SCTE - BROADBAND TELECOMCENTER
SPECIALIST (BTCS) - FACILITIES SECTION
EXAM QUESTIONS AND ANSWERS GRADED A+
2026
1. A telecom center currently operates on a -48 VDC power plant. Due to a load increase,
the maximum current on a 100-foot run will rise from 100 A to 150 A. Using a voltage drop
limit of 2% (0.96 V) and copper conductors with resistivity 10.4 ohms-cmil/ft, what is the
minimum conductor size (in kcmil) required?
A. 250 kcmil
B. 300 kcmil
C. 350 kcmil
D. 400 kcmil
Answer: C
Rationale: Voltage drop Vd = (2 * L * I * R)/1000, where R = resistivity / cmil. Solving for cmil =
(2*L*I*resistivity)/Vd = (2*100*150*10.4)/0.96 = 325,000 cmil 325 kcmil. The next standard
size is 350 kcmil. Options A and B are too small; D is oversized though acceptable but not
minimal.
2. According to TIA-607-B, what is the minimum cross-sectional area of the bonding
conductor between two equipment racks if the total circuit length is 75 feet and the
available fault current is 10 kA?
A. 6 AWG (13.3 mm²)
B. 4 AWG (21.2 mm²)
C. 2 AWG (33.6 mm²)
D. 1/0 AWG (53.5 mm²)
Answer: B
Rationale: TIA-607-B Table 4 requires a minimum bonding conductor size based on the longest
run between racks. For a run of 75 ft and fault current up to 10 kA, the table specifies 4 AWG
(21.2 mm²). Option A is undersized; options C and D exceed the minimum requirement but are
not necessary.
3. You are evaluating fire suppression options for a data center containing sensitive
electronics. For each agent below, determine if it is recommended (Anticipated), not
recommended (Contraindicated), or requires special conditions (Conditional).
Page 1
,A. Correct as shown
B. Incorrect
C. Need re-evaluation
D. Not applicable
Answer: A
Rationale: FM-200 and Novec 1230 are clean agents suitable for electronics (Anticipated). CO ‚
is safe for electronics but poses asphyxiation risk, requiring pre-discharge alarms and delays
(Conditional). Water mist is generally safe for live electronics if deionized, but may cause
shorting (Contraindicated due to risk). Dry chemical leaves corrosive residue, damaging
electronics (Contraindicated).
4. A data center implements hot aisle containment. During commissioning, sensors show
that temperatures in the cold aisle exceed the ASHRAE recommended maximum of 27°C.
What is the most probable cause?
A. Insufficient perforated tile open area
B. Recirculation of hot aisle air through gaps in containment
C. CRAH unit setpoint too high
D. Humidistat setpoint too low
Answer: B
Rationale: Hot aisle containment recirculation occurs when gaps allow hot air to bypass the
return path and mix with cold air, raising cold aisle temperatures. Option A reduces airflow but
does not directly cause mixing. Option C would raise supply temperature, not cold aisle
specifically. Option D affects humidity, not temperature.
5. When installing single-mode fiber optic cable (OS2) and RG-6 coaxial cable in the same
overhead cable tray, what is the primary concern regarding minimum bend radius?
A. Fiber has a stricter bend radius (10 mm) than coax (25 mm), so fiber must be handled more
carefully
B. Coax has a stricter bend radius (10 mm) than fiber (25 mm), so coax must be handled more
carefully
C. Both have identical bend radius limits (20 mm) and can be installed with identical care
D. Bend radius is irrelevant for fiber as it is immune to bending losses
Answer: A
Rationale: Single-mode fiber (OS2) typically has a minimum bend radius of about 10 mm under
tension and 30 mm when installed, while RG-6 coax has a minimum bend radius of about 25 mm
(1 inch) due to dielectric deformation risk. Fiber is more sensitive to bending than coax,
contrary to option B. Option C is false; option D is false because fiber suffers micro-bend losses.
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, 6. A telecom site uses VRLA batteries at 25°C ambient. The manufacturer specifies a float
voltage of 2.27 V/cell at 20°C with a temperature compensation coefficient of -3 mV/°C/cell.
If the actual ambient temperature is 30°C, what float voltage per cell should be applied?
A. 2.27 V
B. 2.24 V
C. 2.30 V
D. 2.21 V
Answer: B
Rationale: Temperature compensation: ”T = 30°C - 20°C = 10°C. Compensation = -3 mV/°C/cell
* 10°C = -30 mV = -0.03 V. Adjusted voltage = 2.27 V - 0.03 V = 2.24 V/cell. Option A ignores
compensation. Option C adds instead of subtracts; option D overcompensates by using 20°C
offset instead of 10°C.
7. A technician suspects an impedance discontinuity in a 50-ohm coaxial trunk line. Using a
TDR, a reflection is observed at 500 feet. The cable has a velocity of propagation (Vp) of
0.85c. What is the round-trip time delay measured by the TDR? (c = 3×10 m/s)
A. 0.60 s
B. 1.20 s
C. 2.40 s
D. 0.30 s
Answer: B
Rationale: Velocity of signal in cable v = Vp * c = 0.85 * 3×10 x m/s = 2.55×10 x m/s. Distance to
fault = 500 ft = 152.4 m. Time one-way = distance/v = 152.4 / (2.55×10) 0.598 s. Round-trip =
2 * 0.598 1.20 s. Option A is one-way, option C double (if used feet instead of meters), option D
uses full speed c.
8. A technician is performing lockout/tagout (LOTO) on a 480 VAC power distribution unit
(PDU) serving multiple racks. After de-energizing the breaker and applying a lock and tag,
what is the NEXT required step before beginning work?
A. Verify zero voltage using a multimeter at the PDU output
B. Activate the emergency power off (EPO) button
C. Notify the network operations center (NOC) of the lock
D. Remove all fuses from backup battery circuits
Answer: A
Rationale: OSHA standard 29 CFR 1910.147 requires verification of zero energy state after
lockout by testing for absence of voltage. Option B (EPO) is redundant and not part of LOTO.
Option C is administrative but not the immediate next step. Option D is unnecessary unless
batteries are part of the PDU, but voltage verification is still primary.
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SPECIALIST (BTCS) - FACILITIES SECTION
EXAM QUESTIONS AND ANSWERS GRADED A+
2026
1. A telecom center currently operates on a -48 VDC power plant. Due to a load increase,
the maximum current on a 100-foot run will rise from 100 A to 150 A. Using a voltage drop
limit of 2% (0.96 V) and copper conductors with resistivity 10.4 ohms-cmil/ft, what is the
minimum conductor size (in kcmil) required?
A. 250 kcmil
B. 300 kcmil
C. 350 kcmil
D. 400 kcmil
Answer: C
Rationale: Voltage drop Vd = (2 * L * I * R)/1000, where R = resistivity / cmil. Solving for cmil =
(2*L*I*resistivity)/Vd = (2*100*150*10.4)/0.96 = 325,000 cmil 325 kcmil. The next standard
size is 350 kcmil. Options A and B are too small; D is oversized though acceptable but not
minimal.
2. According to TIA-607-B, what is the minimum cross-sectional area of the bonding
conductor between two equipment racks if the total circuit length is 75 feet and the
available fault current is 10 kA?
A. 6 AWG (13.3 mm²)
B. 4 AWG (21.2 mm²)
C. 2 AWG (33.6 mm²)
D. 1/0 AWG (53.5 mm²)
Answer: B
Rationale: TIA-607-B Table 4 requires a minimum bonding conductor size based on the longest
run between racks. For a run of 75 ft and fault current up to 10 kA, the table specifies 4 AWG
(21.2 mm²). Option A is undersized; options C and D exceed the minimum requirement but are
not necessary.
3. You are evaluating fire suppression options for a data center containing sensitive
electronics. For each agent below, determine if it is recommended (Anticipated), not
recommended (Contraindicated), or requires special conditions (Conditional).
Page 1
,A. Correct as shown
B. Incorrect
C. Need re-evaluation
D. Not applicable
Answer: A
Rationale: FM-200 and Novec 1230 are clean agents suitable for electronics (Anticipated). CO ‚
is safe for electronics but poses asphyxiation risk, requiring pre-discharge alarms and delays
(Conditional). Water mist is generally safe for live electronics if deionized, but may cause
shorting (Contraindicated due to risk). Dry chemical leaves corrosive residue, damaging
electronics (Contraindicated).
4. A data center implements hot aisle containment. During commissioning, sensors show
that temperatures in the cold aisle exceed the ASHRAE recommended maximum of 27°C.
What is the most probable cause?
A. Insufficient perforated tile open area
B. Recirculation of hot aisle air through gaps in containment
C. CRAH unit setpoint too high
D. Humidistat setpoint too low
Answer: B
Rationale: Hot aisle containment recirculation occurs when gaps allow hot air to bypass the
return path and mix with cold air, raising cold aisle temperatures. Option A reduces airflow but
does not directly cause mixing. Option C would raise supply temperature, not cold aisle
specifically. Option D affects humidity, not temperature.
5. When installing single-mode fiber optic cable (OS2) and RG-6 coaxial cable in the same
overhead cable tray, what is the primary concern regarding minimum bend radius?
A. Fiber has a stricter bend radius (10 mm) than coax (25 mm), so fiber must be handled more
carefully
B. Coax has a stricter bend radius (10 mm) than fiber (25 mm), so coax must be handled more
carefully
C. Both have identical bend radius limits (20 mm) and can be installed with identical care
D. Bend radius is irrelevant for fiber as it is immune to bending losses
Answer: A
Rationale: Single-mode fiber (OS2) typically has a minimum bend radius of about 10 mm under
tension and 30 mm when installed, while RG-6 coax has a minimum bend radius of about 25 mm
(1 inch) due to dielectric deformation risk. Fiber is more sensitive to bending than coax,
contrary to option B. Option C is false; option D is false because fiber suffers micro-bend losses.
Page 2
, 6. A telecom site uses VRLA batteries at 25°C ambient. The manufacturer specifies a float
voltage of 2.27 V/cell at 20°C with a temperature compensation coefficient of -3 mV/°C/cell.
If the actual ambient temperature is 30°C, what float voltage per cell should be applied?
A. 2.27 V
B. 2.24 V
C. 2.30 V
D. 2.21 V
Answer: B
Rationale: Temperature compensation: ”T = 30°C - 20°C = 10°C. Compensation = -3 mV/°C/cell
* 10°C = -30 mV = -0.03 V. Adjusted voltage = 2.27 V - 0.03 V = 2.24 V/cell. Option A ignores
compensation. Option C adds instead of subtracts; option D overcompensates by using 20°C
offset instead of 10°C.
7. A technician suspects an impedance discontinuity in a 50-ohm coaxial trunk line. Using a
TDR, a reflection is observed at 500 feet. The cable has a velocity of propagation (Vp) of
0.85c. What is the round-trip time delay measured by the TDR? (c = 3×10 m/s)
A. 0.60 s
B. 1.20 s
C. 2.40 s
D. 0.30 s
Answer: B
Rationale: Velocity of signal in cable v = Vp * c = 0.85 * 3×10 x m/s = 2.55×10 x m/s. Distance to
fault = 500 ft = 152.4 m. Time one-way = distance/v = 152.4 / (2.55×10) 0.598 s. Round-trip =
2 * 0.598 1.20 s. Option A is one-way, option C double (if used feet instead of meters), option D
uses full speed c.
8. A technician is performing lockout/tagout (LOTO) on a 480 VAC power distribution unit
(PDU) serving multiple racks. After de-energizing the breaker and applying a lock and tag,
what is the NEXT required step before beginning work?
A. Verify zero voltage using a multimeter at the PDU output
B. Activate the emergency power off (EPO) button
C. Notify the network operations center (NOC) of the lock
D. Remove all fuses from backup battery circuits
Answer: A
Rationale: OSHA standard 29 CFR 1910.147 requires verification of zero energy state after
lockout by testing for absence of voltage. Option B (EPO) is redundant and not part of LOTO.
Option C is administrative but not the immediate next step. Option D is unnecessary unless
batteries are part of the PDU, but voltage verification is still primary.
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