Automotive Electronics ASE A6 Practice Test 2026 Complete 60 Questions

AUTOMOTIVE

ELECTRONICS ASE A6

PRACTICE TEST 2026

COMPLETE 60 QUESTIONS

Question 1: A customer reports an intermittent power-lock failure that has occurred several times after rain. After verifying the concern and completing a visual inspection, what should the technician do next?

Choices:

1) Replace the body control module 2) Install a higher-rated door-lock fuse 3) Reprogram every module on the vehicle 4) Research service information and applicable technical service bulletins Correct Answer: Research service information and applicable technical service bulletins Explanation: After verifying the concern and inspecting the vehicle, the technician should check the manufacturer’s service information, history, bulletins, and campaigns for known conditions and prescribed diagnostic steps before replacing parts.Page 1

Question 2: A blower motor is operating while a technician measures 0.95 volt

between the blower motor ground terminal and battery negative. What does this reading most strongly indicate?

Choices:

1) A shorted blower motor armature 2) An open blower fuse 3) Excessive resistance in the ground side of the circuit 4) Normal voltage drop for a loaded ground circuit

Correct Answer: Excessive resistance in the ground side of the circuit

Explanation: A loaded ground circuit should have very little voltage drop. A 0.95-volt drop indicates excessive resistance between the motor ground and battery negative, such as corrosion or a poor connection.Question 3: A 12-volt circuit has 4 ohms of total resistance. Approximately how much current should �ow?

Choices:

1) 3 amps 2) 0.33 amp 3) 48 amps 4) 8 amps

Correct Answer: 3 amps

Explanation: Ohm’s law is I = V/R. Dividing 12 volts by 4 ohms gives 3 amps.

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Question 4: A rear-window defogger circuit consumes 60 watts at 12 volts.

Approximately how much current is the circuit drawing?

Choices:

1) 0.2 amp 2) 720 amps 3) 12 amps 4) 5 amps

Correct Answer: 5 amps

Explanation: Electrical power is P = V × I. Therefore I = P/V = 60 watts ÷ 12 volts = 5 amps.

Question 5: With power removed from a harness, an ohmmeter displays OL when

measuring end-to-end across a single wire that should be continuous. What does OL indicate?

Choices:

1) The wire is open 2) The wire is shorted to power 3) The wire has excessive current �ow 4) The wire has zero resistance

Correct Answer: The wire is open

Explanation: OL on a resistance measurement normally indicates resistance beyond the meter’s range, which in an end-to-end wire check means the circuit is open.Page 3