Analyze simple AC circuits with reactance: Worked examples

The scenario matrix

Before working anything, let us map out every kind of case this topic can produce. Each row is a "cell" — a distinct scenario — and every worked example below is tagged with the cell it covers.

Cell	What makes it special	What you must watch
A. Net inductive	$X_{L} > X_{C}$	$ϕ > 0$ , current lags
B. Net capacitive	$X_{C} > X_{L}$	$ϕ < 0$ , current leads
C. Balanced (resonance)	$X_{L} = X_{C}$	net $X = 0$ , $Z = R$ , $ϕ = 0$
D. Pure R (degenerate)	no L, no C	$Z = R$ , $ϕ = 0$ — same shape as C but different reason
E. Pure reactance ( $R = 0$ )	no resistor	$ϕ = \pm 90°$ , $tan^{- 1}$ blows up
F. Low-freq limit	$f \to 0$ (DC)	$X_{L} \to 0$ , $X_{C} \to \infty$
G. High-freq limit	$f \to \infty$	$X_{L} \to \infty$ , $X_{C} \to 0$
H. Word problem	real device, real units	unit conversions (mH, µF, kHz)
I. Exam twist	given $ϕ$ , find a component	run $tan^{- 1}$ backwards

The sign of $ϕ$ is the spine of everything: positive means inductive (cell A), negative means capacitive (cell B), exactly zero means balanced or resistive (cells C, D). Keep one eye on that sign in every example.

Figure — Analyze simple AC circuits with reactance

The three signs of $ϕ$ (cells A, B, C)

Look at the figure above. We plot the impedance as an arrow in a plane: the horizontal axis is the real part $R$ (always pointing right, since resistance is never negative), and the vertical axis is the net reactance $X_{L} - X_{C}$ . The angle the arrow makes with the horizontal is $ϕ$ .

Arrow tilts up → $X_{L} - X_{C} > 0$ → cell A, inductive, current lags.
Arrow tilts down → $X_{L} - X_{C} < 0$ → cell B, capacitive, current leads.
Arrow lies flat → $X_{L} - X_{C} = 0$ → cell C/D, no phase shift.

Worked examples

Worked example Example 1 — Cell A (net inductive)

Series RLC: $R = 10 Ω$ , $X_{L} = 30 Ω$ , $X_{C} = 22 Ω$ , driven at $V_{r m s} = 120 V$ . Find $∣ Z ∣$ , current $I$ , and phase $ϕ$ .

Forecast: $X_{L}$ beats $X_{C}$ , so guess the arrow tilts up and $ϕ$ is positive (current lags). Guess $∣ Z ∣$ is a bit above $10$ .

Net reactance $X = X_{L} - X_{C} = 30 - 22 = 8 Ω$ . Why this step? $L$ and $C$ push current in opposite timing directions, so their reactances subtract, not add.
$∣ Z ∣ = R^{2} + X^{2} = 1 0^{2} + 8^{2} = 164 = 12.8 Ω$ . Why this step? $R$ and $X$ are perpendicular → Pythagoras gives the arrow length.
$I = \frac{V _{r m s}}{∣ Z ∣} = \frac{120}{12.8} = 9.37 A$ . Why this step? Ohm's Law in AC form: current = voltage ÷ total opposition. Both are RMS, so answer is RMS.
$ϕ = tan^{- 1} \frac{X}{R} = tan^{- 1} \frac{8}{10} = + 38.7°$ . Why this step? The tilt of the arrow; positive because $X > 0$ (inductive).

Verify: $1 0^{2} + 8^{2} = 164$ , $164 \approx 12.81$ ✓. $ϕ > 0$ matches the forecast (inductive, current lags — "ELI"). Units: $Ω$ throughout, $V /Ω = A$ ✓.

Worked example Example 2 — Cell B (net capacitive)

Series RLC: $R = 10 Ω$ , $X_{L} = 15 Ω$ , $X_{C} = 40 Ω$ , $V_{r m s} = 50 V$ . Find $∣ Z ∣$ , $I$ , $ϕ$ .

Forecast: Now $X_{C}$ dominates. Arrow tilts down, $ϕ$ negative, current leads.

Net reactance $X = X_{L} - X_{C} = 15 - 40 = - 25 Ω$ . Why this step? Same subtraction rule; a negative net reactance is the signature of capacitive dominance.
$∣ Z ∣ = 1 0^{2} + (- 25)^{2} = 100 + 625 = 725 = 26.9 Ω$ . Why this step? The square kills the sign — magnitude is always positive. The sign only survives in $ϕ$ .
$I = \frac{50}{26.9} = 1.86 A$ .
$ϕ = tan^{- 1} \frac{- 25}{10} = - 68.2°$ . Why this step? Negative net reactance → negative angle → current leads ("ICE").

Verify: $100 + 625 = 725$ , $725 \approx 26.93$ ✓. Sign of $ϕ$ is negative, forecast confirmed. Notice $∣ Z ∣$ jumped far above $R$ because the reactance ( $25$ ) dwarfs $R$ ( $10$ ).

Worked example Example 3 — Cell C (balanced / resonance)

Series RLC: $R = 10 Ω$ , $X_{L} = 25 Ω$ , $X_{C} = 25 Ω$ , $V_{r m s} = 120 V$ .

Forecast: Equal reactances should cancel. Guess the arrow lies flat, $ϕ = 0$ , and $∣ Z ∣$ collapses to just $R$ — giving the biggest possible current.

Net reactance $X = 25 - 25 = 0 Ω$ . Why this step? At resonance the inductive and capacitive pushes exactly oppose — see Resonance in RLC Circuits.
$∣ Z ∣ = 1 0^{2} + 0^{2} = 10 Ω$ . Why this step? With $X = 0$ the arrow has no vertical part; length equals its horizontal part $R$ .
$I = \frac{120}{10} = 12 A$ . Why this step? Smallest $∣ Z ∣$ possible for this $R$ , so current is at its maximum.
$ϕ = tan^{- 1} (0/10) = 0°$ . Why this step? Flat arrow — voltage and current are perfectly in step.

Verify: Compare with Example 1 (same $R, V$ ): there $∣ Z ∣ = 12.8$ gave $9.37 A$ ; here $∣ Z ∣ = 10$ gives $12 A$ — larger, as forecast. A purely resistive-looking circuit even though $L$ and $C$ are present ✓.

Worked example Example 4 — Cell D (pure resistor) vs Cell E (pure reactance)

(D) Only a resistor: $R = 47 Ω$ , no L, no C, $V_{r m s} = 94 V$ . (E) Only an inductor: $X_{L} = 47 Ω$ , $R = 0$ , $V_{r m s} = 94 V$ .

Forecast: Same $∣ Z ∣$ number for both (47 Ω), but wildly different phase — $0°$ for the resistor, $+ 90°$ for the pure inductor.

(D) $∣ Z ∣ = 4 7^{2} + 0^{2} = 47 Ω$ , $ϕ = tan^{- 1} (0/47) = 0°$ , $I = 94/47 = 2 A$ . Why this step? No reactance → flat arrow → in-phase, pure heat dissipation.
(E) $∣ Z ∣ = 0^{2} + 4 7^{2} = 47 Ω$ , $I = 94/47 = 2 A$ .
(E) phase: $ϕ = tan^{- 1} \frac{47}{0}$ . Why this step? Dividing by zero looks like a crash — but $tan^{- 1}$ of "infinity" is exactly $+ 90°$ . Geometrically the arrow points straight up: all reactance, no resistance.

Verify: Both currents $= 2 A$ ✓. The lesson: $∣ Z ∣$ alone can't tell a resistor from an inductor — you need the phase. That is precisely why we use complex numbers, not just magnitudes.

Worked example Example 5 — Cells F & G (frequency limits, real coil + cap)

A $50 mH$ inductor and a $10 μ F$ capacitor. Evaluate both reactances at $f \to 0$ (DC) and $f = 10 kHz$ .

Forecast (CLiFF): As $f$ rises, $X_{L}$ climbs and $X_{C}$ falls off a cliff. At DC the inductor is a wire ( $X_{L} = 0$ ) and the capacitor is an open gap ( $X_{C} = \infty$ ).

DC ( $f = 0$ ): $X_{L} = 2 π (0) (0.05) = 0 Ω$ (short circuit). Why this step? No changing current → $d i / d t = 0$ → no opposing voltage. See Inductors.
DC ( $f = 0$ ): $X_{C} = \frac{1}{2 π ( 0 ) ( 1 0 ^{- 5} )} \to \infty$ (open circuit). Why this step? Once charged, a capacitor blocks steady current entirely. See Capacitors.
$f = 10 kHz$ : $X_{L} = 2 π (10000) (0.05) = 3141.6 Ω$ . Why this step? Fast $d i / d t$ → large opposing voltage → high reactance.
$f = 10 kHz$ : $X_{C} = \frac{1}{2 π ( 10000 ) ( 10 \times 1 0 ^{- 6} )} = 1.59 Ω$ . Why this step? Rapid charge/discharge → lots of current flows → tiny opposition.

Verify: $2 π \cdot 10000 \cdot 0.05 \approx 3141.6$ ✓. $1/ (2 π \cdot 10000 \cdot 1 0^{- 5}) \approx 1.5915$ ✓. From DC to 10 kHz, $X_{L}$ went $0 \to 3142$ (up) and $X_{C}$ went $\infty \to 1.59$ (down) — opposite trends, exactly CLiFF.

Worked example Example 6 — Cell H (real-world word problem)

A speaker crossover feeds a tweeter through a series capacitor $C = 4.7 μ F$ in series with the tweeter's $R = 8 Ω$ resistance. Audio at $f = 2 kHz$ . What is the impedance the amplifier sees, and does the tweeter get more or less signal at higher frequencies?

Forecast: It's an RC series (cell B territory — capacitive). At 2 kHz, $X_{C}$ should be a handful of ohms, comparable to the $8 Ω$ resistor.

Convert units: $C = 4.7 \times 1 0^{- 6} F$ , $f = 2000 Hz$ . Why this step? Formulas need SI (farads, hertz) or the powers of ten will be wrong.
$X_{C} = \frac{1}{2 π ( 2000 ) ( 4.7 \times 1 0 ^{- 6} )} = 16.93 Ω$ . Why this step? Direct plug-in of $1/ (2 π f C)$ .
$∣ Z ∣ = R^{2} + X_{C}^{2} = 8^{2} + 16.9 3^{2} = 64 + 286.6 = 18.7 Ω$ . Why this step? Series R and C → Pythagoras (net $X = - X_{C}$ ).
$ϕ = tan^{- 1} \frac{- 16.93}{8} = - 64.7°$ (capacitive, current leads). Why this step? Only a capacitor present → sign is negative.

Answer to the design question: as $f$ rises, $X_{C}$ drops, so $∣ Z ∣$ drops and more signal reaches the tweeter — that is why a series capacitor is a high-pass filter. Perfect for a tweeter (which wants highs).

Verify: $1/ (2 π \cdot 2000 \cdot 4.7 \times 1 0^{- 6}) \approx 16.93$ ✓. $64 + 286.6 \approx 18.72$ ✓. Behaviour matches CLiFF (capacitor passes highs).

Worked example Example 7 — Cell I (exam twist: run

$tan^{- 1}$ backwards) A series RL circuit at $f = 60 Hz$ has $R = 100 Ω$ and a measured phase angle $ϕ = + 30°$ (voltage leads current). Find the inductance $L$ .

Forecast: Positive $ϕ$ confirms inductive. Since $tan 30° < 1$ , expect $X_{L} < R$ , so $X_{L}$ around $58 Ω$ , and $L$ a fraction of a henry.

From $ϕ = tan^{- 1} \frac{X _{L}}{R}$ , invert: $X_{L} = R tan ϕ = 100 tan (30°)$ . Why this step? Here we are given the angle and asked for the reactance — the reverse of Examples 1–3. Apply $tan$ to both sides to undo $tan^{- 1}$ .
$X_{L} = 100 \times 0.5774 = 57.74 Ω$ . Why this step? $tan 30° = 1/ 3 \approx 0.5774$ .
Now recover $L$ from $X_{L} = 2 π f L$ : $L = \frac{X _{L}}{2 π f} = \frac{57.74}{2 π ( 60 )}$ . Why this step? Reactance is set; solve the definition of $X_{L}$ for the physical component.
$L = \frac{57.74}{376.99} = 0.1532 H \approx 153 mH$ .

Verify (plug back): With $L = 0.1532 H$ , $X_{L} = 2 π (60) (0.1532) = 57.74 Ω$ , and $tan^{- 1} (57.74/100) = 30.0°$ ✓. Sign positive → inductive, forecast confirmed.

Recall Quick self-test

Which cell has $ϕ = - 90°$ ? ::: Cell E with a pure capacitor ( $R = 0$ ): arrow points straight down. If $X_{L} = X_{C}$ , what is $∣ Z ∣$ ? ::: Just $R$ — cell C (resonance), the current maximum. A series capacitor as a filter passes which frequencies? ::: High frequencies (high-pass), because $X_{C}$ shrinks as $f$ rises. Given $ϕ$ and $R$ , how do you get $X$ ? ::: $X = R tan ϕ$ — invert the arctangent.

Connections

Ohm's Law — every current here is $I = V /∣ Z ∣$ .
Complex Numbers and Phasors — why phase, not just magnitude, is essential (Example 4).
Capacitors and Inductors — the frequency limits in Example 5.
Resonance in RLC Circuits — cell C, the balanced case.
RMS and Peak Values — all voltages/currents above are RMS.
Power in AC Circuits — the same $ϕ$ becomes the power factor $cos ϕ$ .