Title: Module 6: Battery Failure & Diagnosis - Activity
Course: ASE A6 Battery Service > Section: Module 6: Battery Failure & Diagnosis

Q: A 3-year-old battery in a vehicle used only for short 5-minute commutes has gradually lost cranking power. The charging system output is 14.2V and electrolyte level is normal. Which failure mode is most likely?
- Plate shedding from vibration damage
- Sulfation from chronic undercharging
- Cell imbalance from a manufacturing defect
- Parasitic draw from a faulty module
A: Sulfation from chronic undercharging

Q: A 5-year-old work truck battery reads 10.5V open-circuit. One cell has dark, cloudy electrolyte while the other five appear normal. The truck is driven 200+ miles daily on rough roads and runs a liftgate. Which failure mode best explains these symptoms?
- Sulfation from prolonged storage
- Overcharging from a faulty voltage regulator
- Plate shedding leading to an internal short in one cell
- Normal self-discharge from age
A: Plate shedding leading to an internal short in one cell

Q: A customer's battery keeps dying every 3-4 days, but it passes both a load test and a conductance test each time it is recharged. The charging system output is 14.4V. An aftermarket stereo and amplifier were recently installed. What is the most likely cause?
- Sulfation from short-trip driving
- A shorted cell inside the battery
- A weak alternator that cannot keep up with demand
- Parasitic draw from the aftermarket stereo/amplifier installation
A: Parasitic draw from the aftermarket stereo/amplifier installation

Q: What is the normal acceptable range for parasitic draw on a modern vehicle?
- 0-5 mA
- 25-50 mA
- 200-300 mA
- 500-750 mA
A: 25-50 mA

Q: A technician measures a parasitic draw of 0.450A. Convert this reading to milliamps.
- 4.5 mA
- 45 mA
- 450 mA
- 4,500 mA
A: 450 mA

Q: A parasitic draw reading of 450 mA is approximately how many times the upper limit of the acceptable range (50 mA)?
- 3 times
- 5 times
- 9 times
- 15 times
A: 9 times

Q: During a parasitic draw test using the fuse-pulling method, when should the technician pull fuses?
- With the engine running and all accessories on
- Immediately after turning the key off
- With the key removed, all doors closed, and modules in sleep mode (after waiting approximately 45 minutes)
- While a helper cranks the engine
A: With the key removed, all doors closed, and modules in sleep mode (after waiting approximately 45 minutes)

Q: Using the standard battery date code system (A = January, B = February, ... L = December; digit = last digit of year), what month and year does the date code **H3** represent?
- August 2013
- August 2023
- March 2028
- July 2003
A: August 2023

Q: Using the standard date code system, decode **K0**. The current date is June 2026. Is this battery within a 5-year full replacement warranty?
- November 2020 — No, it is beyond the 5-year warranty period
- November 2020 — Yes, it is within the 5-year warranty period
- October 2010 — No, it is beyond the 5-year warranty period
- December 2020 — Yes, it is within the 5-year warranty period
A: November 2020 — No, it is beyond the 5-year warranty period

Q: In a 12V automotive battery with six cells connected in series, what happens if one cell becomes internally shorted?
- The battery voltage increases to compensate
- The remaining five cells share the extra load equally and the battery still functions normally
- The total battery voltage drops to approximately 10.5V because the shorted cell contributes 0V
- The battery voltage remains at 12.6V but the CCA rating drops
A: The total battery voltage drops to approximately 10.5V because the shorted cell contributes 0V

Q: A technician is diagnosing a 3-year-old battery in a vehicle used for daily 5-minute commutes. The battery has gradually lost cranking power over 6 months. It reads 12.4V open-circuit and fails a load test. The electrolyte level is normal and the charging system output is 14.2V. Identify the failure mode and explain your reasoning, including why the driving pattern contributes to this failure. The failure mode is sulfation. The short-trip driving pattern means the alternator never runs long enough to fully recharge the battery. Chronic undercharging leaves lead sulfate on the plates that gradually hardens into permanent crystals. The 12.4V open-circuit voltage suggests slight discharge, and the failed load test indicates reduced capacity from sulfated plates that can no longer deliver adequate current. The charging system is within spec at 14.2V, so the issue is not the alternator — it is the driving pattern preventing the battery from ever reaching a full state of charge. Over 6 months of this cycle, the sulfation has progressed enough to significantly reduce the battery's cranking ability.

Q: A 5-year-old battery in a work truck reads 10.5V open-circuit. The truck runs a liftgate and is driven 200+ miles daily on rough roads. All cells but one have normal electrolyte appearance; one cell has noticeably darker, cloudier electrolyte. Identify the failure mode and explain your reasoning, including how the operating conditions contributed and what the 10.5V reading tells you about the number of functioning cells. The failure mode is plate shedding leading to an internal short circuit. The combination of age (5 years), heavy vibration from rough roads and high daily mileage (200+ miles), and deep cycling from liftgate use has caused active material to separate from the plate grids. The shed material falls to the bottom of the cell and eventually bridges the gap between positive and negative plates, creating an internal short. The 10.5V open-circuit voltage confirms one cell is shorted: five healthy cells at approximately 2.1V each equals 10.5V, while the shorted cell contributes 0V. The dark, cloudy electrolyte in one cell is visual confirmation — the shed plate material is suspended in the electrolyte. The harsh operating conditions (vibration, deep cycling, age) are the classic recipe for plate shedding.

Q: A technician is performing a parasitic draw test. With the vehicle in sleep mode and the DMM connected in series at the negative terminal, the initial reading is 450 mA. The technician begins pulling fuses one at a time and records these readings: |-------------|----------------| Identify which circuit contains the excessive draw, explain how you determined this from the data, and name two possible components on that circuit that could cause the draw. Also address whether the remaining 80 mA reading after isolating the problem circuit warrants further investigation. The Accessory Outlet (15A) circuit contains the excessive draw. When the Accessory Outlet fuse was pulled, the ammeter reading dropped from 450 mA to 80 mA — a decrease of 370 mA. No other fuse caused a significant drop; the other pulls only showed minor fluctuations of 5-20 mA (which are normal for modules waking and sleeping). This large single drop clearly identifies the accessory outlet circuit as the source of the problem. Two possible components that could cause this draw: (1) An aftermarket device such as a dash cam, phone charger, or GPS tracker that is plugged into or hard-wired to the 12V accessory outlet and draws power continuously even with the vehicle off. (2) A faulty accessory outlet relay or wiring that keeps the outlet energized when the vehicle is off. Other acceptable answers include a cigarette lighter accessory left plugged in or an aftermarket USB charger hard-wired to the circuit. The remaining 80 mA after pulling the accessory outlet fuse is still above the normal 25-50 mA range. This may warrant further investigation, though it could be partially explained by modules that were awakened during the fuse-pulling process and need time to return to sleep mode. The technician should note this and recheck after allowing the vehicle to return to full sleep mode.

Q: A vehicle has a measured parasitic draw of 35 mA. The battery is a 48 Ah unit that is 4 years old. The customer complains the battery dies after the vehicle sits for two weeks during vacation. Determine whether the 35 mA draw is acceptable, calculate approximately how many hours it would take to fully drain the battery at this draw rate, and explain why the battery still dies after only two weeks despite the math suggesting it should last longer. The 35 mA draw is within the normal acceptable range of 25-50 mA, so the parasitic draw is not the problem. Calculation: 48 Ah = 48,000 mAh. At 35 mA draw: 48,000 mAh / 35 mA = approximately 1,371 hours, or about 57 days to reach full drain. However, the battery dies after only 14 days (about 336 hours) for two key reasons. First, a 4-year-old battery likely has reduced actual capacity compared to its rated 48 Ah due to normal aging — sulfation, plate degradation, and electrolyte changes reduce the usable capacity well below the original rating. Second, a battery becomes unable to start a vehicle long before it reaches 0% charge. Most vehicles require the battery to be above approximately 50% state of charge to reliably crank the engine. Combining the reduced capacity from aging with the practical starting threshold, 14 days of continuous 35 mA draw is enough to bring the battery below a usable level. The draw is normal; the battery is aging out. A battery maintainer or trickle charger connected during extended parking would solve the problem.

Q: A technician measures individual cell voltages on a fully charged flooded battery and records the following: |------|---------| Calculate the total battery voltage. Identify the weak cell and compare its voltage to the expected range for a healthy, fully charged cell. Predict what would happen to Cell 4 under a load test and how it would affect the overall result. Finally, explain why a series connection of six cells means one bad cell ruins the entire battery's performance. Total battery voltage: 2.13 + 2.11 + 2.12 + 1.88 + 2.12 + 2.10 = 12.46V. Cell 4 is the weak cell at 1.88V. A healthy, fully charged cell should read 2.10-2.15V. Cell 4 is 0.22-0.27V below the expected range, indicating significant degradation from sulfation, shedding, or other internal damage. Under the heavy current demand of a load test, Cell 4 would drop significantly more than the other five cells because it has higher internal resistance due to its degradation. While the five healthy cells might maintain adequate voltage under load, Cell 4 would collapse, pulling the overall battery voltage below the pass/fail threshold. The battery would fail the load test even though five of six cells are healthy. In a series circuit, the total voltage is the sum of all individual cell voltages, and the same current must flow through every cell. If one cell has high internal resistance or is shorted, it becomes a bottleneck for the entire battery. Current flowing through the battery must pass through all six cells sequentially. A weak cell cannot deliver the same current as the other five cells, so it limits the total current output. A shorted cell contributes 0V instead of 2.1V, reducing the total voltage to approximately 10.5V. Either way, the weakest cell determines the battery's overall capability — just like the weakest link determines the strength of a chain. |---------|--------|----------|