Kansas Limited Electrical Contractor
Examination — Local Jurisdiction
Practice Exam 2026 | 100 Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A 120/240-volt single-phase dwelling service supplies a calculated load of
18,000 VA. What is the minimum service current before applying any
permitted demand calculations?
A. 60 A
B. 75 A
C. 150 A
D. 200 A
Answer: 75 A
Rationale: The basic service-load current is calculated as 18,000 VA ÷ 240 V = 75
A. The actual service rating must then satisfy the applicable NEC requirements
and local amendments.
2. A 20-ampere branch circuit supplies several 15-ampere receptacles in a
dwelling. Which statement is correct?
,A. Every receptacle must be rated 20 A
B. 15-A receptacles are generally permitted on a 20-A multioutlet branch circuit
C. Only one receptacle may be installed
D. The breaker must be replaced with a 15-A breaker
Answer: 15-A receptacles are generally permitted on a 20-A multioutlet branch
circuit
Rationale: The NEC permits 15-ampere receptacles on 20-ampere branch circuits
when the circuit supplies multiple receptacle outlets, subject to the applicable
receptacle-rating provisions.
3. A continuous load is calculated at 16 A. What is the minimum branch-circuit
ampacity required when the standard 125% continuous-load factor applies?
A. 16 A
B. 18 A
C. 20 A
D. 25 A
Answer: 20 A
Rationale: 16 A × 125% = 20 A. The conductor and overcurrent protection must
then be selected according to the applicable NEC rules.
4. A 120-volt, 20-ampere circuit supplies a continuous load of 16 A. Which
breaker rating is normally appropriate when applying the 125% continuous-
load rule?
A. 15 A
B. 16 A
C. 20 A
D. 25 A
Answer: 20 A
Rationale: The calculated minimum circuit capacity is 20 A, because 16 A × 1.25
= 20 A.
, 5. A 240-volt single-phase heater draws 20 A. What is its approximate power
consumption?
A. 2,400 W
B. 4,800 W
C. 8,000 W
D. 9,600 W
Answer: 4,800 W
Rationale: For a single-phase resistive load, watts = volts × amperes. Therefore,
240 × 20 = 4,800 W.
6. A 120/240-volt multiwire branch circuit has a 12-A load on one ungrounded
conductor and a 7-A load on the other. What current normally flows in the
shared neutral when the loads are on opposite legs?
A. 19 A
B. 12 A
C. 7 A
D. 5 A
Answer: 5 A
Rationale: Opposite-leg currents are vectorially additive through the neutral by
difference, so 12 A − 7 A = 5 A.
7. What is the primary purpose of the grounded conductor in a typical
120/240-volt single-phase system?
A. To provide overcurrent protection
B. To carry normal unbalanced current
C. To replace the equipment grounding conductor
D. To serve only as a fault-current path
Answer: To carry normal unbalanced current
, Rationale: The grounded conductor normally carries the unbalanced current of
the line-to-neutral loads. It is not interchangeable with the equipment
grounding conductor in branch-circuit wiring.
8. What is the principal function of an equipment grounding conductor?
A. Carry normal load current
B. Increase circuit voltage
C. Provide a low-impedance fault-current path
D. Reduce normal operating current
Answer: Provide a low-impedance fault-current path
Rationale: The equipment grounding path allows sufficient fault current to flow
so that the overcurrent protective device can operate promptly.
9. A copper equipment grounding conductor is installed with a circuit
containing 60-A overcurrent protection. The minimum EGC size is
determined primarily from which NEC table?
A. Table 310.16
B. Table 250.122
C. Table 220.42
D. Table 430.250
Answer: Table 250.122
Rationale: Equipment grounding conductor sizing is based on the rating of the
circuit overcurrent protective device using Table 250.122, subject to applicable
adjustment requirements.
10.An ungrounded conductor is increased in size for voltage-drop purposes.
What must generally happen to the wire-type equipment grounding
conductor installed with it?
A. It may always remain the same size
B. It must be increased proportionately when required by the applicable NEC rule
Examination — Local Jurisdiction
Practice Exam 2026 | 100 Questions &
Answers with Detailed Rationales |
Complete Exam Prep & Study Guide
1. A 120/240-volt single-phase dwelling service supplies a calculated load of
18,000 VA. What is the minimum service current before applying any
permitted demand calculations?
A. 60 A
B. 75 A
C. 150 A
D. 200 A
Answer: 75 A
Rationale: The basic service-load current is calculated as 18,000 VA ÷ 240 V = 75
A. The actual service rating must then satisfy the applicable NEC requirements
and local amendments.
2. A 20-ampere branch circuit supplies several 15-ampere receptacles in a
dwelling. Which statement is correct?
,A. Every receptacle must be rated 20 A
B. 15-A receptacles are generally permitted on a 20-A multioutlet branch circuit
C. Only one receptacle may be installed
D. The breaker must be replaced with a 15-A breaker
Answer: 15-A receptacles are generally permitted on a 20-A multioutlet branch
circuit
Rationale: The NEC permits 15-ampere receptacles on 20-ampere branch circuits
when the circuit supplies multiple receptacle outlets, subject to the applicable
receptacle-rating provisions.
3. A continuous load is calculated at 16 A. What is the minimum branch-circuit
ampacity required when the standard 125% continuous-load factor applies?
A. 16 A
B. 18 A
C. 20 A
D. 25 A
Answer: 20 A
Rationale: 16 A × 125% = 20 A. The conductor and overcurrent protection must
then be selected according to the applicable NEC rules.
4. A 120-volt, 20-ampere circuit supplies a continuous load of 16 A. Which
breaker rating is normally appropriate when applying the 125% continuous-
load rule?
A. 15 A
B. 16 A
C. 20 A
D. 25 A
Answer: 20 A
Rationale: The calculated minimum circuit capacity is 20 A, because 16 A × 1.25
= 20 A.
, 5. A 240-volt single-phase heater draws 20 A. What is its approximate power
consumption?
A. 2,400 W
B. 4,800 W
C. 8,000 W
D. 9,600 W
Answer: 4,800 W
Rationale: For a single-phase resistive load, watts = volts × amperes. Therefore,
240 × 20 = 4,800 W.
6. A 120/240-volt multiwire branch circuit has a 12-A load on one ungrounded
conductor and a 7-A load on the other. What current normally flows in the
shared neutral when the loads are on opposite legs?
A. 19 A
B. 12 A
C. 7 A
D. 5 A
Answer: 5 A
Rationale: Opposite-leg currents are vectorially additive through the neutral by
difference, so 12 A − 7 A = 5 A.
7. What is the primary purpose of the grounded conductor in a typical
120/240-volt single-phase system?
A. To provide overcurrent protection
B. To carry normal unbalanced current
C. To replace the equipment grounding conductor
D. To serve only as a fault-current path
Answer: To carry normal unbalanced current
, Rationale: The grounded conductor normally carries the unbalanced current of
the line-to-neutral loads. It is not interchangeable with the equipment
grounding conductor in branch-circuit wiring.
8. What is the principal function of an equipment grounding conductor?
A. Carry normal load current
B. Increase circuit voltage
C. Provide a low-impedance fault-current path
D. Reduce normal operating current
Answer: Provide a low-impedance fault-current path
Rationale: The equipment grounding path allows sufficient fault current to flow
so that the overcurrent protective device can operate promptly.
9. A copper equipment grounding conductor is installed with a circuit
containing 60-A overcurrent protection. The minimum EGC size is
determined primarily from which NEC table?
A. Table 310.16
B. Table 250.122
C. Table 220.42
D. Table 430.250
Answer: Table 250.122
Rationale: Equipment grounding conductor sizing is based on the rating of the
circuit overcurrent protective device using Table 250.122, subject to applicable
adjustment requirements.
10.An ungrounded conductor is increased in size for voltage-drop purposes.
What must generally happen to the wire-type equipment grounding
conductor installed with it?
A. It may always remain the same size
B. It must be increased proportionately when required by the applicable NEC rule
