All correct, although only when *some* current is flowing, and exactly what
I was saying when I explained how inserting a *known* resistance in a
circuit allows you to use a measured voltage to calculate the current and
resistance in the rest of the circuit as an alternative to measuring current
directly if one doesn't have an ammeter. The whole point is that unless you
do that, or measure current directly, you have no idea what current, if any,
is flowing in the circuit. Measuring the volt-drop across individual
connectors or switches will show zero volts if they are making a good
connection, *but it will also show zero volts if no current is flowing at
all because of an open-circuit elsewhere* which may be the case if an OD
(for example) isn't operating at all and the there could be two or more
faults. You can make all the *voltage* measurements you like on the circuit
as delivered by the factory, and you still won't know whether the solenoid
itself or its earth are high resistance or open-circuit. An open-circuit is
simply a very high resistance fault, it's irrelevant whether one is more
likely than the other. There is no point using a test method that will only
reveal one type of fault but not the other. Measuring the current (or
calculating it using a known resistance in series) will reveal both equally
well.
You might want to revisit your article on the Healey6 site (and incidentally
I object to sites trying to install ActiveX objects on my computer
surreptitiously). You write:
"How an ohmmeter works:
An ohmmeter puts a very small voltage (about .6-.75 Volts at 0.
no amps) out on one lead, and measures what comes back on the
other lead."
This goes completely against Ohms Law (which *is* a Law and not a Theorum
because it always works) which states that "the current through a conductor
between two points is directly proportional to the potential difference or
voltage across the two points, and inversely proportional to the resistance
between them." An ohmmeter applies a voltage to the resistance being
measured which passes a current through it and the ohmmeter, and the
ohmmeter displays that current on a scale calibrated for resistance. You
can test this for yourselves if you have two instruments. Select ohms on
one and volts on the other, and connect the two sets of probes together.
The ohmmeter will show a very high resistance, and the voltmeter will show a
voltage. Now switch the voltmeter to current and the ohmmeter will show a
low resistance and the ammeter a current. My analogue ohmmeter outputs 1.5v
(i.e. the voltage of the battery) at 35mA, a digital 0.2v at 0.8mA, it will
vary from instrument to instrument.
PaulH.
----- Original Message -----
You still don't understand. It is more likely a circuit with high
resistance hot, not an open circuit hot (those are pretty rare)
Assume you have a circuit drawing 8A @12V (assumed, I understand that
the actual voltage will vary, but this is unimportant). Now further
assume that you have a Lucas bullet connector with .3 ohms resistance.
What is going to happen to the voltage across that connector?
Will it stay the same?
Will it go up?
Will it go down?
use Ohms law and solve for voltage E=IR E= 8 X .3 = 2.4V
With an 8 am draw, a .3 ohm resistance will cause a 2.4 volt drop
across that connector.
If you put a voltmeter with the red lead on one side of that
connector, and the black lead on the other side of that connector with
the circuit on, it will register 2.4V. This voltage drop demonstrates
that there is excessive resistance in that connection.
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