Ladwp Electric Distribution Mechanic Trainee
Examination Questions and Verified Answers
Section 1: Electrical Theory, DC/AC Circuits & Fundamentals (Questions 1–35)
Question 1: A distribution mechanic measures 120 volts across a load drawing 10 amperes during
normal operation. Which value represents the power consumption of this resistive load?
A. 12 watts
B. 120 watts
C. 1,200 watts
D. 12,000 watts
CORRECT ANSWER: C. 1,200 watts
Rationale: Power in a single-phase resistive circuit is calculated using the formula P=V×IP = V \times
IP=V×I. Multiplying 120 V by 10 A yields 1,200 W (or 1.2 kW).
Question 2: When two resistors are connected in series within an energized DC test circuit, which
property strictly distinguishes the series connection from a parallel connection?
A. Voltage drop across each resistor is always identical
B. Branch current divides proportionately across components
C. The total circuit resistance is less than the smallest resistor
D. The exact same magnitude of current flows sequentially through every resistor
CORRECT ANSWER: D. The exact same magnitude of current flows sequentially through every resistor
Rationale: In a series circuit, there is only one path for current, making current equal through all series
components. In contrast, parallel circuits divide current across multiple paths while maintaining equal
voltage across each branch.
Question 3: A mechanic needs to calculate the primary current of a single-phase transformer given an
apparent power rating of 50 kVA and a primary voltage of 2,500 V. What is the primary full-load
current?
A. 20 amperes
,B. 50 amperes
C. 125 amperes
D. 200 amperes
CORRECT ANSWER: A. 20 amperes
Rationale: Apparent power S=V×IS = V \times IS=V×I. Rearranging gives I=S/VI = S / VI=S/V.
Converting 50 kVA to 50,000 VA and dividing by 2,500 V results in 50,000/2,500=20 A50,,500 =
20\text{ A}50,000/2,500=20 A.
Question 4: Why does an electrical transformer require alternating current (AC) rather than steady direct
current (DC) to maintain normal transformation action?
A. DC produces excessive magnetic frequency that damages core insulation
B. Continuous electromagnetic induction requires a constantly changing magnetic flux
C. AC eliminates magnetic saturation in the steel laminations
D. DC automatically reverses secondary polarity every half cycle
CORRECT ANSWER: B. Continuous electromagnetic induction requires a constantly changing magnetic
flux
Rationale: Transformer action relies on Faraday's law of induction. Alternating current continuously
changes direction and magnitude, creating a constantly changing magnetic flux in the core that induces
voltage in the secondary winding. Steady DC produces a static magnetic field, which induces no
continuous secondary voltage and causes excessive winding heating.
Question 5: What is the primary operational advantage of maintaining balanced phase loading on a
three-phase, four-wire distribution circuit?
A. It completely eliminates dielectric losses in feeder cables
B. It minimizes current flow in the neutral conductor and balances phase voltages
C. It allows conductors to operate above their thermal rating
D. It prevents line reactance from affecting motor loads
CORRECT ANSWER: B. It minimizes current flow in the neutral conductor and balances phase voltages
Rationale: Balancing loads across all three phases ensures that vectorially, phase currents cancel out in the
neutral wire, minimizing neutral current flow, reducing system line losses (I2RI^2RI2R), and preventing
single-phase voltage unbalance.
,Question 6: If the resistance of an energized circuit loop doubles while the source voltage remains strictly
constant, how does the circuit current respond?
A. Current increases to twice its initial value
B. Current drops to one-half of its initial value
C. Current decreases to one-fourth of its initial value
D. Current remains unchanged because voltage is constant
CORRECT ANSWER: B. Current drops to one-half of its initial value
Rationale: According to Ohm's law (I=V/RI = V / RI=V/R), current is inversely proportional to resistance
when voltage is constant. Doubling the resistance halves the current flow.
Question 7: Which electrical property represents the total opposition to alternating current flow, taking
into account both resistance and reactance?
A. Admittance
B. Inductance
C. Impedance
D. Capacitance
CORRECT ANSWER: C. Impedance
Rationale: Impedance (symbolized by ZZZ and measured in ohms) is the total vector sum of resistance
(RRR) and inductive/capacitive reactance (XXX) in an AC circuit (Z=R2+X2Z = \sqrt{R^2 +
X^2}Z=R2+X2 ).
Question 8: What happens to the inductive reactance (XLX_LXL) of a distribution line reactor when the
frequency of the power supply increases?
A. Inductive reactance increases proportionally
B. Inductive reactance decreases proportionally
C. Inductive reactance drops to zero
D. Inductive reactance remains unaffected
CORRECT ANSWER: A. Inductive reactance increases proportionally
, Rationale: Inductive reactance is calculated by XL=2πfLX_L = 2\pi f LXL=2πfL. Because XLX_LXL is
directly proportional to frequency (fff), an increase in frequency produces a proportional increase in
inductive reactance.
Question 9: What is the power factor of a purely resistive overhead heating element circuit?
A. 0.0
B. 0.5 lagging
C. 0.866 leading
D. 1.0 (Unity)
CORRECT ANSWER: D. 1.0 (Unity)
Rationale: In a purely resistive circuit, voltage and current are exactly in phase (phase angle θ=0∘\theta =
0^\circθ=0∘). Since power factor equals cos(θ)\cos(\theta)cos(θ), cos(0∘)=1.0\cos(0^\circ) =
1.0cos(0∘)=1.0 (unity power factor).
Question 10: How much electrical energy is consumed by a 5 kW overhead space heater running
continuously for 8 hours?
A. 13 kWh
B. 40 kWh
C. 400 kWh
D. 1,000 kWh
CORRECT ANSWER: B. 40 kWh
Rationale: Electrical energy in kilowatt-hours is calculated by multiplying power in kilowatts by time in
hours (E=P×tE = P \times tE=P×t). Here, 5 kW×8 hours=40 kWh5\text{ kW} \times 8\text{ hours} =
40\text{ kWh}5 kW×8 hours=40 kWh.
Question 11: When three 30 Ω30\text{ }\Omega30 Ω resistors are connected in parallel, what is the total
equivalent resistance of the network?
A. 10 Ω10\text{ }\Omega10 Ω
B. 30 Ω30\text{ }\Omega30 Ω
C. 60 Ω60\text{ }\Omega60 Ω
D. 90 Ω90\text{ }\Omega90 Ω
Examination Questions and Verified Answers
Section 1: Electrical Theory, DC/AC Circuits & Fundamentals (Questions 1–35)
Question 1: A distribution mechanic measures 120 volts across a load drawing 10 amperes during
normal operation. Which value represents the power consumption of this resistive load?
A. 12 watts
B. 120 watts
C. 1,200 watts
D. 12,000 watts
CORRECT ANSWER: C. 1,200 watts
Rationale: Power in a single-phase resistive circuit is calculated using the formula P=V×IP = V \times
IP=V×I. Multiplying 120 V by 10 A yields 1,200 W (or 1.2 kW).
Question 2: When two resistors are connected in series within an energized DC test circuit, which
property strictly distinguishes the series connection from a parallel connection?
A. Voltage drop across each resistor is always identical
B. Branch current divides proportionately across components
C. The total circuit resistance is less than the smallest resistor
D. The exact same magnitude of current flows sequentially through every resistor
CORRECT ANSWER: D. The exact same magnitude of current flows sequentially through every resistor
Rationale: In a series circuit, there is only one path for current, making current equal through all series
components. In contrast, parallel circuits divide current across multiple paths while maintaining equal
voltage across each branch.
Question 3: A mechanic needs to calculate the primary current of a single-phase transformer given an
apparent power rating of 50 kVA and a primary voltage of 2,500 V. What is the primary full-load
current?
A. 20 amperes
,B. 50 amperes
C. 125 amperes
D. 200 amperes
CORRECT ANSWER: A. 20 amperes
Rationale: Apparent power S=V×IS = V \times IS=V×I. Rearranging gives I=S/VI = S / VI=S/V.
Converting 50 kVA to 50,000 VA and dividing by 2,500 V results in 50,000/2,500=20 A50,,500 =
20\text{ A}50,000/2,500=20 A.
Question 4: Why does an electrical transformer require alternating current (AC) rather than steady direct
current (DC) to maintain normal transformation action?
A. DC produces excessive magnetic frequency that damages core insulation
B. Continuous electromagnetic induction requires a constantly changing magnetic flux
C. AC eliminates magnetic saturation in the steel laminations
D. DC automatically reverses secondary polarity every half cycle
CORRECT ANSWER: B. Continuous electromagnetic induction requires a constantly changing magnetic
flux
Rationale: Transformer action relies on Faraday's law of induction. Alternating current continuously
changes direction and magnitude, creating a constantly changing magnetic flux in the core that induces
voltage in the secondary winding. Steady DC produces a static magnetic field, which induces no
continuous secondary voltage and causes excessive winding heating.
Question 5: What is the primary operational advantage of maintaining balanced phase loading on a
three-phase, four-wire distribution circuit?
A. It completely eliminates dielectric losses in feeder cables
B. It minimizes current flow in the neutral conductor and balances phase voltages
C. It allows conductors to operate above their thermal rating
D. It prevents line reactance from affecting motor loads
CORRECT ANSWER: B. It minimizes current flow in the neutral conductor and balances phase voltages
Rationale: Balancing loads across all three phases ensures that vectorially, phase currents cancel out in the
neutral wire, minimizing neutral current flow, reducing system line losses (I2RI^2RI2R), and preventing
single-phase voltage unbalance.
,Question 6: If the resistance of an energized circuit loop doubles while the source voltage remains strictly
constant, how does the circuit current respond?
A. Current increases to twice its initial value
B. Current drops to one-half of its initial value
C. Current decreases to one-fourth of its initial value
D. Current remains unchanged because voltage is constant
CORRECT ANSWER: B. Current drops to one-half of its initial value
Rationale: According to Ohm's law (I=V/RI = V / RI=V/R), current is inversely proportional to resistance
when voltage is constant. Doubling the resistance halves the current flow.
Question 7: Which electrical property represents the total opposition to alternating current flow, taking
into account both resistance and reactance?
A. Admittance
B. Inductance
C. Impedance
D. Capacitance
CORRECT ANSWER: C. Impedance
Rationale: Impedance (symbolized by ZZZ and measured in ohms) is the total vector sum of resistance
(RRR) and inductive/capacitive reactance (XXX) in an AC circuit (Z=R2+X2Z = \sqrt{R^2 +
X^2}Z=R2+X2 ).
Question 8: What happens to the inductive reactance (XLX_LXL) of a distribution line reactor when the
frequency of the power supply increases?
A. Inductive reactance increases proportionally
B. Inductive reactance decreases proportionally
C. Inductive reactance drops to zero
D. Inductive reactance remains unaffected
CORRECT ANSWER: A. Inductive reactance increases proportionally
, Rationale: Inductive reactance is calculated by XL=2πfLX_L = 2\pi f LXL=2πfL. Because XLX_LXL is
directly proportional to frequency (fff), an increase in frequency produces a proportional increase in
inductive reactance.
Question 9: What is the power factor of a purely resistive overhead heating element circuit?
A. 0.0
B. 0.5 lagging
C. 0.866 leading
D. 1.0 (Unity)
CORRECT ANSWER: D. 1.0 (Unity)
Rationale: In a purely resistive circuit, voltage and current are exactly in phase (phase angle θ=0∘\theta =
0^\circθ=0∘). Since power factor equals cos(θ)\cos(\theta)cos(θ), cos(0∘)=1.0\cos(0^\circ) =
1.0cos(0∘)=1.0 (unity power factor).
Question 10: How much electrical energy is consumed by a 5 kW overhead space heater running
continuously for 8 hours?
A. 13 kWh
B. 40 kWh
C. 400 kWh
D. 1,000 kWh
CORRECT ANSWER: B. 40 kWh
Rationale: Electrical energy in kilowatt-hours is calculated by multiplying power in kilowatts by time in
hours (E=P×tE = P \times tE=P×t). Here, 5 kW×8 hours=40 kWh5\text{ kW} \times 8\text{ hours} =
40\text{ kWh}5 kW×8 hours=40 kWh.
Question 11: When three 30 Ω30\text{ }\Omega30 Ω resistors are connected in parallel, what is the total
equivalent resistance of the network?
A. 10 Ω10\text{ }\Omega10 Ω
B. 30 Ω30\text{ }\Omega30 Ω
C. 60 Ω60\text{ }\Omega60 Ω
D. 90 Ω90\text{ }\Omega90 Ω
