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CALIFORNIA CSLB C10 ELECTRICAL EXAM WITH VERIFIED
ANSWERS DETAILED RATIONALES GRADED A+
California CSLB C10 Electrical Exam
SECTION 1: PLANNING AND ESTIMATING (Questions 1–45)
1. A contractor is planning the branch circuit wiring for a new 2,400 sq ft single-family
dwelling. Per NEC 220, what is the general lighting and receptacle calculated load in volt-
amperes before applying any demand factor?
A) 2,400 VA
B) 3,000 VA
C) 7,200 VA
D) 9,600 VA
Answer: C) 7,200 VA
Rationale: NEC 220.12 requires a minimum unit load of 3 VA per sq ft for dwelling unit general
lighting and receptacle loads. For a 2,400 sq ft dwelling: 2,400 × 3 = 7,200 VA. This is the
calculated load before demand factors per NEC 220.45 are applied. Option A uses 1 VA/sq ft
which is incorrect. Option B is the first-tier demand threshold, not the calculated load. Option D
would require a 4 VA/sq ft rate which applies to hotel/motel guest rooms, not dwelling units.
2. Using Ohm's Law, if a 240V circuit feeds a resistive heating element that draws 20 amperes,
what is the resistance of the element?
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A) 8 ohms
B) 10 ohms
C) 12 ohms
D) 48 ohms
Answer: C) 12 ohms
Rationale: Ohm's Law: 𝐸 = 𝐼 × 𝑅, rearranged to 𝑅 = 𝐸/𝐼. Given 𝐸 = 240𝑉and 𝐼 = 20𝐴: 𝑅 =
240/20 = 12ohms. Option A (8 ohms) would require 30A at 240V. Option B (10 ohms) would
result in 24A at 240V. Option D (48 ohms) would only draw 5A at 240V. Ohm's Law is
fundamental to circuit analysis per NEC Chapter 9, Table 8.
3. A three-phase 208V wye-connected system supplies a balanced load. What is the phase-to-
neutral voltage?
A) 120V
B) 208V
C) 240V
D) 360V
Answer: A) 120V
Rationale: In a three-phase wye system, the line-to-line voltage equals the phase voltage
multiplied by the square root of 3 (1.732). Therefore, the phase-to-neutral voltage = line voltage
/ √3 = .732 = 120V. Option B is the line-to-line voltage, not phase-to-neutral. Option C
(240V) is the line-to-line voltage of a different common system. Option D (360V) has no
relationship to a 208V wye system. This relationship is fundamental to three-phase system
design per NEC Article 210.
4. What is the minimum size copper service-entrance conductor required for a single-family
dwelling with a calculated load of 200 amperes at 240V? Use THWN conductors rated 75
degrees C.
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A) 2/0 AWG
B) 3/0 AWG
C) 4/0 AWG
D) 250 kcmil
Answer: C) 4/0 AWG
Rationale: Per NEC Table 310.15(B)(16), under the 75°C column, 4/0 AWG copper THWN has an
ampacity of 230A, which is the minimum size that exceeds the 200A calculated load. 2/0 AWG
(175A) and 3/0 AWG (200A) are insufficient when applying the 75°C terminal rating
requirements of NEC 110.14(C). 250 kcmil (255A) would be oversized for this application.
5. A contractor needs 13 receptacles that cost $3.56 each plus sales tax of 7.25%. The supplier
states that for every 10 receptacles that are purchased, the contractor will get one free of
charge. What is the total cost of the receptacles?
A) $42.02
B) $42.72
C) $45.82
D) $49.66
Answer: C) $45.82
Rationale: With the "buy 10, get 1 free" promotion, the contractor purchases 12 receptacles (10
paid + 2 free for the 13th). Cost before tax: 12 × $3.56 = $42.72. Sales tax: $42.72 × 0.0725 =
$3.10. Total cost: $42.72 + $3.10 = $45.82. Option A does not account for the correct number of
paid receptacles. Option B omits sales tax. Option D overcounts the paid receptacles.
6. Which of the following applies to solar modules, metal frames, and metal mounting
equipment on a roof-mounted PV system?
A) They must be weatherproof.
B) They must be wind resistant.
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C) They must be anodized aluminum.
D) They must be bonded and grounded.
Answer: D) They must be bonded and grounded.
Rationale: Per NEC Article 690, all exposed non-current-carrying metal parts of a PV system,
including module frames and mounting equipment, must be bonded and grounded to ensure
safety and proper operation. While weather resistance and wind resistance are important
installation considerations, bonding and grounding are the specific code requirements for
electrical safety.
7. What criteria should be used when selecting a replacement for a defective branch circuit
breaker?
A) The frame type, AIC rating, and ampacity
B) A circuit breaker with higher ampacity to allow for arcing
C) A circuit breaker with a higher AIC rating
D) The frame type, UL rating, and color
Answer: A) The frame type, AIC rating, and ampacity
Rationale: When replacing a circuit breaker, the replacement must match the original in frame
type (physical size and mounting), AIC rating (interrupting capacity), and ampacity (current
rating). Installing a breaker with a higher ampacity than the circuit conductors can allow is a
code violation and safety hazard. Color and UL rating alone are insufficient criteria.
8. A dwelling unit has a general lighting load of 7,200 VA. What is the demand load after
applying the first-tier demand factor per NEC 220.42?
A) 3,000 VA
B) 3,600 VA
C) 4,200 VA
D) 7,200 VA
Answer: A) 3,000 VA
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CALIFORNIA CSLB C10 ELECTRICAL EXAM WITH VERIFIED
ANSWERS DETAILED RATIONALES GRADED A+
California CSLB C10 Electrical Exam
SECTION 1: PLANNING AND ESTIMATING (Questions 1–45)
1. A contractor is planning the branch circuit wiring for a new 2,400 sq ft single-family
dwelling. Per NEC 220, what is the general lighting and receptacle calculated load in volt-
amperes before applying any demand factor?
A) 2,400 VA
B) 3,000 VA
C) 7,200 VA
D) 9,600 VA
Answer: C) 7,200 VA
Rationale: NEC 220.12 requires a minimum unit load of 3 VA per sq ft for dwelling unit general
lighting and receptacle loads. For a 2,400 sq ft dwelling: 2,400 × 3 = 7,200 VA. This is the
calculated load before demand factors per NEC 220.45 are applied. Option A uses 1 VA/sq ft
which is incorrect. Option B is the first-tier demand threshold, not the calculated load. Option D
would require a 4 VA/sq ft rate which applies to hotel/motel guest rooms, not dwelling units.
2. Using Ohm's Law, if a 240V circuit feeds a resistive heating element that draws 20 amperes,
what is the resistance of the element?
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A) 8 ohms
B) 10 ohms
C) 12 ohms
D) 48 ohms
Answer: C) 12 ohms
Rationale: Ohm's Law: 𝐸 = 𝐼 × 𝑅, rearranged to 𝑅 = 𝐸/𝐼. Given 𝐸 = 240𝑉and 𝐼 = 20𝐴: 𝑅 =
240/20 = 12ohms. Option A (8 ohms) would require 30A at 240V. Option B (10 ohms) would
result in 24A at 240V. Option D (48 ohms) would only draw 5A at 240V. Ohm's Law is
fundamental to circuit analysis per NEC Chapter 9, Table 8.
3. A three-phase 208V wye-connected system supplies a balanced load. What is the phase-to-
neutral voltage?
A) 120V
B) 208V
C) 240V
D) 360V
Answer: A) 120V
Rationale: In a three-phase wye system, the line-to-line voltage equals the phase voltage
multiplied by the square root of 3 (1.732). Therefore, the phase-to-neutral voltage = line voltage
/ √3 = .732 = 120V. Option B is the line-to-line voltage, not phase-to-neutral. Option C
(240V) is the line-to-line voltage of a different common system. Option D (360V) has no
relationship to a 208V wye system. This relationship is fundamental to three-phase system
design per NEC Article 210.
4. What is the minimum size copper service-entrance conductor required for a single-family
dwelling with a calculated load of 200 amperes at 240V? Use THWN conductors rated 75
degrees C.
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A) 2/0 AWG
B) 3/0 AWG
C) 4/0 AWG
D) 250 kcmil
Answer: C) 4/0 AWG
Rationale: Per NEC Table 310.15(B)(16), under the 75°C column, 4/0 AWG copper THWN has an
ampacity of 230A, which is the minimum size that exceeds the 200A calculated load. 2/0 AWG
(175A) and 3/0 AWG (200A) are insufficient when applying the 75°C terminal rating
requirements of NEC 110.14(C). 250 kcmil (255A) would be oversized for this application.
5. A contractor needs 13 receptacles that cost $3.56 each plus sales tax of 7.25%. The supplier
states that for every 10 receptacles that are purchased, the contractor will get one free of
charge. What is the total cost of the receptacles?
A) $42.02
B) $42.72
C) $45.82
D) $49.66
Answer: C) $45.82
Rationale: With the "buy 10, get 1 free" promotion, the contractor purchases 12 receptacles (10
paid + 2 free for the 13th). Cost before tax: 12 × $3.56 = $42.72. Sales tax: $42.72 × 0.0725 =
$3.10. Total cost: $42.72 + $3.10 = $45.82. Option A does not account for the correct number of
paid receptacles. Option B omits sales tax. Option D overcounts the paid receptacles.
6. Which of the following applies to solar modules, metal frames, and metal mounting
equipment on a roof-mounted PV system?
A) They must be weatherproof.
B) They must be wind resistant.
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C) They must be anodized aluminum.
D) They must be bonded and grounded.
Answer: D) They must be bonded and grounded.
Rationale: Per NEC Article 690, all exposed non-current-carrying metal parts of a PV system,
including module frames and mounting equipment, must be bonded and grounded to ensure
safety and proper operation. While weather resistance and wind resistance are important
installation considerations, bonding and grounding are the specific code requirements for
electrical safety.
7. What criteria should be used when selecting a replacement for a defective branch circuit
breaker?
A) The frame type, AIC rating, and ampacity
B) A circuit breaker with higher ampacity to allow for arcing
C) A circuit breaker with a higher AIC rating
D) The frame type, UL rating, and color
Answer: A) The frame type, AIC rating, and ampacity
Rationale: When replacing a circuit breaker, the replacement must match the original in frame
type (physical size and mounting), AIC rating (interrupting capacity), and ampacity (current
rating). Installing a breaker with a higher ampacity than the circuit conductors can allow is a
code violation and safety hazard. Color and UL rating alone are insufficient criteria.
8. A dwelling unit has a general lighting load of 7,200 VA. What is the demand load after
applying the first-tier demand factor per NEC 220.42?
A) 3,000 VA
B) 3,600 VA
C) 4,200 VA
D) 7,200 VA
Answer: A) 3,000 VA
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