LADWP Electric Station Operator Exam
2026/2027 Edition
Circuit Breakers, Disconnects, Transformers
Verified Questions with Passed Answers | 100 Questions
Aligned with LADWP Electric Station Operator Certification Standards
Section 1: Electrical Theory and Fundamentals (Q1-Q15)
Q1: An operator monitoring a 34.5 kV bus observes the line-to-line voltage has dropped to 31 kV during a heavy load
period. What is the approximate percentage voltage drop?
A. 8.2%
B. 9.5%
C. 10.1% **[CORRECT]**
D. 11.3%
Correct Answer: C
Rationale: Percentage voltage drop = ((34.5 - 31) / 34.5) x 100 = (3..5) x 100 = 10.14%. Option A uses 34.5 - 31.7
incorrectly, Option B miscalculates the ratio, and Option D uses the wrong base voltage in the denominator. This calculation is
fundamental for LADWP operators assessing bus voltage adequacy during peak loading conditions.
Q2: A three-phase 480V system operates in a wye configuration. What is the phase-to-neutral voltage?
A. 120V
B. 208V
C. 277V **[CORRECT]**
D. 480V
Correct Answer: C
Rationale: In a wye-connected system, phase-to-neutral voltage equals line-to-line voltage divided by the square root of 3. V_phase
= .732 = 277V. Option A (120V) is the phase-to-neutral of a 208V wye system, Option B is the line-to-line voltage of a
120/208V system, and Option D is the line-to-line voltage itself, not the phase-to-neutral value.
Q3: A potential transformer (PT) with a ratio of 2000:1 produces a secondary reading of 115V. What is the primary
voltage?
A. 115 kV
B. 200 kV
C. 230 kV **[CORRECT]**
D. 345 kV
Correct Answer: C
Rationale: The primary voltage equals the PT ratio multiplied by the secondary reading: 2000 x 115V = 230,000V = 230kV.
Option A incorrectly treats the ratio as 1000:1, Option B uses only the ratio number without the secondary reading, and Option D
is an unrelated standard transmission voltage class not derived from this calculation.
Q4: A circuit has a resistance of 6 ohms and a reactance of 8 ohms. What is the total impedance?
A. 7.2 ohms
Page 1
, B. 10 ohms **[CORRECT]**
C. 14 ohms
D. 48 ohms
Correct Answer: B
Rationale: Impedance is calculated using Z = the square root of (R squared + X squared). Z = the square root of (36 + 64) = the
square root of 100 = 10 ohms. Option A represents a simple average, Option C is the arithmetic sum (6 + 8), and Option D is the
product (6 x 8), none of which correctly apply the Pythagorean relationship for impedance.
Q5: A three-phase load draws 500 kW of real power with an apparent power of 600 kVA. What is the power factor?
A. 0.72
B. 0.83 **[CORRECT]**
C. 0.90
D. 1.00
Correct Answer: B
Rationale: Power factor equals real power divided by apparent power: PF = P / S = 500 kW / 600 kVA = 0.833. Option A uses a
reversed calculation, Option C is a common assumption rather than a calculation, and Option D represents a unity power factor
which would require zero reactive power, contradicting the given apparent and real power values.
Q6: In a three-phase wye-connected system, how does the line-to-line voltage relate to the phase voltage?
A. V_line = V_phase
B. V_line = V_phase / the square root of 3
C. V_line = V_phase x the square root of 3 **[CORRECT]**
D. V_line = 3 x V_phase
Correct Answer: C
Rationale: In a wye configuration, the line-to-line voltage is the square root of 3 times the phase voltage. This geometric
relationship arises from the 120-degree phase displacement between phase voltages. Option A describes a delta phase-to-phase
relationship, Option B inverts the ratio, and Option D incorrectly triples the voltage rather than applying the square root of 3
factor.
Q7: A distribution transformer at a LADWP substation has a primary voltage of 4.8 kV. Which of the following is the
most common secondary voltage configuration for this transformer?
A. 120V single-phase only
B. 240V/120V single-phase
C. 480V three-phase
D. 480/240/120V, single or three-phase **[CORRECT]**
Correct Answer: D
Rationale: LADWP distribution transformers with 4.8 kV primaries typically provide 480V, 240V, or 120V on the secondary side,
available in both single-phase and three-phase configurations depending on customer requirements. Option A is too limited,
Option B covers only residential single-phase, and Option C omits the lower voltage classes commonly needed for mixed-load
distribution.
Q8: If a transformer delivers 980 kW from its secondary winding while 1000 kW is supplied to the primary, what is the
approximate efficiency?
A. 96.0%
B. 97.5%
C. 98.0% **[CORRECT]**
D. 99.5%
Correct Answer: C
Page 2
, Rationale: Efficiency equals output divided by input times 100: () x 100 = 98%. This is consistent with the stated
efficiency of large power transformers used at LADWP substations, which typically operate at approximately 98% efficiency.
Option A and B are below the expected range for large transformers, while Option D exceeds typical values for power
transformers of this class.
Q9: What is the relationship between apparent power (S), real power (P), and reactive power (Q) in an AC circuit?
A. S = P + Q
B. S = the square root of (P squared + Q squared) **[CORRECT]**
C. P = S + Q
D. Q = S x P
Correct Answer: B
Rationale: Apparent power is the vector sum of real and reactive power, calculated as S = the square root of (P squared + Q
squared). This follows the Pythagorean theorem because real and reactive power are orthogonal components. Option A incorrectly
adds them algebraically, Option C confuses the relationship, and Option D multiplies rather than combining them as vectors.
Q10: In a delta-connected three-phase system, if the phase current is 20A, what is the line current?
A. 11.5A
B. 20A
C. 34.6A **[CORRECT]**
D. 60A
Correct Answer: C
Rationale: In a delta system, line current equals phase current multiplied by the square root of 3: I_line = 20 x 1.732 = 34.64A.
Option A divides by the square root of 3 (which applies to voltage, not current, in delta), Option B incorrectly assumes they are
equal (which is true for wye phase current, not delta), and Option D multiplies by 3 instead of the square root of 3.
Q11: An LADWP substation bus operates at 138 kV. During a contingency, the voltage drops to 125 kV. What
percentage drop has occurred, and is this within acceptable utility operating limits?
A. 7.2%, within limits
B. 9.4%, within limits **[CORRECT]**
C. 10.2%, outside limits
D. 13.0%, outside limits
Correct Answer: B
Rationale: Percentage drop = ((138 - 125) / 138) x 100 = () x 100 = 9.42%. Utility operating standards typically allow
voltage deviations of up to approximately 5% under normal conditions and up to 10% under emergency conditions. A 9.4% drop is
at the upper boundary but within emergency limits, making Option B correct. Options A and C have calculation errors, and Option
D uses the wrong denominator.
Q12: A load on a 34.5 kV feeder draws 15 A per phase in a balanced three-phase wye system. What is the approximate
three-phase real power if the power factor is 0.85?
A. 443 kW
B. 767 kW **[CORRECT]**
C. 895 kW
D. 1328 kW
Correct Answer: B
Rationale: Three-phase power P = the square root of 3 x V_line x I_line x PF = 1.732 x 34,500 x 15 x 0.85 = 767,288W = 767
kW. Option A omits the power factor, Option C uses a PF of approximately 1.0, and Option D incorrectly multiplies by 3 instead
of the square root of 3. This calculation is essential for operators assessing feeder loading.
Q13: The skin effect in AC conductors causes current to concentrate near the surface of the conductor. What is the
primary operational impact of this phenomenon?
Page 3
2026/2027 Edition
Circuit Breakers, Disconnects, Transformers
Verified Questions with Passed Answers | 100 Questions
Aligned with LADWP Electric Station Operator Certification Standards
Section 1: Electrical Theory and Fundamentals (Q1-Q15)
Q1: An operator monitoring a 34.5 kV bus observes the line-to-line voltage has dropped to 31 kV during a heavy load
period. What is the approximate percentage voltage drop?
A. 8.2%
B. 9.5%
C. 10.1% **[CORRECT]**
D. 11.3%
Correct Answer: C
Rationale: Percentage voltage drop = ((34.5 - 31) / 34.5) x 100 = (3..5) x 100 = 10.14%. Option A uses 34.5 - 31.7
incorrectly, Option B miscalculates the ratio, and Option D uses the wrong base voltage in the denominator. This calculation is
fundamental for LADWP operators assessing bus voltage adequacy during peak loading conditions.
Q2: A three-phase 480V system operates in a wye configuration. What is the phase-to-neutral voltage?
A. 120V
B. 208V
C. 277V **[CORRECT]**
D. 480V
Correct Answer: C
Rationale: In a wye-connected system, phase-to-neutral voltage equals line-to-line voltage divided by the square root of 3. V_phase
= .732 = 277V. Option A (120V) is the phase-to-neutral of a 208V wye system, Option B is the line-to-line voltage of a
120/208V system, and Option D is the line-to-line voltage itself, not the phase-to-neutral value.
Q3: A potential transformer (PT) with a ratio of 2000:1 produces a secondary reading of 115V. What is the primary
voltage?
A. 115 kV
B. 200 kV
C. 230 kV **[CORRECT]**
D. 345 kV
Correct Answer: C
Rationale: The primary voltage equals the PT ratio multiplied by the secondary reading: 2000 x 115V = 230,000V = 230kV.
Option A incorrectly treats the ratio as 1000:1, Option B uses only the ratio number without the secondary reading, and Option D
is an unrelated standard transmission voltage class not derived from this calculation.
Q4: A circuit has a resistance of 6 ohms and a reactance of 8 ohms. What is the total impedance?
A. 7.2 ohms
Page 1
, B. 10 ohms **[CORRECT]**
C. 14 ohms
D. 48 ohms
Correct Answer: B
Rationale: Impedance is calculated using Z = the square root of (R squared + X squared). Z = the square root of (36 + 64) = the
square root of 100 = 10 ohms. Option A represents a simple average, Option C is the arithmetic sum (6 + 8), and Option D is the
product (6 x 8), none of which correctly apply the Pythagorean relationship for impedance.
Q5: A three-phase load draws 500 kW of real power with an apparent power of 600 kVA. What is the power factor?
A. 0.72
B. 0.83 **[CORRECT]**
C. 0.90
D. 1.00
Correct Answer: B
Rationale: Power factor equals real power divided by apparent power: PF = P / S = 500 kW / 600 kVA = 0.833. Option A uses a
reversed calculation, Option C is a common assumption rather than a calculation, and Option D represents a unity power factor
which would require zero reactive power, contradicting the given apparent and real power values.
Q6: In a three-phase wye-connected system, how does the line-to-line voltage relate to the phase voltage?
A. V_line = V_phase
B. V_line = V_phase / the square root of 3
C. V_line = V_phase x the square root of 3 **[CORRECT]**
D. V_line = 3 x V_phase
Correct Answer: C
Rationale: In a wye configuration, the line-to-line voltage is the square root of 3 times the phase voltage. This geometric
relationship arises from the 120-degree phase displacement between phase voltages. Option A describes a delta phase-to-phase
relationship, Option B inverts the ratio, and Option D incorrectly triples the voltage rather than applying the square root of 3
factor.
Q7: A distribution transformer at a LADWP substation has a primary voltage of 4.8 kV. Which of the following is the
most common secondary voltage configuration for this transformer?
A. 120V single-phase only
B. 240V/120V single-phase
C. 480V three-phase
D. 480/240/120V, single or three-phase **[CORRECT]**
Correct Answer: D
Rationale: LADWP distribution transformers with 4.8 kV primaries typically provide 480V, 240V, or 120V on the secondary side,
available in both single-phase and three-phase configurations depending on customer requirements. Option A is too limited,
Option B covers only residential single-phase, and Option C omits the lower voltage classes commonly needed for mixed-load
distribution.
Q8: If a transformer delivers 980 kW from its secondary winding while 1000 kW is supplied to the primary, what is the
approximate efficiency?
A. 96.0%
B. 97.5%
C. 98.0% **[CORRECT]**
D. 99.5%
Correct Answer: C
Page 2
, Rationale: Efficiency equals output divided by input times 100: () x 100 = 98%. This is consistent with the stated
efficiency of large power transformers used at LADWP substations, which typically operate at approximately 98% efficiency.
Option A and B are below the expected range for large transformers, while Option D exceeds typical values for power
transformers of this class.
Q9: What is the relationship between apparent power (S), real power (P), and reactive power (Q) in an AC circuit?
A. S = P + Q
B. S = the square root of (P squared + Q squared) **[CORRECT]**
C. P = S + Q
D. Q = S x P
Correct Answer: B
Rationale: Apparent power is the vector sum of real and reactive power, calculated as S = the square root of (P squared + Q
squared). This follows the Pythagorean theorem because real and reactive power are orthogonal components. Option A incorrectly
adds them algebraically, Option C confuses the relationship, and Option D multiplies rather than combining them as vectors.
Q10: In a delta-connected three-phase system, if the phase current is 20A, what is the line current?
A. 11.5A
B. 20A
C. 34.6A **[CORRECT]**
D. 60A
Correct Answer: C
Rationale: In a delta system, line current equals phase current multiplied by the square root of 3: I_line = 20 x 1.732 = 34.64A.
Option A divides by the square root of 3 (which applies to voltage, not current, in delta), Option B incorrectly assumes they are
equal (which is true for wye phase current, not delta), and Option D multiplies by 3 instead of the square root of 3.
Q11: An LADWP substation bus operates at 138 kV. During a contingency, the voltage drops to 125 kV. What
percentage drop has occurred, and is this within acceptable utility operating limits?
A. 7.2%, within limits
B. 9.4%, within limits **[CORRECT]**
C. 10.2%, outside limits
D. 13.0%, outside limits
Correct Answer: B
Rationale: Percentage drop = ((138 - 125) / 138) x 100 = () x 100 = 9.42%. Utility operating standards typically allow
voltage deviations of up to approximately 5% under normal conditions and up to 10% under emergency conditions. A 9.4% drop is
at the upper boundary but within emergency limits, making Option B correct. Options A and C have calculation errors, and Option
D uses the wrong denominator.
Q12: A load on a 34.5 kV feeder draws 15 A per phase in a balanced three-phase wye system. What is the approximate
three-phase real power if the power factor is 0.85?
A. 443 kW
B. 767 kW **[CORRECT]**
C. 895 kW
D. 1328 kW
Correct Answer: B
Rationale: Three-phase power P = the square root of 3 x V_line x I_line x PF = 1.732 x 34,500 x 15 x 0.85 = 767,288W = 767
kW. Option A omits the power factor, Option C uses a PF of approximately 1.0, and Option D incorrectly multiplies by 3 instead
of the square root of 3. This calculation is essential for operators assessing feeder loading.
Q13: The skin effect in AC conductors causes current to concentrate near the surface of the conductor. What is the
primary operational impact of this phenomenon?
Page 3
