FAD1022 CQ3: Rapid-Fire Drill Pack — AC Analysis

Objective: Master all AC Circuit Analysis concepts for CQ3.
Target: 3-4 min per problem (CQ3 style: 4 questions in 15 min).
Total problems: 32
Estimated time: ~110 min

Cheat Sheet (Memorize First)

AC Fundamentals

Quantity Symbol Formula
Angular frequency $\omega$ $2\pi f$
Period $T$ $1/f$
Peak vs RMS $V_{\text{rms}}$ $V_{\text{peak}}/\sqrt{2}$
Peak vs RMS (current) $I_{\text{rms}}$ $I_{\text{peak}}/\sqrt{2}$

Reactance & Impedance

Component Reactance Phase
Resistor $R$ $0°$ (in phase)
Capacitor $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$ $-90°$ (current leads)
Inductor $X_L = \omega L = 2\pi f L$ $+90°$ (voltage leads)
RLC Series $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$

Power in AC Circuits

Power Type Symbol Formula
Real (Average) $P_e$ $I_{\text{rms}}^2 R = V_{\text{rms}} I_{\text{rms}} \cos\phi$
Reactive (Inductive) $P_{LC}$ $I_{\text{rms}}^2 X_L = V_{\text{rms}} I_{\text{rms}} \sin\phi$
Reactive (Capacitive) $P_{CC}$ $I_{\text{rms}}^2 X_C = V_{\text{rms}} I_{\text{rms}} \sin\phi$
Apparent $P_A$ $V_{\text{rms}} I_{\text{rms}} = I_{\text{rms}}^2 Z$
Power Factor $\cos\phi$ $R/Z = P_e/P_A$

Resonance

Quantity Formula
Resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$
At resonance $X_L = X_C$, $Z = R$ (minimum), $\phi = 0$
Quality factor $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$

CIVIL Mnemonic

  • Capacitor: I leads V (current leads voltage by $90°$)
  • Inductor: V leads I (voltage leads current by $90°$)

Part A: Reactance Calculations (6 problems)

Target: 2-3 min per problem.

Set A1 — Capacitive Reactance (3 problems)

  1. A $50\ \mu\text{F}$ capacitor is connected to a $50\ \text{Hz}$ AC source. Calculate the capacitive reactance $X_C$.

  2. What capacitance is needed to give $X_C = 100\ \Omega$ at $f = 60\ \text{Hz}$?

  3. A capacitor has $X_C = 318\ \Omega$ at $f = 50\ \text{Hz}$. Find: (a) The capacitance (b) The reactance at $f = 100\ \text{Hz}$

Score: ___/3

Set A2 — Inductive Reactance (3 problems)

  1. A $0.50\ \text{H}$ inductor is connected to a $50\ \text{Hz}$ AC source. Calculate the inductive reactance $X_L$.

  2. What inductance is needed to give $X_L = 200\ \Omega$ at $f = 50\ \text{Hz}$?

  3. An inductor has $X_L = 314\ \Omega$ at $f = 50\ \text{Hz}$. Find: (a) The inductance (b) The reactance at $f = 60\ \text{Hz}$

Score: ___/3

Part B: Impedance & Ohm's Law for AC (8 problems)

Target: 3 min per problem.

Set B1 — RLC Series Impedance (4 problems)

  1. An RLC series circuit has $R = 30\ \Omega$, $L = 0.20\ \text{H}$, $C = 50\ \mu\text{F}$, and operates at $f = 50\ \text{Hz}$. Calculate: (a) $X_L$ (b) $X_C$ (c) The impedance $Z$ (d) The phase angle $\phi$

  2. A series circuit has $R = 40\ \Omega$, $X_L = 60\ \Omega$, and $X_C = 25\ \Omega$. Find: (a) The impedance (b) Is the circuit inductive or capacitive? (c) The phase angle

  3. At what frequency will a series RLC circuit with $L = 0.10\ \text{H}$ and $C = 25\ \mu\text{F}$ have $Z = 50\ \Omega$ if $R = 30\ \Omega$?

  4. An RLC circuit has $R = 50\ \Omega$, $L = 0.40\ \text{H}$, and $C = 20\ \mu\text{F}$ at $f = 100\ \text{Hz}$. Calculate the current when connected to $V = 120\ \text{V}_{\text{rms}}$.

Score: ___/4

Set B2 — AC Ohm's Law Applications (4 problems)

  1. A $20\ \Omega$ resistor is connected to $V = 240\ \text{V}_{\text{rms}}$, $50\ \text{Hz}$. Calculate: (a) The RMS current (b) The peak current (c) The power dissipated

  2. A capacitor with $X_C = 40\ \Omega$ is connected to $V = 120\ \text{V}_{\text{rms}}$. Calculate: (a) The RMS current (b) The peak voltage across the capacitor (c) The real power dissipated

  3. An inductor with $X_L = 80\ \Omega$ carries $I = 2.5\ \text{A}_{\text{rms}}$. Calculate: (a) The RMS voltage across the inductor (b) The peak voltage (c) The energy stored in the inductor ($L = 0.25\ \text{H}$)

  4. A series RLC circuit with $Z = 100\ \Omega$ is connected to $V = 200\ \text{V}_{\text{rms}}$. The phase angle is $\phi = 30°$. Calculate: (a) The RMS current (b) The resistance $R$ (c) The reactance $X = X_L - X_C$

Score: ___/4

Part C: Phasor Diagrams & Voltage Relationships (6 problems)

Target: 3-4 min per problem.

Set C1 — Phasor Analysis (3 problems)

  1. An RLC series circuit has $R = 20\ \Omega$, $X_L = 40\ \Omega$, and $X_C = 25\ \Omega$. The current is $I = 3.0\ \text{A}_{\text{rms}}$. (a) Calculate $V_R$, $V_L$, and $V_C$ (b) Draw the phasor diagram (c) Calculate the total voltage $V_T$

  2. In a series circuit, $V_R = 80\ \text{V}$, $V_L = 120\ \text{V}$, and $V_C = 60\ \text{V}$. The current is $I = 2.0\ \text{A}$. (a) Find the impedance $Z$ (b) Find the resistance $R$ (c) Draw the voltage phasor diagram

  3. The voltage phasors in an RLC circuit are: $V_R = 60\ \text{V}$ (horizontal), $V_L = 100\ \text{V}$ (up), and $V_C = 40\ \text{V}$ (down). If $I = 1.5\ \text{A}$, calculate: (a) The impedance (b) The resistance (c) The net reactance

Score: ___/3

Set C2 — Phase Relationships (3 problems)

  1. In a series RLC circuit, $v_R = 100\sin(\omega t)$, $v_L = 150\sin(\omega t + 90°)$, and $v_C = 80\sin(\omega t - 90°)$. Write the expression for the total voltage $v_T(t)$.

  2. The current in a series circuit is $i(t) = 2.0\sin(100\pi t)\ \text{A}$. The voltage across the resistor is $v_R = 60\sin(100\pi t)\ \text{V}$, and across the capacitor is $v_C = 80\sin(100\pi t - 90°)\ \text{V}$. Find: (a) The resistance (b) The capacitive reactance (c) The capacitance

  3. An AC circuit has $V_{\text{rms}} = 120\ \text{V}$ and $I_{\text{rms}} = 3.0\ \text{A}$. The voltage leads the current by $40°$. Express $v(t)$ and $i(t)$ assuming $f = 60\ \text{Hz}$.

Score: ___/3

Part D: Power in AC Circuits (6 problems)

Target: 3-4 min per problem.

Set D1 — Power Calculations (3 problems)

  1. An RLC series circuit has $R = 30\ \Omega$, $X_L = 50\ \Omega$, $X_C = 20\ \Omega$, and $I_{\text{rms}} = 2.0\ \text{A}$. Calculate: (a) The real power $P_e$ (b) The reactive powers $P_{LC}$ and $P_{CC}$ (c) The apparent power $P_A$ (d) The power factor

  2. A circuit draws $I_{\text{rms}} = 5.0\ \text{A}$ from a $240\ \text{V}_{\text{rms}}$ supply. The power factor is $0.80$ lagging. Calculate: (a) The real power (b) The reactive power (c) The apparent power (d) The phase angle

  3. A $100\ \Omega$ resistor, a $0.50\ \text{H}$ inductor, and a $50\ \mu\text{F}$ capacitor are connected in series to a $120\ \text{V}_{\text{rms}}$, $60\ \text{Hz}$ source. Calculate: (a) The impedance (b) The current (c) The power dissipated (d) The power factor

Score: ___/3

Set D2 — Power Factor Correction (3 problems)

  1. An inductive load draws $P_e = 5.0\ \text{kW}$ at $\cos\phi = 0.70$ lagging from a $240\ \text{V}_{\text{rms}}$, $50\ \text{Hz}$ supply. What capacitance must be added in parallel to improve the power factor to unity?

  2. A factory load has $P_e = 100\ \text{kW}$ at $\cos\phi = 0.60$ lagging. Calculate: (a) The apparent power (b) The reactive power (c) The capacitor size needed to correct to $\cos\phi = 0.90$

  3. Explain why adding a capacitor in parallel with an inductive load improves the power factor.

Score: ___/3

Part E: Resonance (6 problems)

Target: 3-4 min per problem.

Set E1 — Resonant Frequency (3 problems)

  1. An RLC series circuit has $L = 0.20\ \text{H}$ and $C = 20\ \mu\text{F}$. Calculate: (a) The resonant frequency $f_0$ (b) The resonant angular frequency $\omega_0$

  2. At what frequency will a circuit with $L = 0.50\ \text{H}$ and $C = 10\ \mu\text{F}$ resonate?

  3. A series circuit resonates at $f_0 = 1.0\ \text{kHz}$. If $C = 0.10\ \mu\text{F}$, what is the inductance?

Score: ___/3

Set E2 — Resonance Behavior (3 problems)

  1. An RLC series circuit has $R = 10\ \Omega$, $L = 0.10\ \text{H}$, and $C = 25\ \mu\text{F}$. It is connected to a variable frequency source with $V = 100\ \text{V}$. (a) Find the resonant frequency (b) Calculate the current at resonance (c) Calculate the voltages across each component at resonance

  2. At resonance in a series RLC circuit with $R = 20\ \Omega$, $L = 0.40\ \text{H}$, and $C = 10\ \mu\text{F}$: (a) What is the impedance? (b) If $V = 50\ \text{V}$, what is the current? (c) What is the Q-factor?

  3. A series circuit has $R = 5.0\ \Omega$, $L = 2.0\ \text{mH}$, and $C = 0.50\ \mu\text{F}$. (a) Find the resonant frequency (b) Calculate the bandwidth (c) What is the maximum current if $V = 10\ \text{V}$?

Score: ___/3

Final Scorecard

Part Topic Problems Raw Score
A — Reactance Sets A1-A2 6 ___/6
B — Impedance Sets B1-B2 8 ___/8
C — Phasors Sets C1-C2 6 ___/6
D — Power Sets D1-D2 6 ___/6
E — Resonance Sets E1-E2 6 ___/6
TOTAL 32 ___/32

Proficiency Benchmarks for CQ3

Level Score Meaning
Exam-Ready $\ge 29/32$ (91%+) Ready for CQ3. Focus on speed.
Solid $25-28/32$ (78-87%) Good grasp. Review missed concepts.
Developing $20-24/32$ (62-75%) Drill weak areas again.
Needs Work $< 20/32$ (<62%) Re-study cheat sheet, retry tutorial.

Speed Benchmarks

  • < 90 min: Excellent speed for CQ3 conditions.
  • 90–110 min: Good. Practice under time pressure.
  • > 110 min: Need more pattern recognition drills.

Error Log Template

After grading, list every wrong problem with a one-word reason:

Problem Reason

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Reactance

  1. $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(50)(50 \times 10^{-6})} = 63.7\ \Omega$

  2. $C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi(60)(100)} = 26.5\ \mu\text{F}$

  3. (a) $C = \frac{1}{2\pi(50)(318)} = 10.0\ \mu\text{F}$ (b) $X_C = \frac{1}{2\pi(100)(10 \times 10^{-6})} = 159\ \Omega$

  4. $X_L = 2\pi f L = 2\pi(50)(0.50) = 157\ \Omega$

  5. $L = \frac{X_L}{2\pi f} = \frac{200}{2\pi(50)} = 0.637\ \text{H}$

  6. (a) $L = \frac{314}{2\pi(50)} = 1.0\ \text{H}$ (b) $X_L = 2\pi(60)(1.0) = 377\ \Omega$

Part B — Impedance & Ohm's Law

  1. (a) $X_L = 2\pi(50)(0.20) = 62.8\ \Omega$ (b) $X_C = \frac{1}{2\pi(50)(50 \times 10^{-6})} = 63.7\ \Omega$ (c) $Z = \sqrt{30^2 + (62.8-63.7)^2} = \sqrt{900 + 0.81} = 30.0\ \Omega$ (d) $\phi = \tan^{-1}\left(\frac{-0.9}{30}\right) = -1.7°$ (slightly capacitive)

  2. (a) $Z = \sqrt{40^2 + (60-25)^2} = \sqrt{1600 + 1225} = 52.2\ \Omega$ (b) Inductive ($X_L > X_C$) (c) $\phi = \tan^{-1}(35/40) = 41.2°$

  3. At $Z = 50\ \Omega$: $50^2 = 30^2 + (X_L - X_C)^2$ → $(X_L - X_C)^2 = 1600$ → $X_L - X_C = \pm 40\ \Omega$ $2\pi f(0.10) - \frac{1}{2\pi f(25 \times 10^{-6})} = 40$ or $-40$ Solving: $f \approx 63\ \text{Hz}$ or $f \approx 100\ \text{Hz}$

  4. $X_L = 2\pi(100)(0.40) = 251\ \Omega$, $X_C = \frac{1}{2\pi(100)(20 \times 10^{-6})} = 79.6\ \Omega$ $$Z = \sqrt{50^2 + (251-79.6)^2} = 178\ \Omega$$ $$I = V/Z = 120/178 = 0.67\ \text{A}$$

  5. (a) $I = V/R = 240/20 = 12\ \text{A}$ (b) $I_{\text{peak}} = 12\sqrt{2} = 17.0\ \text{A}$ (c) $P = I^2 R = (12)^2(20) = 2.88\ \text{kW}$

  6. (a) $I = V/X_C = 120/40 = 3.0\ \text{A}$ (b) $V_{\text{peak}} = 120\sqrt{2} = 170\ \text{V}$ (c) $P = 0$ (pure capacitor)

  7. (a) $V = IX_L = 2.5 \times 80 = 200\ \text{V}$ (b) $V_{\text{peak}} = 200\sqrt{2} = 283\ \text{V}$ (c) $U = \frac{1}{2}LI^2 = 0.5(0.25)(2.5)^2 = 0.781\ \text{J}$

  8. (a) $I = V/Z = 200/100 = 2.0\ \text{A}$ (b) $R = Z\cos\phi = 100\cos(30°) = 86.6\ \Omega$ (c) $X = Z\sin\phi = 100\sin(30°) = 50\ \Omega$

Part C — Phasors

  1. (a) $V_R = IR = 60\ \text{V}$, $V_L = IX_L = 120\ \text{V}$, $V_C = IX_C = 75\ \text{V}$ (c) $V_T = IZ = 3.0\sqrt{20^2 + 15^2} = 75\ \text{V}$

  2. (a) $V_{\text{net}} = \sqrt{80^2 + (120-60)^2} = 100\ \text{V}$, $Z = 100/2 = 50\ \Omega$ (b) $R = V_R/I = 80/2 = 40\ \Omega$

  3. Net vertical: $100 - 40 = 60\ \text{V}$ up. $V_T = \sqrt{60^2 + 60^2} = 84.9\ \text{V}$ (a) $Z = 84.9/1.5 = 56.6\ \Omega$ (b) $R = 60/1.5 = 40\ \Omega$ (c) $X = 60/1.5 = 40\ \Omega$

  4. Net: $100\sin(\omega t) + 150\cos(\omega t) - 80\cos(\omega t) = 100\sin(\omega t) + 70\cos(\omega t)$ $$v_T = \sqrt{100^2 + 70^2}\sin(\omega t + \phi) = 122\sin(\omega t + 35°)\ \text{V}$$

  5. (a) $R = V_R/I = 60/2 = 30\ \Omega$ (b) $X_C = V_C/I = 80/2 = 40\ \Omega$ (c) $C = \frac{1}{\omega X_C} = \frac{1}{100\pi \times 40} = 79.6\ \mu\text{F}$

  6. $i(t) = 3.0\sqrt{2}\sin(120\pi t) = 4.24\sin(120\pi t)\ \text{A}$ $$v(t) = 120\sqrt{2}\sin(120\pi t + 40°) = 170\sin(120\pi t + 40°)\ \text{V}$$

Part D — Power

  1. (a) $P_e = I^2 R = (2.0)^2(30) = 120\ \text{W}$ (b) $P_{LC} = I^2 X_L = 200\ \text{VAR}$, $P_{CC} = I^2 X_C = 80\ \text{VAR}$ (c) $P_A = I^2 Z = (2.0)^2\sqrt{30^2 + 30^2} = 170\ \text{VA}$ (d) $\cos\phi = R/Z = 30/42.4 = 0.707$

  2. (a) $P_e = VI\cos\phi = 240 \times 5.0 \times 0.80 = 960\ \text{W}$ (b) $\sin\phi = 0.60$, $P_{\text{reactive}} = 240 \times 5.0 \times 0.60 = 720\ \text{VAR}$ (c) $P_A = 240 \times 5.0 = 1200\ \text{VA}$ (d) $\phi = \cos^{-1}(0.80) = 36.9°$

  3. $X_L = 2\pi(60)(0.50) = 188.5\ \Omega$, $X_C = \frac{1}{2\pi(60)(50 \times 10^{-6})} = 53.1\ \Omega$ (a) $Z = \sqrt{100^2 + (188.5-53.1)^2} = 168\ \Omega$ (b) $I = 120/168 = 0.714\ \text{A}$ (c) $P = I^2 R = (0.714)^2(100) = 51.0\ \text{W}$ (d) $\cos\phi = 100/168 = 0.595$

  4. $Q_{\text{old}} = P\tan\phi = 5000 \times \tan(\cos^{-1}0.70) = 5102\ \text{VAR}$ Need $Q_C = 5102\ \text{VAR}$ to cancel $$X_C = V^2/Q_C = 240^2/5102 = 11.3\ \Omega$$ $$C = \frac{1}{2\pi(50)(11.3)} = 282\ \mu\text{F}$$

  5. (a) $P_A = P_e/\cos\phi = 100/0.60 = 167\ \text{kVA}$ (b) $Q = P\tan\phi = 100 \times 1.33 = 133\ \text{kVAR}$ (c) New $\phi = \cos^{-1}(0.90) = 26°$, new $Q = 100 \times \tan(26°) = 48.4\ \text{kVAR}$ $$Q_C = 133 - 48.4 = 84.6\ \text{kVAR}$$

  6. The capacitor supplies reactive power locally, reducing the reactive power the source must provide. This reduces the total current drawn from the source for the same real power, improving the power factor.

Part E — Resonance

  1. (a) $f_0 = \frac{1}{2\pi\sqrt{0.20 \times 20 \times 10^{-6}}} = 79.6\ \text{Hz}$ (b) $\omega_0 = 2\pi f_0 = 500\ \text{rad/s}$

  2. $f_0 = \frac{1}{2\pi\sqrt{0.50 \times 10 \times 10^{-6}}} = 71.2\ \text{Hz}$

  3. $L = \frac{1}{\omega_0^2 C} = \frac{1}{(2\pi \times 1000)^2 \times 0.10 \times 10^{-6}} = 0.253\ \text{H}$

  4. (a) $f_0 = \frac{1}{2\pi\sqrt{0.10 \times 25 \times 10^{-6}}} = 100.7\ \text{Hz}$ (b) $I = V/R = 100/10 = 10\ \text{A}$ (c) $V_R = 100\ \text{V}$, $V_L = V_C = I\omega_0 L = 10 \times 2\pi(100.7)(0.10) = 632\ \text{V}$

  5. (a) $Z = R = 20\ \Omega$ (b) $I = 50/20 = 2.5\ \text{A}$ (c) $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{20}\sqrt{\frac{0.40}{10 \times 10^{-6}}} = 10$

  6. (a) $f_0 = \frac{1}{2\pi\sqrt{2 \times 10^{-3} \times 0.5 \times 10^{-6}}} = 5033\ \text{Hz}$ (b) $\Delta f = f_0/Q = f_0 \times \frac{R}{\sqrt{L/C}} = 5033 \times \frac{5}{63.2} = 398\ \text{Hz}$ (c) $I_{\text{max}} = V/R = 10/5 = 2.0\ \text{A}$

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