VEHICLE COMMUNICATION SYSTEM
(LATEST) QUESTIONS AND ANSWERS
100% CORRECT
Automotive Network & Diagnostic Communication
Comprehensive Practice Review
50 Questions | 2026/2027 Aligned
Aligned with SAE J1939/J2534, ISO 11898/17987, IEEE 802.3bw/802.1AS,
ISO 14229 UDS & ISO/SAE 21434 Cybersecurity Standards
May 2026
, Vehicle Communication System Exam Review | 2026/2027 Standards
Table of Contents
CAN Bus Fundamentals: Architecture, Message Structure & Q1 -
Section 1
Arbitration Logic Q10
Secondary Network Protocols: LIN, FlexRay, MOST & Q11 -
Section 2
Application-Specific Use Cases Q20
Automotive Ethernet, Gateway Modules & Network Topology Q21 -
Section 3
Integration Q30
Diagnostic Communication: UDS, OBD-II, J1939 & Scan Tool Q31 -
Section 4
Protocols Q40
Cybersecurity, OTA Updates, ADAS Communication & Q41 -
Section 5
Troubleshooting Q50
Standards Alignment: SAE J1939/J2534 | ISO 11898 (CAN) / ISO 17987 (LIN) / ISO 17458 (FlexRay) | IEEE 802.3bw (100BASE-T1) /
802.3bp (1000BASE-T1) / 802.1AS (TSN gPTP) | ISO 14229 (UDS) / ISO 15765 (ISO-TP) / ISO 13400 (DoIP) | ISO/SAE 21434 (Cybersecurity)
| AUTOSAR SecOC
Page 2
, Vehicle Communication System Exam Review | 2026/2027 Standards
Section 1: CAN Bus Fundamentals: Architecture, Message Structure &
Arbitration Logic
Q1-Q10
Q1: A technician measures 2.5V on CAN_H and 2.5V on CAN_L with the ignition on but the engine off.
The oscilloscope shows a flat line on both traces. What is the most likely cause?
A. Normal recessive bus state with no active communication
B. CAN_H shorted to battery voltage through a faulty ECU
C. Both CAN lines are shorted together at a connector, preventing differential signaling [CORRECT]
D. The 120 ohm termination resistor has failed open
Correct Answer: C
Rationale: When both CAN lines are shorted together, no voltage differential exists, resulting in identical 2.5V readings and a
flat oscilloscope trace. A normal recessive state shows a flat line at 2.5V differential but would show activity when
communication occurs; shorted-to-battery would show ~12V on CAN_H; open termination would cause signal reflection, not
identical voltages.
Q2: Two ECUs on the same CAN bus attempt to transmit simultaneously. ECU-A sends an 11-bit
identifier of 0x0A3, and ECU-B sends 0x1F4. Which statement correctly describes the arbitration
outcome?
A. ECU-B wins arbitration because its identifier has a higher numeric value and higher priority
B. ECU-A wins arbitration because it transmits more dominant bits at the beginning of the identifier field
[CORRECT]
C. Neither ECU wins; the bus enters a bus-off state due to simultaneous transmission
D. The gateway module intervenes and assigns transmission priority based on network load
Correct Answer: B
Rationale: CAN arbitration is bitwise: the identifier with more leading dominant (logic 0) bits wins. 0x0A3 (000 1010 0011) has
a dominant bit in position 7 while 0x1F4 (001 1111 0100) has a recessive bit, so ECU-A wins. Lower numeric values in
standard 11-bit CAN have higher priority. Bus-off occurs only after repeated errors, not during normal arbitration.
Q3: During CAN bus troubleshooting, a technician discovers that removing either termination resistor
results in normal communication, but with both resistors installed, the oscilloscope shows severely
distorted waveforms. What is the fault?
A. The bus is overloaded with excessive message traffic
B. There is an extra (third) termination resistor somewhere on the network, creating 80 ohms total instead
of 60 ohms [CORRECT]
C. One ECU is transmitting extended 29-bit frames on a network configured for standard 11-bit only
D. The CAN transceivers in all ECUs have incompatible differential voltage thresholds
Correct Answer: B
Rationale: A properly terminated CAN bus measures approximately 60 ohms (two 120-ohm resistors in parallel). An extra
termination resistor creates approximately 40 ohms, distorting signals. Removing one of the two correct resistors restores
near-normal impedance (~80 ohms with the extra), allowing marginal communication. Bus overload, frame format mismatch,
and transceiver incompatibility produce different symptom patterns.
Page 3
(LATEST) QUESTIONS AND ANSWERS
100% CORRECT
Automotive Network & Diagnostic Communication
Comprehensive Practice Review
50 Questions | 2026/2027 Aligned
Aligned with SAE J1939/J2534, ISO 11898/17987, IEEE 802.3bw/802.1AS,
ISO 14229 UDS & ISO/SAE 21434 Cybersecurity Standards
May 2026
, Vehicle Communication System Exam Review | 2026/2027 Standards
Table of Contents
CAN Bus Fundamentals: Architecture, Message Structure & Q1 -
Section 1
Arbitration Logic Q10
Secondary Network Protocols: LIN, FlexRay, MOST & Q11 -
Section 2
Application-Specific Use Cases Q20
Automotive Ethernet, Gateway Modules & Network Topology Q21 -
Section 3
Integration Q30
Diagnostic Communication: UDS, OBD-II, J1939 & Scan Tool Q31 -
Section 4
Protocols Q40
Cybersecurity, OTA Updates, ADAS Communication & Q41 -
Section 5
Troubleshooting Q50
Standards Alignment: SAE J1939/J2534 | ISO 11898 (CAN) / ISO 17987 (LIN) / ISO 17458 (FlexRay) | IEEE 802.3bw (100BASE-T1) /
802.3bp (1000BASE-T1) / 802.1AS (TSN gPTP) | ISO 14229 (UDS) / ISO 15765 (ISO-TP) / ISO 13400 (DoIP) | ISO/SAE 21434 (Cybersecurity)
| AUTOSAR SecOC
Page 2
, Vehicle Communication System Exam Review | 2026/2027 Standards
Section 1: CAN Bus Fundamentals: Architecture, Message Structure &
Arbitration Logic
Q1-Q10
Q1: A technician measures 2.5V on CAN_H and 2.5V on CAN_L with the ignition on but the engine off.
The oscilloscope shows a flat line on both traces. What is the most likely cause?
A. Normal recessive bus state with no active communication
B. CAN_H shorted to battery voltage through a faulty ECU
C. Both CAN lines are shorted together at a connector, preventing differential signaling [CORRECT]
D. The 120 ohm termination resistor has failed open
Correct Answer: C
Rationale: When both CAN lines are shorted together, no voltage differential exists, resulting in identical 2.5V readings and a
flat oscilloscope trace. A normal recessive state shows a flat line at 2.5V differential but would show activity when
communication occurs; shorted-to-battery would show ~12V on CAN_H; open termination would cause signal reflection, not
identical voltages.
Q2: Two ECUs on the same CAN bus attempt to transmit simultaneously. ECU-A sends an 11-bit
identifier of 0x0A3, and ECU-B sends 0x1F4. Which statement correctly describes the arbitration
outcome?
A. ECU-B wins arbitration because its identifier has a higher numeric value and higher priority
B. ECU-A wins arbitration because it transmits more dominant bits at the beginning of the identifier field
[CORRECT]
C. Neither ECU wins; the bus enters a bus-off state due to simultaneous transmission
D. The gateway module intervenes and assigns transmission priority based on network load
Correct Answer: B
Rationale: CAN arbitration is bitwise: the identifier with more leading dominant (logic 0) bits wins. 0x0A3 (000 1010 0011) has
a dominant bit in position 7 while 0x1F4 (001 1111 0100) has a recessive bit, so ECU-A wins. Lower numeric values in
standard 11-bit CAN have higher priority. Bus-off occurs only after repeated errors, not during normal arbitration.
Q3: During CAN bus troubleshooting, a technician discovers that removing either termination resistor
results in normal communication, but with both resistors installed, the oscilloscope shows severely
distorted waveforms. What is the fault?
A. The bus is overloaded with excessive message traffic
B. There is an extra (third) termination resistor somewhere on the network, creating 80 ohms total instead
of 60 ohms [CORRECT]
C. One ECU is transmitting extended 29-bit frames on a network configured for standard 11-bit only
D. The CAN transceivers in all ECUs have incompatible differential voltage thresholds
Correct Answer: B
Rationale: A properly terminated CAN bus measures approximately 60 ohms (two 120-ohm resistors in parallel). An extra
termination resistor creates approximately 40 ohms, distorting signals. Removing one of the two correct resistors restores
near-normal impedance (~80 ohms with the extra), allowing marginal communication. Bus overload, frame format mismatch,
and transceiver incompatibility produce different symptom patterns.
Page 3
