LADWP ELECTRICAL MECHANIC EXAM
WITH ACTUAL QUESTIONS AND VERIFIED
ANSWERS, PLUS EXPLAINED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED 100% PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
1. Electrical Fundamentals
An electrical mechanic is troubleshooting a 240-V resistive heating
element that draws 20 A when operating normally. The mechanic wants
to determine the approximate resistance of the heating element before
removing it from service. Assuming the element behaves as a purely
resistive load, which resistance should be expected?
A. 4.8 Ω
B. 12 Ω
C. 24 Ω
D. 48 Ω
Answer: C. 12 Ω
Rationale: Ohm’s law states that R = V/I. Substituting 240 V and 20 A
gives R = 240/20 = 12 Ω. A resistive heating element converts electrical
energy primarily into heat, so its operating resistance can be
approximated directly from the applied voltage and current.
2. Electrical Power
A three-phase motor operates from a 480-V line-to-line supply and
draws 30 A at a power factor of 0.80. Assuming the motor is reasonably
balanced, what is its approximate real input power?
1
,A. 11.52 kW
B. 19.92 kW
C. 27.65 kW
D. 33.26 kW
Answer: C. 19.92 kW
Rationale: Three-phase real power is P = √3 × VL × IL × PF.
Therefore, P ≈ 1.732 × 480 × 30 × 0.80 = 19,958 W, or approximately
19.96 kW. The important distinction is that apparent power alone
would be √3VI, while real power additionally incorporates power
factor.
3. Series Circuit
Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series to a 60-V
DC source. What current flows through the circuit?
A. 1.0 A
B. 2.0 A
C. 3.0 A
D. 6.0 A
Answer: B. 2.0 A
Rationale: In a series circuit, resistances add directly: RT = 5 + 10 +
15 = 30 Ω. Using Ohm’s law, I = V/R = 60/30 = 2 A. The same current
flows through every component in a series circuit.
4. Parallel Circuit
A 12-V control circuit contains two parallel resistors, one rated at 6 Ω
and the other at 3 Ω. What total current is supplied by the source?
2
,A. 2 A
B. 4 A
C. 6 A
D. 8 A
Answer: C. 6 A
Rationale: The equivalent resistance of parallel resistors is found from
1/RT = 1/6 + 1/3 = 1/2, giving RT = 2 Ω. Therefore, total current is I =
12/2 = 6 A. Each branch has the full 12 V across it, with 2 A through
the 6-Ω branch and 4 A through the 3-Ω branch.
5. Kirchhoff’s Current Law
At a junction inside a control cabinet, 12 A enters from one conductor.
Two outgoing conductors carry 5 A and 4 A respectively. Assuming the
circuit is operating normally, what current must leave through the
remaining conductor?
A. 2 A
B. 3 A
C. 7 A
D. 9 A
Answer: B. 3 A
Rationale: Kirchhoff’s Current Law states that the total current
entering a node must equal the total current leaving it. Therefore, 12 A
= 5 A + 4 A + I, giving I = 3 A. This principle is particularly useful
when troubleshooting branches of electrical control circuits.
6. AC Frequency
An alternating-current waveform completes 60 complete cycles every
second. What is its frequency?
3
, A. 30 Hz
B. 60 Hz
C. 120 Hz
D. 360 Hz
Answer: B. 60 Hz
Rationale: Frequency is the number of complete cycles occurring per
second and is measured in hertz. Therefore, 60 cycles per second
corresponds to 60 Hz. Frequency affects motors, transformers,
inductive reactance, capacitive reactance, and many other AC devices.
7. Transformer Operation
A transformer has 2,400 turns on its primary winding and 600 turns on
its secondary winding. If 2,400 V AC is applied to the primary, what
ideal secondary voltage should be expected?
A. 300 V
B. 600 V
C. 1,200 V
D. 9,600 V
Answer: B. 600 V
Rationale: For an ideal transformer, Vp/Vs = Np/Ns. Thus Vs = 2,400
× (600/2,400) = 600 V. Because the secondary has fewer turns than the
primary, this is a step-down transformer.
8. Transformer and DC
Why should a conventional transformer not be connected directly to a
steady DC source?
4
WITH ACTUAL QUESTIONS AND VERIFIED
ANSWERS, PLUS EXPLAINED
RATIONALES/EXPERT VERIFIED FOR
GUARANTEED 100% PASS 2026/LATEST
UPDATE/INSTANT DOWNLOAD PDF
1. Electrical Fundamentals
An electrical mechanic is troubleshooting a 240-V resistive heating
element that draws 20 A when operating normally. The mechanic wants
to determine the approximate resistance of the heating element before
removing it from service. Assuming the element behaves as a purely
resistive load, which resistance should be expected?
A. 4.8 Ω
B. 12 Ω
C. 24 Ω
D. 48 Ω
Answer: C. 12 Ω
Rationale: Ohm’s law states that R = V/I. Substituting 240 V and 20 A
gives R = 240/20 = 12 Ω. A resistive heating element converts electrical
energy primarily into heat, so its operating resistance can be
approximated directly from the applied voltage and current.
2. Electrical Power
A three-phase motor operates from a 480-V line-to-line supply and
draws 30 A at a power factor of 0.80. Assuming the motor is reasonably
balanced, what is its approximate real input power?
1
,A. 11.52 kW
B. 19.92 kW
C. 27.65 kW
D. 33.26 kW
Answer: C. 19.92 kW
Rationale: Three-phase real power is P = √3 × VL × IL × PF.
Therefore, P ≈ 1.732 × 480 × 30 × 0.80 = 19,958 W, or approximately
19.96 kW. The important distinction is that apparent power alone
would be √3VI, while real power additionally incorporates power
factor.
3. Series Circuit
Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series to a 60-V
DC source. What current flows through the circuit?
A. 1.0 A
B. 2.0 A
C. 3.0 A
D. 6.0 A
Answer: B. 2.0 A
Rationale: In a series circuit, resistances add directly: RT = 5 + 10 +
15 = 30 Ω. Using Ohm’s law, I = V/R = 60/30 = 2 A. The same current
flows through every component in a series circuit.
4. Parallel Circuit
A 12-V control circuit contains two parallel resistors, one rated at 6 Ω
and the other at 3 Ω. What total current is supplied by the source?
2
,A. 2 A
B. 4 A
C. 6 A
D. 8 A
Answer: C. 6 A
Rationale: The equivalent resistance of parallel resistors is found from
1/RT = 1/6 + 1/3 = 1/2, giving RT = 2 Ω. Therefore, total current is I =
12/2 = 6 A. Each branch has the full 12 V across it, with 2 A through
the 6-Ω branch and 4 A through the 3-Ω branch.
5. Kirchhoff’s Current Law
At a junction inside a control cabinet, 12 A enters from one conductor.
Two outgoing conductors carry 5 A and 4 A respectively. Assuming the
circuit is operating normally, what current must leave through the
remaining conductor?
A. 2 A
B. 3 A
C. 7 A
D. 9 A
Answer: B. 3 A
Rationale: Kirchhoff’s Current Law states that the total current
entering a node must equal the total current leaving it. Therefore, 12 A
= 5 A + 4 A + I, giving I = 3 A. This principle is particularly useful
when troubleshooting branches of electrical control circuits.
6. AC Frequency
An alternating-current waveform completes 60 complete cycles every
second. What is its frequency?
3
, A. 30 Hz
B. 60 Hz
C. 120 Hz
D. 360 Hz
Answer: B. 60 Hz
Rationale: Frequency is the number of complete cycles occurring per
second and is measured in hertz. Therefore, 60 cycles per second
corresponds to 60 Hz. Frequency affects motors, transformers,
inductive reactance, capacitive reactance, and many other AC devices.
7. Transformer Operation
A transformer has 2,400 turns on its primary winding and 600 turns on
its secondary winding. If 2,400 V AC is applied to the primary, what
ideal secondary voltage should be expected?
A. 300 V
B. 600 V
C. 1,200 V
D. 9,600 V
Answer: B. 600 V
Rationale: For an ideal transformer, Vp/Vs = Np/Ns. Thus Vs = 2,400
× (600/2,400) = 600 V. Because the secondary has fewer turns than the
primary, this is a step-down transformer.
8. Transformer and DC
Why should a conventional transformer not be connected directly to a
steady DC source?
4
