VEHICLE COMMUNICATION SYSTEM (LATEST)
QUESTIONS AND ANSWERS 100% CORRECT
Q1: On a high-speed CAN bus with the ignition on and no faults present, what approximate
voltages should be measured on CAN-H and CAN-L during dominant bit transmission?
A. CAN-H 2.5 V, CAN-L 2.5 V
B. CAN-H 5.0 V, CAN-L 0 V
C. CAN-H 3.5 V, CAN-L 1.5 V [CORRECT]
D. CAN-H 12 V, CAN-L 0 V
Correct Answer: C
Rationale: During a dominant bit the CAN transceiver actively drives CAN-H toward 3.5 V and
CAN-L toward 1.5 V, producing a differential of approximately 2 V. Recessive bits float both
lines near the 2.5 V common-mode voltage; 5 V/0 V or 12 V levels are characteristic of other
protocols or faults.
Q2: What is the expected total resistance measured across the DLC pins 6 and 14 on a
properly terminated high-speed CAN bus with the battery disconnected?
A. 120 ohms
B. 0 ohms
C. 60 ohms [CORRECT]
D. 240 ohms
Correct Answer: C
Rationale: Two 120-ohm terminating resistors are placed in parallel at the ends of the bus,
yielding a total resistance of 60 ohms. A reading of 120 ohms indicates one resistor is open; 0
ohms indicates a short.
Q3: Which statement correctly describes a recessive bit on a CAN bus?
A. CAN-H is driven high and CAN-L is driven low
B. Both CAN-H and CAN-L are actively driven to opposite voltages
C. Both lines rest near the common-mode voltage of approximately 2.5 V [CORRECT]
D. The differential voltage is approximately 2 V
,Correct Answer: C
Rationale: In the recessive state the transmitters are off and the bus is passively biased to ~2.5
V on both lines, producing near-zero differential voltage. Dominant bits produce the 2 V
differential.
Q4: A technician measures 236 ohms between CAN-H and CAN-L at the DLC with the battery
disconnected. What does this indicate?
A. Both terminating resistors are good
B. The bus is shorted to ground
C. One terminating resistor is open [CORRECT]
D. The network is operating normally
Correct Answer: C
Rationale: With one 120-ohm resistor open, only the remaining 120-ohm resistor is in the
circuit; wiring resistance and meter accuracy can produce a reading near 236 ohms. Normal is
60 ohms.
Q5: What is the standard data rate for a high-speed CAN network used for powertrain
communication?
A. 10 kbps
B. 125 kbps
C. 500 kbps [CORRECT]
D. 10 Mbps
Correct Answer: C
Rationale: Classical high-speed CAN (CAN-C) operates at 500 kbps. Lower rates are used for
low-speed/body CAN; 10 Mbps is typical of Automotive Ethernet.
Q6: During arbitration on a CAN bus, which message wins the bus?
A. The message with the highest identifier
B. The message with the longest data field
C. The message with the lowest identifier [CORRECT]
D. The message transmitted by the module closest to the DLC
,Correct Answer: C
Rationale: CAN uses non-destructive bitwise arbitration; a dominant (0) bit overwrites a
recessive (1) bit. The lowest identifier therefore remains dominant and wins arbitration.
Q7: What is the primary purpose of the two 120-ohm terminating resistors on a high-speed
CAN bus?
A. To limit current flow into the modules
B. To provide a voltage reference for the transceivers
C. To prevent signal reflections and stabilize the bus [CORRECT]
D. To allow the bus to operate at 12 V
Correct Answer: C
Rationale: The resistors match the characteristic impedance of the twisted-pair cable,
absorbing energy at the ends of the bus and preventing reflections that would corrupt the
differential signal.
Q8: On a CAN bus that is active and free of faults, approximate resting (recessive) voltages
measured with a digital multimeter are:
A. CAN-H 0 V, CAN-L 5 V
B. CAN-H 3.5 V, CAN-L 1.5 V
C. CAN-H and CAN-L both near 2.5 V [CORRECT]
D. CAN-H 12 V, CAN-L 0 V
Correct Answer: C
Rationale: With no dominant bits being transmitted, both lines float to the common-mode
bias voltage of approximately 2.5 V. Dominant voltages appear only while a module is actively
transmitting.
Q9: Which of the following is a key difference between classical CAN and CAN-FD?
A. CAN-FD uses a single wire
B. Classical CAN supports up to 64 data bytes
C. CAN-FD allows flexible data rates and up to 64 data bytes [CORRECT]
D. CAN-FD eliminates the need for terminating resistors
, Correct Answer: C
Rationale: CAN-FD retains the same arbitration phase but permits a higher bit rate and up to
64 bytes in the data phase, increasing throughput while remaining compatible with classical
CAN physical layers.
Q10: A technician observes CAN-H at 3.2 V and CAN-L at 1.8 V while the bus is active. This
indicates:
A. A short between CAN-H and CAN-L
B. An open terminating resistor
C. Normal dominant/recessive activity on a high-speed CAN bus [CORRECT]
D. A short of CAN-H to battery voltage
Correct Answer: C
Rationale: Average voltages of approximately 3.2 V on CAN-H and 1.8 V on CAN-L are typical
when the bus is actively communicating (mix of dominant and recessive bits). Faults produce
significantly different readings.
Q11: What voltage levels are characteristic of a low-speed (fault-tolerant) CAN-B network
when active and healthy?
A. CAN-H ≈ 3.5 V, CAN-L ≈ 1.5 V
B. CAN-H ≈ 4.8 V, CAN-L ≈ 0.4 V [CORRECT]
C. Both lines ≈ 2.5 V
D. CAN-H ≈ 12 V, CAN-L ≈ 0 V
Correct Answer: B
Rationale: Low-speed CAN uses different common-mode levels; dominant states drive CAN-H
near battery voltage and CAN-L near ground, producing the approximate 4.8 V / 0.4 V
readings when averaged.
Q12: Message arbitration on CAN is possible because:
A. Modules transmit on different frequencies
B. Dominant bits override recessive bits [CORRECT]
C. Each module has a unique clock
QUESTIONS AND ANSWERS 100% CORRECT
Q1: On a high-speed CAN bus with the ignition on and no faults present, what approximate
voltages should be measured on CAN-H and CAN-L during dominant bit transmission?
A. CAN-H 2.5 V, CAN-L 2.5 V
B. CAN-H 5.0 V, CAN-L 0 V
C. CAN-H 3.5 V, CAN-L 1.5 V [CORRECT]
D. CAN-H 12 V, CAN-L 0 V
Correct Answer: C
Rationale: During a dominant bit the CAN transceiver actively drives CAN-H toward 3.5 V and
CAN-L toward 1.5 V, producing a differential of approximately 2 V. Recessive bits float both
lines near the 2.5 V common-mode voltage; 5 V/0 V or 12 V levels are characteristic of other
protocols or faults.
Q2: What is the expected total resistance measured across the DLC pins 6 and 14 on a
properly terminated high-speed CAN bus with the battery disconnected?
A. 120 ohms
B. 0 ohms
C. 60 ohms [CORRECT]
D. 240 ohms
Correct Answer: C
Rationale: Two 120-ohm terminating resistors are placed in parallel at the ends of the bus,
yielding a total resistance of 60 ohms. A reading of 120 ohms indicates one resistor is open; 0
ohms indicates a short.
Q3: Which statement correctly describes a recessive bit on a CAN bus?
A. CAN-H is driven high and CAN-L is driven low
B. Both CAN-H and CAN-L are actively driven to opposite voltages
C. Both lines rest near the common-mode voltage of approximately 2.5 V [CORRECT]
D. The differential voltage is approximately 2 V
,Correct Answer: C
Rationale: In the recessive state the transmitters are off and the bus is passively biased to ~2.5
V on both lines, producing near-zero differential voltage. Dominant bits produce the 2 V
differential.
Q4: A technician measures 236 ohms between CAN-H and CAN-L at the DLC with the battery
disconnected. What does this indicate?
A. Both terminating resistors are good
B. The bus is shorted to ground
C. One terminating resistor is open [CORRECT]
D. The network is operating normally
Correct Answer: C
Rationale: With one 120-ohm resistor open, only the remaining 120-ohm resistor is in the
circuit; wiring resistance and meter accuracy can produce a reading near 236 ohms. Normal is
60 ohms.
Q5: What is the standard data rate for a high-speed CAN network used for powertrain
communication?
A. 10 kbps
B. 125 kbps
C. 500 kbps [CORRECT]
D. 10 Mbps
Correct Answer: C
Rationale: Classical high-speed CAN (CAN-C) operates at 500 kbps. Lower rates are used for
low-speed/body CAN; 10 Mbps is typical of Automotive Ethernet.
Q6: During arbitration on a CAN bus, which message wins the bus?
A. The message with the highest identifier
B. The message with the longest data field
C. The message with the lowest identifier [CORRECT]
D. The message transmitted by the module closest to the DLC
,Correct Answer: C
Rationale: CAN uses non-destructive bitwise arbitration; a dominant (0) bit overwrites a
recessive (1) bit. The lowest identifier therefore remains dominant and wins arbitration.
Q7: What is the primary purpose of the two 120-ohm terminating resistors on a high-speed
CAN bus?
A. To limit current flow into the modules
B. To provide a voltage reference for the transceivers
C. To prevent signal reflections and stabilize the bus [CORRECT]
D. To allow the bus to operate at 12 V
Correct Answer: C
Rationale: The resistors match the characteristic impedance of the twisted-pair cable,
absorbing energy at the ends of the bus and preventing reflections that would corrupt the
differential signal.
Q8: On a CAN bus that is active and free of faults, approximate resting (recessive) voltages
measured with a digital multimeter are:
A. CAN-H 0 V, CAN-L 5 V
B. CAN-H 3.5 V, CAN-L 1.5 V
C. CAN-H and CAN-L both near 2.5 V [CORRECT]
D. CAN-H 12 V, CAN-L 0 V
Correct Answer: C
Rationale: With no dominant bits being transmitted, both lines float to the common-mode
bias voltage of approximately 2.5 V. Dominant voltages appear only while a module is actively
transmitting.
Q9: Which of the following is a key difference between classical CAN and CAN-FD?
A. CAN-FD uses a single wire
B. Classical CAN supports up to 64 data bytes
C. CAN-FD allows flexible data rates and up to 64 data bytes [CORRECT]
D. CAN-FD eliminates the need for terminating resistors
, Correct Answer: C
Rationale: CAN-FD retains the same arbitration phase but permits a higher bit rate and up to
64 bytes in the data phase, increasing throughput while remaining compatible with classical
CAN physical layers.
Q10: A technician observes CAN-H at 3.2 V and CAN-L at 1.8 V while the bus is active. This
indicates:
A. A short between CAN-H and CAN-L
B. An open terminating resistor
C. Normal dominant/recessive activity on a high-speed CAN bus [CORRECT]
D. A short of CAN-H to battery voltage
Correct Answer: C
Rationale: Average voltages of approximately 3.2 V on CAN-H and 1.8 V on CAN-L are typical
when the bus is actively communicating (mix of dominant and recessive bits). Faults produce
significantly different readings.
Q11: What voltage levels are characteristic of a low-speed (fault-tolerant) CAN-B network
when active and healthy?
A. CAN-H ≈ 3.5 V, CAN-L ≈ 1.5 V
B. CAN-H ≈ 4.8 V, CAN-L ≈ 0.4 V [CORRECT]
C. Both lines ≈ 2.5 V
D. CAN-H ≈ 12 V, CAN-L ≈ 0 V
Correct Answer: B
Rationale: Low-speed CAN uses different common-mode levels; dominant states drive CAN-H
near battery voltage and CAN-L near ground, producing the approximate 4.8 V / 0.4 V
readings when averaged.
Q12: Message arbitration on CAN is possible because:
A. Modules transmit on different frequencies
B. Dominant bits override recessive bits [CORRECT]
C. Each module has a unique clock
