Complex Numbers in AC Circuits – Impedance and Phasors

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Everything in the first three parts of this series was preparation for this. The four forms, the operations, De Moivre’s theorem — all of it points here. In real engineering board exams, complex numbers almost never appear as pure algebra problems. They appear dressed as AC circuit problems. Impedance, phasors, power factor, apparent power — these are all complex numbers with engineering labels attached. Once you see that clearly, entire sections of the board exam become straightforward.

This is Part 4 of the Complete Complex Numbers ECE and EE Board Exam Reviewer Series on PinoyBIX.org. Part 1 covered the four forms. Part 2 covered operations. Part 3 covered De Moivre’s theorem. This final part connects all of it to real circuit problems with worked examples and every major board exam question type in this topic.

📋 BOARD EXAM RELEVANCE

ECE (Electronics Engineer) — AC circuit analysis using complex numbers is one of the highest-weight topics on the ECE board exam. Impedance, phasor voltage and current, Ohm’s law in phasor form, complex power, and power factor all appear regularly. Expect 6 to 12 items across Engineering Mathematics and Electronics subjects. This is a critical topic.
EE (Electrical Engineer) — This is the core application topic of the entire EE board exam. Phasor analysis, impedance of RLC circuits, series and parallel impedance combinations, complex power, power factor correction — all are tested heavily. Very high frequency. Mastery is non-negotiable.
ME (Mechanical Engineer) — Complex numbers in AC circuits appear in Engineering Mathematics and occasionally in Electrical Technology subjects. Impedance of series RLC circuits and basic phasor representation are the most tested. Moderate frequency.
CE (Civil Engineer) — Basic AC circuit analysis appears in Engineering Mathematics. Impedance and power factor calculations appear occasionally. Low to moderate frequency.
ChE (Chemical Engineer) — Complex numbers in process control transfer functions use the same mathematics as AC circuit analysis. The variable $s = \sigma + j\omega$ in Laplace transforms is a direct extension of phasor concepts. Moderate frequency in Engineering Mathematics and Chemical Engineering subjects.
GeE (Geodetic Engineer) — Basic AC circuit concepts appear in Engineering Mathematics. Low frequency.
Naval Architect and Marine Engineer — AC circuit analysis and vibration analysis both use complex numbers. Impedance and phasor problems appear in Engineering Mathematics and Marine Electrical subjects. Moderate frequency.

Bottom line: ECE and EE examinees must master every section of this post completely. ME, ChE, and Naval Architecture examinees must master impedance and basic phasor analysis. All other boards need a working understanding of the impedance formula and power factor.

Why AC Circuits Use Complex Numbers

In a DC circuit, resistance is the only opposition to current flow. It is a single real number measured in ohms. In an AC circuit, inductors and capacitors also oppose current — but in a frequency-dependent way that shifts the phase between voltage and current. A single real number cannot capture both the magnitude and the phase shift simultaneously.

Complex numbers solve this problem exactly. The real part carries the resistive component. The imaginary part carries the reactive component. The angle between them is the phase shift. One complex number describes everything about how a component or circuit behaves under AC conditions.

KEY CONCEPT — Why Impedance is a Complex Number

In DC circuits: $V = IR$ where $R$ is a real number.

In AC circuits: $\mathbf{V} = \mathbf{I} \cdot \mathbf{Z}$ where $\mathbf{Z} = R + jX$ is a complex number.

The real part $R$ is resistance. The imaginary part $X$ is reactance. The angle $\theta = \arctan(X/R)$ is the phase difference between voltage and current.

Impedance — $Z = R + jX$

Impedance $\mathbf{Z}$ is the total opposition to AC current flow. It is a complex number in rectangular form where the real part is resistance and the imaginary part is reactance.

KEY FORMULAS — Impedance

$\mathbf{Z} = R + jX \quad (\Omega)$

$|\mathbf{Z}| = \sqrt{R^2 + X^2} \quad (\Omega)$

$\theta = \arctan\!\left(\dfrac{X}{R}\right)$

where $R$ is resistance in ohms, $X$ is reactance in ohms, $|\mathbf{Z}|$ is the magnitude of impedance, and $\theta$ is the phase angle.

The sign of the reactance $X$ tells you whether the circuit is inductive or capacitive.

Positive $X$ means inductive reactance. The circuit is inductive. Voltage leads current.
Negative $X$ means capacitive reactance. The circuit is capacitive. Current leads voltage.
$X = 0$ means purely resistive. Voltage and current are in phase.

Impedance of Individual Components

KEY FORMULAS — Component Impedances

Component	Impedance	Phase (V vs I)	Reactance
Resistor $R$	$\mathbf{Z}_R = R$	$0°$ — in phase	No reactance
Inductor $L$	$\mathbf{Z}_L = j\omega L = jX_L$	$+90°$ — V leads I	$X_L = \omega L$
Capacitor $C$	$\mathbf{Z}_C = \dfrac{1}{j\omega C} = -j\dfrac{1}{\omega C} = -jX_C$	$-90°$ — I leads V	$X_C = \dfrac{1}{\omega C}$

where $\omega = 2\pi f$ is the angular frequency in radians per second.

Series RLC Circuit Impedance

For a series RLC circuit, the total impedance is the sum of the individual impedances.

KEY FORMULA — Series RLC Impedance

$\mathbf{Z}_{total} = R + j(X_L - X_C) = R + j\!\left(\omega L - \dfrac{1}{\omega C}\right)$

$|\mathbf{Z}_{total}| = \sqrt{R^2 + (X_L - X_C)^2}$

$\theta = \arctan\!\left(\dfrac{X_L - X_C}{R}\right)$

When $X_L > X_C$ , the net reactance is positive and the circuit is inductive. When $X_L < X_C$ , the net reactance is negative and the circuit is capacitive. When $X_L = X_C$ , the circuit is at resonance and $\mathbf{Z} = R$ .

Phasors — Representing Sinusoidal Signals as Complex Numbers

A sinusoidal voltage or current varies with time. Writing it as a time function in every calculation is tedious. Phasor representation solves this by converting the time domain signal into a complex number — a fixed vector in the complex plane whose magnitude is the amplitude and whose angle is the phase.

KEY FORMULAS — Phasor Conversion

Time domain to phasor:

$v(t) = V_m\cos(\omega t + \phi) \quad \longrightarrow \quad \mathbf{V} = V_m\angle\phi$

Drop the $\cos$ and $\omega t$ . Keep the amplitude and phase angle.

Phasor to time domain:

$\mathbf{V} = V_m\angle\phi \quad \longrightarrow \quad v(t) = V_m\cos(\omega t + \phi)$

Add back the $\cos$ and $\omega t$ . The magnitude becomes the amplitude.

Important: phasors assume all signals in the circuit share the same frequency $\omega$ . The phasor captures only the amplitude and phase — not the frequency. This is why phasor analysis works only for single-frequency AC circuits.

Ohm’s Law in Phasor Form

KEY FORMULA — Ohm’s Law for AC Circuits

$\mathbf{V} = \mathbf{I} \cdot \mathbf{Z}$

$\mathbf{I} = \dfrac{\mathbf{V}}{\mathbf{Z}}$

$\mathbf{Z} = \dfrac{\mathbf{V}}{\mathbf{I}}$

These are exactly the same as DC Ohm’s law except that $\mathbf{V}$ , $\mathbf{I}$ , and $\mathbf{Z}$ are all complex numbers. Multiply and divide them using polar form: multiply magnitudes and add or subtract angles.

The memory device “ELI the ICE man” tells you the phase relationship without calculation.

ELI — In an inductor (L), voltage (E) leads current (I). The current lags.
ICE — In a capacitor (C), current (I) leads voltage (E). The current leads.

Complex Power — $S$ , $P$ , $Q$ , and Power Factor

Power in AC circuits has three components, each representing a different aspect of energy behavior. Together they form the power triangle, which is geometrically identical to the impedance triangle.

KEY FORMULAS — Complex Power

$\mathbf{S} = P + jQ \quad \text{(complex power, VA)}$

$P = |\mathbf{S}|\cos\theta = I_{rms}^2 R \quad \text{(real power, W)}$

$Q = |\mathbf{S}|\sin\theta = I_{rms}^2 X \quad \text{(reactive power, VAR)}$

$|\mathbf{S}| = \sqrt{P^2 + Q^2} = V_{rms} \cdot I_{rms} \quad \text{(apparent power, VA)}$

$\text{pf} = \cos\theta = \dfrac{P}{|\mathbf{S}|} \quad \text{(power factor)}$

Real power $P$ is the actual energy consumed per second. It is what your electric meter measures. Reactive power $Q$ is energy stored and returned by inductors and capacitors — it does no net work but increases the current demand. Apparent power $|\mathbf{S}|$ is the product of rms voltage and rms current regardless of phase.

Power factor ranges from $0$ to $1$ . A power factor of $1$ means the load is purely resistive — all power delivered is consumed. A low power factor means significant reactive power is present, which wastes transmission capacity.

Inductive loads have lagging power factor — current lags voltage.
Capacitive loads have leading power factor — current leads voltage.

Parallel Impedance

For two impedances connected in parallel, the equivalent impedance uses the product over sum formula — the same as parallel resistance in DC circuits, but with complex number arithmetic.

KEY FORMULA — Two Impedances in Parallel

$\mathbf{Z}_{eq} = \dfrac{\mathbf{Z}_1 \cdot \mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2}$

Multiply the numerator in polar form. Add the denominator in rectangular form. Convert the denominator to polar for the final division.

Worked Problems — Board Exam Type Questions

The following 10 problems represent actual ECE, EE, ME, CE, and ChE board exam question types on complex numbers in AC circuits. Work each problem completely before reading the solution.

Problem 1 — ECE Board Exam Type

A series RLC circuit has $R = 6\,\Omega$ , $X_L = 10\,\Omega$ , and $X_C = 2\,\Omega$ . Find the total impedance in both rectangular and polar form.

Given: $R = 6\,\Omega$ , $X_L = 10\,\Omega$ , $X_C = 2\,\Omega$

Find: $\mathbf{Z}$ in rectangular and polar form.

Solution:

Step 1: Write the total impedance in rectangular form.

$\mathbf{Z} = R + j(X_L - X_C) = 6 + j(10 - 2) = 6 + j8\,\Omega$

Step 2: Find the magnitude.

$|\mathbf{Z}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\Omega$

Step 3: Find the phase angle. Since both $R$ and $X$ are positive, the angle is in Quadrant I.

$\theta = \arctan\!\left(\dfrac{8}{6}\right) = \arctan(1.333) = 53.13°$

Step 4: Write the polar form.

$\mathbf{Z} = 10\angle 53.13°\,\Omega$

✓ ANSWER: $\mathbf{Z} = 6 + j8\,\Omega$

(rectangular) and $\mathbf{Z} = 10\angle 53.13°\,\Omega$

(polar)

Examiner note: The 6-8-10 right triangle is a scaled version of the 3-4-5 triangle. Recognize it immediately. $|\mathbf{Z}| = 10\,\Omega$ and $\theta = 53.13°$ without a calculator. This specific combination appears frequently on the board exam.

Problem 2 — EE Board Exam Type

A source voltage $\mathbf{V} = 100\angle 0°$ V is applied to a circuit with impedance $\mathbf{Z} = 10\angle 53.13°\,\Omega$ . Find the phasor current $\mathbf{I}$ .

Given: $\mathbf{V} = 100\angle 0°$ V, $\mathbf{Z} = 10\angle 53.13°\,\Omega$

Find: $\mathbf{I}$

Solution:

Step 1: Apply Ohm’s law in phasor form.

$\mathbf{I} = \dfrac{\mathbf{V}}{\mathbf{Z}} = \dfrac{100\angle 0°}{10\angle 53.13°}$

Step 2: Divide the magnitudes and subtract the angles.

$|\mathbf{I}| = \dfrac{100}{10} = 10\,\text{A}$

$\angle\mathbf{I} = 0° - 53.13° = -53.13°$

Step 3: Write the result.

$\mathbf{I} = 10\angle{-53.13°}\,\text{A}$

✓ ANSWER: $\mathbf{I} = 10\angle{-53.13°}$

Examiner note: The negative angle on the current confirms the circuit is inductive — current lags voltage by $53.13°$ . This matches the positive reactance $X = +8\,\Omega$ from Problem 1. Always check that the sign of the current angle is consistent with the nature of the circuit reactance.

Problem 3 — ECE Board Exam Type

Convert the time domain voltage $v(t) = 311\cos(\omega t + 30°)$ V to phasor form. Then convert the phasor $\mathbf{I} = 5\angle{-45°}$ A back to time domain.

Given: $v(t) = 311\cos(\omega t + 30°)$ V and $\mathbf{I} = 5\angle{-45°}$ A

Find: Phasor $\mathbf{V}$ and time domain $i(t)$ .

Solution:

Step 1: Convert $v(t)$ to phasor form. Drop $\cos$ and $\omega t$ . Keep amplitude and phase angle.

$\mathbf{V} = 311\angle 30°\,\text{V}$

Step 2: Convert $\mathbf{I} = 5\angle{-45°}$ A back to time domain. Add $\cos(\omega t + \phi)$ .

$i(t) = 5\cos(\omega t - 45°)\,\text{A}$

✓ ANSWER: $\mathbf{V} = 311\angle 30°$

V and $i(t) = 5\cos(\omega t - 45°)$

Examiner note: The amplitude in phasor form is the peak value $V_m$ , not the rms value. If the problem gives rms voltage and asks for the phasor, use the peak value $V_m = V_{rms}\sqrt{2}$ . If it asks for rms phasor directly, use the rms value as the magnitude. Read the problem statement carefully to determine which convention is required.

Problem 4 — ME Board Exam Type

A coil has resistance $R = 8\,\Omega$ and inductive reactance $X_L = 6\,\Omega$ . Find the impedance, the phase angle, and state whether current leads or lags voltage.

Given: $R = 8\,\Omega$ , $X_L = 6\,\Omega$

Find: $\mathbf{Z}$ , $\theta$ , and the phase relationship.

Solution:

Step 1: Write the impedance in rectangular form. There is no capacitor so $X_C = 0$ .

$\mathbf{Z} = R + jX_L = 8 + j6\,\Omega$

Step 2: Find the magnitude.

$|\mathbf{Z}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\,\Omega$

Step 3: Find the phase angle.

$\theta = \arctan\!\left(\dfrac{6}{8}\right) = \arctan(0.75) = 36.87°$

Step 4: Determine the phase relationship. The reactance is positive (inductive), so the circuit is inductive. By ELI: voltage leads current, or equivalently, current lags voltage by $36.87°$ .

✓ ANSWER: $\mathbf{Z} = 10\angle 36.87°\,\Omega$

. Current lags voltage by $36.87°$

(inductive circuit).

Examiner note: The 6-8-10 triangle (a scaled 3-4-5 triangle) gives $|\mathbf{Z}| = 10\,\Omega$ directly. Recognize common Pythagorean triples in impedance problems to save computation time: 3-4-5, 5-12-13, 8-15-17, and their multiples.

Problem 5 — EE Board Exam Type

Two impedances $\mathbf{Z}_1 = 3 + j4\,\Omega$ and $\mathbf{Z}_2 = 4 - j3\,\Omega$ are connected in series. Find the total impedance and the total current if the source voltage is $\mathbf{V} = 50\angle 0°$ V.

Given: $\mathbf{Z}_1 = 3 + j4\,\Omega$ , $\mathbf{Z}_2 = 4 - j3\,\Omega$ , $\mathbf{V} = 50\angle 0°$ V

Find: $\mathbf{Z}_{total}$ and $\mathbf{I}$ .

Solution:

Step 1: Add the impedances in rectangular form. Series impedances add directly.

$\mathbf{Z}_{total} = \mathbf{Z}_1 + \mathbf{Z}_2 = (3 + 4) + j(4 - 3) = 7 + j1\,\Omega$

Step 2: Convert to polar form.

$|\mathbf{Z}_{total}| = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \approx 7.07\,\Omega$

$\theta = \arctan\!\left(\dfrac{1}{7}\right) = 8.13°$

$\mathbf{Z}_{total} = 7.07\angle 8.13°\,\Omega$

Step 3: Find the current using Ohm’s law.

$\mathbf{I} = \dfrac{\mathbf{V}}{\mathbf{Z}_{total}} = \dfrac{50\angle 0°}{7.07\angle 8.13°} = \dfrac{50}{7.07}\angle(0° - 8.13°) = 7.07\angle{-8.13°}\,\text{A}$

✓ ANSWER: $\mathbf{Z}_{total} = 7.07\angle 8.13°\,\Omega$

and $\mathbf{I} = 7.07\angle{-8.13°}$

Examiner note: Series impedances are always added in rectangular form. The result $7 + j1$ has a small positive reactance, which means the combined circuit is slightly inductive. The small phase angle of $8.13°$ confirms this — current lags voltage by only a small amount.

Problem 6 — ECE Board Exam Type

Two impedances $\mathbf{Z}_1 = 3 + j4\,\Omega$ and $\mathbf{Z}_2 = 4 - j3\,\Omega$ are connected in parallel. Find the equivalent impedance.

Given: $\mathbf{Z}_1 = 3 + j4\,\Omega$ , $\mathbf{Z}_2 = 4 - j3\,\Omega$

Find: $\mathbf{Z}_{eq}$

Solution:

Step 1: Convert both impedances to polar form for multiplication.

$|\mathbf{Z}_1| = \sqrt{3^2 + 4^2} = 5\,\Omega \quad \theta_1 = \arctan\!\left(\dfrac{4}{3}\right) = 53.13° \quad \Rightarrow \quad \mathbf{Z}_1 = 5\angle 53.13°$

$|\mathbf{Z}_2| = \sqrt{4^2 + (-3)^2} = 5\,\Omega \quad \theta_2 = \arctan\!\left(\dfrac{-3}{4}\right) = -36.87° \quad \Rightarrow \quad \mathbf{Z}_2 = 5\angle{-36.87°}$

Step 2: Multiply $\mathbf{Z}_1 \cdot \mathbf{Z}_2$ in polar form.

$\mathbf{Z}_1 \cdot \mathbf{Z}_2 = (5 \times 5)\angle(53.13° + (-36.87°)) = 25\angle 16.26°$

Step 3: Add $\mathbf{Z}_1 + \mathbf{Z}_2$ in rectangular form.

$\mathbf{Z}_1 + \mathbf{Z}_2 = (3 + 4) + j(4 - 3) = 7 + j1\,\Omega$

Step 4: Convert the denominator to polar form.

$|7 + j1| = \sqrt{49 + 1} = \sqrt{50} \approx 7.07\,\Omega \quad \theta = \arctan\!\left(\dfrac{1}{7}\right) = 8.13°$

$\mathbf{Z}_1 + \mathbf{Z}_2 = 7.07\angle 8.13°$

Step 5: Divide numerator by denominator in polar form.

$\mathbf{Z}_{eq} = \dfrac{25\angle 16.26°}{7.07\angle 8.13°} = \dfrac{25}{7.07}\angle(16.26° - 8.13°) = 3.54\angle 8.13°\,\Omega$

Step 6: Convert to rectangular form.

$R = 3.54\cos 8.13° = 3.54(0.990) = 3.50\,\Omega$

$X = 3.54\sin 8.13° = 3.54(0.141) = 0.50\,\Omega$

✓ ANSWER: $\mathbf{Z}_{eq} = 3.5 + j0.5\,\Omega$

or $3.54\angle 8.13°\,\Omega$

Examiner note: Parallel impedance always requires the product over sum formula. Multiply the numerator in polar form and add the denominator in rectangular form. Never add impedances in polar form directly. This mixed form approach is the correct and efficient method.

Problem 7 — EE Board Exam Type

A single-phase load draws $P = 1{,}200$ W at a power factor of $0.8$ lagging from a $240$ V rms source. Find the apparent power $|\mathbf{S}|$ , reactive power $Q$ , and the load current $I_{rms}$ .

Given: $P = 1{,}200$ W, $\text{pf} = 0.8$ lagging, $V_{rms} = 240$ V

Find: $|\mathbf{S}|$ , $Q$ , and $I_{rms}$

Solution:

Step 1: Find the apparent power using the power factor definition.

$\text{pf} = \dfrac{P}{|\mathbf{S}|} \quad \Rightarrow \quad |\mathbf{S}| = \dfrac{P}{\text{pf}} = \dfrac{1{,}200}{0.8} = 1{,}500\,\text{VA}$

Step 2: Find the reactive power using the power triangle.

$|\mathbf{S}|^2 = P^2 + Q^2 \quad \Rightarrow \quad Q = \sqrt{|\mathbf{S}|^2 - P^2} = \sqrt{1{,}500^2 - 1{,}200^2}$

$Q = \sqrt{2{,}250{,}000 - 1{,}440{,}000} = \sqrt{810{,}000} = 900\,\text{VAR}$

Step 3: Find the rms current.

$|\mathbf{S}| = V_{rms} \cdot I_{rms} \quad \Rightarrow \quad I_{rms} = \dfrac{|\mathbf{S}|}{V_{rms}} = \dfrac{1{,}500}{240} = 6.25\,\text{A}$

✓ ANSWER: $|\mathbf{S}| = 1{,}500$

VA, $Q = 900$

VAR, $I_{rms} = 6.25$

Examiner note: The power triangle relationship $P^2 + Q^2 = |\mathbf{S}|^2$ is the Pythagorean theorem applied to power. It is structurally identical to $R^2 + X^2 = |\mathbf{Z}|^2$ . A lagging power factor means the load is inductive, so $Q$ is positive.

Problem 8 — CE Board Exam Type

A circuit has impedance $\mathbf{Z} = 5 - j12\,\Omega$ . Find the power factor and state whether it is leading or lagging.

Given: $\mathbf{Z} = 5 - j12\,\Omega$

Find: Power factor and its nature.

Solution:

Step 1: Find the magnitude of the impedance.

$|\mathbf{Z}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\,\Omega$

Step 2: Find the phase angle.

$\theta = \arctan\!\left(\dfrac{-12}{5}\right) = -67.38°$

Step 3: Compute the power factor.

$\text{pf} = \cos\theta = \cos(-67.38°) = 0.385$

Step 4: Determine the nature. The reactance is negative ( $X = -12\,\Omega$ ), which means the circuit is capacitive. Capacitive circuits have leading power factor — current leads voltage.

✓ ANSWER: Power factor $= 0.385$

leading (capacitive circuit)

Examiner note: The 5-12-13 Pythagorean triple gives $|\mathbf{Z}| = 13\,\Omega$ directly. Negative reactance means capacitive means leading power factor. Positive reactance means inductive means lagging power factor. These three-part connections must be automatic on the board exam.

Problem 9 — ECE Board Exam Type

A series circuit has $R = 10\,\Omega$ and $L = 31.83$ mH. The source frequency is $f = 60$ Hz. Find the impedance at this frequency and the current if $\mathbf{V} = 120\angle 0°$ V.

Given: $R = 10\,\Omega$ , $L = 31.83$ mH, $f = 60$ Hz, $\mathbf{V} = 120\angle 0°$ V

Find: $\mathbf{Z}$ and $\mathbf{I}$

Solution:

Step 1: Compute the angular frequency.

$\omega = 2\pi f = 2\pi(60) = 376.99\,\text{rad/s}$

Step 2: Compute the inductive reactance.

$X_L = \omega L = 376.99 \times 0.03183 = 12\,\Omega$

Step 3: Write the total impedance.

$\mathbf{Z} = R + jX_L = 10 + j12\,\Omega$

Step 4: Convert to polar form.

$|\mathbf{Z}| = \sqrt{10^2 + 12^2} = \sqrt{100 + 144} = \sqrt{244} \approx 15.62\,\Omega$

$\theta = \arctan\!\left(\dfrac{12}{10}\right) = \arctan(1.2) = 50.19°$

$\mathbf{Z} = 15.62\angle 50.19°\,\Omega$

Step 5: Find the current.

$\mathbf{I} = \dfrac{\mathbf{V}}{\mathbf{Z}} = \dfrac{120\angle 0°}{15.62\angle 50.19°} = 7.68\angle{-50.19°}\,\text{A}$

✓ ANSWER: $\mathbf{Z} = 15.62\angle 50.19°\,\Omega$

and $\mathbf{I} = 7.68\angle{-50.19°}$

Examiner note: When a problem gives inductance in mH and frequency in Hz, always compute $X_L = \omega L = 2\pi f L$ as the first step. Substituting directly without computing $\omega$ first is a common source of error. Also note that $L = 31.83$ mH is specifically chosen to give $X_L = 12\,\Omega$ at $60$ Hz — a deliberate board exam construction.

Problem 10 — EE Board Exam Type

A load draws complex power $\mathbf{S} = 800 + j600$ VA from a $100$ V rms source. Find the power factor, the load impedance, and the load current.

Given: $\mathbf{S} = 800 + j600$ VA, $V_{rms} = 100$ V

Find: pf, $\mathbf{Z}_{load}$ , and $I_{rms}$

Solution:

Step 1: Identify $P = 800$ W and $Q = 600$ VAR from the complex power.

Step 2: Find the apparent power.

$|\mathbf{S}| = \sqrt{P^2 + Q^2} = \sqrt{800^2 + 600^2} = \sqrt{640{,}000 + 360{,}000} = \sqrt{1{,}000{,}000} = 1{,}000\,\text{VA}$

Step 3: Find the power factor. The reactive power $Q$ is positive, which means the load is inductive (lagging).

$\text{pf} = \dfrac{P}{|\mathbf{S}|} = \dfrac{800}{1{,}000} = 0.8 \text{ lagging}$

Step 4: Find the rms current.

$I_{rms} = \dfrac{|\mathbf{S}|}{V_{rms}} = \dfrac{1{,}000}{100} = 10\,\text{A}$

Step 5: Find the load impedance.

$|\mathbf{Z}_{load}| = \dfrac{V_{rms}}{I_{rms}} = \dfrac{100}{10} = 10\,\Omega$

$\theta = \arctan\!\left(\dfrac{Q}{P}\right) = \arctan\!\left(\dfrac{600}{800}\right) = \arctan(0.75) = 36.87°$

$\mathbf{Z}_{load} = 10\angle 36.87°\,\Omega = 8 + j6\,\Omega$

✓ ANSWER: pf $= 0.8$

lagging, $\mathbf{Z}_{load} = 8 + j6\,\Omega$

, $I_{rms} = 10$

Examiner note: The power triangle and the impedance triangle share the same angle $\theta$ . The power factor angle, the impedance angle, and the voltage-current phase difference are all the same angle. Recognizing this connection allows you to extract the load impedance directly from the complex power without additional computation.

Common Mistakes and Examiner Traps

These are the most consistent error patterns in board exam solutions for AC circuit problems involving complex numbers.

❌ Common Mistake	✅ Correct Approach
Writing $\mathbf{Z}_C = +jX_C$ (positive sign for capacitive reactance). Treating capacitive reactance as positive imaginary, same as inductive reactance.	Capacitive impedance is $\mathbf{Z}_C = -jX_C$ where $X_C = \dfrac{1}{\omega C}$ . The imaginary part is negative. This places capacitive impedance below the real axis in the impedance diagram, confirming that current leads voltage in a capacitive circuit.
Adding impedances in polar form for series combinations. Writing $\mathbf{Z}_{total} = r_1\angle\theta_1 + r_2\angle\theta_2$ directly without converting to rectangular form.	Series impedances are always added in rectangular form: $\mathbf{Z}_{total} = (R_1 + R_2) + j(X_1 + X_2)$ . Convert to polar only after completing the addition, if polar form is needed for the final answer.
Confusing peak value with rms value in power calculations. Using peak voltage $V_m$ instead of rms voltage $V_{rms}$ in formulas like $\|\mathbf{S}\| = V \cdot I$ and $P = I^2 R$ .	Power formulas use rms values: $\|\mathbf{S}\| = V_{rms} \cdot I_{rms}$ and $P = I_{rms}^2 R$ . The relationship between peak and rms for sinusoidal signals is $V_{rms} = \dfrac{V_m}{\sqrt{2}}$ . Always identify which value the problem provides before substituting.
Stating power factor as leading when the circuit is inductive, and vice versa. Mixing up the ELI and ICE conventions.	Use ELI the ICE man: in an inductive circuit (L), voltage (E) leads current (I), so current lags — lagging power factor. In a capacitive circuit (C), current (I) leads voltage (E) — leading power factor. Positive reactance means inductive means lagging. Negative reactance means capacitive means leading.
Using the wrong formula for parallel impedance. Adding reciprocals and forgetting to take the reciprocal of the sum, or attempting to add polar impedances directly.	For two parallel impedances: $\mathbf{Z}_{eq} = \dfrac{\mathbf{Z}_1 \cdot \mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2}$ . Multiply the numerator in polar form. Add the denominator in rectangular form. Then divide in polar form.
Forgetting to compute $\omega = 2\pi f$ before finding $X_L$ or $X_C$ . Substituting frequency $f$ directly into $X_L = \omega L$ instead of angular frequency $\omega$ .	Always compute $\omega = 2\pi f$ first when inductance or capacitance values are given with frequency in Hz. Then compute $X_L = \omega L$ and $X_C = \dfrac{1}{\omega C}$ . Substituting $f$ directly for $\omega$ gives an answer that is off by a factor of $2\pi$ .
Ignoring the phase angle when stating the final current or voltage. Writing $I = 10$ A instead of $\mathbf{I} = 10\angle{-53.13°}$ A after computing the phasor current.	Phasor quantities require both magnitude and angle. Always write the complete phasor form as $r\angle\theta$ in the final answer. The magnitude alone is incomplete and loses the phase information that most board exam problems specifically ask for.

Board Exam Quick Tips

Write $\mathbf{Z} = R + j(X_L - X_C)$ as your first line for any series RLC problem. This single formula sets up the entire solution — magnitude, phase angle, current, and power factor all follow directly from it. Writing this line first also prevents you from mixing up the signs of $X_L$ and $X_C$ .
Apply Ohm’s law as $\mathbf{I} = \mathbf{V}/\mathbf{Z}$ in polar form every time. Divide the voltage magnitude by the impedance magnitude. Subtract the impedance angle from the voltage angle. Two arithmetic steps. This is the fastest method and the one least prone to sign errors.
ELI the ICE man is a required memory item, not optional. Positive reactance means inductive means current lags means lagging power factor. Negative reactance means capacitive means current leads means leading power factor. Know all three links in that chain for both cases.
The power triangle and the impedance triangle share the same angle. The power factor angle $\theta$ , the impedance phase angle, and the voltage-current phase difference are all the same quantity. Use this connection to move between power and impedance problems efficiently.
Always compute $\omega = 2\pi f$ before substituting into $X_L = \omega L$ or $X_C = 1/(\omega C)$ . The board exam gives you $f$ in Hz and expects you to convert to $\omega$ first. Skipping this step gives an answer that is off by a factor of $2\pi$ — a systematic error that eliminates the item.

Frequently Asked Questions

Q1. Why is capacitive reactance written as $-jX_C$ instead of $+jX_C$ ?

The impedance of a capacitor comes from its voltage-current relationship in the frequency domain: $\mathbf{Z}_C = \dfrac{1}{j\omega C}$ . To simplify, multiply numerator and denominator by $-j$ : $\dfrac{1}{j\omega C} \cdot \dfrac{-j}{-j} = \dfrac{-j}{\omega C} = -j\dfrac{1}{\omega C} = -jX_C$ . The negative sign arises naturally from the mathematics of the capacitor’s impedance. It is not a convention — it is a derived result.

Q2. What happens to impedance at resonance in a series RLC circuit?

At resonance, $X_L = X_C$ , which means $\omega L = \dfrac{1}{\omega C}$ . The reactive components cancel and the total impedance becomes $\mathbf{Z} = R + j(X_L - X_C) = R + j(0) = R$ . The impedance is purely resistive, the phase angle is $0°$ , the power factor is $1$ , and the current is at its maximum value $I = V/R$ . This is the resonance condition.

Q3. What is the difference between real power, reactive power, and apparent power?

Real power $P$ in watts is the actual energy consumed per second by resistive elements. It does useful work — heating, lighting, mechanical output. Reactive power $Q$ in VAR is energy stored and returned alternately by inductors and capacitors. It does no net work but increases the current demand on the source. Apparent power $|\mathbf{S}|$ in VA is the total power that the source must supply, equal to $V_{rms} \times I_{rms}$ regardless of phase. The three are related by $|\mathbf{S}|^2 = P^2 + Q^2$ .

Q4. Can a circuit have both inductive and capacitive elements and still have a unity power factor?

Yes. If $X_L = X_C$ , the inductive and capacitive reactances cancel exactly. The net reactance is zero, making the impedance purely resistive. The phase angle is $0°$ and the power factor is $\cos 0° = 1$ . This is the resonance condition. In practice, power factor correction capacitors are deliberately added to inductive loads to reduce the reactive component and bring the power factor closer to unity.

Q5. How is complex power used in the board exam?

Complex power problems typically give you two of the three quantities — $P$ , $Q$ , $|\mathbf{S}|$ , or power factor — and ask for the rest. The solution always uses $|\mathbf{S}|^2 = P^2 + Q^2$ and $\text{pf} = P/|\mathbf{S}|$ . A second common type gives the complex power $\mathbf{S} = P + jQ$ directly and asks for the load impedance, which requires computing $I_{rms} = |\mathbf{S}|/V_{rms}$ and then $\mathbf{Z} = V_{rms}/I_{rms}$ at the appropriate angle.

Series Complete — What You Now Know

This post completes the Complex Numbers ECE and EE Board Exam Reviewer Series on PinoyBIX.org. Across four posts, the series covered every major subtopic that appears in engineering board exams under this heading.

Part	Topic	Key Concepts
1	Forms and the j-Operator	Four forms, Argand diagram, modulus, argument, powers of $j$
2	Operations	Addition, subtraction, multiplication, division, conjugate method
3	De Moivre’s Theorem	Integer powers, all nth roots, root spacing, principal root
4	AC Circuits and Applications	Impedance, phasors, Ohm’s law, complex power, power factor

→ Back to the Complete Complex Numbers ECE and EE Board Exam Reviewer Series

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Complex Numbers in AC Circuits – Impedance and Phasors | PinoyBIX

Why AC Circuits Use Complex Numbers

Impedance — $Z = R + jX$

Impedance of Individual Components

Series RLC Circuit Impedance

Phasors — Representing Sinusoidal Signals as Complex Numbers

Ohm’s Law in Phasor Form

Complex Power — $S$ , $P$ , $Q$ , and Power Factor

Parallel Impedance

Worked Problems — Board Exam Type Questions

Problem 1 — ECE Board Exam Type

Problem 2 — EE Board Exam Type

Problem 3 — ECE Board Exam Type

Problem 4 — ME Board Exam Type

Problem 5 — EE Board Exam Type

Problem 6 — ECE Board Exam Type

Problem 7 — EE Board Exam Type

Problem 8 — CE Board Exam Type

Problem 9 — ECE Board Exam Type

Problem 10 — EE Board Exam Type

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

Series Complete — What You Now Know

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Why AC Circuits Use Complex Numbers

Impedance of Individual Components

Series RLC Circuit Impedance

Phasors — Representing Sinusoidal Signals as Complex Numbers

Ohm’s Law in Phasor Form

Parallel Impedance

Worked Problems — Board Exam Type Questions

Problem 1 — ECE Board Exam Type

Problem 2 — EE Board Exam Type

Problem 3 — ECE Board Exam Type

Problem 4 — ME Board Exam Type

Problem 5 — EE Board Exam Type

Problem 6 — ECE Board Exam Type

Problem 7 — EE Board Exam Type

Problem 8 — CE Board Exam Type

Problem 9 — ECE Board Exam Type

Problem 10 — EE Board Exam Type

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

Series Complete — What You Now Know

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