Analyze simple AC circuits with reactance: Question bank

Before you start, one shared vocabulary reminder so no symbol is unearned:

Definition Symbols used on this page

$R$ — resistance in ohms $Ω$ ; the "in-phase" opposition that turns current into heat.
$X$ — reactance in $Ω$ ; frequency-dependent opposition that only stores energy. $X_{L} = ω L$ (inductor), $X_{C} = 1/ (ω C)$ (capacitor). Both formulas give a positive number, but the sign each carries inside $Z$ differs (see below).
$X_{net}$ — net reactance $X_{L} - X_{C}$ , the single signed number that survives after the inductor and capacitor partly cancel. Positive = net inductive, negative = net capacitive. Whenever we mean the combined value (not one component alone) we write $X_{net}$ .
$V_{m}$ — peak voltage: the maximum value the AC voltage reaches at the top of its wave. The RMS value is $V_{r m s} = V_{m} / 2$ ; likewise $I_{m}$ is the peak current.
$Z$ — impedance, written $Z = R + j X_{net}$ . Its length is total opposition and its angle $ϕ$ is the voltage-vs-current timing shift.
$ω = 2 π f$ — angular frequency ( $f$ in hertz).
$j = - 1$ — the unit that points " $90°$ sideways" so timing can be stored as an angle.
"lead / lag" — current reaching its peak before / after the voltage peak.

Two pictures anchor every card on this page — study them first, then reveal the cards against them:

Figure — Analyze simple AC circuits with reactance

Figure 1 — Phasor triangle. A blue arrow $R$ points right along the horizontal (resistance, in phase). From its tip an orange arrow $X_{net} = X_{L} - X_{C}$ points straight up (net reactance; it would point down if net capacitive). The green diagonal from the origin to the top of that arrow is $Z$ , whose length is $R^{2} + X_{net}^{2}$ and whose red angle $ϕ$ from the horizontal is the phase. A small square marks the right angle between $R$ and $X_{net}$ — the reason we use Pythagoras, not plain addition.

Figure 2 — Time-domain waveforms. A gray voltage cosine is the reference. The orange curve (capacitor current) peaks a quarter-cycle before the voltage — it leads (ICE). The blue curve (inductor current) peaks a quarter-cycle after the voltage — it lags (ELI). Same voltage, opposite timing.

The first figure is the phasor triangle: $R$ along the horizontal, the net reactance $X_{net} = X_{L} - X_{C}$ straight up (or down if capacitive), and $Z$ as the diagonal — the picture behind "use Pythagoras, not addition". The second is the time-domain waveform: it shows current leading (capacitor) and lagging (inductor) the same voltage by a quarter-cycle, the picture behind ELI-the-ICE-man.

True or false — justify

Reactance and resistance have the same unit, so a reactance also dissipates power as heat.

False. Both are measured in

Ω

, but reactance only stores energy in a field and returns it each cycle; the average power in a pure

X

is zero, so no heat — see Power in AC Circuits.

An ideal inductor connected to a DC source (

f = 0

) behaves like an open circuit.

False — it's the opposite.

X_{L} = ω L = 2 π f L \to 0

f \to 0

, so at DC an ideal inductor is a short (a plain wire), not an open.

An ideal capacitor at DC (

f = 0

) acts like an open circuit.

True.

X_{C} = 1/ (ω C) \to \infty

f \to 0

, so it blocks steady current entirely — matching the picture of a fully-charged capacitor passing no current.

For a series R-L-C circuit you can find total opposition by adding

R + X_{L} + X_{C}

because they're all in ohms.

False.

R

and

X

sit

90°

apart, and

X_{L}

X_{C}

point opposite ways, so you combine as

∣ Z ∣ = R^{2} + X_{net}^{2}

with

X_{net} = X_{L} - X_{C}

— the diagonal in the phasor figure, never a plain sum.

In a purely capacitive circuit the current leads the voltage by

90°

True. From

i = C d v / d t

, differentiating

cos

gives

+ 90°

: the current peaks a quarter-cycle before the voltage (the "ICE" of ELI-the-ICE-man, orange curve in the waveform figure).

Doubling the frequency doubles the reactance of any component.

False.

X_{L} = ω L

is linear in

ω

, so

2 ω

doubles it;

X_{C} = 1/ (ω C)

is inverse in

ω

, so

2 ω

halves it. Same input, opposite algebra — that is the whole asymmetry.

At resonance the circuit's impedance is zero.

False. At resonance

X_{L} = X_{C}

so the reactive part

X_{net}

cancels, leaving

∣ Z ∣ = R

— the minimum, not zero (unless

R = 0

). See Resonance in RLC Circuits.

The phase angle

ϕ

of the impedance is also the phase angle used in the power factor.

True. The power factor is

cos ϕ

using the same

ϕ = atan2 (X_{net}, R)

; the geometry that sets timing also sets how much power is real.

A negative phase angle

ϕ

means the circuit is net inductive.

False. Negative

ϕ

means

X_{net} = X_{L} - X_{C} < 0

, i.e. net capacitive (current leads). Net inductive gives positive

ϕ

You may plug a peak voltage

V_{m}

and an RMS current into

Z = V / I

together.

False. Mixing conventions corrupts the ratio; use peak-with-peak or RMS-with-RMS since

V_{r m s} = V_{m} / 2

— see RMS and Peak Values.

Spot the error

X_{C} = \frac{1}{2 π f C}

, and since

C

is in the denominator, a bigger capacitor gives more reactance."

Error: bigger

C

means smaller

X_{C}

. A larger capacitor stores more charge per volt, so more current flows for the same voltage swing — less opposition.

"Impedance is

Z = R + j X_{net}

, so its magnitude is

R + X_{net}

Error: the magnitude of

R + j X_{net}

R^{2} + X_{net}^{2}

, not

R + X_{net}

. Adding the parts linearly ignores that the real and imaginary axes are perpendicular — see the diagonal of the phasor triangle.

Z_{C} = 1/ (j ω C)

, so its impedance is

+ j / (ω C)

Error:

1/ j = - j

, so

Z_{C} = - j / (ω C)

. That minus sign is exactly why capacitive reactance enters

Z

negatively and lowers the phase angle.

"The current lags in a capacitor because the capacitor 'resists' the voltage building up."

Error: in a capacitor current leads, not lags. Current must flow first to deposit charge before voltage can rise, so current peaks earlier.

"Since

X_{L} = ω L

grows without limit, a real inductor blocks all high frequencies completely."

Error: ideal

X_{L} \to \infty

, but real inductors have parasitic winding capacitance and resistance that eventually let high frequencies through — the ideal model is a limiting case, not literal truth.

"In a series RLC circuit both

X_{L}

and

X_{C}

oppose current, so the net reactance is

X_{L} + X_{C}

Error: they oppose current at opposite moments (inductor lags, capacitor leads), so their reactances partly cancel:

X_{net} = X_{L} - X_{C}

Why questions

Why do we use complex numbers for AC opposition instead of a single real number?

Because AC opposition carries two facts — how much (magnitude) and when (phase timing). One real number can't store both; a complex number's length and angle store both at once. See Complex Numbers and Phasors.

Why does

X_{L}

grow with frequency while

X_{C}

shrinks?

An inductor opposes change in current (

v = L d i / d t

), so faster wiggling means bigger opposing voltage → more opposition. A capacitor passes current whenever voltage changes (

i = C d v / d t

), so faster wiggling means more current flows → less opposition.

Why is Pythagoras the right tool for combining

R

and

X_{net}

Because the resistive drop is in phase with current while the reactive drop is

90°

out of phase; treating them as perpendicular legs of a right triangle, the total is the hypotenuse

R^{2} + X_{net}^{2}

— the diagonal in the phasor figure.

Why do we subtract

X_{C}

from

X_{L}

rather than add them?

Inductive voltage leads the current and capacitive voltage lags it — they point in opposite directions on the imaginary axis, so their signed sum is a difference

X_{net} = X_{L} - X_{C}

Why does an inductor's current lag the voltage?

Voltage is proportional to how fast current changes (

v = L d i / d t

), which is largest when current is crossing zero; so the current peak arrives a quarter-cycle after the voltage peak — a

90°

lag (the "ELI", blue curve in the waveform figure).

Why is reactance the same value whether you use peak or RMS quantities?

Reactance is a ratio

V / I

, and both voltage and current shrink by the same factor

2

going from peak to RMS, so the factor cancels and

X

is unchanged.

Why prefer

atan2 (X_{net}, R)

over

tan^{- 1} (X_{net} / R)

for the phase?

Plain arctan only spans

\pm 90°

and the ratio blows up as

R \to 0

;

atan2

keeps

X_{net}

and

R

separate so it tracks the correct sign and returns

\pm 90°

cleanly at the

R = 0

boundary.

Why does the impedance angle equal the power-factor angle?

The angle of

Z

measures how far current timing is offset from voltage, and power delivered depends on that same offset — so

cos ϕ

of the impedance angle is exactly the power factor. See Power in AC Circuits.

Edge cases

What is

∣ Z ∣

for a circuit with

R = 0

and

X_{L} = X_{C}

(pure LC at resonance)?

∣ Z ∣ = 0 + 0^{2} = 0

: an ideal lossless LC at resonance is a short circuit and current is limited only by the source.

f \to 0

(approaching DC) in a series RLC, what does the circuit look like?

X_{L} \to 0

(short) and

X_{C} \to \infty

(open), so the capacitor dominates and blocks all current — the impedance magnitude blows up.

f \to \infty

in a series RLC, what dominates?

X_{C} \to 0

(short) and

X_{L} \to \infty

(open), so the inductor dominates and blocks the current — again impedance grows without bound.

For a purely resistive circuit, what is the phase angle

ϕ

ϕ = atan2 (0, R) = 0°

: with no reactance, voltage and current peak together — the boundary case where AC behaves just like DC in timing.

R = 0

but

X_{L} \neq = X_{C}

, what is

ϕ

ϕ = atan2 (X_{net}, 0) = \pm 90°

(sign follows

X_{net}

): with no resistance the impedance is purely reactive, so current is exactly a quarter-cycle out of phase and no real power is dissipated. This is exactly the boundary where plain arctan fails and

atan2

still works.

What happens to the phase angle exactly at resonance (

X_{L} = X_{C}

R \neq = 0

ϕ = atan2 (0, R) = 0°

: the reactances cancel, the circuit looks purely resistive, and current is maximal and in phase with voltage.

Can the magnitude

∣ Z ∣

ever be smaller than

R

in a series RLC?

No. Since

∣ Z ∣ = R^{2} + X_{net}^{2}

and the squared term is

\geq 0

, the smallest possible value is

R

itself, reached only at resonance.

Recall One-line summary of every trap

Reactance shares the ohm but not the heat; $X_{L}$ climbs and $X_{C}$ falls with frequency; opposites cancel into $X_{net} = X_{L} - X_{C}$ ; $R$ and $X_{net}$ combine by Pythagoras; negative $ϕ$ is capacitive; resonance leaves $R$ , not zero; use $atan2$ so the sign and the $R \to 0$ boundary stay correct; and peak/RMS must never be mixed in one Ohm's-law step.