac1{\omega_0CR}$

\omega_0=2\pi f_0

is angular frequency at resonance. With

\omega_0L=50\,\Omega

and

R=5\,\Omega

Q_s=10

Parallel RLC: fixed current amplitude

Hold the source's RMS current fixed while changing frequency. Voltage is greatest at resonance, where the inductor and capacitor branch currents cancel at the input. Away from resonance, the circuit's impedance magnitude falls, so the same source current produces less voltage.

Average power is $P=V_{\mathrm{rms}}^2/R$ . It falls to half its resonant value when voltage falls to $1/\sqrt2\approx0.707$ of its resonant magnitude. These two frequencies mark the bandwidth.

The parallel resistor is also called a shunt resistor. At a given voltage, a larger shunt resistance converts less energy into heat, so it raises the parallel quality factor, $Q_p$ :

$Q_p=\frac R{\omega_0L}=\omega_0CR$ With

\omega_0L=50\,\Omega

and a 500 Ω shunt resistor,

Q_p=10

A fixed voltage across this parallel circuit would keep its average power constant as frequency changes. That is why we use a fixed current to show its power peak and bandwidth.

Both examples give Q = 10 at 2.45 GHz, so both have a bandwidth of 245 MHz, between approximately 2.331 GHz and 2.576 GHz. Compare them in the plot below: change Q and watch the half-power frequencies move. Each curve is divided by its own resonant amplitude, so equal drawn peaks do not mean equal currents or voltages.

Higher Q gives a smaller bandwidth

Continuous sinusoidal driveFrequency sweep ·

f_0=2.45\,\mathrm{GHz}

The source stays connected. These plots use the same L and C as the ring-down animation, with a separate Q setting.

Drive

Fixed RMS voltage across series RLC: compare current. Each curve is divided by its own resonant value. The frequency scale stays fixed.

Selected Q = 10
Reference Q = 2

Quality factor Q 10 · series R = 5 Ω

Sweep frequency 2.45 GHz

Impedance magnitude: 5 Ω
Impedance phase: 0°
Bandwidth: 245 MHz

Half-power frequencies: 2.3306–2.5756 GHz
Resonance frequency: 2.45 GHz

At Q = 10, the half-power frequencies are closer together than at Q = 2: the bandwidth is smaller. The 0.707 line marks half the average power at resonance.

Think about itIn the 2.45 GHz series circuit, with 0.7071 V RMS held across it, what happens at resonance if resistance falls from 50 Ω to 5 Ω?

Answer

At resonance the net reactance is zero, so current is set by the remaining resistance. Reducing it tenfold raises current tenfold, from about 14.1 to 141 mA RMS under the same imposed voltage. Each reactance still has magnitude 50 Ω, so each component voltage is about $0.1414\,\mathrm A\times50\,\Omega\approx7.07\,\mathrm V$ RMS. Their opposite phases make these larger voltages cancel in the input sum. The ratio $\frac{\omega_0L}{R}$ also rises tenfold, from $Q_s=1$ to $Q_s=10$ .

Think about itA parallel resonator with ideal L and C resonates at 2.45 GHz. Its only loss is a 500 Ω shunt resistor, giving $Q_p=10$ . If the resistance increases to 1,000 Ω while L and C stay fixed, what happens to Q, bandwidth, and ring-down?

Answer

Q doubles to 20 because $Q_p=\frac R{\omega_0L}$ . The resonant frequency stays fixed, so the half-power bandwidth halves from 245 MHz to 122.5 MHz. At the same stored energy, the larger shunt resistance dissipates less power, so the oscillation decays more slowly after the drive stops. This is the opposite resistance change from reducing loss in the series circuit.

Connecting equipment can lower Q

A resonator already loses energy as heat in its own resistance. Now connect a measuring instrument that also absorbs some of its energy. The oscillations fade faster and Q becomes lower, even though the resonator's components have not changed. This gives us two ways to describe its Q:

Unloaded Q: includes only the resonator's own losses.
Loaded Q: includes its own losses plus energy taken by connected equipment, such as a source, load, or measuring instrument.

For the same resonant frequency, the lower loaded Q also means a larger bandwidth. The measurement connection can therefore change the behavior you are trying to measure.

Two other labels describe just one part of the picture: component Q applies to a single part, such as an inductor; external Q counts only the losses through connections to other equipment.

Common misconceptionHigher Q is always better.

At the same resonance frequency, higher Q reduces bandwidth but makes oscillations take longer to fade after the drive stops or changes. At resonance, component voltages or branch currents can also become larger relative to the input. Choose Q for the required bandwidth, response time and component ratings. The $\frac{f_0}{\mathrm{BW}}$ shortcut applies to the stated circuit connections, loading and half-power response; it does not describe every network.

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11 / 13

Real RF components are not ideal components

Can a capacitor become inductive—or an inductor capacitive?

Yes. We have just seen how inductance and capacitance can resonate together. A real component can contain both effects, even when its label says only “capacitor” or “inductor.” These unwanted electrical effects are called parasitics. They come from the component's materials and construction.

A real capacitor also has resistance and inductance

A capacitor's conductors add inductance, and its materials dissipate some energy. We represent these effects with equivalent series inductance (ESL) and equivalent series resistance (ESR), alongside its capacitance. The drawing is an electrical model of one component, not three separate parts hidden inside it.

Capacitor model: resistance, inductance and capacitance in series.

$Z_{\mathrm{capacitor}} \approx R_{\mathrm{ESR}} + j(\omega L_{\mathrm{ESL}} - \frac{1}{\omega C})$ Add the series resistance and the two opposing reactances, just as in the series RLC circuit.

At low frequencies, capacitive reactance dominates. As frequency rises, its magnitude falls while inductive reactance grows. At the self-resonant frequency (SRF), the two cancel. This model's impedance reaches its minimum: only ESR remains. Above SRF, inductance dominates, so the component's terminal behavior is inductive.

A real inductor also has resistance and capacitance

The coil's wire has resistance. Its nearby conductors also store energy in an electric field, creating unwanted capacitance. A simple model puts this capacitance, $C_{\mathrm p}$ , in parallel with the coil and its wire resistance.

Inductor model: unwanted capacitance across the coil and its wire resistance.

$Z_{\mathrm{inductor}} \approx \left(\frac{1}{R+j\omega L}+j\omega C_{\mathrm p}\right)^{-1}$

R

is the wire resistance and

L

is the coil's inductance. Add the admittances of the coil branch and the unwanted capacitance, then invert the sum.

Below its first self-resonance, the component behaves inductively. Near self-resonance, the opposing reactive currents in the two branches largely cancel at the input. Only a small source current is needed for a given voltage, so impedance reaches a high peak. Loss keeps that peak finite. Above this resonance, the unwanted capacitance dominates, and the terminal behavior becomes capacitive.

When the real component stops following the ideal one

Illustrative modelsSame fixed axes · component terminals only

Compare

The ideal curve keeps falling. With parasitics, the curve reaches a minimum and then rises as inductance takes over.

Real model · solidIdeal · dashed

Both axes use logarithmic spacing: equal distances represent equal ratios, such as 1 to 10 and 10 to 100.

Frequency 1 GHz

Real model · |Z|: 117.4 Ω
Ideal component · |Z|: 122.4 Ω

Model values: ESR 0.3 Ω and ESL 0.8 nH.

The part has not physically changed into a different component: its overall voltage–current relationship has changed with frequency. Check impedance at your operating frequency, not just the printed pF or nH value. Use manufacturer curves or models to check behavior, loss and self-resonance. These simple models explain the first resonance; real losses vary with frequency, and more complex behavior can occur at higher frequencies. Next, we will see how the connections used to measure the part can also affect the result.

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12 / 13

Measuring impedance

Are we measuring the component itself, or the component together with its connections?

Take the 1.3 pF capacitor model from Section 11 and examine it at 4.5 GHz. Connecting an instrument to the part requires leads and mounting pads. This connection structure is called a fixture. Its own inductance and capacitance contribute to the impedance seen by the instrument.

Choose where the impedance is reported

Recall that the reference plane identifies the measurement location. Here we compare two terminal pairs: the component's terminals and the fixture input. The first result describes the component; the second includes its mounting connections. The component stays the same in both cases.

Start with the capacitor at 4.5 GHz below. Its device-terminal impedance is capacitive, but the fixture-input impedance is inductive. Follow the two curves as frequency changes. The explorer calculates both results from known models; obtaining a component-only result from a real instrument also requires accounting for the fixture.

Where is impedance measured?

Calculated model · fixed scales

Component

Device terminals
Fixture input

Impedance magnitude [Ω]

Impedance phase [°]

Logarithmic frequency and magnitude; linear phase.

Frequency 4.5 GHz

Device terminalsCapacitive: 0.30 − j4.59 Ω
Fixture inputInductive: 0.29 + j5.38 Ω

Use an ideal component model

Model values and plot limits

Real capacitor: ESR 0.3 Ω and ESL 0.8 nH. Real inductor: wire resistance 0.25 Ω and unwanted capacitance 0.5 pF. Fixture: series lead inductance 0.35 nH and pad capacitance 0.12 pF in parallel with the component.

These illustrative models are not manufacturer measurements. Curves can leave the fixed 0.01 Ω–100 kΩ window. An ideal lossless network can reach zero impedance or an unbounded peak; phase is undefined there and the trace breaks.

Correct the measurement at the intended location

Consider a vector network analyzer (VNA). It measures the size and phase of the signal reflected back from the connected circuit, relative to the signal sent toward it, and can use that information to calculate impedance. That conversion uses a reference impedance: the VNA's 50 Ω system reference in this example. We will keep the measurement at one port.

Before measuring the capacitor, we connect three reference devices to the cable end, one at a time. These are calibration standards: an open leaves the cable's two conductors unconnected; a short joins them; and a load connects them through a known impedance, usually 50 Ω in this system.

The VNA compares the reflection from each standard with its known response. From the differences, it calculates corrections for measurement errors. This process is calibration. We then remove the final standard and connect the fixture and capacitor at the same cable end. The setup below shows both reference planes.

The fixture stays connected between the calibration plane and the component terminals. SMA connectors shown schematically in cutaway.

Calibration at the cable end leaves the fixture beyond it included in the result. To report at the component terminals, we need measured data or a suitable model describing the fixture's electrical behavior. De-embedding uses that information to remove the fixture's effect mathematically. The fixture remains physically connected.

Choose an instrument for the frequency and component

The VNA is one option. Use an instrument and measurement method that support both your test frequency and the impedance range of the component.

Three common ways to characterize a component
Instrument	What it tells you
LCR meter	Reports inductance, capacitance, resistance and loss at selected test frequencies.
Impedance analyzer	Shows how impedance magnitude and phase change across a supported frequency range.
Vector network analyzer (VNA)	Measures reflected and transmitted signals at RF and microwave frequencies. A calibrated reflection measurement can be converted to impedance.

Report the result with its conditions

Signal level and temperature can change a real component's response. So can DC bias: a steady voltage or current applied along with the alternating test signal. Record the conditions used, especially when comparing different tests.

A useful report lets someone else understand what the number describes. The example below uses the capacitor model's rounded impedance and fictional test conditions. It is an illustration of reporting, not measured component data.

Illustrative report

Component: capacitor C1, nominally 1.3 pF.

Result: $Z=(0.3-j4.59)\,\Omega$ at 4.5 GHz, reported at the component terminals.

Method: VNA reflection measurement using a 50 Ω reference impedance. Calibration with open, short and load standards at the cable end; fixture effects removed using the lead-and-pad model shown above.

Test conditions: source setting −20 dBm, 0 V DC bias, 23 °C. The source setting is not the power absorbed by the capacitor.

What was measured? Identify the component, frequency and complex impedance, or magnitude and phase.
Where was it reported? Name the terminals, calibration plane and treatment of the fixture.
Under what conditions? Record the instrument and method, signal level, and relevant bias and temperature.

An LCR meter measures impedance and can calculate a capacitance or inductance value from it. To do that, it uses an equivalent circuit: a simple model of the component.

If the component behaves capacitively at the test frequency, the meter can model it as an ideal capacitor with a resistor in series or in parallel. Both models can match the same measured impedance, but can use different resistance and capacitance values. Record the frequency and whether the meter used its series or parallel model. Changing mode changes the calculation; the physical component stays the same.

Think about itYou calibrate at the cable end, then connect a fixture and capacitor. Does the instrument now report only the capacitor's impedance?

Answer

No. The fixture beyond the calibration plane remains included. To report the capacitor alone, remove the fixture's effect using suitable measured data or a model, or calibrate directly at the component terminals if the setup supports it.

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13 / 13

Put it together: a series RLC circuit

How much current flows, how much power is absorbed, and what changes with frequency?

Use what you have learned to study the circuit below. A sine-wave source applies 0.5 V RMS across a resistor, inductor and capacitor connected in series. Start at 200 MHz, with the source voltage as the 0° phase reference. All three components are ideal, and we consider the response after switching effects have settled.

At 200 MHz,

X_{\mathrm L}

and

X_{\mathrm C}

are the inductor and capacitor reactances. Use the exact +50 Ω and −20 Ω values for the calculations; the nH and pF values are rounded. The arrow and +/− marks define references for the alternating current and voltage.

When frequency changes, keep the same components and the same 0.5 V RMS across the whole circuit. Try these three questions before opening the solution.

At 200 MHz: find the total impedance as $R+jX$ and as magnitude and angle. Then find the RMS current, whether it leads or lags the source voltage, and the average power absorbed by the circuit.
At 400 MHz: first predict whether current will increase or decrease. Then calculate the new impedance, RMS current and average power to check your prediction.
Explain the result: write a short report for 200 MHz, including the circuit, frequency, applied voltage and your results. What extra information would a real measurement report need?

Reveal the solution

1. Find impedance, current and power at 200 MHz

In series, add the resistance and the signed reactances separately. The +50 Ω from the inductor and −20 Ω from the capacitor leave +30 Ω of reactance, so the circuit is inductive:

Z=(40+j30)\,\Omega

|Z|=\sqrt{40^2+30^2}\,\Omega=50\,\Omega

\phi=\tan^{-1}\!\left(\frac{30}{40}\right)\approx+36.9^\circ

In magnitude and angle form, this is approximately $50\angle36.9^\circ\,\Omega$ . The source voltage is across all three components, so use the whole impedance to find current:

I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{|Z|}=\frac{0.5\,\mathrm V}{50\,\Omega}=10\,\mathrm{mA}

Current phase is voltage phase minus impedance angle: $0^\circ-36.9^\circ=-36.9^\circ$ . Current therefore lags the source voltage by 36.9°.

The ideal inductor and capacitor store and return energy. Only the resistor absorbs energy on average, so calculate power using the circuit current and the 40 Ω resistance:

P=I_{\mathrm{rms}}^2R=(0.01\,\mathrm A)^2(40\,\Omega)=4\,\mathrm{mW}

2. Check what changes at 400 MHz

Doubling frequency doubles the inductor reactance to +100 Ω and halves the capacitor reactance magnitude to 10 Ω. Its sign stays negative. The resistance stays 40 Ω, while the net reactance grows from +30 Ω to +90 Ω:

Z=(40+j90)\,\Omega\approx98.5\angle66.0^\circ\,\Omega

I_{\mathrm{rms}}=\frac{0.5\,\mathrm V}{|Z|}\approx5.08\,\mathrm{mA}

P=I_{\mathrm{rms}}^2R\approx1.03\,\mathrm{mW}

Current now lags voltage by approximately 66°. Current and power both decrease: the voltage stayed fixed, but the impedance magnitude increased.

3. Write a short calculation report

At 200 MHz, the ideal series RLC circuit has an input impedance of $(40+j30)\,\Omega$ . With 0.5 V RMS applied across the whole circuit, the steady sine-wave current is 10 mA RMS and lags the source voltage by approximately 36.9°. The circuit absorbs 4 mW of average power.

This reports a calculation. A real measurement report would also identify the actual parts, measurement terminals, instrument and method, calibration and fixture treatment, and test conditions such as signal level, DC bias and temperature. Record those details from the test; do not assume them from this ideal circuit.

Think about itAt 200 MHz, what is the RMS voltage across each component? Why do those magnitudes not add to 0.5 V?

Answer

Each component carries the same 10 mA RMS. Multiplying that current by each component's impedance magnitude gives 0.4 V across R, 0.5 V across L, and 0.2 V across C, all RMS values.

The inductor and capacitor voltages are 180° apart, so their combined RMS magnitude is $0.5-0.2=0.3\,\mathrm V$ . The resistor voltage is 90° from that combined voltage. Their phasors form a right triangle:

V_{\mathrm{rms}}=\sqrt{0.4^2+0.3^2}\,\mathrm V=0.5\,\mathrm V

The voltages add at every instant. Their RMS magnitudes cannot simply be added because the voltages are out of phase.

Think about itAt what frequency does this circuit resonate, and how much current flows then? Does it match a 50 Ω reference? A match requires $Z=(50+j0)\,\Omega$ .

Answer

Use the resonance formula from Section 09 with the rounded L and C values shown in the drawing:

f_0=\frac{1}{2\pi\sqrt{LC}}\approx126.5\,\mathrm{MHz}

At resonance, the reactances cancel and only the 40 Ω resistance remains. With the same 0.5 V RMS source, current reaches its maximum:

I_{\mathrm{rms}}=\frac{0.5\,\mathrm V}{40\,\Omega}=12.5\,\mathrm{mA}

Average power is then 6.25 mW. The circuit is not matched to a 50 Ω reference: its resistance is 40 Ω. At 200 MHz it was not matched either: its impedance magnitude was 50 Ω, but it still had +30 Ω of reactance.

When you apply these calculations to a real circuit, check the component models at your operating frequency and state where the impedance is reported.

Fields & Waves builds on this picture of voltage, current and stored energy to explain how RF signals travel through space and structures.

Ungraded review

Check your understanding

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

01Two loads each draw 10 mA RMS from 1 V RMS. One current is in phase with voltage; the other lags by 60°. Do they absorb equal average power?
Model answer
No. Their impedance magnitudes are both 100 Ω, but phase matters: $P=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi$ gives 10 mW and 5 mW. The second load stores and returns some energy during each cycle while still absorbing positive average real power.
02An ideal inductor and capacitor are connected in parallel across 1 V RMS. At one frequency their impedances are $j100\,\Omega$ and $-j100\,\Omega$ . Does zero net source current mean neither branch carries current?
Model answer
No. Each branch carries 10 mA RMS. The branch-current phasors are opposite, so they cancel at the source connection while energy is exchanged between the fields. Any real loss or added parallel load requires additional source current.
03A simple resonator has a 100 MHz center frequency and a 5 MHz half-power bandwidth. An added load widens the bandwidth to 10 MHz without moving the center appreciably. What happened to loaded Q and ring-down?
Model answer
Loaded Q falls from $\frac{100}{5}=20$ to $\frac{100}{10}=10$ . The load provides another route for energy to leave the resonator, so the amplitude decays faster after excitation stops. The component's internal loss need not have changed.
04A capacitor looks inductive in a fixture measurement. Is that enough to conclude that the capacitor itself is above self-resonance?
Model answer
No. The part may be above self-resonance, but fixture inductance can also change the reported impedance. Check the measurement frequency, reference plane and fixture removal, then compare a device-terminal result with an appropriate component model or component data at the measurement frequency.
05Turn “impedance: 68 Ω” into a useful report template. Which missing facts must be obtained rather than guessed?
Model answer
Record complex impedance ( $R+jX$ , or magnitude and phase), frequency, device/port and reference plane, instrument and method, signal level, calibration and fixture treatment. Add bias, temperature and equivalent-model choice when relevant. Establish whether 68 Ω was a magnitude or a resistance; the original statement does not supply the missing angle or test conditions.

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

Something unclear or incorrect? Suggest a correction

Impedance & Reactance

Why resistance is no longer enough

Resistance and real power transfer

Reading the direction of energy flow

What resistance can represent in RF

Capacitance: energy in an electric field

Capacitance links voltage and charge

Current depends on how quickly voltage changes

Example — a 1 μF capacitor

1 Rising zero crossing

2 Positive voltage peak

Stored energy depends on voltage

Inductance: energy in a magnetic field

Voltage sets how quickly current changes

Example — a 10 mH inductor

1 Positive voltage peak

2 Following voltage zero crossing

Stored energy depends on current

Two ways to build the coil

Wirewound

Multilayer

Same RMS values. Different average power.

Where does the energy go?

Reactance: how inductors and capacitors respond to frequency

Reactance puts a number on this opposition to current

Same voltage. Change the frequency.

Calculate reactance at any frequency

Complex impedance: magnitude and phase together

1. Find the magnitude: how much current?

2. Find the angle: does current lead or lag?

3. Keep magnitude and phase in the calculation

Write the three component impedances

From impedance to current

Combining impedances without losing phase

Series: the same current flows through every element

Parallel: every branch has the same voltage

Admittance makes the parallel calculation shorter

Follow the connection, then add

One current through R, L and C

Add ZZZ directly

A compact complex voltage divider

Resonance: when L and C cancel at the input

Frequency changes the balance

Parallel resonance cancels branch currents instead

Q, bandwidth, and ring-down

A charged capacitor sets the loop in motion

Q compares stored energy with loss

Bandwidth measures the frequency range around resonance

Series RLC: fixed voltage amplitude

Parallel RLC: fixed current amplitude

Higher Q gives a smaller bandwidth

Connecting equipment can lower Q

Real RF components are not ideal components

A real capacitor also has resistance and inductance

A real inductor also has resistance and capacitance

When the real component stops following the ideal one

Measuring impedance

Choose where the impedance is reported

Where is impedance measured?

Correct the measurement at the intended location

Choose an instrument for the frequency and component

Report the result with its conditions

Put it together: a series RLC circuit

1. Find impedance, current and power at 200 MHz

2. Check what changes at 400 MHz

3. Write a short calculation report

Check your understanding

Try 6 quick questions.

Sources and further study

Core theory

Components, measurement, and next steps

Add $Z$ directly