LADWP ELECTRICAL MATHEMATICS
COMPREHENSIVE PRACTICE EXAM WITH
QUESTIONS AND VERIFIED ANSWERS,
PLUS DETAILED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1.
A 120 V resistive heating element draws 15 A under normal operating
conditions. What is the resistance of the heating element?
A. 6 Ω
B. 8 Ω
C. 10 Ω
D. 12 Ω
Answer: B. 8 Ω
Rationale: Ohm’s law states that R=V/I. Substituting 120 V and 15 A
gives R=120/15=8 Ω. This calculation is fundamental in electrical
troubleshooting because knowing any two of voltage, current, and
resistance allows the technician to determine the third.
2.
A circuit has a resistance of 24 Ω and is connected to a 480 V source.
What current will flow through the circuit?
A. 10 A
B. 15 A
C. 20 A
D. 24 A
1
,Answer: C. 20 A
Rationale: Using Ohm’s law, I=V/R. Therefore, I=480/24=20 A. In an
ideal resistive circuit, the current is directly proportional to applied
voltage and inversely proportional to resistance.
3.
A 240 V motor circuit draws 18 A at a power factor of 0.82. What is the
approximate real power consumed by the motor?
A. 2.95 kW
B. 3.54 kW
C. 4.32 kW
D. 5.27 kW
Answer: B. 3.54 kW
Rationale: For a single-phase AC circuit, P=VI(PF). Thus
P=240×18×0.82=3,542.4 W, or approximately 3.54 kW. Power factor is
important because apparent power alone does not represent the actual
rate at which electrical energy is converted into useful work.
4.
A 10 Ω resistor carries 6 A continuously. How much power is dissipated
by the resistor?
A. 36 W
B. 60 W
C. 240 W
D. 360 W
Answer: D. 360 W
2
,Rationale: Resistor power can be calculated using P=I2R. Therefore,
P=62×10=36×10=360 W. The same result can be obtained from P=VI
after determining that the voltage across the resistor is 60 V.
5.
Three 12 Ω resistors are connected in series. What is the total resistance?
A. 4 Ω
B. 12 Ω
C. 24 Ω
D. 36 Ω
Answer: D. 36 Ω
Rationale: Series resistances add directly: RT=R1+R2+R3. Thus
12+12+12=36 Ω. Series circuits have the same current through every
component while the source voltage divides among the resistances.
6.
Three 12 Ω resistors are connected in parallel. What is their equivalent
resistance?
A. 4 Ω
B. 12 Ω
C. 24 Ω
D. 36 Ω
Answer: A. 4 Ω
Rationale: For three equal resistors in parallel, the equivalent
resistance is the resistance of one resistor divided by the number of
resistors: 12/3=4 Ω. Parallel circuits provide multiple current paths, so
3
, their equivalent resistance is always lower than the smallest individual
branch resistance.
7.
A 600 V three-phase system supplies a balanced load drawing 50 A at a
power factor of 0.90. What is the approximate real power?
A. 27.0 kW
B. 38.97 kW
C. 46.76 kW
D. 54.0 kW
Answer: C. 46.76 kW
Rationale: Three-phase real power is P=3VLILPF. Therefore,
P=1.732×600×50×0.90≈46,764 W, or 46.76 kW. This equation is one
of the most important calculations for three-phase distribution and
industrial loads.
8.
A 10 kW load operates for 6 hours. How much electrical energy does it
consume?
A. 1.67 kWh
B. 16 kWh
C. 60 kWh
D. 600 kWh
Answer: C. 60 kWh
Rationale: Electrical energy is calculated as E=Pt. Therefore,
E=10 kW×6 h=60 kWh. Power describes the rate of energy use, while
kilowatt-hours measure accumulated energy consumption.
4
COMPREHENSIVE PRACTICE EXAM WITH
QUESTIONS AND VERIFIED ANSWERS,
PLUS DETAILED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1.
A 120 V resistive heating element draws 15 A under normal operating
conditions. What is the resistance of the heating element?
A. 6 Ω
B. 8 Ω
C. 10 Ω
D. 12 Ω
Answer: B. 8 Ω
Rationale: Ohm’s law states that R=V/I. Substituting 120 V and 15 A
gives R=120/15=8 Ω. This calculation is fundamental in electrical
troubleshooting because knowing any two of voltage, current, and
resistance allows the technician to determine the third.
2.
A circuit has a resistance of 24 Ω and is connected to a 480 V source.
What current will flow through the circuit?
A. 10 A
B. 15 A
C. 20 A
D. 24 A
1
,Answer: C. 20 A
Rationale: Using Ohm’s law, I=V/R. Therefore, I=480/24=20 A. In an
ideal resistive circuit, the current is directly proportional to applied
voltage and inversely proportional to resistance.
3.
A 240 V motor circuit draws 18 A at a power factor of 0.82. What is the
approximate real power consumed by the motor?
A. 2.95 kW
B. 3.54 kW
C. 4.32 kW
D. 5.27 kW
Answer: B. 3.54 kW
Rationale: For a single-phase AC circuit, P=VI(PF). Thus
P=240×18×0.82=3,542.4 W, or approximately 3.54 kW. Power factor is
important because apparent power alone does not represent the actual
rate at which electrical energy is converted into useful work.
4.
A 10 Ω resistor carries 6 A continuously. How much power is dissipated
by the resistor?
A. 36 W
B. 60 W
C. 240 W
D. 360 W
Answer: D. 360 W
2
,Rationale: Resistor power can be calculated using P=I2R. Therefore,
P=62×10=36×10=360 W. The same result can be obtained from P=VI
after determining that the voltage across the resistor is 60 V.
5.
Three 12 Ω resistors are connected in series. What is the total resistance?
A. 4 Ω
B. 12 Ω
C. 24 Ω
D. 36 Ω
Answer: D. 36 Ω
Rationale: Series resistances add directly: RT=R1+R2+R3. Thus
12+12+12=36 Ω. Series circuits have the same current through every
component while the source voltage divides among the resistances.
6.
Three 12 Ω resistors are connected in parallel. What is their equivalent
resistance?
A. 4 Ω
B. 12 Ω
C. 24 Ω
D. 36 Ω
Answer: A. 4 Ω
Rationale: For three equal resistors in parallel, the equivalent
resistance is the resistance of one resistor divided by the number of
resistors: 12/3=4 Ω. Parallel circuits provide multiple current paths, so
3
, their equivalent resistance is always lower than the smallest individual
branch resistance.
7.
A 600 V three-phase system supplies a balanced load drawing 50 A at a
power factor of 0.90. What is the approximate real power?
A. 27.0 kW
B. 38.97 kW
C. 46.76 kW
D. 54.0 kW
Answer: C. 46.76 kW
Rationale: Three-phase real power is P=3VLILPF. Therefore,
P=1.732×600×50×0.90≈46,764 W, or 46.76 kW. This equation is one
of the most important calculations for three-phase distribution and
industrial loads.
8.
A 10 kW load operates for 6 hours. How much electrical energy does it
consume?
A. 1.67 kWh
B. 16 kWh
C. 60 kWh
D. 600 kWh
Answer: C. 60 kWh
Rationale: Electrical energy is calculated as E=Pt. Therefore,
E=10 kW×6 h=60 kWh. Power describes the rate of energy use, while
kilowatt-hours measure accumulated energy consumption.
4
