CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
CSLB C-10 ELECTRICAL CONTRACTOR LICENSE EXAM
TEST 1
2026/2027 Actual Exam – 120 Verified Questions and Correct Answers
California State Licensing Board (CSLB)
Aligned with: CSLB C-10 Examination Content Outline • California Electrical Code (Title 24, Part 3) • National Electrical
Code (NEC) • OSHA Safety Regulations 29 CFR 1910 & 1926
Cognitive Distribution: 25% Recall • 50% Application • 25% Analysis (including calculations, code interpretation, and project
management scenarios)
Section 1: General Electrical Knowledge & Theory
20 Questions
Q1: A 240-volt circuit supplies a resistive heating load drawing 18 amperes. Using Ohm's Law,
what is the resistance of the heating element?
A. 13.3 ohms *[CORRECT]
B. 10.5 ohms
C. 4,320 ohms
D. 0.075 ohm
Correct Answer: A
Rationale: Ohm's Law states R = V / I. Substituting 240 V / 18 A = 13.33 ohms. Option B miscalculates by
inverting current; option C incorrectly multiplies V x I (yielding power in watts); option D divides I / V. The CSLB
C-10 exam routinely tests this fundamental relationship, so candidates must apply V = IR fluently in both
directions.
Q2: A 120-volt circuit feeds a 1,500-watt resistive load. What current will the branch circuit
conductors carry?
A. 8.0 A
B. 12.5 A *[CORRECT]
C. 15.0 A
D. 180,000 A
Correct Answer: B
Rationale: Power formula P = V x I rearranged to I = P / V gives 1, = 12.5 A. Option A is the result of
using 240 V; option C confuses the load with a typical 15 A breaker rating; option D multiplies P x V instead of
dividing. NEC 210.19(A) requires branch-circuit conductors sized to carry the calculated load.
California State Licensing Board (CSLB) Official Practice Examination Page 1
,CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q3: Three resistors (10 ohms, 20 ohms, 30 ohms) are connected in series across a 120-V source.
What is the voltage drop across the 20-ohm resistor?
A. 20 V
B. 40 V *[CORRECT]
C. 60 V
D. 120 V
Correct Answer: B
Rationale: Total series resistance = 60 ohms, so circuit current I = = 2 A. Voltage across the 20-ohm
resistor V = IR = 2 x 20 = 40 V. Option A assumes 1 A; option C incorrectly allocates half the supply; option D
would only occur if all voltage dropped across one element, which violates Kirchhoff's Voltage Law.
Q4: Two resistors, 6 ohms and 3 ohms, are connected in parallel across a 12-V battery. What total
current does the source supply?
A. 1.5 A
B. 3.0 A
C. 6.0 A *[CORRECT]
D. 9.0 A
Correct Answer: C
Rationale: Parallel equivalent resistance R_T = (6 x 3) / (6 + 3) = = 2 ohms. Total current I = V / R =
= 6 A. Option A calculates only one branch; option B finds only one branch's current; option D adds the
resistances as if in series. CSLB C-10 candidates must distinguish series vs. parallel behavior for service-load
calculations.
Q5: Which statement correctly distinguishes alternating current (AC) from direct current (DC)?
A. AC cannot be transformed; DC can be transformed at any voltage.
B. AC periodically reverses direction; DC flows in one direction only. *[CORRECT]
C. AC requires two conductors; DC requires four conductors.
D. AC is used only at voltages below 50 V; DC is used above 50 V.
Correct Answer: B
Rationale: By definition, AC reverses direction periodically (60 Hz in North America) while DC maintains constant
polarity. Option A is reversed - AC is easily transformed via mutual induction; DC requires electronic conversion.
Option C is false; both can use various conductor arrangements. Option D is unsupported by any code or theory
reference.
California State Licensing Board (CSLB) Official Practice Examination Page 2
,CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q6: A single-phase step-down transformer has a primary voltage of 4,800 V and a turns ratio of
20:1. What is the secondary voltage?
A. 240 V *[CORRECT]
B. 120 V
C. 96 V
D. 4,800 V
Correct Answer: A
Rationale: Turns ratio N_p / N_s = V_p / V_s. With a 20:1 ratio, V_s = V_p / 20 = 4, = 240 V. Option B
assumes a 40:1 ratio; option C miscalculates (4,); option D ignores the ratio entirely. NEC Article 450
governs transformer installations, and ratio calculations are routinely tested on the CSLB C-10 exam.
Q7: A 240-V single-phase circuit supplies a load drawing 30 A at a 0.80 power factor. What is the
true (real) power consumed by the load?
A. 7,200 W
B. 5,760 W *[CORRECT]
C. 9,000 W
D. 4,800 W
Correct Answer: B
Rationale: True power P = V x I x PF = 240 x 30 x 0.80 = 5,760 W. Option A is apparent power (V x I, ignoring
PF); option C uses PF = 1.25 which is impossible; option D halves the apparent power without basis. Power factor
correction is a recurring C-10 topic because motors and fluorescent ballasts draw reactive current that does no
useful work.
Q8: An AC circuit draws 20 A at 480 V with a 0.60 lagging power factor. What is the reactive power
(VAR) of the load?
A. 9,600 VAR
B. 4,800 VAR
C. 7,680 VAR *[CORRECT]
D. 12,800 VAR
Correct Answer: C
Rationale: Apparent power S = V x I = 480 x 20 = 9,600 VA. Reactive power Q = S x sin(theta) where cos(theta)
= 0.60, so sin(theta) = 0.80. Q = 9,600 x 0.80 = 7,680 VAR. Option A confuses reactive power with apparent
power; option B is the true power (S x cos theta); option D is unrelated. CSLB candidates must distinguish W, VA,
and VAR.
California State Licensing Board (CSLB) Official Practice Examination Page 3
, CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q9: What is the relationship between true power (P), reactive power (Q), and apparent power (S) in
an AC circuit?
A. P = S + Q
B. S = P x Q
C. S = sqrt(P^2 + Q^2) *[CORRECT]
D. P = S / Q
Correct Answer: C
Rationale: The power triangle gives S = sqrt(P^2 + Q^2). Option A treats them as additive scalars (incorrect -
they are phasors 90 degrees apart); option B multiplies them (dimensionally wrong); option D divides them. The
power-triangle relationship is foundational for power factor correction sizing and is regularly tested on the CSLB
C-10 exam.
Q10: In a balanced three-phase wye-connected system, what is the relationship between line
voltage and phase voltage?
A. Line voltage = phase voltage
B. Line voltage = phase voltage x sqrt(3) *[CORRECT]
C. Line voltage = phase voltage / sqrt(3)
D. Line voltage = phase voltage x 3
Correct Answer: B
Rationale: In a wye system, line voltage = sqrt(3) x phase voltage (about 1.732 x V_phase). For example, 277 V
phase-to-neutral yields 480 V phase-to-phase. Option A describes a delta system; option C inverts the
relationship; option D incorrectly uses a factor of 3. Three-phase calculations appear throughout NEC Article 220
load calculations and motor circuits (Article 430).
Q11: How does an ideal inductor behave in a steady-state DC circuit after the transient charging
period?
A. It acts as an open circuit (infinite resistance).
B. It acts as a short circuit (zero resistance). *[CORRECT]
C. It produces a continuous alternating voltage.
D. It blocks DC entirely while passing AC.
Correct Answer: B
Rationale: An ideal inductor's reactance X_L = 2*pi*f*L. With DC, f = 0, so X_L = 0 - the inductor acts as a short.
Option A describes capacitor behavior in DC steady state; option C contradicts the steady-state condition; option
D reverses inductor behavior (inductors pass DC, block high-frequency AC). Understanding transient vs.
steady-state behavior is essential for motor and transformer analysis.
California State Licensing Board (CSLB) Official Practice Examination Page 4
CSLB C-10 ELECTRICAL CONTRACTOR LICENSE EXAM
TEST 1
2026/2027 Actual Exam – 120 Verified Questions and Correct Answers
California State Licensing Board (CSLB)
Aligned with: CSLB C-10 Examination Content Outline • California Electrical Code (Title 24, Part 3) • National Electrical
Code (NEC) • OSHA Safety Regulations 29 CFR 1910 & 1926
Cognitive Distribution: 25% Recall • 50% Application • 25% Analysis (including calculations, code interpretation, and project
management scenarios)
Section 1: General Electrical Knowledge & Theory
20 Questions
Q1: A 240-volt circuit supplies a resistive heating load drawing 18 amperes. Using Ohm's Law,
what is the resistance of the heating element?
A. 13.3 ohms *[CORRECT]
B. 10.5 ohms
C. 4,320 ohms
D. 0.075 ohm
Correct Answer: A
Rationale: Ohm's Law states R = V / I. Substituting 240 V / 18 A = 13.33 ohms. Option B miscalculates by
inverting current; option C incorrectly multiplies V x I (yielding power in watts); option D divides I / V. The CSLB
C-10 exam routinely tests this fundamental relationship, so candidates must apply V = IR fluently in both
directions.
Q2: A 120-volt circuit feeds a 1,500-watt resistive load. What current will the branch circuit
conductors carry?
A. 8.0 A
B. 12.5 A *[CORRECT]
C. 15.0 A
D. 180,000 A
Correct Answer: B
Rationale: Power formula P = V x I rearranged to I = P / V gives 1, = 12.5 A. Option A is the result of
using 240 V; option C confuses the load with a typical 15 A breaker rating; option D multiplies P x V instead of
dividing. NEC 210.19(A) requires branch-circuit conductors sized to carry the calculated load.
California State Licensing Board (CSLB) Official Practice Examination Page 1
,CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q3: Three resistors (10 ohms, 20 ohms, 30 ohms) are connected in series across a 120-V source.
What is the voltage drop across the 20-ohm resistor?
A. 20 V
B. 40 V *[CORRECT]
C. 60 V
D. 120 V
Correct Answer: B
Rationale: Total series resistance = 60 ohms, so circuit current I = = 2 A. Voltage across the 20-ohm
resistor V = IR = 2 x 20 = 40 V. Option A assumes 1 A; option C incorrectly allocates half the supply; option D
would only occur if all voltage dropped across one element, which violates Kirchhoff's Voltage Law.
Q4: Two resistors, 6 ohms and 3 ohms, are connected in parallel across a 12-V battery. What total
current does the source supply?
A. 1.5 A
B. 3.0 A
C. 6.0 A *[CORRECT]
D. 9.0 A
Correct Answer: C
Rationale: Parallel equivalent resistance R_T = (6 x 3) / (6 + 3) = = 2 ohms. Total current I = V / R =
= 6 A. Option A calculates only one branch; option B finds only one branch's current; option D adds the
resistances as if in series. CSLB C-10 candidates must distinguish series vs. parallel behavior for service-load
calculations.
Q5: Which statement correctly distinguishes alternating current (AC) from direct current (DC)?
A. AC cannot be transformed; DC can be transformed at any voltage.
B. AC periodically reverses direction; DC flows in one direction only. *[CORRECT]
C. AC requires two conductors; DC requires four conductors.
D. AC is used only at voltages below 50 V; DC is used above 50 V.
Correct Answer: B
Rationale: By definition, AC reverses direction periodically (60 Hz in North America) while DC maintains constant
polarity. Option A is reversed - AC is easily transformed via mutual induction; DC requires electronic conversion.
Option C is false; both can use various conductor arrangements. Option D is unsupported by any code or theory
reference.
California State Licensing Board (CSLB) Official Practice Examination Page 2
,CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q6: A single-phase step-down transformer has a primary voltage of 4,800 V and a turns ratio of
20:1. What is the secondary voltage?
A. 240 V *[CORRECT]
B. 120 V
C. 96 V
D. 4,800 V
Correct Answer: A
Rationale: Turns ratio N_p / N_s = V_p / V_s. With a 20:1 ratio, V_s = V_p / 20 = 4, = 240 V. Option B
assumes a 40:1 ratio; option C miscalculates (4,); option D ignores the ratio entirely. NEC Article 450
governs transformer installations, and ratio calculations are routinely tested on the CSLB C-10 exam.
Q7: A 240-V single-phase circuit supplies a load drawing 30 A at a 0.80 power factor. What is the
true (real) power consumed by the load?
A. 7,200 W
B. 5,760 W *[CORRECT]
C. 9,000 W
D. 4,800 W
Correct Answer: B
Rationale: True power P = V x I x PF = 240 x 30 x 0.80 = 5,760 W. Option A is apparent power (V x I, ignoring
PF); option C uses PF = 1.25 which is impossible; option D halves the apparent power without basis. Power factor
correction is a recurring C-10 topic because motors and fluorescent ballasts draw reactive current that does no
useful work.
Q8: An AC circuit draws 20 A at 480 V with a 0.60 lagging power factor. What is the reactive power
(VAR) of the load?
A. 9,600 VAR
B. 4,800 VAR
C. 7,680 VAR *[CORRECT]
D. 12,800 VAR
Correct Answer: C
Rationale: Apparent power S = V x I = 480 x 20 = 9,600 VA. Reactive power Q = S x sin(theta) where cos(theta)
= 0.60, so sin(theta) = 0.80. Q = 9,600 x 0.80 = 7,680 VAR. Option A confuses reactive power with apparent
power; option B is the true power (S x cos theta); option D is unrelated. CSLB candidates must distinguish W, VA,
and VAR.
California State Licensing Board (CSLB) Official Practice Examination Page 3
, CSLB C-10 Electrical Contractor License Exam Test 1 | 2026/2027 Actual Exam 120 Verified Questions
Q9: What is the relationship between true power (P), reactive power (Q), and apparent power (S) in
an AC circuit?
A. P = S + Q
B. S = P x Q
C. S = sqrt(P^2 + Q^2) *[CORRECT]
D. P = S / Q
Correct Answer: C
Rationale: The power triangle gives S = sqrt(P^2 + Q^2). Option A treats them as additive scalars (incorrect -
they are phasors 90 degrees apart); option B multiplies them (dimensionally wrong); option D divides them. The
power-triangle relationship is foundational for power factor correction sizing and is regularly tested on the CSLB
C-10 exam.
Q10: In a balanced three-phase wye-connected system, what is the relationship between line
voltage and phase voltage?
A. Line voltage = phase voltage
B. Line voltage = phase voltage x sqrt(3) *[CORRECT]
C. Line voltage = phase voltage / sqrt(3)
D. Line voltage = phase voltage x 3
Correct Answer: B
Rationale: In a wye system, line voltage = sqrt(3) x phase voltage (about 1.732 x V_phase). For example, 277 V
phase-to-neutral yields 480 V phase-to-phase. Option A describes a delta system; option C inverts the
relationship; option D incorrectly uses a factor of 3. Three-phase calculations appear throughout NEC Article 220
load calculations and motor circuits (Article 430).
Q11: How does an ideal inductor behave in a steady-state DC circuit after the transient charging
period?
A. It acts as an open circuit (infinite resistance).
B. It acts as a short circuit (zero resistance). *[CORRECT]
C. It produces a continuous alternating voltage.
D. It blocks DC entirely while passing AC.
Correct Answer: B
Rationale: An ideal inductor's reactance X_L = 2*pi*f*L. With DC, f = 0, so X_L = 0 - the inductor acts as a short.
Option A describes capacitor behavior in DC steady state; option C contradicts the steady-state condition; option
D reverses inductor behavior (inductors pass DC, block high-frequency AC). Understanding transient vs.
steady-state behavior is essential for motor and transformer analysis.
California State Licensing Board (CSLB) Official Practice Examination Page 4
