Slow crank every third morning: battery warranty claim or parasitic drain?
A customer brings a 2019 petrol hatchback (1.2 L engine, 45 Ah, 12 V lead-acid battery, 14 months old) to your Guwahati workshop. Complaint: the engine cranks slowly or not at all every second or third morning, but starts immediately after a jump start and runs normally all day. Battery open-circuit voltage is 12.55 V after a 12-hour rest; charging voltage is 14.1 V at 2000 rpm with headlamps and blower on; a load test at half CCA passes; battery terminals are clean and tight. You connect a clamp meter around the negative cable, lock the vehicle, and after the body-control-module sleep period of 40 minutes the reading settles at 340 mA. Manufacturer specification is under 50 mA after sleep. Pulling fuses one at a time, the current drops to 40 mA only when the 10 A interior-lamp fuse is removed. Further checking shows the glovebox lamp stays lit with the glovebox shut because the switch striker is bent. The customer insists the 14-month-old battery is defective and demands a free warranty replacement today.
1. Using the readings, decide whether the battery or the vehicle is at fault and show the arithmetic that explains why the car fails on about the third morning.
Read the worked answer
Fault is on the vehicle, not in the battery.
- Rest voltage 12.55 V after 12 hours is roughly 80-90% state of charge for a 12 V lead-acid battery - it is holding charge.
- Charging voltage 14.1 V at 2000 rpm with loads on sits inside the normal 13.8-14.4 V window, so the alternator and regulator are delivering.
- The load test passed, so capacity is broadly intact.
- 340 mA against a 50 mA limit is 6.8 times specification, and removing the interior-lamp fuse takes away 300 mA of it. That points at a circuit, not at the battery.
Why the third morning fails. Drain per day = 0.340 A x 24 h = 8.16 Ah. The battery is 45 Ah, and a car needs roughly half its capacity plus a cranking margin to start reliably. Usable margin = about 45 x 0.4 = 18 Ah. 18 Ah / 8.16 Ah per day = about 2.2 days. So on mornings one and two the engine still cranks briskly; by morning three the state of charge has fallen far enough that the 150-200 A cranking draw collapses terminal voltage and the starter turns slowly. The complaint pattern and the measurement agree.
Documented sequence: battery condition (rest voltage, load test) → charging system (voltage, AC ripple) → sleep-mode current measured with doors locked, key away, after the full 40-minute sleep period → fuse-pull or fuse voltage-drop isolation → circuit-level check. Record every value with units and note engine/ambient temperature on the job card.
2. The fuse-pull test pointed at the interior-lamp circuit. Describe two alternative methods to narrow the drain without pulling fuses, and list what must be verified before you declare the repair finished.
Read the worked answer
Narrowing without pulling fuses
- Millivolt drop across fuses. With current flowing, measure DC millivolts across each fuse (probe the two test points on the fuse top). A fuse carrying current shows a small drop; the fuse with the drain shows the largest. A millivolt-to-current chart for the fuse type converts the reading to amps. This is faster and avoids waking modules by removing and refitting fuses.
- Progressive disconnection and scan data. Unplug relays and module connectors on the suspect circuit one at a time, re-measuring after each sleep cycle, and read module sleep/bus-activity status with a scan tool. A module that will not go to sleep, or a stuck door/glovebox switch input visible in live data, shows up here.
The 300 mA step is far too big for an LED glovebox lamp alone in most cars - a bulb-type lamp is plausible, an LED is not. Confirm the actual current of that lamp with a meter in series, and check whether the same fuse also feeds a boot lamp, vanity lamp or an aftermarket dash camera or tracker spliced into that circuit.
Before signing off:
- Sleep current back under 50 mA after a full 40-minute sleep, measured twice.
- Alternator diode leak ruled out (AC ripple typically under 0.5 V; disconnect B+ and re-measure).
- No aftermarket accessory left draining.
- Overnight recheck or a 24-hour retest agreed with the customer, because a drain that only appears when a module fails to sleep may not show in a 10-minute test.
3. The customer still wants a free warranty battery. Write how you would handle the conversation and what you would put in writing.
Read the worked answer
Conversation: Show the numbers rather than argue. "Your battery holds 12.55 V after standing all night, takes 14.1 V from the alternator and passed a load test. If I fit a new battery today under warranty, the claim will be rejected because the battery tests good - and more importantly, the car will still be flat by the third morning, because something is drawing 340 mA when it should draw under 50. We found it: the glovebox lamp is staying on."
Then give the choice plainly: repair the striker/switch (small job), or leave it and keep jump-starting. Point out that a warranty replacement would not be honest, would not fix the symptom, and would cost the customer more in the long run as the new battery is repeatedly flattened - deep discharge also shortens battery life.
In writing, on the job card:
- All test values: rest voltage, charging voltage, load-test result, sleep current 340 mA against a 50 mA limit.
- The isolation result: fuse removed, current fell to 40 mA; glovebox lamp found lit with the glovebox shut; striker bent.
- Recommendation and the customer's decision, with signature if the repair is declined.
- A note that the battery is not faulty and a warranty claim is not supported by the test data.
Keep the tone factual and let the evidence carry the refusal. Never disable the lamp and return the car without telling the customer what was found and what was done.