FAD1022 CQ — AC Circuits Quiz Set

Course quiz problems covering inductive reactance, RLC series circuits, RL/RC series circuit analysis, and AC voltage fundamentals.

[!tip] Quick Answer Key

Q Answer Q Answer Q Answer
1 $1.00 \times 10^2\ \Omega$ 5 $5.66$ V 9 $157.98$ V
2 $155.11$ V 6 $172.70$ V 10 $2.99 \times 10^{-5}$ F
3 $0.0358$ H 7 $3.76 \times 10^{-5}$ F 11 $2.86$ A
4 $1.30$ A 8 $2.16$ A

Question 1: Inductive Reactance with Frequency Change

Problem Statement: An ideal inductor is connected to an AC supply of frequency 50 Hz. The rms voltage across the inductor is 100 V, and the rms current is 2.0 A. The frequency of the supply is then increased to 100 Hz while the voltage remains constant. Determine the new inductive reactance.

Given:

  • Initial frequency: $f_1 = 50$ Hz
  • Final frequency: $f_2 = 100$ Hz
  • RMS voltage: $V_{rms} = 100$ V (constant)
  • Initial RMS current: $I_1 = 2.0$ A

Solution:

Step 1: Calculate initial inductive reactance $$X_{L1} = \frac{V_{rms}}{I_1} = \frac{100}{2.0} = 50\ \Omega$$

Step 2: Apply frequency relationship Since $X_L = 2\pi f L$ and inductance $L$ is constant: $$\frac{X_{L2}}{X_{L1}} = \frac{f_2}{f_1} = \frac{100}{50} = 2$$

Step 3: Calculate new inductive reactance $$X_{L2} = 2 \times X_{L1} = 2 \times 50 = 100\ \Omega$$

Answer: $1.00 \times 10^2\ \Omega$

Key Concept: Inductive Reactance — $X_L \propto f$. Doubling frequency doubles inductive reactance.

Question 2: Capacitive RLC Circuit — Voltage Across Capacitor

Problem Statement: In a capacitive RLC series circuit consists of 85 Ω resistance, inductive reactance of 76 Ω and achieved a circuit impedance of 101 Ω. The circuit is driven by 60 Hz with 120 V (Vrms) signal generator. Calculate voltage across the capacitor.

Given:

  • Resistance: $R = 85\ \Omega$
  • Inductive reactance: $X_L = 76\ \Omega$
  • Impedance: $Z = 101\ \Omega$
  • Frequency: $f = 60$ Hz
  • RMS voltage: $V_{rms} = 120$ V

Solution:

Step 1: Find capacitive reactance using impedance formula $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Squaring both sides: $$Z^2 = R^2 + (X_L - X_C)^2$$

$$101^2 = 85^2 + (76 - X_C)^2$$

$$10201 = 7225 + (76 - X_C)^2$$

$$(76 - X_C)^2 = 2976$$

$$76 - X_C = \pm 54.55$$

Since the circuit is capacitive, $X_C > X_L$. We use the positive magnitude: $$X_C = 76 + 54.55 = 130.55\ \Omega$$

Step 2: Calculate RMS current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{120}{101} = 1.188\ \text{A}$$

Step 3: Calculate voltage across capacitor $$V_C = I_{rms} \times X_C = 1.188 \times 130.55 = 155.1\ \text{V}$$

Answer: $155.11$ V

Key Concepts:

  • RLC Series Circuit — Impedance formula for combined R, L, C elements
  • Capacitive Reactance — $X_C = \frac{1}{2\pi f C}$
  • Impedance — Opposition to AC current flow

Question 3: RL Series Circuit — Calculate Inductance

Problem Statement: In an RL series circuit supply by 5.2V (Vrms), 60 Hz, given the phase angle is 32.7°. Calculate the inductance if the resistance is 21 Ω.

Given:

  • RMS voltage: $V_{rms} = 5.2$ V
  • Frequency: $f = 60$ Hz
  • Phase angle: $\phi = 32.7°$
  • Resistance: $R = 21\ \Omega$

Solution:

Step 1: Use phase angle formula for RL circuit $$\tan\phi = \frac{X_L}{R}$$

Step 2: Solve for inductive reactance $$X_L = R \times \tan\phi = 21 \times \tan(32.7°) = 21 \times 0.642 = 13.48\ \Omega$$

Step 3: Calculate inductance $$X_L = 2\pi f L$$

$$L = \frac{X_L}{2\pi f} = \frac{13.48}{2\pi \times 60} = \frac{13.48}{376.99} = 0.0358\ \text{H}$$

Answer: $0.0358$ H

Key Concepts:

  • RL Series Circuit — Phase angle relationship $\tan\phi = X_L/R$
  • Inductance — $L$ calculated from reactance and frequency
  • Phase Angle — Positive angle indicates voltage leads current (inductive)

Question 4: RL Series Circuit — Calculate Current

Problem Statement: An RL series circuit is connected to 8.7 V(Vrms). Given the phase angle between the current and voltage is 61.38° and the resistance is 3.2 Ω. Calculate the current flows in circuit.

Given:

  • RMS voltage: $V_{rms} = 8.7$ V
  • Phase angle: $\phi = 61.38°$
  • Resistance: $R = 3.2\ \Omega$

Solution:

Step 1: Use power factor relationship $$\cos\phi = \frac{R}{Z}$$

Step 2: Solve for impedance $$Z = \frac{R}{\cos\phi} = \frac{3.2}{\cos(61.38°)} = \frac{3.2}{0.479} = 6.68\ \Omega$$

Step 3: Calculate current using Ohm's law for AC $$I_{rms} = \frac{V_{rms}}{Z} = \frac{8.7}{6.68} = 1.30\ \text{A}$$

Answer: $1.30$ A

Key Concepts:

  • Impedance — Total opposition in AC circuit
  • Power Factor — $\cos\phi = R/Z$
  • RL Series Circuit — Current calculation using impedance triangle

Question 5: RMS Value from AC Voltage Equation

Problem Statement: An alternating voltage is given by $V(t) = 8\sin(120\pi t + \frac{\pi}{6})$. Determine the r.m.s. value of the voltage.

Given:

  • Voltage equation: $V(t) = 8\sin(120\pi t + \frac{\pi}{6})$

Solution:

Step 1: Identify peak voltage from equation The general form is $V(t) = V_0 \sin(\omega t + \phi)$ $$V_0 = 8\ \text{V}$$

Step 2: Calculate RMS value $$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 5.66\ \text{V}$$

Answer: $5.66$ V

[!warning] Common Mistake The phase angle $\frac{\pi}{6}$ in the equation does NOT affect the RMS calculation. RMS only depends on the peak value.

Key Concept: RMS Values — $V_{rms} = V_0/\sqrt{2}$ for sinusoidal AC signals.

Question 6: Inductive RLC Circuit — Voltage Across Inductor

Problem Statement: In an inductive RLC series circuit consists of 76 Ω resistance, capacitive reactance of 79 Ω and achieved a circuit impedance of 113 Ω. The circuit is driven by 60 Hz with 120 V (Vrms) signal generator. Calculate voltage across the inductor.

Given:

  • Resistance: $R = 76\ \Omega$
  • Capacitive reactance: $X_C = 79\ \Omega$
  • Impedance: $Z = 113\ \Omega$
  • Frequency: $f = 60$ Hz
  • RMS voltage: $V_{rms} = 120$ V

Solution:

Step 1: Find inductive reactance using impedance formula $$Z^2 = R^2 + (X_L - X_C)^2$$

$$113^2 = 76^2 + (X_L - 79)^2$$

$$12769 = 5776 + (X_L - 79)^2$$

$$(X_L - 79)^2 = 6993$$

$$X_L - 79 = \pm 83.6$$

Since the circuit is inductive, $X_L > X_C$. We take the positive root: $$X_L = 79 + 83.6 = 162.6\ \Omega$$

[!note] Why positive root? When solving $(X_L - X_C)^2 = \text{value}$ for reactance, always use the positive square root. The magnitude $|X_L - X_C| = +\sqrt{Z^2 - R^2}$ gives the correct reactance value. For inductive circuits: $X_L = X_C + \sqrt{Z^2 - R^2}$.

Step 2: Calculate RMS current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{120}{113} = 1.062\ \text{A}$$

Step 3: Calculate voltage across inductor $$V_L = I_{rms} \times X_L = 1.062 \times 162.6 = 172.7\ \text{V}$$

Answer: $172.70$ V

Key Concepts:

  • RLC Series Circuit — Impedance formula
  • Inductive Reactance — When $X_L > X_C$, circuit is inductive
  • Voltage Across Components — $V_L = I \times X_L$

Question 7: RC Series Circuit — Calculate Capacitance

Problem Statement: In an RC series circuit supply by 5.7 V (Vrms), 60 Hz, given the phase angle is -32.7°. Calculate the capacitance of the source if the resistance is 110 Ω.

Given:

  • RMS voltage: $V_{rms} = 5.7$ V
  • Frequency: $f = 60$ Hz
  • Phase angle: $\phi = -32.7°$ (negative = capacitive)
  • Resistance: $R = 110\ \Omega$

Solution:

Step 1: Use phase angle formula for RC circuit $$\tan\phi = \frac{-X_C}{R}$$

Step 2: Solve for capacitive reactance $$\tan(-32.7°) = \frac{-X_C}{110}$$

$$-0.642 = \frac{-X_C}{110}$$

$$X_C = 70.6\ \Omega$$

Step 3: Calculate capacitance $$X_C = \frac{1}{2\pi f C}$$

$$C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 60 \times 70.6} = \frac{1}{26619} = 3.76 \times 10^{-5}\ \text{F}$$

Answer: $3.76 \times 10^{-5}$ F

Key Concepts:

  • RC Series Circuit — Negative phase angle indicates current leads voltage
  • Capacitive Reactance — $X_C = 1/(2\pi f C)$
  • Capacitance — Calculated from reactance and frequency

Question 8: RL Series Circuit — Calculate Current (Variation)

Problem Statement: An RL series circuit is connected to 101 V(Vrms). Given the phase angle between the current and voltage is 61.38° and the resistance is 22.4 Ω. Calculate the current flows in circuit.

Given:

  • RMS voltage: $V_{rms} = 101$ V
  • Phase angle: $\phi = 61.38°$
  • Resistance: $R = 22.4\ \Omega$

Solution:

Step 1: Use power factor relationship $$\cos\phi = \frac{R}{Z}$$

Step 2: Solve for impedance $$Z = \frac{R}{\cos\phi} = \frac{22.4}{\cos(61.38°)} = \frac{22.4}{0.479} = 46.76\ \Omega$$

Step 3: Calculate current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{101}{46.76} = 2.16\ \text{A}$$

Answer: $2.16$ A

Key Concepts:

  • RL Series Circuit — Same approach as Question 4 with different values
  • Impedance Triangle — $Z = R/\cos\phi$
  • AC Ohm's Law — $I = V/Z$

Question 9: Inductive RLC Circuit — Voltage Across Inductor (Variation)

Problem Statement: In an inductive RLC series circuit consists of 89 Ω resistance, capacitive reactance of 81 Ω and achieved a circuit impedance of 108 Ω. The circuit is driven by 60 Hz with 120 V (Vrms) signal generator. Calculate voltage across the inductor.

Given:

  • Resistance: $R = 89\ \Omega$
  • Capacitive reactance: $X_C = 81\ \Omega$
  • Impedance: $Z = 108\ \Omega$
  • Frequency: $f = 60$ Hz
  • RMS voltage: $V_{rms} = 120$ V

Solution:

Step 1: Find inductive reactance $$Z^2 = R^2 + (X_L - X_C)^2$$

$$108^2 = 89^2 + (X_L - 81)^2$$

$$11664 = 7921 + (X_L - 81)^2$$

$$(X_L - 81)^2 = 3743$$

$$X_L - 81 = +61.18$$ (taking positive root)

$$X_L = 81 + 61.18 = 142.18\ \Omega$$

Step 2: Calculate RMS current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{120}{108} = 1.111\ \text{A}$$

Step 3: Calculate voltage across inductor $$V_L = I_{rms} \times X_L = 1.111 \times 142.18 = 158.0\ \text{V}$$

Answer: $157.98$ V

Key Concepts:

  • RLC Series Circuit — Using positive square root for reactance
  • Voltage Across Inductor — $V_L = I \times X_L$

Question 10: RC Series Circuit — Calculate Capacitance (Variation)

Problem Statement: In an RC series circuit supply by 7.3 V (Vrms), 60 Hz, given the phase angle is -32.7°. Calculate the capacitance of the source if the resistance is 138 Ω.

Given:

  • RMS voltage: $V_{rms} = 7.3$ V
  • Frequency: $f = 60$ Hz
  • Phase angle: $\phi = -32.7°$
  • Resistance: $R = 138\ \Omega$

Solution:

Step 1: Find capacitive reactance $$\tan\phi = \frac{-X_C}{R}$$

$$\tan(-32.7°) = \frac{-X_C}{138}$$

$$-0.642 = \frac{-X_C}{138}$$

$$X_C = 88.6\ \Omega$$

Step 2: Calculate capacitance $$C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 60 \times 88.6} = 2.99 \times 10^{-5}\ \text{F}$$

Answer: $2.99 \times 10^{-5}$ F

Key Concepts:

  • RC Series Circuit — Negative phase angle
  • Capacitance Calculation — From $X_C$ and frequency

Question 11: RL Series Circuit — Calculate Current (Variation 2)

Problem Statement: An RL series circuit is connected to 104 V(Vrms). Given the phase angle between the current and voltage is 61.38° and the resistance is 17.4 Ω. Calculate the current flows in circuit.

Given:

  • RMS voltage: $V_{rms} = 104$ V
  • Phase angle: $\phi = 61.38°$
  • Resistance: $R = 17.4\ \Omega$

Solution:

Step 1: Calculate impedance $$Z = \frac{R}{\cos\phi} = \frac{17.4}{\cos(61.38°)} = \frac{17.4}{0.479} = 36.33\ \Omega$$

Step 2: Calculate current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{104}{36.33} = 2.86\ \text{A}$$

Answer: $2.86$ A

Key Concepts:

  • RL Series Circuit — Impedance from power factor
  • AC Current — Ohm's law for AC circuits

Question 12: Inductive Reactance with Frequency Change (Variation)

Problem Statement: An ideal inductor is connected to an AC supply of frequency 50 Hz. The rms voltage across the inductor is 100 V, and the rms current is 2.0 A. The frequency of the supply is then increased to 100 Hz while the voltage remains constant. Determine the new inductive reactance.

Given:

  • Initial frequency: $f_1 = 50$ Hz
  • Final frequency: $f_2 = 100$ Hz
  • RMS voltage: $V_{rms} = 100$ V
  • Initial RMS current: $I_1 = 2.0$ A

Solution:

Step 1: Calculate initial inductive reactance $$X_{L1} = \frac{V_{rms}}{I_1} = \frac{100}{2.0} = 50\ \Omega$$

Step 2: Apply frequency relationship $$\frac{X_{L2}}{X_{L1}} = \frac{f_2}{f_1} = \frac{100}{50} = 2$$

Step 3: Calculate new inductive reactance $$X_{L2} = 2 \times 50 = 100\ \Omega$$

Answer: $1.00 \times 10^2\ \Omega$

Key Concept: Inductive Reactance — $X_L \propto f$. Same problem type as Question 1 with identical values.

Question 13: Capacitive RLC Circuit — Voltage Across Capacitor (Variation)

Problem Statement: In a capacitive RLC series circuit consists of 79 Ω resistance, inductive reactance of 75 Ω and achieved a circuit impedance of 118 Ω. The circuit is driven by 60 Hz with 120 V (Vrms) signal generator. Calculate voltage across the capacitor.

Given:

  • Resistance: $R = 79\ \Omega$
  • Inductive reactance: $X_L = 75\ \Omega$
  • Impedance: $Z = 118\ \Omega$
  • Frequency: $f = 60$ Hz
  • RMS voltage: $V_{rms} = 120$ V

Solution:

Step 1: Find capacitive reactance $$Z^2 = R^2 + (X_L - X_C)^2$$

$$118^2 = 79^2 + (75 - X_C)^2$$

$$13924 = 6241 + (75 - X_C)^2$$

$$(75 - X_C)^2 = 7683$$

$$|75 - X_C| = 87.65$$

Since circuit is capacitive ($X_C > X_L$): $$X_C = 75 + 87.65 = 162.65\ \Omega$$

Step 2: Calculate RMS current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{120}{118} = 1.017\ \text{A}$$

Step 3: Calculate voltage across capacitor $$V_C = I_{rms} \times X_C = 1.017 \times 162.65 = 165.4\ \text{V}$$

Answer: $165.41$ V

Key Concepts:

  • RLC Series Circuit — Using positive magnitude for reactance
  • Voltage Across Capacitor — $V_C = I \times X_C$

Question 14: RL Series Circuit — Calculate Inductance (Variation)

Problem Statement: In an RL series circuit supply by 2.1V (Vrms), 60 Hz, given the phase angle is 32.7°. Calculate the inductance if the resistance is 29 Ω.

Given:

  • RMS voltage: $V_{rms} = 2.1$ V
  • Frequency: $f = 60$ Hz
  • Phase angle: $\phi = 32.7°$
  • Resistance: $R = 29\ \Omega$

Solution:

Step 1: Find inductive reactance $$\tan\phi = \frac{X_L}{R}$$

$$X_L = R \times \tan\phi = 29 \times \tan(32.7°) = 29 \times 0.642 = 18.62\ \Omega$$

Step 2: Calculate inductance $$L = \frac{X_L}{2\pi f} = \frac{18.62}{2\pi \times 60} = 0.0494\ \text{H}$$

Answer: $0.0494$ H

Key Concepts:

  • RL Series Circuit — Phase angle relationship
  • Inductance Calculation — From $X_L$ and frequency

Question 15: RL Series Circuit — Calculate Current (Variation 3)

Problem Statement: An RL series circuit is connected to 1.7 V(Vrms). Given the phase angle between the current and voltage is 61.38° and the resistance is 4.9 Ω. Calculate the current flows in circuit.

Given:

  • RMS voltage: $V_{rms} = 1.7$ V
  • Phase angle: $\phi = 61.38°$
  • Resistance: $R = 4.9\ \Omega$

Solution:

Step 1: Calculate impedance from power factor $$Z = \frac{R}{\cos\phi} = \frac{4.9}{\cos(61.38°)} = \frac{4.9}{0.479} = 10.23\ \Omega$$

Step 2: Calculate current $$I_{rms} = \frac{V_{rms}}{Z} = \frac{1.7}{10.23} = 0.166\ \text{A}$$

Answer: $0.17$ A

Key Concepts:

  • RL Series Circuit — Same approach as Questions 4, 8, 11
  • Power Factor — $\cos\phi = R/Z$

Summary Table

Question Circuit Type Key Formula Answer
1 Pure Inductor $X_L \propto f$ $1.00 \times 10^2\ \Omega$
2 Capacitive RLC $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $155.11$ V
3 RL Series $\tan\phi = X_L/R$, $X_L = 2\pi f L$ $0.0358$ H
4 RL Series $\cos\phi = R/Z$, $I = V/Z$ $1.30$ A
5 AC Voltage Equation $V_{rms} = V_0/\sqrt{2}$ $5.66$ V
6 Inductive RLC $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $172.70$ V
7 RC Series $\tan\phi = -X_C/R$, $C = 1/(2\pi f X_C)$ $3.76 \times 10^{-5}$ F
8 RL Series $\cos\phi = R/Z$, $I = V/Z$ $2.16$ A
9 Inductive RLC $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $157.98$ V
10 RC Series $\tan\phi = -X_C/R$, $C = 1/(2\pi f X_C)$ $2.99 \times 10^{-5}$ F
11 RL Series $\cos\phi = R/Z$, $I = V/Z$ $2.86$ A
12 Pure Inductor $X_L \propto f$ $1.00 \times 10^2\ \Omega$
13 Capacitive RLC $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $165.41$ V
14 RL Series $\tan\phi = X_L/R$, $L = X_L/(2\pi f)$ $0.0494$ H
15 RL Series $\cos\phi = R/Z$, $I = V/Z$ $0.17$ A

Key Formulas Used

Reactance: $$X_L = 2\pi f L \quad\text{(inductive)}$$ $$X_C = \frac{1}{2\pi f C} \quad\text{(capacitive)}$$

Impedance: $$Z = \frac{V_{rms}}{I_{rms}} = \sqrt{R^2 + (X_L - X_C)^2}$$

Phase Angle: $$\tan\phi = \frac{X_L - X_C}{R}$$ $$\cos\phi = \frac{R}{Z}$$

Voltage Across Components: $$V_C = I \times X_C$$ $$V_L = I \times X_L$$ $$V_R = I \times R$$

Related