VEHICLE COMMUNICATION SYSTEMS CERTIFICATION
EXAM 2026/2027 | CAN/LIN/FlexRay/MOST/Ethernet |
100% Correct Answers | Pass Guaranteed - A+ Graded
Section 1: CAN (Controller Area Network) - Architecture & Protocols (Q1-20)
Q1. What is the nominal differential voltage between CAN_H and CAN_L during a
dominant bit state on a standard ISO 11898 CAN bus?
A. 0V
B. 1.0V
C. 2.0V
D. 5.0V
Correct Answer: C. 2.0V [CORRECT]
Rationale: In dominant state, CAN_H ≈ 3.5V and CAN_L ≈ 1.5V, creating a 2.0V
differential. 0V represents recessive state, while 1.0V and 5.0V are incorrect values.
Q2. A technician measures 60 ohms between CAN_H and CAN_L at the DLC connector
with the vehicle powered down. What does this indicate?
A. One termination resistor is open circuit
B. The bus has a short to ground
C. Both 120-ohm termination resistors are properly installed
D. The bus has a short between CAN_H and CAN_L
Correct Answer: C. Both 120-ohm termination resistors are properly installed [CORRECT]
Rationale: Two 120-ohm resistors in parallel at opposite bus ends yield 60 ohms. An
open resistor would show 120 ohms, a short would show near 0 ohms, and a short
between lines would also show near 0 ohms.
Q3. A vehicle has three ECUs on a CAN bus with 120-ohm termination resistors at the
two end nodes. A fourth ECU is added in the middle with an internal 120-ohm
,termination resistor enabled. What total resistance will now be measured across CAN_H
and CAN_L?
A. 60 ohms
B. 40 ohms
C. 30 ohms
D. 120 ohms
Correct Answer: B. 40 ohms [CORRECT]
Rationale: Three 120-ohm resistors in parallel equal 40 ohms (1/(1/120+1/120+1/120)).
This improper termination causes signal reflections and communication errors.
Q4. During a resistance check, a technician measures 0.3 ohms between CAN_H and
CAN_L with the vehicle off. What is the most likely fault?
A. Open termination resistor
B. Short circuit between CAN_H and CAN_L
C. Missing termination resistor
D. CAN_H short to battery
Correct Answer: B. Short circuit between CAN_H and CAN_L [CORRECT]
Rationale: A near-zero ohm reading indicates a direct short between the lines. An open
or missing resistor would show high resistance, while a short to battery would not affect
line-to-line resistance.
Q5. Which CAN protocol specification supports a 29-bit identifier?
A. CAN 2.0A
B. CAN 2.0B
C. ISO 11898-1:2015 classic only
D. LIN 2.2
Correct Answer: B. CAN 2.0B [CORRECT]
Rationale: CAN 2.0B uses 29-bit identifiers for extended addressing, while CAN 2.0A
uses 11-bit identifiers. LIN is a completely different protocol.
Q6. A technician measures 120 ohms between CAN_H and CAN_L on a two-node CAN
bus. Both nodes are supposed to have internal termination. What is the most likely
fault?
, A. Both termination resistors are properly installed
B. One termination resistor is open circuit
C. The bus is shorted to ground
D. The data rate is too high for the bus length
Correct Answer: B. One termination resistor is open circuit [CORRECT]
Rationale: Two 120-ohm resistors in parallel should measure 60 ohms. A 120-ohm
reading indicates one resistor is open or missing. A short would show near 0 ohms, and
data rate does not affect DC resistance.
Q7. A vehicle has a CAN bus network with termination resistors of 120 ohms at each
end. If one resistor fails open circuit, what resistance will be measured between CAN_H
and CAN_L at the DLC?
A. 0 ohms
B. 30 ohms
C. 60 ohms
D. 120 ohms
Correct Answer: D. 120 ohms [CORRECT]
Rationale: With one open termination, only the remaining 120-ohm resistor is measured.
60 ohms indicates both are good, 0 ohms indicates a short, and 30 ohms would require
additional parallel resistors.
Q8. In a non-destructive arbitration scenario, two CAN nodes attempt to transmit
simultaneously. Node A has identifier 0x123 and Node B has identifier 0x122. Which
node wins arbitration?
A. Node A wins because it has the higher numeric identifier
B. Node B wins because it has the lower numeric identifier
C. Both nodes transmit simultaneously without collision
D. Arbitration is determined by message priority field, not identifier
Correct Answer: B. Node B wins because it has the lower numeric identifier [CORRECT]
Rationale: CAN arbitration uses CSMA/CR where the lower identifier value (0x122 <
0x123) wins because it transmits more dominant bits. This is non-destructive
arbitration.
EXAM 2026/2027 | CAN/LIN/FlexRay/MOST/Ethernet |
100% Correct Answers | Pass Guaranteed - A+ Graded
Section 1: CAN (Controller Area Network) - Architecture & Protocols (Q1-20)
Q1. What is the nominal differential voltage between CAN_H and CAN_L during a
dominant bit state on a standard ISO 11898 CAN bus?
A. 0V
B. 1.0V
C. 2.0V
D. 5.0V
Correct Answer: C. 2.0V [CORRECT]
Rationale: In dominant state, CAN_H ≈ 3.5V and CAN_L ≈ 1.5V, creating a 2.0V
differential. 0V represents recessive state, while 1.0V and 5.0V are incorrect values.
Q2. A technician measures 60 ohms between CAN_H and CAN_L at the DLC connector
with the vehicle powered down. What does this indicate?
A. One termination resistor is open circuit
B. The bus has a short to ground
C. Both 120-ohm termination resistors are properly installed
D. The bus has a short between CAN_H and CAN_L
Correct Answer: C. Both 120-ohm termination resistors are properly installed [CORRECT]
Rationale: Two 120-ohm resistors in parallel at opposite bus ends yield 60 ohms. An
open resistor would show 120 ohms, a short would show near 0 ohms, and a short
between lines would also show near 0 ohms.
Q3. A vehicle has three ECUs on a CAN bus with 120-ohm termination resistors at the
two end nodes. A fourth ECU is added in the middle with an internal 120-ohm
,termination resistor enabled. What total resistance will now be measured across CAN_H
and CAN_L?
A. 60 ohms
B. 40 ohms
C. 30 ohms
D. 120 ohms
Correct Answer: B. 40 ohms [CORRECT]
Rationale: Three 120-ohm resistors in parallel equal 40 ohms (1/(1/120+1/120+1/120)).
This improper termination causes signal reflections and communication errors.
Q4. During a resistance check, a technician measures 0.3 ohms between CAN_H and
CAN_L with the vehicle off. What is the most likely fault?
A. Open termination resistor
B. Short circuit between CAN_H and CAN_L
C. Missing termination resistor
D. CAN_H short to battery
Correct Answer: B. Short circuit between CAN_H and CAN_L [CORRECT]
Rationale: A near-zero ohm reading indicates a direct short between the lines. An open
or missing resistor would show high resistance, while a short to battery would not affect
line-to-line resistance.
Q5. Which CAN protocol specification supports a 29-bit identifier?
A. CAN 2.0A
B. CAN 2.0B
C. ISO 11898-1:2015 classic only
D. LIN 2.2
Correct Answer: B. CAN 2.0B [CORRECT]
Rationale: CAN 2.0B uses 29-bit identifiers for extended addressing, while CAN 2.0A
uses 11-bit identifiers. LIN is a completely different protocol.
Q6. A technician measures 120 ohms between CAN_H and CAN_L on a two-node CAN
bus. Both nodes are supposed to have internal termination. What is the most likely
fault?
, A. Both termination resistors are properly installed
B. One termination resistor is open circuit
C. The bus is shorted to ground
D. The data rate is too high for the bus length
Correct Answer: B. One termination resistor is open circuit [CORRECT]
Rationale: Two 120-ohm resistors in parallel should measure 60 ohms. A 120-ohm
reading indicates one resistor is open or missing. A short would show near 0 ohms, and
data rate does not affect DC resistance.
Q7. A vehicle has a CAN bus network with termination resistors of 120 ohms at each
end. If one resistor fails open circuit, what resistance will be measured between CAN_H
and CAN_L at the DLC?
A. 0 ohms
B. 30 ohms
C. 60 ohms
D. 120 ohms
Correct Answer: D. 120 ohms [CORRECT]
Rationale: With one open termination, only the remaining 120-ohm resistor is measured.
60 ohms indicates both are good, 0 ohms indicates a short, and 30 ohms would require
additional parallel resistors.
Q8. In a non-destructive arbitration scenario, two CAN nodes attempt to transmit
simultaneously. Node A has identifier 0x123 and Node B has identifier 0x122. Which
node wins arbitration?
A. Node A wins because it has the higher numeric identifier
B. Node B wins because it has the lower numeric identifier
C. Both nodes transmit simultaneously without collision
D. Arbitration is determined by message priority field, not identifier
Correct Answer: B. Node B wins because it has the lower numeric identifier [CORRECT]
Rationale: CAN arbitration uses CSMA/CR where the lower identifier value (0x122 <
0x123) wins because it transmits more dominant bits. This is non-destructive
arbitration.
