Two ammeters in parallel

[williamwright]

I wanted to measure to consumption of a 12V pump. It was obviously going
to be more than 10A when working hard, but I only had 10A multimeters. I
used two meters in parallel, and added the readings. The two meters gave
different readings, so obviously there was some imbalance. But will the
summed readings I got be valid?
Examples:
5.15A and 10.30A
3.32A and 7.31A

Bill

[Custos Custodum]

Kirchhoff is your friend.

[Java Jive]

It's a long time ago that I did any serious measurement in anger, as it
were, but the principle is correct. However, there is clearly a
significant discrepancy between the performance of your ammeters, and I
think that needs to be explained and allowed for before putting any
critical faith in the result.

AIR, it's all about internal resistance, when measuring voltage you want
it to be as high as possible, but when measuring current as low as
possible. Clearly there seems to be a significant difference between
your ammeters in this respect. But, also AIR, these things can also
have shunts, that either can be physically attached or else more
conveniently switched in and out of circuit, for situations when the
load is expected to be beyond the tolerance of the meter. Could the
difference between the two be related to a shunt in the lower reading one?

--

Fake news kills!

I may be contacted via the contact address given on my website:
www.macfh.co.uk

[Theo]

I think it's a bit more complicated than that.

Moving-coil ammeters are actually ammeters - the movement is proportional to
the current. The current is split, so each will show directly the current
passing through the coil and you can sum them.

However digital ammeters aren't really ammeters, they're voltmeters. They
measure the voltage across a fixed resistance and, by ohm's law, determine
the current. The problem comes when you put two in parallel, becuase the
two resistances are effectively in parallel too. So you change the
effective resistance the voltmeter sees and thus the ohm's law calculation
is wrong. Effectively both voltmeters are across the same pair of parallel
resistors, but each meter doesn't know the resistance has changed.

Now a decent digital ammeter will compensate for this effect (eg add
components in series to isolate the measurement resistance), but that would
depend on how they were constructed.

I don't know if this is a real problem, or one that they all compensate for.
I think I'd want to do some calibration runs to find out.

Theo

[Fredxx]

Yes, all sounds good to me.

[Clive Arthur]

On 04/09/2020 16:20, williamwright wrote:

Yes that's fine, the current has nowhere else to go.

Even if the meters are the same type, small differences in wiring and
contact resistance would probably make them read differently, but the
total is accurate assuming the meters are accurate enough individually.

--
Cheers
Clive

[alan_m]

On 04/09/2020 16:20, williamwright wrote:

Were they both the same model?

--
mailto : news {at} admac {dot} myzen {dot} co {dot} uk

[Kellerman]

On 04/09/2020 16:20, williamwright wrote:

If you want to check that the meters read the same - put them in series.
Any error will them show up :-)


--
Ask how to email me.

[David Wade]

It doesn't need to. It measures voltage. The voltage across a resistor
depends on the current flowing through it. You put 1 amp through a 1 ohm
meter you get one volt. It doesn't matter what's in parallel with it, so
long as 1 amp is flowing through the meter you get 1 volt across its
resistor. So the meter reads 1amp for every volt.

You put a resistor in parallel, say another 1 ohm, then the current
through each resistor is now 0.5 amp so the voltage is now 0.5volt.

The meter now reads 0.5 amp because that is what is flowing through it.

>
> Now a decent digital ammeter will compensate for this effect (eg add
> components in series to isolate the measurement resistance), but that would
> depend on how they were constructed.

It does not need to.

>
> I don't know if this is a real problem, or one that they all compensate for.
> I think I'd want to do some calibration runs to find out.
>


> Theo
>

Dave

[John Williamson]

On 04/09/2020 16:20, williamwright wrote:

Possibly.

If the meters are moving coil or moving iron movements, rated at 10 amps
each, with no shunts in place, then yes.

Any meter, digital or analogue, that uses a shunt is actually measuring
the voltage across the shunt. Putting two shunts in parallel will result
in both meters reading the same voltage across the resistance of both
shunts in parallel, so the readings will not be trustworthy.

A better way to do the job is to add a known resistor in series with the
supply, and measure the voltage across it. It needs to be a very low
value, though, to get an accurate result not affected by the voltage drop.

This is why I recently spent about eighty quid on a DC capable clamp meter.


--
Tciao for Now!

John.

[Jimk]

BattyPuss & D i m?
--
Jimk


----Android NewsGroup Reader----
http://usenet.sinaapp.com/

[Clive Arthur]

You'd have been better off spending your money on a copy of 'Electricity
for Beginners'.

--
Cheers
Clive

[John Williamson]

On 04/09/2020 19:18, Clive Arthur wrote:

> You'd have been better off spending your money on a copy of 'Electricity
> for Beginners'.
>

And in what way do you think I'm wrong?

Just use ohm's law and draw the circuit out to verify.

[The Natural Philosopher]

On 04/09/2020 17:52, Bob Latham wrote:
> In article <hrf497...@mid.individual.net>,

> Assuming we're talking DC, I think so yes.
>
No need to think. Kirchoff's law.*

> The difference in the two readings is probably caused by the
> different internal resistance of each meter used to measure the
> voltage drop and hence derive the current.
>
It is

> That would mean that each meter would be calibrated to assume a
> different voltage drop for full scale deflection at 10 amps.
>
yes

> Never the less, the current measurement of each meter is still
> correct for that meter and I believe they can be summed
>
No belief. it's a fact.

ob.
>

*Kirchhoff's current law (1st Law) states that current flowing into a
node (or a junction) must be equal to current flowing out of it. This is
a consequence of charge conservation.

--
How fortunate for governments that the people they administer don't think.

Adolf Hitler

[The Natural Philosopher]

On 04/09/2020 17:05, Theo wrote:

> In uk.d-i-y Custos Custodum <m...@privacy.net> wrote:
>> On Fri, 4 Sep 2020 16:20:09 +0100, williamwright
>> <wrights...@f2s.com> wrote:
>>
>>> I wanted to measure to consumption of a 12V pump. It was obviously going
>>> to be more than 10A when working hard, but I only had 10A multimeters. I
>>> used two meters in parallel, and added the readings. The two meters gave
>>> different readings, so obviously there was some imbalance. But will the
>>> summed readings I got be valid?
>>> Examples:
>>> 5.15A and 10.30A
>>> 3.32A and 7.31A
>>>
>>> Bill
>>
>> Kirchhoff is your friend.
>
> I think it's a bit more complicated than that.
>

No, it isn't


--
Microsoft : the best reason to go to Linux that ever existed.

[The Natural Philosopher]

On 04/09/2020 18:17, John Williamson wrote:

> On 04/09/2020 16:20, williamwright wrote:
>> I wanted to measure to consumption of a 12V pump. It was obviously going
>> to be more than 10A when working hard, but I only had 10A multimeters. I
>> used two meters in parallel, and added the readings. The two meters gave
>> different readings, so obviously there was some imbalance. But will the
>> summed readings I got be valid?
>> Examples:
>> 5.15A and 10.30A
>> 3.32A and 7.31A
>>
> Possibly.
>
> If the meters are moving coil or moving iron movements, rated at 10 amps
> each, with no shunts in place, then yes.
>
> Any meter, digital or analogue, that uses a shunt is actually measuring
> the voltage across the shunt. Putting two shunts in parallel will result
> in both meters reading the same voltage across the resistance of both
> shunts in parallel, so the readings will not be trustworthy.
>
> A better way to do the job

is to study electrical circuit theory, in particular Kirchoff, and STFU




--
"The most difficult subjects can be explained to the most slow witted
man if he has not formed any idea of them already; but the simplest
thing cannot be made clear to the most intelligent man if he is firmly
persuaded that he knows already, without a shadow of doubt, what is laid
before him."

- Leo Tolstoy

[The Natural Philosopher]

On 04/09/2020 19:18, Clive Arthur wrote:

+100

[The Natural Philosopher]

On 04/09/2020 19:23, John Williamson wrote:
> On 04/09/2020 19:18, Clive Arthur wrote:
>
>> You'd have been better off spending your money on a copy of 'Electricity
>> for Beginners'.
>>
> And in what way do you think I'm wrong?
>
> Just use ohm's law and draw the circuit out to verify.
>

Dont need to. Just use Kirchoff's law and add the two ammeters together

I mean if the sum of the currents through the meters isn't all going
through the load where the fuck is it going?

[alan_m]

On 04/09/2020 18:17, John Williamson wrote:

> Any meter, digital or analogue, that uses a shunt is actually measuring
> the voltage across the shunt. Putting two shunts in parallel will result
> in both meters reading the same voltage across the resistance of both
> shunts in parallel, so the readings will not be trustworthy.
>

Another vote here for saying you are wrong.

> A better way to do the job is to add a known resistor in series with the
> supply, and measure the voltage across it.

That is what the shunt in the meter is effectively doing. It's measuring
the current across a low value series resistor.

>
> This is why I recently spent about eighty quid on a DC capable clamp meter.
>

How does a DC clamp meter add a series resistor in series with the
supply? Hint: it doesn't.

[John Williamson]

On 04/09/2020 19:36, The Natural Philosopher wrote:
> On 04/09/2020 19:23, John Williamson wrote:

>> Just use ohm's law and draw the circuit out to verify.
>>
> Dont need to. Just use Kirchoff's law and add the two ammeters together
>
> I mean if the sum of the currents through the meters isn't all going
> through the load where the fuck is it going?
>

All the current is going through the shunts in a ratio determined by
their values, then through the load. This generates a voltage across the
pair of shunts. Each meter will be measuring the voltage across the
combined resistances, and displaying that as a current. Basically, any
multimeter will always be operating as a low voltage voltmeter, with
additional components such as a current shunt for measuring amps, a high
impedance network to measure higher voltages, or a constant current
power supply to measure resistance.

The calibration of each of the meters varies according to the value of
their shunt, so each meter is measuring the voltage across the shunts,
and displaying what it thinks is the current, but as the shunt
resistance is not as calibrated by the maker in the case given, they do
not read correctly. Two identical meters will read within the limits of
their accuracy and tolerances, but as these meters do not read the same,
given that there is a 2:1 ratio between the readings, near enough, the
readings are not to be trusted.

[John Williamson]

On 04/09/2020 19:34, The Natural Philosopher wrote:

>> A better way to do the job
> is to study electrical circuit theory, in particular Kirchoff, and STFU
>

Back at you. Kirchoffs law is modified by Ohms law and the manufacture
and design of the meters in this case.

In this case, Kirchoff's law applies only if the meter movements measure
current, not voltage, in the same way as the ammeter in my car works.

[alan_m]

On 04/09/2020 19:23, John Williamson wrote:

> On 04/09/2020 19:18, Clive Arthur wrote:
>
>> You'd have been better off spending your money on a copy of 'Electricity
>> for Beginners'.
>>
> And in what way do you think I'm wrong?
>
> Just use ohm's law and draw the circuit out to verify.
>

I suggest that you do the same.

You are making the mistake in assuming that two meter shunts in parallel
change the value of the shunts in each meter and the one meter has to
compensate for this.

If you have one meter with a 0.01 ohm shunt and there is 10 amps flowing
through it the meter reads 100mV across it. The meter scales this 100mV
= 10 Amps. This is exactly your suggested implementation for measuring
current of having a known resistor in series with the load and measuring
the voltage across it.

You now introduce another identical meter. You have two shunts in
parallel so the effective shunt resistance is now 0.005 ohms and 10A
amps flowing through 0.005 Ohms gives 50mV. Each meter scales this to 5Amps

BUT you have two meters both saying 5 Amps and 2 x 5 Amps = 10 Amps
flowing into the load.

As others have pointed out Kirchhoff says when you have two identical
resistors in parallel half the current will flow in each. So 5 Amps
flows through the 0.01 Ohm shunt in one meter and 5 Amps flows through
the 0.01 Ohm shunt in the other meter. The shunts don't physically
change their 0.01 Ohm value so 5 Amps through 0.01 Ohms gives 50mV BUT
again the total current is shared between tow measuring devices and
their reading have to be added to obtain the total.

It also doesn't matter if the shunt values are different in the two
meters as long as the scaling (measured volts across the shunt to amps
through the shunt) is calibrated to be correct.

[alan_m]

Doesn't the ammeter in your car have a natural full range of say +/-
micro Amps and it's the shunt connected to the terminals and current
sharing as defined by Kirchoff that allows it to report 10s of Amps?

[John Williamson]

On 04/09/2020 20:04, alan_m wrote:
> On 04/09/2020 18:17, John Williamson wrote:
>
>> A better way to do the job is to add a known resistor in series with
>> the supply, and measure the voltage across it.
>
> That is what the shunt in the meter is effectively doing. It's measuring
> the current across a low value series resistor.
>

And the voltage varies according to both the value of the shunt
resistance and the current. The meter is calibrated to show the correct
current only with the correct value of shunt resistance. If two shunts
are connected in parallel, this voltage will not correspond to the
current shown by either meter.

> How does a DC clamp meter add a series resistor in series with the
> supply? Hint: it doesn't.
>

I know this, which is why I bought one. It measures the magnetic field
round the cable, and works the current out from that. The oldest version
I own has a spring mounted lump of iron near a magnet, with the iron
attached to a needle. As the current varies, the net magnetic field
varies, and the needle moves. Made in the 1960s and still accurate at
currents between 5 and 300 amps.

[John Williamson]

On 04/09/2020 20:48, alan_m wrote:

> Doesn't the ammeter in your car have a natural full range of say +/-
> micro Amps and it's the shunt connected to the terminals and current
> sharing as defined by Kirchoff that allows it to report 10s of Amps?
>

No, it has a native sensitivity of +-60 Amps. I checked, there is no
shunt, and the coil wire is rather thick...

[NY]

"alan_m" <ju...@admac.myzen.co.uk> wrote in message
news:hrfk0e...@mid.individual.net...

The problem with using a shunt in parallel with an ammeter is that you need
very reproducible resistance which is of the order of an ohm or so.

When I was in my teens I built an analogue multimeter with several different
shunts, and a rotary switch to choose which one was connected across the
meter movement. I quickly learned that I could get a variety of readings
depending on how I wiggled the rotary switch, evidently because its contacts
had a resistance that was a similar order of magnitude to the high-current
range (lowest resistance) shunt.

I'm not sure how high-precision ammeters (eg Avo) get around that problem
while still presenting a very low in-series resistance to the circuit being
measured.

And how do digital meters do it? Still with shunts?

[Vir Campestris]

The problem Bill has is that he has two different meters. I thought Mr
Williamson had it right - but then I did the calculations. Even for
shunt meters that measure voltage, and will be reading the same voltage
across their different shunts, they'll be working out that the current
at that voltage. And that is the current running through.

Bill, I hope your meters were fused... because if the difference between
them had been bigger you'd have sent almost all the current through one
meter. Look slike you got away with it though!

Andy

[williamwright]

On 04/09/2020 17:38, alan_m wrote:
> On 04/09/2020 16:20, williamwright wrote:
>> I wanted to measure to consumption of a 12V pump. It was obviously
>> going to be more than 10A when working hard, but I only had 10A
>> multimeters. I used two meters in parallel, and added the readings.
>> The two meters gave different readings, so obviously there was some
>> imbalance. But will the summed readings I got be valid?
>> Examples:
>> 5.15A and 10.30A
>> 3.32A and 7.31A
>>
>> Bill
>
>
> Were they both the same model?
>

No. I have two identical ones but one of them only works on resistance
for some reason.
So one is an iso-tech IDM91E and the other is an Extech EX410A.

Bill

[RayL12]

On 04/09/2020 21:20, Bob Latham wrote:
> In article <hrfk2a...@mid.individual.net>,

> John Williamson <johnwil...@btinternet.com> wrote:
>> On 04/09/2020 20:04, alan_m wrote:
>>> On 04/09/2020 18:17, John Williamson wrote:
>>>
>>>> A better way to do the job is to add a known resistor in series with
>>>> the supply, and measure the voltage across it.
>>>
>>> That is what the shunt in the meter is effectively doing. It's measuring
>>> the current across a low value series resistor.
>
>>>
>> And the voltage varies according to both the value of the shunt
>> resistance and the current. The meter is calibrated to show the
>> correct current only with the correct value of shunt resistance.
>> If two shunts are connected in parallel, this voltage will not
>> correspond to the current shown by either meter.
>

> You are looking too hard it is much simpler than you are looking.
>
> Assume you have a 0.1 ohm shunt resistor used in the meter.
> The meter would be calibrated to show 10 amps when there was 1 volt
> across the meter.
> When 7 amps is flowing through that shunt the voltage drop across it
> will be 7 x 0.1 = 0.7 volts that will be calibrated to show 7 amps on
> the meter.
>
> Placing a parallel resistor across the multimeter will change the
> overall shunt resistance that is true, however, if the meter has 0.7
> v across it then I guarantee the meter is passing 7amps through its
> shunt regardless of any parallel circuit.
>
> The other meter is using a different value shunt resitor maybe 0.2
> ohms and may be calibrated to read 10 amps with 2 volts across the
> meter.
> Figures just to make sums easy, not real values!
>
> That's how both meters don't read the same when in parallel because
> of their calibration but are both accurate. The meter with the lower
> shunt resistance will hog more of the current.
>
> This is very basic ohms law.
>
>
> Bob.
>


How do 2 shunt resistors that are COMMONED at the source have different
voltages?

[alan_m]

On 04/09/2020 22:51, RayL12 wrote:

>
>
> How do 2 shunt resistors that are COMMONED at the source have different
> voltages?

They don't.

If the meter have different values of shunt resistors the conversion
between the voltage across the shunt and reported current will be
different.

Example
The load takes 10 Amps.

Meter 1 has a shunt resistor of 0.01 ohms
Meter 2 has a shunt resistor of 0.02 ohm

Using meter 1 alone to measure the current results in 10 Amps across
0.01 ohms giving a voltage across the shunt of 100mV. The meter is
calibrated to give a conversion factor of 100mV = 10 Amps - this never
changes.

Using meter 2 alone to measure the current results in 10 Amps across
0.02 ohms giving a voltage across the shunt of 200mV. The meter is
calibrated to give a conversion factor of 200mV = 10 Amps - again, this
never changes.


Put the two meters in parallel to measure the current.
2/3 or the current flows through meter 1.
1/3 of the current flows through meter 2.

The volts measured across the shunt in meter 1 = 66.666 mV
The volts measured across the shunt in meter 2 = 66.666 mV

The volts drop between the red and black leads on both meters is the
same @ 66.666mV

BUT
Meter 1 calibrated conversion is 100mV = 10A so 66.666mV = 6.6666A
and this is the value meter 1 reports on its display

Meter 2 calibrated conversion is 200mV = 10 A so 66,666mV = 3.3333A
and this is the value meter 2 reports on its display

Add the two together and you are back to the 10 Amps.

When two meter are connected in parallel the volts drop across the two
parallel shunt resistors is identical BUT it's NOT the same volts drop
if the meters are used individually.

[Theo]

In uk.d-i-y David Wade <g4...@dave.invalid> wrote:
> It doesn't need to. It measures voltage. The voltage across a resistor
> depends on the current flowing through it. You put 1 amp through a 1 ohm
> meter you get one volt. It doesn't matter what's in parallel with it, so
> long as 1 amp is flowing through the meter you get 1 volt across its
> resistor. So the meter reads 1amp for every volt.
>
> You put a resistor in parallel, say another 1 ohm, then the current
> through each resistor is now 0.5 amp so the voltage is now 0.5volt.
>
> The meter now reads 0.5 amp because that is what is flowing through it.

Ah yes, you're right. While the voltage each one reads is affected by the
other resistor, when I do an example the scale factors cancel out. So they
do actually display the right thing.

Theo

[alan_m]

On 04/09/2020 20:39, John Williamson wrote:

Kirchoff law still applies. At the shunt junction some of the current
goes through the shunt and some of the current goes to the voltage
measuring circuit BUT as the shunt is likely to be less than 1 ohm and
the input impedance of the voltage measuring circuit many tens of
MegaOhms the error introduced as a result is much less than 0.0001%

[Java Jive]

On 04/09/2020 20:55, Bob Latham wrote:
>
> In article <ritmpd$1v6m$1...@gioia.aioe.org>,

> Java Jive <ja...@evij.com.invalid> wrote:
>>
>> On 04/09/2020 16:20, williamwright wrote:
>>>
>>> I wanted to measure to consumption of a 12V pump. It was
>>> obviously going to be more than 10A when working hard, but I
>>> only had 10A multimeters. I used two meters in parallel, and
>>> added the readings. The two meters gave different readings, so
>>> obviously there was some imbalance. But will the summed readings
>>> I got be valid?
>>>
>>> Examples:
>>> 5.15A and 10.30A
>>> 3.32A and 7.31A
>>

>> It's a long time ago that I did any serious measurement in anger,
>> as it were, but the principle is correct. However, there is
>> clearly a significant discrepancy between the performance of your
>> ammeters, and I think that needs to be explained and allowed for
>> before putting any critical faith in the result.
>
> Dear me, JJ is wrong again and doesn't understand ohms law.

Reread the end of the first sentence that I wrote above. You might also
care to consider that I got a first in an electronics unit at
university, which I find rather more convincing than your infamous
self-confessed two hours' homework during the whole of your time at school.

>> AIR, it's all about internal resistance, when measuring voltage you
>> want it to be as high as possible, but when measuring current as
>> low as possible. Clearly there seems to be a significant
>> difference between your ammeters in this respect.
>
> That's because they use different current measuring voltage drop
> resistor values and different calibration, neither is faulty or
> unreliable.

Initially Bill didn't make it clear that they weren't identical meters,
which he now has, and, obviously wrongly in retrospect, for some reason
I was assuming they were identical, and therefore was worried by the
difference between them.

>> But, also AIR, these things can also have shunts, that either can
>> be physically attached or else more conveniently switched in and
>> out of circuit, for situations when the load is expected to be
>> beyond the tolerance of the meter. Could the difference between
>> the two be related to a shunt in the lower reading one?

So, it now seems this assumption about the differing shunt values was
entirely correct.

> I've explained above.

You've explained bollocks as usual. However, fortunately for Bill,
others have given very good and detailed explanations to back up what I
was merely the first to confirm, that the principle he was following was
correct.

--

Fake news kills!

I may be contacted via the contact address given on my website:
www.macfh.co.uk

[John Rumm]

Something that you can do with a moving coil meter designed for a single
measurement range. Multimeters (as their name would imply) can't be of
such a basic design.

In the days of analogue multimeters the galvanometer was a sensitive
current measuring device where typically only a few 10s of uA would
result in FSD. So combinations of series and parallel resistances would
be switched in to set the "range". The series resistance to both lower
the sensitivity of the movement, and to ensure that the load of the
galvanometer itself was negligible, and the parallel resistance to
bypass most of the current to keep the insertion resistance of the
multimeter to a low value, while also providing a small potential
difference proportional to the current being passed for the galvanometer
to measure. With modern digital meters, the measurement element is now
nearer to being is ideal voltage sensitive device, but the rest of the
principle remains the same.



--
Cheers,

John.

/=================================================================\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\=================================================================/

[John Rumm]

+1

I was about to go through a similar exercise - but you saved me the
bother. :-)

Ohms law is all you really need - looking at just one meter, if you see
a known voltage across a known value of shunt, then the sums tell you
the current passing through that resistor, and hence the meter as a
whole. There is nothing you can do outside of the meter that will change
that basic relationship. Additional parallel paths will just lower the
voltage seen across *your* shunt, and hence the proportion of the
current your meter will measure[1]

[1] In fact (all other things being equal) multiple meters will give a
slightly more accurate reading overall since the total insertion
resistance of the multiple meters will be lower than that of a single one.

[J. P. Gilliver (John)]

On Fri, 4 Sep 2020 at 20:56:19, NY <m...@privacy.invalid> wrote:
>"alan_m" <ju...@admac.myzen.co.uk> wrote in message
>news:hrfk0e...@mid.individual.net...
>> On 04/09/2020 20:39, John Williamson wrote:
>>> On 04/09/2020 19:34, The Natural Philosopher wrote:
>>>
>>>>> A better way to do the job
>>>> is to study electrical circuit theory, in particular Kirchoff, and STFU
>>>>
>>> Back at you. Kirchoffs law is modified by Ohms law and the
>>>manufacture and design of the meters in this case.

I've forgotten who is on which side, but yes, you can add the readings
of two current meters connected in parallel; each is reporting the
current going through its own shunt. The fact that you've connected
another meter's shunt in parallel will change the total shunt
resistance, and thus the reading on the first meter, but doesn't affect
the principle.

Say 8A is flowing, and you only have one meter in series with it: it
will show 8A.

Now you connect another meter in parallel with the first; the combined
resistance of the shunt falls, so the voltage measured across it will
fall; our meter might now show 6A - but the other meter will in that
case show 2A. (If you now disconnect the first meter, the second - now
only - meter will show 8A.)

The only slight theoretical effect, as John Rumm has said, is the one
due to the series resistance of the meter/meters itself/themselves; as
(again as someone has said) the combined resistance is lower with more
meters, this will actually be less significant with two or more meters -
but with modern digital meters (which Bill is using, from the number of
digits he's quoted), any variation here will be far less than the
accuracy with which the 12V supply voltage can be set anyway. (The
discrepancy is that you're supplying the 12V across the motor/meter
combination: what's actually across the motor will be _fractionally_
lower. But, as I said, the drop across the meter(s) is going to be far
less than the setting precision of the supply.) [A supply with sense
connections would get round even that concern, but would be way overkill
when the load is a pump motor!]
[]

>When I was in my teens I built an analogue multimeter with several
>different shunts, and a rotary switch to choose which one was connected
>across the meter movement. I quickly learned that I could get a variety
>of readings depending on how I wiggled the rotary switch, evidently
>because its contacts had a resistance that was a similar order of
>magnitude to the high-current range (lowest resistance) shunt.
>
>I'm not sure how high-precision ammeters (eg Avo) get around that
>problem while still presenting a very low in-series resistance to the
>circuit being measured.

They usually have a different terminal for the high-current ranges, thus
bypassing the switch altogether - for that reason, mainly.

>
>And how do digital meters do it? Still with shunts?

Yes, more so, as the basic module in them is a true voltmeter. But -
even (I'd say especially) low and medium-price ones - they too use a
different socket for high-current measurement.
--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

The breathtaking wonders of nature revealed to the soothing tones of Sir David
Attenborough. Life doesn't get much better than that.
- Ben Preston, Radio Times editor (2016/11/26-12/2)

[J. P. Gilliver (John)]

On Fri, 4 Sep 2020 at 20:49:29, John Williamson
<johnwil...@btinternet.com> wrote:
>On 04/09/2020 20:04, alan_m wrote:

[]

>> How does a DC clamp meter add a series resistor in series with the
>> supply? Hint: it doesn't.
>>
>I know this, which is why I bought one. It measures the magnetic field
>round the cable, and works the current out from that. The oldest
>version I own has a spring mounted lump of iron near a magnet, with the
>iron attached to a needle. As the current varies, the net magnetic
>field varies, and the needle moves. Made in the 1960s and still
>accurate at currents between 5 and 300 amps.
>

The older ones use the transformer effect - they make the through wire
the primary (one turn, or more like half a turn!) of a transformer, and
are thus capable of measuring AC only (and only for relatively large
currents). [I presume their calibration is only valid over a certain
range of mains frequencies, too.] Ones that can measure direct current
use a Hall effect sensor, requiring electronics to amplify its output -
i. e. if your current meter doesn't need a battery, it measures AC only.

The AC-only types are used in the electricity supply industry, are known
as "current transformers" or CTs, and are often built into things like
switchgear (usually with a "bar primary", i. e. the current being
measured is carried by a straight conductor); in the same way as you
should never short the output of a normal transformer, you should never
leave the output of a CT _open_ circuit, or it will damage itself; very
counter-intuitive for anyone brought up in electronics!

[John Williamson]

On 05/09/2020 03:41, J. P. Gilliver (John) wrote:

> The older ones use the transformer effect - they make the through wire
> the primary (one turn, or more like half a turn!) of a transformer, and
> are thus capable of measuring AC only (and only for relatively large
> currents). [I presume their calibration is only valid over a certain
> range of mains frequencies, too.] Ones that can measure direct current
> use a Hall effect sensor, requiring electronics to amplify its output -
> i. e. if your current meter doesn't need a battery, it measures AC only.
>

<Grin> This one was made by Lucas for use on vehicles, so it works only
on DC. AC just vibrates the needle and makes it buzz. Designed to
non-intrusively measure the starter and charging currents without
disturbing the vehicle wiring.

No hall effect sensors, which didn't exist in the 1950s when it was
designed, and no electronics. Like this one, except that mine has two
ranges, and two positions for the wire, one for starter current (0 - 300
Amps) and one for dynamo current (0 - 30 Amps).

http://www.stuttgartperformanceengineering.com/IMAGES/inductiveammeter/inducammpx2web.jpg

[Brian Gaff]

How do these ammeters work? Ie are they just shunted meters? Presumably not
clamp meters.

I'd have thought it would have been better to make up a shunt by
calculation so the meter could read the current and adjust it afterwards
using the ohms law and all that clever stuff my brain at this time of day
can't figure out. Nurse the screens.

Brian

--
--
This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...
bri...@blueyonder.co.uk
Blind user, so no pictures please
Note this Signature is meaningless.!
"williamwright" <wrights...@f2s.com> wrote in message
news:hrf497...@mid.individual.net...

>I wanted to measure to consumption of a 12V pump. It was obviously going to
>be more than 10A when working hard, but I only had 10A multimeters. I used
>two meters in parallel, and added the readings. The two meters gave
>different readings, so obviously there was some imbalance. But will the
>summed readings I got be valid?
> Examples:
> 5.15A and 10.30A
> 3.32A and 7.31A
>

> Bill

[Brian Gaff]

This is where analogue meters score of course.

Brian

--
--
This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...
bri...@blueyonder.co.uk
Blind user, so no pictures please
Note this Signature is meaningless.!

"Theo" <theom...@chiark.greenend.org.uk> wrote in message
news:KpF*YL...@news.chiark.greenend.org.uk...

> In uk.d-i-y Custos Custodum <m...@privacy.net> wrote:
>> On Fri, 4 Sep 2020 16:20:09 +0100, williamwright
>> <wrights...@f2s.com> wrote:
>>

>> >I wanted to measure to consumption of a 12V pump. It was obviously going
>> >to be more than 10A when working hard, but I only had 10A multimeters. I
>> >used two meters in parallel, and added the readings. The two meters gave
>> >different readings, so obviously there was some imbalance. But will the
>> >summed readings I got be valid?
>> >Examples:
>> >5.15A and 10.30A
>> >3.32A and 7.31A
>> >
>> >Bill
>>

>> Kirchhoff is your friend.
>
> I think it's a bit more complicated than that.
>

> Moving-coil ammeters are actually ammeters - the movement is proportional
> to
> the current. The current is split, so each will show directly the current
> passing through the coil and you can sum them.
>
> However digital ammeters aren't really ammeters, they're voltmeters. They
> measure the voltage across a fixed resistance and, by ohm's law, determine
> the current. The problem comes when you put two in parallel, becuase the
> two resistances are effectively in parallel too. So you change the
> effective resistance the voltmeter sees and thus the ohm's law calculation
> is wrong. Effectively both voltmeters are across the same pair of
> parallel
> resistors, but each meter doesn't know the resistance has changed.
>
> Now a decent digital ammeter will compensate for this effect (eg add
> components in series to isolate the measurement resistance), but that
> would
> depend on how they were constructed.
>
> I don't know if this is a real problem, or one that they all compensate
> for.
> I think I'd want to do some calibration runs to find out.
>
> Theo

[Paul]

williamwright wrote:
> I wanted to measure to consumption of a 12V pump. It was obviously going
> to be more than 10A when working hard, but I only had 10A multimeters. I
> used two meters in parallel, and added the readings. The two meters gave
> different readings, so obviously there was some imbalance. But will the
> summed readings I got be valid?
> Examples:
> 5.15A and 10.30A
> 3.32A and 7.31A
>
> Bill

Seeing as this is so difficult, this notion of adding two
numbers, try the following. Hook up three meters.
There's no shame in doing this, as it's all part of
the neat tricks you can do with meters. I like having
two meters handy, because of the things you can have
one meter do to another meter.

3.32
+---+
+----| A |----+
Src | +---+ | +---+
----+ +----| A |---- Load -----+
| | +---+ |
| +---+ | 10.63 |
+----| A |----+ |
+---+ |
Return 7.31 |
-------------------------------------------+

Already, I can hear plaintive bleating "you'll blow
the fuse on my third meter". OK then. Time for
another diagram. Use a piece of 14 gauge wire
for the bypass. Then check the "now wrong" readings
on the meter, to see that addition indeed happens
and is correct. In my made up example, roughly
half the current is flowing down my Rbypass,
which just magically is the correct equivalent
value to the network below it. If the wire I used
is less than the resistance of the network below,
the readings on all three meters will be less
than before, but the addition property still holds.
That's why I'm using a 14 gauge wire, to ensure the
ammeters don't pop a fuse when the pump starts
and the stall current is highest. The fuse in the
meter, if memory serves, isn't a slowblow, and
would likely pop at the least provocation.

+---------- Rbypass -----+ (bypass lots of the current)
| | (avoid blowing meters...)
| 1.66 | (check addition property)
| +---+ |
+----| A |----+ |
Src | +---+ | +---+ |
----+ +----| A |-+-- Load -----+
| | +---+ |
| +---+ | 5.27 |
+----| A |----+ |
+---+ |
Return 3.61 |
-------------------------------------------+

14 gauge wire (bypasses lots and lots)
+------------------------+
| |
| 0.7 | (addition property still holding...)
| +---+ |
+----| A |----+ |
Src | +---+ | +---+ |
----+ +----| A |-+-- Load -----+
| | +---+ |
| +---+ | 2.1 |
+----| A |----+ |
+---+ |
Return 1.4 |
-------------------------------------------+

Once you have proven to your satisfaction, the
additive property, the third meter can be removed,
knowing that addition still holds.

Now, you're excited. "Can I keep the bypass ?
That looks useful." Well, no, because copper wire
has a high thermal coefficient. If your meters
are good, the shunt inside uses manganin, with
a low tempco. If you use copper for a shunt, as
the wire gets hot, it throws off the divider property
a little bit and screws up your calibration. If you
were to use a bypass, the bypass should have a good
tempco property like the shunt component inside each
ammeter.

But considering the accuracy of this measurement,
it's a pump, sure, leave the copper wire and
use it to "scale" what you're doing.

In fact, using ye olde Ohms law, you can build
your own ammeter. Buy yourself a shunt and go to it.
The worst you can do, is burn up your new shunt
(Rmeasure). Work out the power dissipation using
I^2*R and ensure you're not exceeding the power
limit of your new shunt.

------------ Rmeasure --------------
^ ^
| |
+---- Volts -----+

It helps if you can find a four point shunt.
The top terminals are where the current flows.
The front terminals are for your voltmeter connection.
You can use screw-down lugs, such that with everything
screwed down, your setup won't fall apart while
you're juggling the motor and hoses.

http://www.deltecco.com/imageMKA_5_150.jpg

A typical device for sale might say

50A 75mV

Then you would know that your multimeter could
profitably have a 100mV full scale voltage readout,
and because they didn't make the numbers align
nicely, you'll have to do maths (scaling) to
work out the amps flowing. If it measured 20mV,
then about 12A would be flowing (I didn't use
a calculator, just a guesstimate). It's pretty
hard to measure 20mV with cheap meters. That's
why the meter didn't have a 50A scale in the
first place.

For power, 50A * 0.075V is 3.75W, and a thing
that size is going to get burning hot. But not
"glowing red" hot. And that's why the full scale
voltage they designed for, is so low and hard to
measure. If they made it develop 200mV to match
your 200mV scale on the meter, the power would be
three times higher, and we're talking serious
skin burns for a surface temperature on the shunt.

If the device has a calibration, well, you would
really appreciate that. Otherwise, if you
"just buy a hunk" of manganin, you'll need a
constant current source to calibrate with.
And that's a tall order at these current flow
levels. For a big ass shunt, you need that
calibration.

You would think making shunts would be a dull business, and
you wouldn't be wrong...

https://res.mdpi.com/d_attachment/materials/materials-10-00876/article_deploy/materials-10-00876.pdf

Paul

[Roger Hayter]

On 5 Sep 2020 at 09:46:01 BST, "Bob Latham" <b...@sick-of-spam.invalid> wrote:

> In article <riujej$1g4m$1...@gioia.aioe.org>,

> Java Jive <ja...@evij.com.invalid> wrote:
>> On 04/09/2020 20:55, Bob Latham wrote:
>> >
>> > In article <ritmpd$1v6m$1...@gioia.aioe.org>,
>> > Java Jive <ja...@evij.com.invalid> wrote:
>> >>
>> >> On 04/09/2020 16:20, williamwright wrote:
>> >>>
>> >>> I wanted to measure to consumption of a 12V pump. It was
>> >>> obviously going to be more than 10A when working hard, but I
>> >>> only had 10A multimeters. I used two meters in parallel, and
>> >>> added the readings. The two meters gave different readings, so
>> >>> obviously there was some imbalance. But will the summed readings
>> >>> I got be valid?
>> >>>
>> >>> Examples:
>> >>> 5.15A and 10.30A
>> >>> 3.32A and 7.31A
>> >>
>> >> It's a long time ago that I did any serious measurement in anger,
>> >> as it were, but the principle is correct. However, there is
>> >> clearly a significant discrepancy between the performance of your
>> >> ammeters, and I think that needs to be explained and allowed for
>> >> before putting any critical faith in the result.
>> >
>> > Dear me, JJ is wrong again and doesn't understand ohms law.
>
>> Reread the end of the first sentence that I wrote above.
>

> Ha, you were all at sea, have the grace to admit it, you couldn't see
> the simple issue as just ohms law.

>
>> You might also care to consider that I got a first in an
>> electronics unit at university,
>

> Did they teach ohms law, apparently not. It completely explains your
> left wing brain washing though.
>
Empirical evidence throws extreme doubt on the suggestion that engineering
degrees generally result in left wing brain washing.


--
Roger Hayter

[J. P. Gilliver (John)]

On Sat, 5 Sep 2020 at 09:42:51, Brian Gaff
<mildew...@blueyonder.co.uk> wrote:
>This is where analogue meters score of course.
> Brian
>

Not really; the principle of adding the reading from two current meters
in parallel applies whatever technology they use. (Probably easier to
add digits.)

--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

Science is magic that works.
- James Thompson @JamesPsychol 2020-3-17

[J. P. Gilliver (John)]

Interesting! In your earlier post you said "The oldest version I own has

a spring mounted lump of iron near a magnet, with the iron attached to a
needle. As the current varies, the net magnetic field varies, and the

needle moves.", so I assumed it was a clamp meter of some sort (and you
saying "non-intrusively" and "without disturbing the vehicle's wiring"
this time suggests it still is, though I can't see a clamp in the above
picture).

If it is a clamp meter, I'm interested to know what principle - if not
the Hall effect - it does work by.

Ah, I've just re-read what you say - "As the current varies" - so does
it only indicate a _change_ of current, which might be sufficient for
measuring starting current (though I'm not sure about "dynamo"). Or are
you saying it still indicates - i. e. the needle remains away from the
centre position - for a _steady_ current?

--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

[Java Jive]

On 05/09/2020 10:45, Roger Hayter wrote:
>
> On 5 Sep 2020 at 09:46:01 BST, "Bob Latham" <b...@sick-of-spam.invalid> wrote:
>>

>> [snip Bob's usual diarrhoea]

>>
>> Did they teach ohms law, apparently not. It completely explains your
>> left wing brain washing though.
>>
> Empirical evidence throws extreme doubt on the suggestion that engineering
> degrees generally result in left wing brain washing.

LOL!

Unfortunately, however, as his posting history repeatedly demonstrates,
Bob is useless at understanding empirical evidence!

[John Williamson]

It doesn't clamp on, you just lay the wire into the groove. And, yes, it
does give a steady reading for a steady direct current. There are no
electrical parts inside the meter, just a carefully positioned and
calibrated magnet, a piece of iron, a small spiral spring such as those
used in watch and clock escapements and a needle mounted on jewelled
bearings. Zero calibration is by moving the outer end of the spring
using a small screwdriver in a slot in the pivot.

The principle is that every wire carrying a current is surrounded by a
magnetic field whose strength varies with the current, and this field
interferes with the field generated by the magnet inside the meter, so
allowing the spring to move the needle one way as the field decreases,
or the increase in field moves it the other way. At zero current, the
field and spring hold the needle in the zero position, and a current in
the wire allows either the spring or the magnetic field to move the
needle in the appropriate direction. No batteries, no electronics, and
no current actually flowing inside the meter.

By careful design of the positions of the magnet and the two grooves
which hold the wire, this has two different ranges of measurement.

Oh, and when this was made, cars and commercial vehicles had dynamos,
not alternators.

[alan_m]

On 05/09/2020 11:31, J. P. Gilliver (John) wrote:

> Ah, I've just re-read what you say - "As the current varies" - so does
> it only indicate a _change_ of current, which might be sufficient for
> measuring starting current (though I'm not sure about "dynamo"). Or are
> you saying it still indicates - i. e. the needle remains away from the
> centre position - for a _steady_ current?

http://www.stuttgartperformanceengineering.com/inductiveammater.html

suggests how to use a similar device but also indicates that it may not
be too accurate (possibly in error by 20%)

[Ian Jackson]

In message <hrhc15...@mid.individual.net>, alan_m
<ju...@admac.myzen.co.uk> writes

I guess such a device will use what I was taught, in school physics
lessons, as the 'SNOW rule' (south to north over west) and/or the 'SNEW
rule' (south to north under west). [I feel a Google coming on.]
--
Ian

[Ian Jackson]

In message <riu0v0$f15$4...@dont-email.me>, The Natural Philosopher
<t...@invalid.invalid> writes



>
>*Kirchhoff's current law (1st Law) states that current flowing into a
>node (or a junction) must be equal to current flowing out of it. This
>is a consequence of charge conservation.
>
While this simply must be true, it does not, in itself, totally allay
the sneaky doubt as to whether the sum of the currents shown on the two
meters really does add up to the total going in and out. [It has been
explained why this doubt was misplaced - but it does no harm to
occasionally question the bleedin' obvious!]
--
Ian

[The Natural Philosopher]

On 04/09/2020 20:39, John Williamson wrote:
> On 04/09/2020 19:34, The Natural Philosopher wrote:
>
>>> A better way to do the job
>> is to study electrical circuit theory, in particular Kirchoff, and STFU
>>
> Back at you. Kirchoffs law is modified by Ohms law and the manufacture
> and design of the meters in this case.
>

> In this case, Kirchoff's law applies only if the meter movements measure
> current, not voltage, in the same way as the ammeter in my car works.
>
>

It does not matter in what ratio the current splits between the two
resistors. Each resistor will show the correct voltage according to its
v. drop

If one resistor is say twice the size of te other, it will take half the
current for the same v drop, and the meter will show half the current of
the other meter.

you have missed the wood for the trees.


--
Labour - a bunch of rich people convincing poor people to vote for rich
people by telling poor people that "other" rich people are the reason
they are poor.

Peter Thompson

[The Natural Philosopher]

On 04/09/2020 20:39, alan_m wrote:

> On 04/09/2020 19:23, John Williamson wrote:
>> On 04/09/2020 19:18, Clive Arthur wrote:
>>
>>> You'd have been better off spending your money on a copy of 'Electricity
>>> for Beginners'.
>>>
>> And in what way do you think I'm wrong?
>>
>> Just use ohm's law and draw the circuit out to verify.

>>
>
> I suggest that you do the same.
>
> You are making the mistake in assuming that two meter shunts in parallel
> change the value of the shunts in each meter and the one meter has to
> compensate for this.
>

No, you are making a mistake

The v drop is of course identical. Each meter will be calibrated to show
a current corresponding to its internal resistor at that v drop. A
little calculation which you are too lazy to do shows that it all works
correctly.


--
"It is an established fact to 97% confidence limits that left wing
conspirators see right wing conspiracies everywhere"

[The Natural Philosopher]

sigh. Ohms laws and kirchoffs laws are basic . So basic you shoudldn't
need to think about it.




--
If I had all the money I've spent on drink...
..I'd spend it on drink.

Sir Henry (at Rawlinson's End)

[John Williamson]

They are not accurate, and are only designed to be a tool to diagnose
major problems and narrow down the area of search. Checking whether a
motor is drawing current when trying to sort out a non working demister,
or whether the dynamo is actually producing any current at all when the
battery keeps going flat, that sort of thing.

Once they show you a problem, then it's worth getting a proper meter out
and starting to disconnect and test items individually.

[Roger Hayter]

On 5 Sep 2020 at 13:41:57 BST, "The Natural Philosopher" <t...@invalid.invalid>
wrote:

> On 05/09/2020 13:14, Ian Jackson wrote:
>> In message <riu0v0$f15$4...@dont-email.me>, The Natural Philosopher
>> <t...@invalid.invalid> writes
>>
>>
>>
>>>
>>> *Kirchhoff's current law (1st Law) states that current flowing into a
>>> node (or a junction) must be equal to current flowing out of it. This
>>> is a consequence of charge conservation.
>>>
>> While this simply must be true, it does not, in itself, totally allay
>> the sneaky doubt as to whether the sum of the currents shown on the two
>> meters really does add up to the total going in and out. [It has been
>> explained why this doubt was misplaced - but it does no harm to
>> occasionally question the bleedin' obvious!]
>
> sigh. Ohms laws and kirchoffs laws are basic . So basic you shoudldn't
> need to think about it.

To play devil's advocate on this, although you are correct, the accuracy of
the answer to the implied question: "What would be the current if the meters
were replaced by a wire?" is based on the assumption that the meters have
negligible effect. So one has to go a degree more sophisticated than Ohm's
law to decide, a) that in this case a typical multimeter-type current meter
has negligible effect, b) two current meters in parallel will have less effect
and c) neither meter contain a voltage or current generator of significance in
the context. None of these would affect the truth of Kirchoff's law, but any
could affect the quantity you are trying to measure before it shared itself
between the meters.

So a moment's thought about the validity of the measurement setup is, as
usual, well in order.




--
Roger Hayter

[Brian Gregory]

On 04/09/2020 17:05, Theo wrote:
> However digital ammeters aren't really ammeters, they're voltmeters. They
> measure the voltage across a fixed resistance and, by ohm's law, determine
> the current. The problem comes when you put two in parallel, becuase the
> two resistances are effectively in parallel too. So you change the
> effective resistance the voltmeter sees and thus the ohm's law calculation
> is wrong. Effectively both voltmeters are across the same pair of parallel
> resistors, but each meter doesn't know the resistance has changed.

You're thinking too hard and getting it wrong.

Of course adding the two results gives the correct answer.

--
Brian Gregory (in England).

[Brian Gregory]

On 04/09/2020 20:17, John Williamson wrote:

> On 04/09/2020 19:36, The Natural Philosopher wrote:
>> On 04/09/2020 19:23, John Williamson wrote:
>
>>> Just use ohm's law and draw the circuit out to verify.
>>>

>> Dont need to. Just use Kirchoff's law and add the two ammeters together
>>
>> I mean if the sum of the currents through the meters isn't all going
>> through the load where the fuck is it going?
>>
> All the current is going through the shunts in a ratio determined by
> their values, then through the load. This generates a voltage across the
> pair of shunts. Each meter will be measuring the voltage across the
> combined resistances, and displaying that as a current. Basically, any
> multimeter will always be operating as a low voltage voltmeter, with
> additional components such as a current shunt for measuring amps, a high
> impedance network to measure higher voltages, or a constant current
> power supply to measure resistance.
>
> The calibration of each of the meters varies according to the value of
> their shunt, so each meter is measuring the voltage across the shunts,
> and displaying what it thinks is the current, but as the shunt
> resistance is not as calibrated by the maker in the case given, they do
> not read correctly. Two identical meters will read within the limits of
> their accuracy and tolerances, but as these meters do not read the same,
> given that there is a 2:1 ratio between the readings, near enough, the
> readings are not to be trusted.
>

Idiot.

[Brian Gregory]

On 05/09/2020 20:11, Tim Streater wrote:
>
> While correct, you're a bit late to the party.
>

I can assure you that in fact everybody else was early.

I checked my invitation.

[Indy Jess John]

On 05/09/2020 11:17, J. P. Gilliver (John) wrote:
> On Sat, 5 Sep 2020 at 09:42:51, Brian Gaff
> <mildew...@blueyonder.co.uk> wrote:
>> This is where analogue meters score of course.
>> Brian
>>
> Not really; the principle of adding the reading from two current meters
> in parallel applies whatever technology they use. (Probably easier to
> add digits.)

A similar approach was taken for weighing clinically obese people where
the readings of a set of scales beneath each foot were added together
when the estimated weight exceeded the capacity of one set.

Jim

[J. P. Gilliver (John)]

On Sat, 5 Sep 2020 at 12:44:28, John Williamson
<johnwil...@btinternet.com> wrote:
[]

>It doesn't clamp on, you just lay the wire into the groove. And, yes,
>it does give a steady reading for a steady direct current. There are no
>electrical parts inside the meter, just a carefully positioned and
>calibrated magnet, a piece of iron, a small spiral spring such as those
>used in watch and clock escapements and a needle mounted on jewelled
>bearings. Zero calibration is by moving the outer end of the spring
>using a small screwdriver in a slot in the pivot.
>
>The principle is that every wire carrying a current is surrounded by a
>magnetic field whose strength varies with the current, and this field
>interferes with the field generated by the magnet inside the meter, so
>allowing the spring to move the needle one way as the field decreases,
>or the increase in field moves it the other way. At zero current, the
>field and spring hold the needle in the zero position, and a current in
>the wire allows either the spring or the magnetic field to move the
>needle in the appropriate direction. No batteries, no electronics, and
>no current actually flowing inside the meter.

Fascinating, thanks.

>
>By careful design of the positions of the magnet and the two grooves
>which hold the wire, this has two different ranges of measurement.
>
>Oh, and when this was made, cars and commercial vehicles had dynamos,
>not alternators.
>

Indeed. A development (to alternators) for which I'm very thankful!

--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

And on the question of authorship, I subscribe to the view that the plays were
not in fact written by Shakespeare but by someone of the same name.
- Hugh Bonneville (RT 2014/10/11-17)

[RayL12]

On 04/09/2020 23:38, alan_m wrote:
> On 04/09/2020 22:51, RayL12 wrote:
>
>>
>>
>> How do 2 shunt resistors that are COMMONED at the source have
>> different voltages?
>
>
> They don't.

<snip>

Cheers Alan. I didn't bother reading the rest as it was not my concern.

My concern was with the idea that 2 resistors in parallel can be put
into a single path and display 2 different voltage readings across them.
If both shunts are tied to the source and the ground they share the same
voltage.

[Indy Jess John]

On 05/09/2020 23:06, J. P. Gilliver (John) wrote:
> On Sat, 5 Sep 2020 at 12:44:28, John Williamson
> <johnwil...@btinternet.com> wrote:

>> Oh, and when this was made, cars and commercial vehicles had dynamos,
>> not alternators.
>>
> Indeed. A development (to alternators) for which I'm very thankful!

There are some benefits from a dynamo too.

I have an old car with a dynamo. A couple of winters ago a neighbour who
was going away for Christmas knocked on my door and asked if I had any
jump leads (and I had). Apparently her diesel car had failed to start
and she had run the battery down trying.

I put my old car next to hers and linked the batteries, and as a
precaution I left my car idling. I tried to start hers and despite the
light coming on showing the glow plug circuit was operating and then
going out to show that it was ready for starting, the engine showed no
sign of starting when I turned the key. It was obvious that the glow
plugs weren't heating.

An alternator is current limited, and there is no way it can deliver a
continuous 100+ amps to turn a starter motor, but a dynamo wired through
an original control box has no such limitation. So I propped my car's
throttle giving somewhere around 2500 revs and then went to the
neighbours car, gave it a quarter throttle and turned the key and kept
it turned and I watched the rev counter as it started at just under 500
revs and as the compression warmed things up it gradually climbed the
scale until just about a minute and a half later it got to 1500 revs and
the engine suddenly fired up.

By this time I had smoke coming out of my starter motor, but it
obviously wasn't badly damaged because it has started my car ever since.
I told my neighbour that she wouldn't have any trouble restarting once
the engine had warmed up, but she needed the glow plugs repaired before
it would start from cold again.

Jim

[williamwright]

On 06/09/2020 10:38, Indy Jess John wrote:

> I put my old car next to hers and linked the batteries, and as a
> precaution I left my car idling. I tried to start hers and despite the
> light coming on showing the glow plug circuit was operating and then
> going out to show that it was ready for starting, the engine showed no
> sign of starting when I turned the key. It was obvious that the glow
> plugs weren't heating.
>
> An alternator is current limited, and there is no way it can deliver a
> continuous 100+ amps to turn a starter motor, but a dynamo wired through
> an original control box has no such limitation. So I propped my car's
> throttle giving somewhere around 2500 revs and then went to the
> neighbours car, gave it a quarter throttle and turned the key and kept
> it turned and I watched the rev counter as it started at just under 500
> revs and as the compression warmed things up it gradually climbed the
> scale until just about a minute and a half later it got to 1500 revs and
> the engine suddenly fired up.
>
> By this time I had smoke coming out of my starter motor, but it
> obviously wasn't badly damaged because it has started my car ever since.
> I told my neighbour that she wouldn't have any trouble restarting once
> the engine had warmed up, but she needed the glow plugs repaired before
> it would start from cold again.
>
> Jim

I don't understand what you did.
1. I thought dynamo control boxes limited the output to 22A
2. Why had all four glow plugs failed on the same occasion?
3. Why did your starter motor smoke?

Bill

[tony sayer]

In article <n625H.302661$Dg6.1...@fx01.am4>, Indy Jess John <jimwarren
@OMITblueyonder.co.uk> scribeth thus

Errmm.. Why did you have smoke coming from your starter motor don't you
mean your dynamo?...

>obviously wasn't badly damaged because it has started my car ever since.
>I told my neighbour that she wouldn't have any trouble restarting once
>the engine had warmed up, but she needed the glow plugs repaired before
>it would start from cold again.
>
>Jim

--
Tony Sayer


Man is least himself when he talks in his own person.

Give him a keyboard, and he will reveal himself.

[J. P. Gilliver (John)]

On Sun, 6 Sep 2020 at 13:36:20, williamwright <wrights...@f2s.com>
wrote:

>On 06/09/2020 10:38, Indy Jess John wrote:
>
>> I put my old car next to hers and linked the batteries, and as a
>>precaution I left my car idling.

(I'd always do that, I think.)

>> I tried to start hers and despite the light coming on showing the
>>glow plug circuit was operating and then going out to show that it
>>was ready for starting, the engine showed no sign of starting when I
>>turned the key. It was obvious that the glow plugs weren't heating.

Yes, the going out is often just a timer, rather than detecting warming.

>> An alternator is current limited, and there is no way it can deliver
>>a continuous 100+ amps to turn a starter motor, but a dynamo wired
>>through an original control box has no such limitation. So I propped
>>my car's throttle giving somewhere around 2500 revs and then went to
>>the neighbours car, gave it a quarter throttle and turned the key and
>>kept it turned and I watched the rev counter as it started at just
>>under 500 revs and as the compression warmed things up it gradually
>>climbed the scale until just about a minute and a half later it got
>>to 1500 revs and the engine suddenly fired up.
>> By this time I had smoke coming out of my starter motor, but it
>>obviously wasn't badly damaged because it has started my car ever
>>since. I told my neighbour that she wouldn't have any trouble
>>restarting once the engine had warmed up, but she needed the glow
>>plugs repaired before it would start from cold again.
>> Jim
>
>I don't understand what you did.
>1. I thought dynamo control boxes limited the output to 22A

(I don't know anything about that.)

>2. Why had all four glow plugs failed on the same occasion?

They'd probably failed gradually over time; depending on weather etc., a
diesel engine may start from cold with only three, two, or possibly even
only one glowplug still operating (I think in particularly warm weather,
it can with none). What probably happened here was they'd been
deteriorating, and the failure (to start) happened on the first really
cold morning. (The owner would perhaps have been experiencing needing
more and more cranking to start, and would be pleased at how well it
started after she'd had them all changed.)

>3. Why did your starter motor smoke?

I wondered that, then thought: is it possible that Indy's car uses the
starter motor _as_ the dynamo?
>
>Bill
John

--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

"Offer me a drink." "It's six o'clock in the morning!" "Then float a Cheerio
in it." - Franks and Gibbs, NCIS

[Paul Ratcliffe]

On Sun, 6 Sep 2020 13:36:20 +0100, williamwright <wrights...@f2s.com> wrote:

> 3. Why did your starter motor smoke?

Peer pressure?

[Indy Jess John]

On 06/09/2020 13:46, tony sayer wrote:

> Errmm.. Why did you have smoke coming from your starter motor don't you
> mean your dynamo?...

Yes I did mean dynamo.

Brain fade!

Jim

[Indy Jess John]

On 06/09/2020 13:36, williamwright wrote:

> I don't understand what you did.
> 1. I thought dynamo control boxes limited the output to 22A

Maybe it is supposed to. Mine clearly didn't.
The workshop manual describes the operation of the control box as a
voltage regulator "incorporating constant voltage control for the
charging circuit", with the contacts opening to prevent over-voltage
being output. The implication is that if I am taking a lot of amps out
of the dynamo, the voltage is unlikely to exceed the contact opening
voltage and everything the dynamo produces goes down my jump leads.

> 2. Why had all four glow plugs failed on the same occasion?

I assumed it was a circuit board failure or a blown fuse that stopped
them all working at the same time, but I have never looked at the way
they work to see the possibilities. I have seen a diesel tractor started
from cold using smouldering blotting paper as a glow plug, so it
obviously doesn't need anything complicated in a car to provide heat,
but it probably has associated computer linkage for the fault diagnostic
system to report faults, and circuit boards do fail.

Elsewhere in this thread someone has suggested that perhaps the glow
plugs failed progressively and when the last one failed the engine
wouldn't start. That is a possible alternative.

I have no idea which is correct.

> 3. Why did your starter motor smoke?

It was the dynamo not the starter motor. I typed the wrong thing.
It was driving another car's starter motor (in parallel with my own
battery) for a lot longer than would normally be the case, and the
windings will have heated from the number of watts being generated.
Dynamos don't have much of an integral fan for cooling.

Jim

[williamwright]

So re-reading your text, you use the dynamo to start the engine?

Bill

[Indy Jess John]

I used the power from my dynamo to drive the starter motor in the other car.

Jim

[tony sayer]

In article <ksc5H.3087857$aGl.2...@fx36.am4>, Indy Jess John
<jimw...@OMITblueyonder.co.uk> scribeth thus

Ever thought of converting to an Alternator?.

Did this with several old cars over time and it worked very well:)...

[Indy Jess John]

On 07/09/2020 09:21, tony sayer wrote:

> Ever thought of converting to an Alternator?.
>
> Did this with several old cars over time and it worked very well:)...

I have kept the car configured as it was when it left the factory, and
the dynamo and control box combination works OK (and I have a spare
dynamo in my garage in case it is needed).

I am also aware that I could convert the electrics to negative earth,
but I haven't done that either, though I have installed a cigarette
lighter socket delivering a negative earth connection so that I can use
a car battery powered tyre pump or other gadgets like a Satnav which
require that polarity.

Jim

[John Williamson]

On 07/09/2020 10:22, Indy Jess John wrote:

> I have kept the car configured as it was when it left the factory, and
> the dynamo and control box combination works OK (and I have a spare
> dynamo in my garage in case it is needed).
>
> I am also aware that I could convert the electrics to negative earth,
> but I haven't done that either, though I have installed a cigarette
> lighter socket delivering a negative earth connection so that I can use
> a car battery powered tyre pump or other gadgets like a Satnav which
> require that polarity.
>
> Jim
>

Converting to negative earth will reduce corrosion, and is worth doing
for that reason alone. The only reason not to do it is of the existing
factory fitted radio needs a positive earthed supply.

It does not affect the originality, and is easily reversed.

[J. P. Gilliver (John)]

On Sun, 6 Sep 2020 at 22:24:31, Indy Jess John
<jimw...@OMITblueyonder.co.uk> wrote:
>On 06/09/2020 13:36, williamwright wrote:

[]

>> 2. Why had all four glow plugs failed on the same occasion?

[]

>it obviously doesn't need anything complicated in a car to provide
>heat, but it probably has associated computer linkage for the fault
>diagnostic system to report faults, and circuit boards do fail.
>
>Elsewhere in this thread someone has suggested that perhaps the glow
>plugs failed progressively and when the last one failed the engine
>wouldn't start. That is a possible alternative.

That was me. When I first got my (Diesel) Škoda Felicia (far from new!
But I presume it had recently had the 'plugs replaced), it seemed to
start within a quarter of a turn, if I waited for the light to go out
before cranking; I'm not exaggerating, it really did seem to fire as
soon as the engine was turned. It progressively got harder to start, so
I eventually had them replaced, which brought it back to similar (though
not quite as good). I don't think it had any fancy diagnostics, though
if your neighbour's is modern, that might have.

>
>I have no idea which is correct.
>
>> 3. Why did your starter motor smoke?

(I loved the "peer pressure" response!)

>It was the dynamo not the starter motor. I typed the wrong thing.

Ah, so I guessed wrong.

It always struck me as odd - especially in the days of dynamos - that
cars had the two devices; surely one could serve as both. Presumably the
design differences needed for operation at different speeds/durations
made it less bother just to fit two devices.
[]

--
J. P. Gilliver. UMRA: 1960/<1985 MB++G()AL-IS-Ch++(p)Ar@T+H+Sh0!:`)DNAf

User Error: Replace user, hit any key to continue.

[charles]

In article <Ktz6AvZp...@255soft.uk>,

There was a car in the 60's that had a unit doing both jobs. Was it the
Beetle?

--
from KT24 in Surrey, England
"I'd rather die of exhaustion than die of boredom" Thomas Carlyle

[Paul_news]

On 07/09/2020 17:12, J. P. Gilliver (John) wrote:
> It always struck me as odd - especially in the days of dynamos - that
> cars had the two devices; surely one could serve as both. Presumably the
> design differences needed for operation at different speeds/durations
> made it less bother just to fit two devices.

Trouble is trying to do 2 jobs it likely does neither well.
Starters are quite low geared but dynamos about engine speed.
At least 1 cheap (eastern-european?) car had a combined starter-dynamo.
I can't remember the make.
As a car tinkerer I was asked to look at 1 by a co-worker, took 1 look
at the manual and declined.

Paul.

[Indy Jess John]

My brother bought an old Gogomobile in the 1960s, dirt cheap because it
was advertised as a non-runner and he had to get a friend with a car to
tow it round to his house. It was a Gogomobile Royal, like this
https://www.super-hobby.de/zdjecia/4/5/1/2255_rd.jpg

It looked like a Daf 33 in body shape, but it had a BMW horizontally
opposed twin air cooled engine in the front to power it via the rear
axle, and it came with a handbook written in German which nobody could
understand. However it had pictures and diagrams so it was possible to
work out most things, and when I went round to help him work out why it
wouldn't go, there was a diagram in the middle of a lot of German words
of something that could have been a fuel filter. So we followed the
fuel pipe back from the carburettor and eventually found this fuel
filter behind the rear axle, and when we removed the very mucky glass
bowl the filter inside was completely clogged with muck. The quick
answer was to throw away the filter and reassemble it without one.

Then the way the engine started was a puzzle, because there was no sign
of a starter motor, or any kick start mechanism but there was a diagram
showing that there was a "lichtsmaschin" (I can't remember exactly how
it was spelt) on the front of the engine underneath some airflow cowling
for cooling. That was wired through some complicated relay arrangement
to the ignition switch and the battery. When the key was turned it
operated a relay and connected the battery to one set of wires going
into the cowling, and when the key was released it dropped the relay and
connected another set of wires to the battery. It was a single device
attached to the crankshaft that turned the engine to start it and one
started it charged the battery.

After charging the battery and winding the engine over for a while for
the mechanical pump to get the petrol from the tank to the engine it did
start.

Gogomobiles were made in a Bavarian factory, so not quite eastern
European, and BMW bought the company in 1967 to gain ownership of some
useful patents.

Jim

[williamwright]

Ah! I'm not sure if my skills of comprehension are failing or if your
account was slightly unclear!

Bill

[williamwright]

On 07/09/2020 10:22, Indy Jess John wrote:

> I have kept the car configured as it was when it left the factory, and
> the dynamo and control box combination works OK (and I have a spare
> dynamo in my garage in case it is needed).
>
> I am also aware that I could convert the electrics to negative earth,
> but I haven't done that either, though I have installed a cigarette
> lighter socket delivering a negative earth connection so that I can use
> a car battery powered tyre pump or other gadgets like a Satnav which
> require that polarity.
>
> Jim
>

Sounds risky!

Bill

[Indy Jess John]

It is in an non-conducting box. It works perfectly.

Jim

[Indy Jess John]

1 Heavy duty[1] jump leads connect my battery to neighbour's flat battery.
2 I intend to keep the neighbour's starter motor turning for as long as
it takes to get an engine designed to be assisted by glow plugs to start
from compression alone on a frosty morning.
3 I expect that to take more Amp Hours than my battery holds so I run my
engine at enough revs to get plenty of supplementary power from my dynamo.
4 I get in neighbour's car and turn the ignition key and hold it in the
Start position. It does take quite a while but neighbour's engine
finally starts, by which time the windings in my dynamo have got hot
enough to start smoking.
5 Disconnect jump leads, so that neighbour's alternator charges her
battery and my dynamo recharges mine.
6 Neighbour drives off. I go back to my parking place and then run the
engine for another 5 minutes to get a reasonable level of charge in my
battery before switching off.

[1] Home made jump leads using cables intended for an arc welder.

Jim

[Java Jive]

On 08/09/2020 14:41, williamwright wrote:
>

But potentially good for lighting cigarettes!

--

Fake news kills!

I may be contacted via the contact address given on my website:
www.macfh.co.uk

[No Name]

On 08/09/2020 14:41, williamwright wrote:

That reminds me.... When I was doing work experience with the
electricity boards in the very late 1980's, all their works vehicles all
had signs on the dashboard saying

WARNING THIS VEHICLE IS POSITIVE EARTH

and negative earth became commonplace in the 1970's.....

I enquired around other vans for the water and gas companies then and
they all had the same warning sticker

Does anyone know why?

S

[R. Mark Clayton]

On Friday, 4 September 2020 16:20:11 UTC+1, wrights...@aol.com wrote:
> I wanted to measure to consumption of a 12V pump. It was obviously going
> to be more than 10A when working hard, but I only had 10A multimeters. I
> used two meters in parallel, and added the readings. The two meters gave
> different readings, so obviously there was some imbalance. But will the
> summed readings I got be valid?
> Examples:
> 5.15A and 10.30A
> 3.32A and 7.31A
>
> Bill

The readings will be different because of the different sensitivity of the meters (usually measured in k ohms per volt).

If the meters are in parallel and 4A is going through one and 8A is going through the other then the total current is of course 12A, so yes you can just sum them.

[R. Mark Clayton]

On Friday, 4 September 2020 17:05:32 UTC+1, Theo wrote:
> In uk.d-i-y Custos Custodum <m...@privacy.net> wrote:

> > On Fri, 4 Sep 2020 16:20:09 +0100, williamwright
> > <wrights...@f2s.com> wrote:
> >
> > >I wanted to measure to consumption of a 12V pump. It was obviously going
> > >to be more than 10A when working hard, but I only had 10A multimeters. I
> > >used two meters in parallel, and added the readings. The two meters gave
> > >different readings, so obviously there was some imbalance. But will the
> > >summed readings I got be valid?
> > >Examples:
> > >5.15A and 10.30A
> > >3.32A and 7.31A
> > >
> > >Bill
> >

> > Kirchhoff is your friend.
>
> I think it's a bit more complicated than that.
>
> Moving-coil ammeters are actually ammeters - the movement is proportional to
> the current. The current is split, so each will show directly the current
> passing through the coil and you can sum them.
>
> However digital ammeters aren't really ammeters, they're voltmeters. They
> measure the voltage across a fixed resistance and, by ohm's law, determine
> the current. The problem comes when you put two in parallel, becuase the
> two resistances are effectively in parallel too. So you change the
> effective resistance the voltmeter sees and thus the ohm's law calculation
> is wrong. Effectively both voltmeters are across the same pair of parallel
> resistors, but each meter doesn't know the resistance has changed.
>
> Now a decent digital ammeter will compensate for this effect (eg add
> components in series to isolate the measurement resistance), but that would
> depend on how they were constructed.
>
> I don't know if this is a real problem, or one that they all compensate for.
> I think I'd want to do some calibration runs to find out.
>
> Theo

Afraid not the current will divide between the two meters in inverse proportion to their shunt resistance (the meter measures the tiny voltage across it) as in V = IR, so each meter will tell you how much current is flowing through it. It neither knows nor cares what current is flowing in the other arm, and the shunt resistance in each meter won't change (unless it is auto-ranging) so you can just sum them.

[williamwright]

On 08/09/2020 15:52, Indy Jess John wrote:

>> Sounds risky!
>>
>> Bill
>
> It is in an non-conducting box. It works perfectly.
>
> Jim
>

I'd forget and connect up something with exposed metal parts.

Bill

[Indy Jess John]

But to be a problem the exposed metal parts have to make a short circuit.
Inside the car there is only the (painted) seat frame or the door
handles that your "something" might short against, because the seats are
upholstered and the floor is carpeted and the parcel shelf is hardboard.
Also the wire to the new cigarette lighter socket is protected by its
own in-line fuse, so a short would blow it.

Jim

[williamwright]

On 08/09/2020 20:42, Indy Jess John wrote:

>> I'd forget and connect up something with exposed metal parts.
>>
>> Bill
>
> But to be a problem the exposed metal parts have to make a short circuit.
> Inside the car there is only the (painted) seat frame or the door
> handles that your "something" might short against, because the seats are
> upholstered and the floor is carpeted and the parcel shelf is hardboard.
> Also the wire to the new cigarette lighter socket is protected by its
> own in-line fuse, so a short would blow it.
>
> Jim
>

Don't worry, I'd fuck it up.

Bill

[Paul]

If running appliance loads off the lighter socket, you
might want to know that.

Paul

[Indy Jess John]

I won't be offering you a lift then :-)

Jim

[RayL12]

On 08/09/2020 16:47, Java Jive wrote:
> On 08/09/2020 14:41, williamwright wrote:
>>
>> On 07/09/2020 10:22, Indy Jess John wrote:
>>>
>>> I am also aware that I could convert the electrics to negative earth,
>>> but I haven't done that either, though I have installed a cigarette
>>> lighter socket delivering a negative earth connection so that I can
>>> use a car battery powered tyre pump or other gadgets like a Satnav
>>> which require that polarity.
>>>
>>> Jim
>>
>> Sounds risky!
>
> But potentially good for lighting cigarettes!
>

LOL. Excellent picture :-)

[williamwright]

Best not.

Bill