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CALIFORNIA CSLB C-10 ELECTRICAL EXAMINATION
2026/2027 COMPLETE (150) CURRENT TESTING
QUESTIONS AND CORRECT ANSWERS WITH DETAILED
RATIONALES.
ELECTRICAL
Prepare effectively for the California CSLB C-10 Electrical Examination with this
focused study resource. It supports review of electrical systems, wiring methods,
installation practices, safety requirements, electrical codes, and contractor
responsibilities. Use the material to reinforce your technical knowledge, review key
topics, and identify areas that may require additional study. This resource is suited for
electricians, electrical contractors, and candidates preparing for the California CSLB
C-10 Electrical examination.
MULTIPLE CHOICE.
SECTION 1: ELECTRICAL THEORY & FUNDAMENTALS (Questions 1–25)
Question 1:
According to Ohm's Law, if the voltage in a circuit is 240 volts and the
resistance is 30 ohms, what is the current flowing in the circuit?
A) 2 amperes
B) 8 amperes
C) 12 amperes
D) 60 amperes
Answer: B
Rationale: Ohm's Law states that I = V/R (Current = Voltage / Resistance). I =
240V / 30Ω = 8 amperes. The three fundamental forms are: I = V/R, V = I × R,
and R = V/I.
Question 2:
A 1,500-watt, 120-volt electric heater will draw how many amperes when
operating at rated voltage?
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A) 8 amperes
B) 10 amperes
C) 12.5 amperes
D) 15 amperes
Answer: C
Rationale: Power (P) = Voltage (V) × Current (I). Therefore, I = P/V = 1,500W /
120V = 12.5 amperes. This is a fundamental application of the power formula
in electrical calculations.
Question 3:
Voltage drop in a wire is a function of which of the following factors?
A) Wire gauge only
B) Length of wire only
C) Wire gauge, length of wire, and current flow
D) The ambient temperature only
Answer: C
Rationale: Voltage drop is directly influenced by the resistance of the wire
(determined by its gauge and length) and the amount of current (amperage)
flowing through it, according to Ohm's Law (V = I × R).
Question 4:
What is the power factor of a circuit that draws 10A at 240V with a true power
of 1,920W?
A) 0.60
B) 0.70
C) 0.80
D) 0.90
Answer: C
Rationale: Power Factor = True Power ÷ Apparent Power. Apparent Power = V ×
I = 240V × 10A = 2,400VA. Power Factor = 1,920W / 2,400VA = 0.80.
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Question 5:
A shunt is used in parallel with a d.c. ammeter measuring large currents to:
A) Dampen the meter pointer
B) Reduce the meter current
C) Protect the meter against shorts
D) Increase the sensitivity of the meter
Answer: B
Rationale: A shunt is connected in parallel with an ammeter to bypass the
majority of the current, allowing the meter to measure large currents without
being damaged. The shunt "reduces the meter current".
Question 6:
To measure the voltage across a load, you would use a(n):
A) Voltmeter in series with the load
B) Voltmeter across the load
C) Ammeter across the load
D) Ammeter in series with the load
Answer: B
Rationale: A voltmeter must be connected in parallel (across) the load to
measure voltage. An ammeter is connected in series to measure current.
Question 7:
A 12-volt, 13-watt fluorescent light operates for an average of 12 hours per
day. How many watt-hours does it consume daily?
A) 144 Wh
B) 1.08 Wh
C) 156 Wh
D) 1.56 Wh
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Answer: C
Rationale: Watt-hours = Power (watts) × Time (hours) = 13W × 12h = 156 Wh.
Question 8:
What is the minimum size THWN copper conductor permitted for a 20-
ampere branch circuit in a dry location?
A) 14 AWG
B) 12 AWG
C) 10 AWG
D) 8 AWG
Answer: B
Rationale: NEC Table 310.16 lists 12 AWG THWN copper with an ampacity of
25 amperes at 75°C. While 14 AWG is rated for 20 amperes, 210.19(A)(1)
requires branch circuit conductors to have an ampacity not less than the
maximum load to be served, and 240.4(D) limits 14 AWG overcurrent
protection to 15 amperes.
Question 9:
What is the maximum number of quarter bends (90-degree turns) allowed in a
single conduit run between pull points or outlet boxes?
A) Two quarter bends (180 degrees total)
B) Three quarter bends (270 degrees total)
C) Four quarter bends (360 degrees total)
D) Five quarter bends (450 degrees total)
Answer: C
Rationale: NEC 358.26 limits the total number of bends in a conduit run
(including offset bends) to not more than 360 degrees between pull points,
junction boxes, or outlet boxes.
CALIFORNIA CSLB C-10 ELECTRICAL EXAMINATION
2026/2027 COMPLETE (150) CURRENT TESTING
QUESTIONS AND CORRECT ANSWERS WITH DETAILED
RATIONALES.
ELECTRICAL
Prepare effectively for the California CSLB C-10 Electrical Examination with this
focused study resource. It supports review of electrical systems, wiring methods,
installation practices, safety requirements, electrical codes, and contractor
responsibilities. Use the material to reinforce your technical knowledge, review key
topics, and identify areas that may require additional study. This resource is suited for
electricians, electrical contractors, and candidates preparing for the California CSLB
C-10 Electrical examination.
MULTIPLE CHOICE.
SECTION 1: ELECTRICAL THEORY & FUNDAMENTALS (Questions 1–25)
Question 1:
According to Ohm's Law, if the voltage in a circuit is 240 volts and the
resistance is 30 ohms, what is the current flowing in the circuit?
A) 2 amperes
B) 8 amperes
C) 12 amperes
D) 60 amperes
Answer: B
Rationale: Ohm's Law states that I = V/R (Current = Voltage / Resistance). I =
240V / 30Ω = 8 amperes. The three fundamental forms are: I = V/R, V = I × R,
and R = V/I.
Question 2:
A 1,500-watt, 120-volt electric heater will draw how many amperes when
operating at rated voltage?
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A) 8 amperes
B) 10 amperes
C) 12.5 amperes
D) 15 amperes
Answer: C
Rationale: Power (P) = Voltage (V) × Current (I). Therefore, I = P/V = 1,500W /
120V = 12.5 amperes. This is a fundamental application of the power formula
in electrical calculations.
Question 3:
Voltage drop in a wire is a function of which of the following factors?
A) Wire gauge only
B) Length of wire only
C) Wire gauge, length of wire, and current flow
D) The ambient temperature only
Answer: C
Rationale: Voltage drop is directly influenced by the resistance of the wire
(determined by its gauge and length) and the amount of current (amperage)
flowing through it, according to Ohm's Law (V = I × R).
Question 4:
What is the power factor of a circuit that draws 10A at 240V with a true power
of 1,920W?
A) 0.60
B) 0.70
C) 0.80
D) 0.90
Answer: C
Rationale: Power Factor = True Power ÷ Apparent Power. Apparent Power = V ×
I = 240V × 10A = 2,400VA. Power Factor = 1,920W / 2,400VA = 0.80.
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Question 5:
A shunt is used in parallel with a d.c. ammeter measuring large currents to:
A) Dampen the meter pointer
B) Reduce the meter current
C) Protect the meter against shorts
D) Increase the sensitivity of the meter
Answer: B
Rationale: A shunt is connected in parallel with an ammeter to bypass the
majority of the current, allowing the meter to measure large currents without
being damaged. The shunt "reduces the meter current".
Question 6:
To measure the voltage across a load, you would use a(n):
A) Voltmeter in series with the load
B) Voltmeter across the load
C) Ammeter across the load
D) Ammeter in series with the load
Answer: B
Rationale: A voltmeter must be connected in parallel (across) the load to
measure voltage. An ammeter is connected in series to measure current.
Question 7:
A 12-volt, 13-watt fluorescent light operates for an average of 12 hours per
day. How many watt-hours does it consume daily?
A) 144 Wh
B) 1.08 Wh
C) 156 Wh
D) 1.56 Wh
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Answer: C
Rationale: Watt-hours = Power (watts) × Time (hours) = 13W × 12h = 156 Wh.
Question 8:
What is the minimum size THWN copper conductor permitted for a 20-
ampere branch circuit in a dry location?
A) 14 AWG
B) 12 AWG
C) 10 AWG
D) 8 AWG
Answer: B
Rationale: NEC Table 310.16 lists 12 AWG THWN copper with an ampacity of
25 amperes at 75°C. While 14 AWG is rated for 20 amperes, 210.19(A)(1)
requires branch circuit conductors to have an ampacity not less than the
maximum load to be served, and 240.4(D) limits 14 AWG overcurrent
protection to 15 amperes.
Question 9:
What is the maximum number of quarter bends (90-degree turns) allowed in a
single conduit run between pull points or outlet boxes?
A) Two quarter bends (180 degrees total)
B) Three quarter bends (270 degrees total)
C) Four quarter bends (360 degrees total)
D) Five quarter bends (450 degrees total)
Answer: C
Rationale: NEC 358.26 limits the total number of bends in a conduit run
(including offset bends) to not more than 360 degrees between pull points,
junction boxes, or outlet boxes.
