AUTOMOTIVE TECHNOLOGY
VEHICLE COMMUNICATION SYSTEM (LATEST)
QUESTIONS AND ANSWERS 100% CORRECT
**2026/2027**
A+
Complete Domain Coverage
A+ 5 100%
QUESTIONS VERIFIED EXAM DOMAINS COVERED RATIONALES INCLUDED
CATEGORIES
1. CAN Bus Fundamentals Physical Layer and Frame Structure
2. Higher-Layer Protocols Diagnostics OBD-II and UDS
3. LIN FlexRay MOST and Automotive Ethernet
4. Network Architecture Gateways and Cybersecurity
5. Troubleshooting Testing Tools and System Integration
STUVIAACTUALEXAM
, SECTION 1: CAN BUS FUNDAMENTALS PHYSICAL LAYER AND FRAME STRUCTURE
Q1
A technician measuring a high-speed CAN bus at the DLC finds recessive voltage near 2.5 V on both CAN-H and CAN-L
when the bus is idle. The differential voltage is essentially zero. This observation indicates that the bus is in the:
A. Recessive state with no dominant bit currently being driven
B. Dominant state with active transmission
C. Shorted condition between CAN-H and CAN-L
D. Open circuit on the termination resistors
Correct Answer: A
Rationale:
In the recessive state both lines sit near 2.5 V so the differential is approximately 0 V. A dominant bit drives CAN-H high and CAN-L low,
producing a differential of about 2 V. Zero differential at idle is normal recessive behavior, not a short or open.
Q2
While diagnosing intermittent communication loss, a scope capture shows the CAN waveform amplitude collapsing when
a particular module is connected. The most likely physical-layer cause is:
A. The termination resistors are both 120 Ω as specified
B. The bit rate is set 10 % higher than the network design
C. The twisted-pair cable has the correct characteristic impedance
D. That module has an internal short or excessive loading on the bus
Correct Answer: D
Rationale:
A single module that collapses the bus amplitude usually presents a low-impedance fault (partial short or failed transceiver). Correct
termination and proper impedance support normal amplitude; an incorrect bit rate produces timing errors rather than amplitude collapse.
Q3
A CAN frame is observed with the IDE bit set to 1. This bit setting means the frame is using:
A. Remote-frame request format only
B. Error-frame format
C. An extended 29-bit identifier
D. A standard 11-bit identifier
Correct Answer: C
Rationale:
The Identifier Extension (IDE) bit distinguishes standard (IDE = 0, 11-bit ID) from extended (IDE = 1, 29-bit ID) frames. It does not by itself
define remote or error frames.
VEHICLE COMMUNICATION SYSTEM (LATEST) QUESTIONS AND ANSWERS 100% CORRECT **2026/2027*... Page 2
, Q4
During arbitration on a CAN network two nodes begin transmitting simultaneously. Node A's identifier is numerically
lower than Node B's. The outcome is that:
A. An error frame is immediately generated by both nodes
B. Node A wins arbitration and continues transmitting while Node B becomes a receiver
C. Both nodes stop and retry after a random delay
D. Node B wins because it has the higher identifier
Correct Answer: B
Rationale:
CAN uses non-destructive bitwise arbitration: the dominant bit (0) wins over recessive (1). A lower numerical ID contains earlier dominant bits
and therefore wins, allowing the higher-priority message to continue without collision.
Q5
A technician finds only one 120 Ω termination resistor on a high-speed CAN backbone that should have two. The
expected consequence is:
A. Increased reflections and potential communication errors, especially at higher bit rates
B. No effect because a single resistor is sufficient for most vehicles
C. Complete inability of any node to transmit dominant bits
D. Automatic reduction of the bus speed to 125 kbit/s
Correct Answer: A
Rationale:
Two 60 Ω equivalent terminations (two 120 Ω resistors) are required to match the cable impedance and absorb reflections. A single resistor
leaves the line improperly terminated, producing reflections that corrupt edges at higher speeds.
Q6
The CRC field in a CAN data frame is used primarily to:
A. Identify the source module address
B. Indicate the number of data bytes that follow
C. Request a remote transmission from another node
D. Detect bit errors that occurred during transmission of the frame
Correct Answer: D
Rationale:
The Cyclic Redundancy Check allows every receiver to verify that the received bits match what the transmitter sent. Address and DLC
information are carried in other fields; remote requests use the RTR bit.
Q7
A scope shows dominant bits lasting longer than the nominal bit time on a CAN bus. This stretch is characteristic of:
A. Correct operation of the acknowledge slot
B. A properly terminated bus at idle
C. A node that has lost arbitration or is generating an error frame with extra dominant bits
D. Normal recessive-to-dominant transitions
Correct Answer: C
Rationale:
Error frames and overload frames deliberately insert sequences of dominant bits. A node that detects an error or loses arbitration may also
contribute dominant bits, lengthening the observed dominant period beyond a single bit time.
VEHICLE COMMUNICATION SYSTEM (LATEST) QUESTIONS AND ANSWERS 100% CORRECT **2026/2027*... Page 3
VEHICLE COMMUNICATION SYSTEM (LATEST)
QUESTIONS AND ANSWERS 100% CORRECT
**2026/2027**
A+
Complete Domain Coverage
A+ 5 100%
QUESTIONS VERIFIED EXAM DOMAINS COVERED RATIONALES INCLUDED
CATEGORIES
1. CAN Bus Fundamentals Physical Layer and Frame Structure
2. Higher-Layer Protocols Diagnostics OBD-II and UDS
3. LIN FlexRay MOST and Automotive Ethernet
4. Network Architecture Gateways and Cybersecurity
5. Troubleshooting Testing Tools and System Integration
STUVIAACTUALEXAM
, SECTION 1: CAN BUS FUNDAMENTALS PHYSICAL LAYER AND FRAME STRUCTURE
Q1
A technician measuring a high-speed CAN bus at the DLC finds recessive voltage near 2.5 V on both CAN-H and CAN-L
when the bus is idle. The differential voltage is essentially zero. This observation indicates that the bus is in the:
A. Recessive state with no dominant bit currently being driven
B. Dominant state with active transmission
C. Shorted condition between CAN-H and CAN-L
D. Open circuit on the termination resistors
Correct Answer: A
Rationale:
In the recessive state both lines sit near 2.5 V so the differential is approximately 0 V. A dominant bit drives CAN-H high and CAN-L low,
producing a differential of about 2 V. Zero differential at idle is normal recessive behavior, not a short or open.
Q2
While diagnosing intermittent communication loss, a scope capture shows the CAN waveform amplitude collapsing when
a particular module is connected. The most likely physical-layer cause is:
A. The termination resistors are both 120 Ω as specified
B. The bit rate is set 10 % higher than the network design
C. The twisted-pair cable has the correct characteristic impedance
D. That module has an internal short or excessive loading on the bus
Correct Answer: D
Rationale:
A single module that collapses the bus amplitude usually presents a low-impedance fault (partial short or failed transceiver). Correct
termination and proper impedance support normal amplitude; an incorrect bit rate produces timing errors rather than amplitude collapse.
Q3
A CAN frame is observed with the IDE bit set to 1. This bit setting means the frame is using:
A. Remote-frame request format only
B. Error-frame format
C. An extended 29-bit identifier
D. A standard 11-bit identifier
Correct Answer: C
Rationale:
The Identifier Extension (IDE) bit distinguishes standard (IDE = 0, 11-bit ID) from extended (IDE = 1, 29-bit ID) frames. It does not by itself
define remote or error frames.
VEHICLE COMMUNICATION SYSTEM (LATEST) QUESTIONS AND ANSWERS 100% CORRECT **2026/2027*... Page 2
, Q4
During arbitration on a CAN network two nodes begin transmitting simultaneously. Node A's identifier is numerically
lower than Node B's. The outcome is that:
A. An error frame is immediately generated by both nodes
B. Node A wins arbitration and continues transmitting while Node B becomes a receiver
C. Both nodes stop and retry after a random delay
D. Node B wins because it has the higher identifier
Correct Answer: B
Rationale:
CAN uses non-destructive bitwise arbitration: the dominant bit (0) wins over recessive (1). A lower numerical ID contains earlier dominant bits
and therefore wins, allowing the higher-priority message to continue without collision.
Q5
A technician finds only one 120 Ω termination resistor on a high-speed CAN backbone that should have two. The
expected consequence is:
A. Increased reflections and potential communication errors, especially at higher bit rates
B. No effect because a single resistor is sufficient for most vehicles
C. Complete inability of any node to transmit dominant bits
D. Automatic reduction of the bus speed to 125 kbit/s
Correct Answer: A
Rationale:
Two 60 Ω equivalent terminations (two 120 Ω resistors) are required to match the cable impedance and absorb reflections. A single resistor
leaves the line improperly terminated, producing reflections that corrupt edges at higher speeds.
Q6
The CRC field in a CAN data frame is used primarily to:
A. Identify the source module address
B. Indicate the number of data bytes that follow
C. Request a remote transmission from another node
D. Detect bit errors that occurred during transmission of the frame
Correct Answer: D
Rationale:
The Cyclic Redundancy Check allows every receiver to verify that the received bits match what the transmitter sent. Address and DLC
information are carried in other fields; remote requests use the RTR bit.
Q7
A scope shows dominant bits lasting longer than the nominal bit time on a CAN bus. This stretch is characteristic of:
A. Correct operation of the acknowledge slot
B. A properly terminated bus at idle
C. A node that has lost arbitration or is generating an error frame with extra dominant bits
D. Normal recessive-to-dominant transitions
Correct Answer: C
Rationale:
Error frames and overload frames deliberately insert sequences of dominant bits. A node that detects an error or loses arbitration may also
contribute dominant bits, lengthening the observed dominant period beyond a single bit time.
VEHICLE COMMUNICATION SYSTEM (LATEST) QUESTIONS AND ANSWERS 100% CORRECT **2026/2027*... Page 3
