ALARM COURSE CERTIFICATION EXAM 2026/2027 | Fire
& Burglar Alarm Complete Solutions | Pass Guaranteed -
A+ Graded
SECTION 1: ELECTRICAL FUNDAMENTALS & OHM'S LAW (15
Questions)
Q1: A Notification Appliance Circuit (NAC) operates at 24 VDC and powers two
horn-strobe devices wired in series, each with an internal resistance of 12 ohms. What
is the total current flowing through the circuit?
A. 0.5 A, because current divides equally in a series circuit regardless of total resistance
B. 2.0 A, calculated by dividing 24 V by the parallel equivalent of 12 ohms
C. 1.0 A, determined by Ohm's Law (I = E ÷ R) where total resistance equals 24 ohms
[CORRECT]
D. 4.0 A, because each device draws 2 A independently in series
Correct Answer: C
Rationale: In a series circuit, total resistance is the sum of individual resistances (12 Ω +
12 Ω = 24 Ω). Applying Ohm's Law (I = E ÷ R), current equals 24 V ÷ 24 Ω = 1.0 A. Current
remains constant throughout a series path, so it does not divide or multiply per device.
VERIFIED ✓
Q2: An alarm panel supervises a zone using a 4.7 kΩ End-of-Line (EOL) resistor on a 12
VDC circuit. What is the standby current through the EOL resistor?
A. 2.55 mA, calculated as 12 V ÷ 4,700 Ω [CORRECT]
B. 12.0 mA, because EOL resistors draw 1 mA per volt of panel voltage
C. 0.47 mA, calculated by dividing voltage by resistance in kilohms directly
D. 4.7 mA, matching the resistor value to current numerically
Correct Answer: A
Rationale: Ohm's Law states I = E ÷ R. For a 12 VDC panel supervising a 4.7 kΩ (4,700 Ω)
EOL resistor, current equals 12 ÷ 4,700 = 0.00255 A or 2.55 mA. This supervisory current
,allows the panel to detect opens and shorts by monitoring the voltage drop across the
EOL. VERIFIED ✓
Q3: A fire alarm circuit uses 18 AWG solid copper wire (0.0065 ohms per foot) to supply
power to a device 250 feet from the panel. The device draws 0.5 A. What is the
approximate voltage drop using the formula 2 × I × R × Length?
A. 0.81 V, calculated as 2 × 0.5 A × 0.0065 Ω/ft × 250 ft × 2 conductors
B. 1.63 V, calculated as 2 × 0.5 A × 0.0065 Ω/ft × 250 ft [CORRECT]
C. 3.25 V, using only the one-way distance without the multiplier for the return path
D. 0.33 V, forgetting to multiply by the 2-factor for both supply and return conductors
Correct Answer: B
Rationale: The voltage drop formula 2 × I × R × L accounts for both supply and return
conductors. Substituting values: 2 × 0.5 A × 0.0065 Ω/ft × 250 ft = 1.625 V
(approximately 1.63 V). NEC recommends limiting voltage drop to 3% for power-limited
fire alarm circuits to ensure reliable device operation. VERIFIED ✓
Q4: Two identical 2.2 kΩ EOL resistors are accidentally installed in parallel at the end of
a supervised alarm zone. What is the total resistance the panel sees?
A. 4.4 kΩ, incorrectly adding resistances as if wired in series
B. 2.2 kΩ, assuming the second resistor has no effect on total resistance
C. 1.1 kΩ, calculated using the parallel resistance formula (R1 × R2) ÷ (R1 + R2)
[CORRECT]
D. 0.55 kΩ, dividing the single resistor value by four instead of two
Correct Answer: C
Rationale: For parallel resistors, total resistance equals (R1 × R2) ÷ (R1 + R2). With two
2.2 kΩ resistors: (2,200 × 2,200) ÷ (2,200 + 2,200) = 1,100 Ω or 1.1 kΩ. The panel would
interpret this lower resistance as a fault condition since it expects the single 2.2 kΩ EOL
value. VERIFIED ✓
Q5: A strobe notification appliance is rated at 24 VDC and draws 0.1 A during alarm.
How much power does it consume?
A. 0.24 W, incorrectly dividing voltage by current
B. 2.4 W, calculated using the power formula P = E × I [CORRECT]
C. 240 W, multiplying voltage by current and adding a decimal error
D. 24 W, using only the voltage value as the power rating
,Correct Answer: B
Rationale: Electrical power is calculated as P = E × I (watts = volts × amps). For this
strobe: 24 V × 0.1 A = 2.4 W. Accurate power calculations are essential when sizing NAC
power supplies and secondary battery capacity to ensure the system meets NFPA 72
secondary power requirements. VERIFIED ✓
Q6: Three resistors (100 Ω, 200 Ω, and 300 Ω) are connected in series across a 24 VDC
fire alarm power supply. What is the voltage drop across the 200 Ω resistor?
A. 4.0 V, calculated by dividing 24 V by the number of resistors
B. 8.0 V, determined by finding the current (0.04 A) and multiplying by 200 Ω [CORRECT]
C. 12.0 V, using half the supply voltage regardless of resistance proportion
D. 16.0 V, incorrectly applying the voltage divider formula in reverse
Correct Answer: B
Rationale: Total series resistance equals 600 Ω. Circuit current is I = 24 V ÷ 600 Ω = 0.04
A. Voltage drop across the 200 Ω resistor is V = I × R = 0.04 A × 200 Ω = 8.0 V. This
demonstrates Kirchhoff's Voltage Law, where the sum of voltage drops equals the
source voltage. VERIFIED ✓
Q7: At a junction box in an alarm circuit, 2.0 A enters on one conductor and 3.0 A enters
on another. If 4.0 A leaves on one branch, what is the current on the remaining branch
per Kirchhoff's Current Law?
A. 1.0 A leaving the node, violating conservation of charge
B. 1.0 A entering the node, balancing total incoming and outgoing current [CORRECT]
C. 5.0 A leaving the node, adding all entering currents together
D. 9.0 A entering the node, incorrectly summing all values regardless of direction
Correct Answer: B
Rationale: Kirchhoff's Current Law states that the sum of currents entering a node
equals the sum leaving. Total entering current is 2.0 A + 3.0 A = 5.0 A. With 4.0 A
leaving, the remaining branch must carry 1.0 A entering the node to maintain balance
(5.0 A in = 5.0 A out). VERIFIED ✓
Q8: A step-down transformer has 500 primary turns and 100 secondary turns connected
to a 120 VAC source. What is the secondary voltage?
A. 600 V, incorrectly multiplying instead of dividing by the turns ratio
B. 24 V, calculated by the turns ratio (100 ÷ 500 × 120 V) [CORRECT]
, C. 120 V, assuming transformers do not change voltage levels
D. 12 V, using a 10:1 ratio instead of the actual 5:1 ratio
Correct Answer: B
Rationale: Transformer voltage ratio equals the turns ratio: V_secondary = (N_secondary
÷ N_primary) × V_primary. Here, (100 ÷ 500) × 120 V = 0.2 × 120 = 24 V. Step-down
transformers are the most common power supply component in alarm systems,
converting line voltage to low-voltage DC for panel and device operation. VERIFIED ✓
Q9: A fire alarm control panel requires 24 hours of standby power at 0.5 A and 15
minutes of alarm power at 2.0 A. What is the minimum battery capacity required?
A. 12.0 Ah, accounting only for standby current without alarm load
B. 12.5 Ah, calculated as (0.5 A × 24 h) + (2.0 A × 0.25 h) [CORRECT]
C. 15.0 Ah, using 24 hours at the alarm current draw
D. 48.0 Ah, multiplying standby current by alarm current incorrectly
Correct Answer: B
Rationale: Battery capacity equals the sum of standby and alarm loads: (0.5 A × 24 h) +
(2.0 A × 0.25 h) = 12 Ah + 0.5 Ah = 12.5 Ah. NFPA 72 requires secondary power to
support 24 hours of standby plus 5 minutes of alarm for non-voice systems (15 minutes
for voice systems). Commercial systems often use 30 Ah or larger batteries. VERIFIED
✓
Q10: An 18 AWG copper conductor has a resistance of approximately 6.5 ohms per
1,000 feet. What is the total resistance of a 500-foot loop (out and back) of this wire?
A. 3.25 Ω, using only the one-way length in the calculation
B. 6.5 Ω, calculated as (6.5 Ω/1,000 ft × 500 ft × 2 conductors) [CORRECT]
C. 13.0 Ω, forgetting to divide by 1,000 and using 6.5 directly
D. 1.63 Ω, dividing the correct answer by four incorrectly
Correct Answer: B
Rationale: Total conductor length for a 500-foot run includes both supply and return
paths (1,000 feet total). Resistance = (6.5 Ω ÷ 1,000 ft) × 1,000 ft = 6.5 Ω. Accurate wire
resistance calculations are critical for voltage drop analysis in NAC and power circuits
to ensure devices receive adequate operating voltage. VERIFIED ✓
& Burglar Alarm Complete Solutions | Pass Guaranteed -
A+ Graded
SECTION 1: ELECTRICAL FUNDAMENTALS & OHM'S LAW (15
Questions)
Q1: A Notification Appliance Circuit (NAC) operates at 24 VDC and powers two
horn-strobe devices wired in series, each with an internal resistance of 12 ohms. What
is the total current flowing through the circuit?
A. 0.5 A, because current divides equally in a series circuit regardless of total resistance
B. 2.0 A, calculated by dividing 24 V by the parallel equivalent of 12 ohms
C. 1.0 A, determined by Ohm's Law (I = E ÷ R) where total resistance equals 24 ohms
[CORRECT]
D. 4.0 A, because each device draws 2 A independently in series
Correct Answer: C
Rationale: In a series circuit, total resistance is the sum of individual resistances (12 Ω +
12 Ω = 24 Ω). Applying Ohm's Law (I = E ÷ R), current equals 24 V ÷ 24 Ω = 1.0 A. Current
remains constant throughout a series path, so it does not divide or multiply per device.
VERIFIED ✓
Q2: An alarm panel supervises a zone using a 4.7 kΩ End-of-Line (EOL) resistor on a 12
VDC circuit. What is the standby current through the EOL resistor?
A. 2.55 mA, calculated as 12 V ÷ 4,700 Ω [CORRECT]
B. 12.0 mA, because EOL resistors draw 1 mA per volt of panel voltage
C. 0.47 mA, calculated by dividing voltage by resistance in kilohms directly
D. 4.7 mA, matching the resistor value to current numerically
Correct Answer: A
Rationale: Ohm's Law states I = E ÷ R. For a 12 VDC panel supervising a 4.7 kΩ (4,700 Ω)
EOL resistor, current equals 12 ÷ 4,700 = 0.00255 A or 2.55 mA. This supervisory current
,allows the panel to detect opens and shorts by monitoring the voltage drop across the
EOL. VERIFIED ✓
Q3: A fire alarm circuit uses 18 AWG solid copper wire (0.0065 ohms per foot) to supply
power to a device 250 feet from the panel. The device draws 0.5 A. What is the
approximate voltage drop using the formula 2 × I × R × Length?
A. 0.81 V, calculated as 2 × 0.5 A × 0.0065 Ω/ft × 250 ft × 2 conductors
B. 1.63 V, calculated as 2 × 0.5 A × 0.0065 Ω/ft × 250 ft [CORRECT]
C. 3.25 V, using only the one-way distance without the multiplier for the return path
D. 0.33 V, forgetting to multiply by the 2-factor for both supply and return conductors
Correct Answer: B
Rationale: The voltage drop formula 2 × I × R × L accounts for both supply and return
conductors. Substituting values: 2 × 0.5 A × 0.0065 Ω/ft × 250 ft = 1.625 V
(approximately 1.63 V). NEC recommends limiting voltage drop to 3% for power-limited
fire alarm circuits to ensure reliable device operation. VERIFIED ✓
Q4: Two identical 2.2 kΩ EOL resistors are accidentally installed in parallel at the end of
a supervised alarm zone. What is the total resistance the panel sees?
A. 4.4 kΩ, incorrectly adding resistances as if wired in series
B. 2.2 kΩ, assuming the second resistor has no effect on total resistance
C. 1.1 kΩ, calculated using the parallel resistance formula (R1 × R2) ÷ (R1 + R2)
[CORRECT]
D. 0.55 kΩ, dividing the single resistor value by four instead of two
Correct Answer: C
Rationale: For parallel resistors, total resistance equals (R1 × R2) ÷ (R1 + R2). With two
2.2 kΩ resistors: (2,200 × 2,200) ÷ (2,200 + 2,200) = 1,100 Ω or 1.1 kΩ. The panel would
interpret this lower resistance as a fault condition since it expects the single 2.2 kΩ EOL
value. VERIFIED ✓
Q5: A strobe notification appliance is rated at 24 VDC and draws 0.1 A during alarm.
How much power does it consume?
A. 0.24 W, incorrectly dividing voltage by current
B. 2.4 W, calculated using the power formula P = E × I [CORRECT]
C. 240 W, multiplying voltage by current and adding a decimal error
D. 24 W, using only the voltage value as the power rating
,Correct Answer: B
Rationale: Electrical power is calculated as P = E × I (watts = volts × amps). For this
strobe: 24 V × 0.1 A = 2.4 W. Accurate power calculations are essential when sizing NAC
power supplies and secondary battery capacity to ensure the system meets NFPA 72
secondary power requirements. VERIFIED ✓
Q6: Three resistors (100 Ω, 200 Ω, and 300 Ω) are connected in series across a 24 VDC
fire alarm power supply. What is the voltage drop across the 200 Ω resistor?
A. 4.0 V, calculated by dividing 24 V by the number of resistors
B. 8.0 V, determined by finding the current (0.04 A) and multiplying by 200 Ω [CORRECT]
C. 12.0 V, using half the supply voltage regardless of resistance proportion
D. 16.0 V, incorrectly applying the voltage divider formula in reverse
Correct Answer: B
Rationale: Total series resistance equals 600 Ω. Circuit current is I = 24 V ÷ 600 Ω = 0.04
A. Voltage drop across the 200 Ω resistor is V = I × R = 0.04 A × 200 Ω = 8.0 V. This
demonstrates Kirchhoff's Voltage Law, where the sum of voltage drops equals the
source voltage. VERIFIED ✓
Q7: At a junction box in an alarm circuit, 2.0 A enters on one conductor and 3.0 A enters
on another. If 4.0 A leaves on one branch, what is the current on the remaining branch
per Kirchhoff's Current Law?
A. 1.0 A leaving the node, violating conservation of charge
B. 1.0 A entering the node, balancing total incoming and outgoing current [CORRECT]
C. 5.0 A leaving the node, adding all entering currents together
D. 9.0 A entering the node, incorrectly summing all values regardless of direction
Correct Answer: B
Rationale: Kirchhoff's Current Law states that the sum of currents entering a node
equals the sum leaving. Total entering current is 2.0 A + 3.0 A = 5.0 A. With 4.0 A
leaving, the remaining branch must carry 1.0 A entering the node to maintain balance
(5.0 A in = 5.0 A out). VERIFIED ✓
Q8: A step-down transformer has 500 primary turns and 100 secondary turns connected
to a 120 VAC source. What is the secondary voltage?
A. 600 V, incorrectly multiplying instead of dividing by the turns ratio
B. 24 V, calculated by the turns ratio (100 ÷ 500 × 120 V) [CORRECT]
, C. 120 V, assuming transformers do not change voltage levels
D. 12 V, using a 10:1 ratio instead of the actual 5:1 ratio
Correct Answer: B
Rationale: Transformer voltage ratio equals the turns ratio: V_secondary = (N_secondary
÷ N_primary) × V_primary. Here, (100 ÷ 500) × 120 V = 0.2 × 120 = 24 V. Step-down
transformers are the most common power supply component in alarm systems,
converting line voltage to low-voltage DC for panel and device operation. VERIFIED ✓
Q9: A fire alarm control panel requires 24 hours of standby power at 0.5 A and 15
minutes of alarm power at 2.0 A. What is the minimum battery capacity required?
A. 12.0 Ah, accounting only for standby current without alarm load
B. 12.5 Ah, calculated as (0.5 A × 24 h) + (2.0 A × 0.25 h) [CORRECT]
C. 15.0 Ah, using 24 hours at the alarm current draw
D. 48.0 Ah, multiplying standby current by alarm current incorrectly
Correct Answer: B
Rationale: Battery capacity equals the sum of standby and alarm loads: (0.5 A × 24 h) +
(2.0 A × 0.25 h) = 12 Ah + 0.5 Ah = 12.5 Ah. NFPA 72 requires secondary power to
support 24 hours of standby plus 5 minutes of alarm for non-voice systems (15 minutes
for voice systems). Commercial systems often use 30 Ah or larger batteries. VERIFIED
✓
Q10: An 18 AWG copper conductor has a resistance of approximately 6.5 ohms per
1,000 feet. What is the total resistance of a 500-foot loop (out and back) of this wire?
A. 3.25 Ω, using only the one-way length in the calculation
B. 6.5 Ω, calculated as (6.5 Ω/1,000 ft × 500 ft × 2 conductors) [CORRECT]
C. 13.0 Ω, forgetting to divide by 1,000 and using 6.5 directly
D. 1.63 Ω, dividing the correct answer by four incorrectly
Correct Answer: B
Rationale: Total conductor length for a 500-foot run includes both supply and return
paths (1,000 feet total). Resistance = (6.5 Ω ÷ 1,000 ft) × 1,000 ft = 6.5 Ω. Accurate wire
resistance calculations are critical for voltage drop analysis in NAC and power circuits
to ensure devices receive adequate operating voltage. VERIFIED ✓
