This tip is useful in a situation where you suspect some electrical item is being left on and slowly flattening the battery over a couple of days. The boot or bonnet light is a typical case. You would think that you might have to disconnect one of the battery leads to measure the current with an ammeter but here is an easy method with a cheap multimeter.

Depending on the model of the vehicle, there is usually a heavy tinned copper braid or a solid 'earth' cable connecting one of the battery posts to the chassis. If you connect a multimeter (on the mV millivolt DC range) across this lead (ie one probe on the chassis end and the other on the battery post) with everything turned off (engine, lights etc) there should only be a small current passing through the cable to drive low powered items like the radio memory or the alarm in standby mode.

A typical voltage drop across a short earth cable will probably be less than a volt eg 450 mV DC (0.45 Volts). Any more and you need to try to get a sense of what might be causing the extra drain. For instance, if you turn on the sidelights you will see a large increase in the voltage reading; if you open the boot you should only see a small increase on the meter (if you see no increase then the boot light was probably already on and you have found the problem - fix the switch, make a brew and look smug).

This is based on the Ohm's Law principle that a small current (say 0.5 Amps) flowing through a cable of very low resistance (eg 0.9 Ohm) will have a measurable voltage across it (Voltage=Current X Resistance) eg 0.5 X 0.9 = 0.45 V or 450 mV.

UPDATE 2 August 2015: Real world test on a 2007 SE 2.2 TD4
The chassis end of the earth cable is not easy to access so I have measured some voltage drops across the red positive cable going from the battery to the fusebox. The resistance is obviously very low but one can still read small voltage changes on a cheap meter. Mine is a rapitest MAS830 that has a low DC voltage setting of 200mV. Without clips you should be able to insert one probe under the heat-shrink insulation on the battery lead and the other, horizontally, between the terminals in the fuse box. I have a photo if anyone needs more help.

The quiescent voltage after everything is stable after either a single lock or a double lock is about 0.2mV. It seems to take up to a minute for the system to go into standby after any event. However, contrary to some reports, this is the same as when the car is unlocked. It may be that less current is drawn after a longer delay; perhaps someone else may be able to elaborate.

Here are my figures from a 2007 SE 2.2 TD4 with Satnav/heated screen/seats/Bluetooth/6CD and other extras (yours may very depending on options). The meter was connected and then rested on the headlamp washers because the hood must be down for the alarm to work (unless you can beat the hood switch with a screwdriver).

After unlocking: 2.2mV (system awoken with interior light on) the reading peaks as the mirrors swing out.
Autolock after about 1 minute: 0.2mV
Unlocked, door opened and closed (to prevent autolock) after 1 minute: 0.2mV
Open tailgate: 1.0mV
Tailgate closed, sidelights on: 2.4mV
Sidelights off, open driver's door (with interior and puddle light on) 1.9mV
Close door, interior light on, puddle light off: 1.4mV dropping to 0.2 after interior light goes out.

Find the fault by pulling fuses
Assuming you are testing the battery drain because you suspect a fault, you might also try pulling out fuses one at a time to see if there is a sudden drop on the meter. The measurements above are taken on a fault-free vehicle so if you are seeing 20mV instead of 0.2, then you almost certainly have a problem. It's difficult to say exactly what the current drain is without disconnecting the battery lead and inserting an ammeter but let's suppose that 20mV might represent a current drain of a few amps - that would almost certainly put undue load on the battery when the car is parked up.

If you are seeing a randomly fluctuating reading, it could indicate an intermittent short circuit and I would err on the side of caution and not drive the car until it is properly fixed.

After the tests, with the engine off, I measured the nominal battery voltage (at the rear cigar lighter socket) to be 12.15V; this dropped to 12.09 with the sidelights on. I think this might indicate that the battery is near the end of its useful life and will need monitoring or replacing before the winter. With the engine running, the voltage was 14.14.

Please be aware that low voltage batteries can generate extremely dangerous heavy currents and you should not attempt any tests or repairs if you are not competent. This simple test with a cheap multimeter is a safer way to test the current drain without disconnecting the battery cables. Always remove wrist watches, bracelets and rings etc before undertaking electrical tests or repairs.

I hope this tip helps someone avoid an expensive diagnostic bill or replacement battery.

Last edited by Mando on 3rd Aug 2015 4:50 pm. Edited 3 times in total

Top tip

Excellent tip.
I'll keep that in my note book.
Thanks  Now at the point when I learn something new something old is lost out the other side !
Now retired so it doesn't matter anymore.
Freelander now gone.

great application of ohm's law. Thanks for that

Interesting way of detecting if you have a high quiescent drain.
As you say, the only way of measuring the actual current is to use an ammeter between the battery and battery pos lead.
I've always done it this way with a shunt (don't want to either damage my Fluke or blow it's fuse!) Another member of the failed FL2 clutch/DMF club, twice.

...surely a volt meter in series on the batt earth is more accurate?any more than 0.25 after ecu shutting down is excessive and will flatten a battery if not used after a couple of days.i might have this wrong, but it seem s that you are in fact measuring volt drop by going from batt earth to chassis in parallel ,so the measurement if more than 0.5 will show the earth lead is not carring full current ie if meter shows 1 volt the earth lead is only carring 11 volts and 1 volt takes the easy route through the meter,thr higher the voltage on the meter the worse the connection.for example completely disconnect earth lead and the meter will read batt voltage when load is applied.this also applys to engine earths,ie volt meter from block to body, crank or attempt to crank meter should not read more than 0.5,if 0.5 or below no earthing problem exists

Unnecessary quote removed

Hi Sid. Ideally, one would measure the actual DC current with an in-circuit ammeter but, as explained, this can be hazardous and will probably fry your meter's fuse or wipe your radio code. However, the tip is about measuring the DC voltage across any current-carrying component. This is an accepted and accurate way to calculate the DC current flowing in the component, assuming you know its DC resistance. You then simply calculate V/R (where V is in Volts and R is in Ohms) and you have the current in Amps.

In the case of a battery lead (+ or -) we don't care about the actual resistance (which will typically be less than 1 Ohm). We are just looking for ANY small voltage across it which indicates that some current is flowing. Of course, if you disconnected one end of the battery lead and measured its actual resistance, you could then reconnect it, measure the voltage drop and calculate the actual current. For example, if the cable is 1 Ohm and there is a drop of 2 V across it, the real current is 2 Amps.

In the case of a connected battery lead supplying a current of say 0.25A, from V=I*R, you will only see a voltage across a 1 Ohm cable of 0.25 Volts.

The tip related to residual current drains so please bear in mind that when the starter motor is energised, the battery voltage always drops a good deal. The above method still holds true; it's just that the voltage readings will be much lower. In the case of a failed battery, you may see the battery's terminal voltage (ie voltmeter across both battery posts) drop to almost nothing under heavy load.

Sid, you mention a voltage drop of 0.5 V between the end of the cable and the chassis. Using the same rule (ignoring the rest of the cable) this indicates that the resistance of the earth joint is high: because V=I*R, for any current, if R is high then V would also be high. A perfect chassis connection would be zero Ohms so, even with 100A flowing to the starter, the voltage drop across the joint would be zero. It's worth mentioning that my tip assumes that the earth connection is not being included in the test. Of course, if the chassis connection was poor, there probably wouldn't be much drain anyway!

I hope that explains the theory a little better. Feel free to raise any questions.