Maryland Master Electrician Exam – 300 Practice
Questions with Verified Answers & Rationales 2026
NEC All Core Domains Covered A+ Graded
Exam Authority: Maryland State Board of
Electricians
Code Edition: Based on the 2026 National Electrical
Code (NEC)
Exam Format: 90 questions (100 points), 4-hour
time limit, open-book
Passing Score: 70% minimum
1
,SECTION 1: ELECTRICAL THEORY (60
Questions)
Exam Domain Weight: 20%
Question 1
In a series circuit consisting of three resistors of 10
Ω, 20 Ω, and 30 Ω connected to a 120 V source,
what is the current flowing through the circuit?
A) 1.0 A
B) 1.5 A
C) 2.0 A
2
,D) 2.5 A
Correct Answer: C) 2.0 A
Rationale: Total resistance Rtotal = R1 + R2 + R3 =
10 + 20 + 30 = 60 Ω. Current I = V / R = 120 V / 60
Ω=2 A.
Question 2
According to Ohm's Law, if a conductor has a
voltage of 240 V across it and a resistance of 12 Ω,
what is the power dissipated as heat?
A) 4 kW
3
, B) 4.8 kW
C) 1 kW
D) 480 W
Correct Answer: B) 4.8 kW
Rationale: First find current I = V/R = 240 V / 12 Ω
= 20 A. Power P = V × I = 240 V × 20 A = 4800 W =
4.8 kW.
Question 3
4
Questions with Verified Answers & Rationales 2026
NEC All Core Domains Covered A+ Graded
Exam Authority: Maryland State Board of
Electricians
Code Edition: Based on the 2026 National Electrical
Code (NEC)
Exam Format: 90 questions (100 points), 4-hour
time limit, open-book
Passing Score: 70% minimum
1
,SECTION 1: ELECTRICAL THEORY (60
Questions)
Exam Domain Weight: 20%
Question 1
In a series circuit consisting of three resistors of 10
Ω, 20 Ω, and 30 Ω connected to a 120 V source,
what is the current flowing through the circuit?
A) 1.0 A
B) 1.5 A
C) 2.0 A
2
,D) 2.5 A
Correct Answer: C) 2.0 A
Rationale: Total resistance Rtotal = R1 + R2 + R3 =
10 + 20 + 30 = 60 Ω. Current I = V / R = 120 V / 60
Ω=2 A.
Question 2
According to Ohm's Law, if a conductor has a
voltage of 240 V across it and a resistance of 12 Ω,
what is the power dissipated as heat?
A) 4 kW
3
, B) 4.8 kW
C) 1 kW
D) 480 W
Correct Answer: B) 4.8 kW
Rationale: First find current I = V/R = 240 V / 12 Ω
= 20 A. Power P = V × I = 240 V × 20 A = 4800 W =
4.8 kW.
Question 3
4
