Vehicle Communication System Certification Exam – 2026/2027 Complete Practice Question Bank

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