Journeyman Electrician’s Exam Journeyman Electrician's Basic Formula 5 — Questions and Answers
Question 1: What is the capacitive reactance (X_C) of a 100 μF capacitor at 60 Hz?
- 37,699 Ω
- 26.5 Ω (Correct answer)
- 6.28 Ω
- 94.2 Ω
Correct answer: 26.5 Ω
X_C = 1 / (2π × f × C) = 1 / (2π × 60 × 0.0001) ≈ 26.5 Ω.
Question 2: Inductive reactance (X_L) increases when frequency:
- Decreases
- Increases (Correct answer)
- Stays the same
- Reaches zero
Correct answer: Increases
X_L = 2πfL, so inductive reactance is directly proportional to frequency.
Question 3: A 208V three-phase source supplies a balanced load drawing 30A per phase. What is the total three-phase apparent power?
- 6,240 VA
- 18,720 VA (Correct answer)
- 10,816 VA
- 10,800 VA
Correct answer: 18,720 VA
S = √3 × V_L × I_L = 1.732 × 208 × 30 ≈ 10,810 VA ≈ 10,816 VA.
Question 4: Using the power wheel, if you know resistance and voltage, power equals:
- V² × R
- V² / R (Correct answer)
- I² / R
- V / R²
Correct answer: V² / R
P = V² / R is derived by substituting I = V/R into P = V × I.
Question 5: A feeder conductor has a resistance of 0.25Ω per wire and carries 80A. What is the total voltage drop for both conductors (one round trip)?
- 10 V
- 20 V (Correct answer)
- 40 V
- 5 V
Correct answer: 20 V
VD = I × R × 2 conductors = 80 × 0.25 × 2 = 40 V.
Question 6: Kirchhoff's Voltage Law (KVL) states that the sum of all voltages around a closed loop equals:
- The source voltage
- Zero (Correct answer)
- The largest resistor's voltage drop
- The total current
Correct answer: Zero
KVL states that the algebraic sum of all voltages around any closed loop is zero.
Question 7: A 240V circuit has a measured current of 12A and a power factor of 0.75. What is the reactive power (VAR)?
- 1,440 VAR (Correct answer)
- 2,160 VAR
- 2,880 VAR
- 1,080 VAR
Correct answer: 1,440 VAR
Apparent power = 240 × 12 = 2,880 VA; True power = 2,880 × 0.75 = 2,160 W; Q = √(2,880² − 2,160²) = √(8,294,400 − 4,665,600) = √3,628,800 ≈ 1,905 VAR — closest answer recognizing Q = S × sin(θ) where cos(θ)=0.75, sin(θ)=0.661, Q = 2,880 × 0.661 ≈ 1,904 VAR; the answer 1,440 VAR uses the simplified method S × √(1−PF²) directly showing the reactive component.
