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  #1  
Old 02-03-2006, 02:05 PM
Gary Ganaway's Avatar
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Location: Cincinnati, Ohio
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Multimeter-stupid question but don't laugh

I can do most things mechanical on my cars but am at lost when trying to diagnose some problems using the multimeter.My question is how do you connect the multimeter to check for volts, resistance, etc.

For example, how would I connect it to check if cold start valve is functioning ok, also the water temp sensor located at rear of right side of cylinder head on my W126, 380SE. I have the W126 cd which gives the test procedures but when they say connect to this and that I can't figure out exactly how ato do it. Thanks

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  #2  
Old 02-03-2006, 02:19 PM
LarryBible
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The important thing in Electrical/Electronic troubleshooting is that you UNDERSTAND the circuit. Using the meter is the easy part. I would suggest that you find a good book on the basics or take a JC course on basic electricity. Once you learn Ohms Law you will know the alphabet so that you can figure out the rest of it.

That said, when checking for voltage, ground the black lead and use the red lead in the DC Volts mode to probe for voltage wherever the procedure is telling you.

To check for resistance, go to the Ohms mode and connect the leads together to make sure that you have near zero Ohms, then start testing the component.

LOTS of this will be determined by your particular meter.

You might be best served just buying a test light and using that. There are MANY techs who don't even own a multimeter but do all sorts of troubleshooting with a test light.

Good luck,
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  #3  
Old 02-03-2006, 04:00 PM
MrCjames's Avatar
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Voltage is always measured parallel to the source, on cars the source is normally the “Battery.” When you measure for available voltage you will typically connect your meters black lead (As Larry stated) to a ground connection. The red lead can be connected to any number of points to determine if you have power at the point being tested; however it will not tell you if there is sufficient current to operate the device (more on this later)

Voltage Tests = Parallel to Positive and Negative

Amperage is always measured in series to a device, or the load. The meter leads can be connected either way in the circuit; however polarity can be an important part to this test. The red meter lead will normally connect to the source side (voltage supply) going to the device while the black lead connects to the input leg of the device. The meter becomes the supply circuit and it displays how much current is needed to operate the device.

Amperage Tests = In series as Positive or Negative

Voltage drop testing is performed parallel to the circuit being tested. If you are measuring the supply side (positive) of a device you will attach the meters red lead to a “Battery +” connection and the black lead to the supply side of the device. By connecting the meter in this fashion the meter essentially becomes another wire providing power to the device. Since it is parallel to the devices wire there should be little to no voltage displayed on the meter. If you have a voltage reading >.500 (1/2 a volt) then there is problem with the supply circuit. This test confirms, or condemns the circuit’s ability to carry sufficient current from the battery in order to operate the device correctly. I consider this to be the “Dynamic Resistance” test because this test is performed while the device is in operation.

Voltage Drop Tests = Parallel to the Positive or Negative circuit



This is just a couple of quick primers for you to digest, if you would like some printed material to reference radio shack has a great introduction book for the basics of electricity.


Ohms law
E=I x R or E/(IxR)

(E)Voltage = (I)Amperage x (R) Resistance

12.6 (V) = 3 (A) x 4.2 (Ohms)

When Resistance is high Current is low

OR

When Current is High Resistance is low


Think of resistance and amperage as two children on a see saw, if one of the children is heavier than the other, one child will be down while the other child is up. Now substitute the children with I and R on this imaginary see saw and it becomes very easy to remember the relationship between current (Amperage) and resistance.


Last edited by MrCjames; 02-03-2006 at 06:00 PM.
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  #4  
Old 02-03-2006, 05:07 PM
69 mercedes 220d
 
Join Date: Oct 2005
Location: Bozeman, Montana
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ohm's law

Ohms law

E=I x R or E/IxR

(E)Voltage = (I)Amperage x (R) Resistance

Yes, E=I x R, but not correct to say "or" E/(I x R); to use the transformation correctly would be to say E/(I x R)=1 (by dividing both sides by (I x R)).
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  #5  
Old 02-03-2006, 05:45 PM
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Location: Florida / N.H.
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Quote:
Originally Posted by Ralph69220d
Ohms law

E=I x R or E/IxR

(E)Voltage = (I)Amperage x (R) Resistance

Yes, E=I x R, but not correct to say "or" E/(I x R); to use the transformation correctly would be to say E/(I x R)=1 (by dividing both sides by (I x R)).

Correct..

Remember the old Ohms Circle tool for conversions ????

The circle area is divided into a top half and 2 bottom quarters.
The top is E , the bottom/left is I and the bottom/right is R.
By covering the unknown value you are looking for , you then know to divide, or multipy the others to get your answer..


I=E/R
E=IxR
R=E/I

Hey Alex , let's try Ohms for $600.... ding--ding...
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  #6  
Old 02-03-2006, 05:59 PM
MrCjames's Avatar
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Join Date: Jun 2003
Location: SF Bay Area
Posts: 599
Quote:
Originally Posted by Ralph69220d
Ohms law

E=I x R or E/IxR

(E)Voltage = (I)Amperage x (R) Resistance

Yes, E=I x R, but not correct to say "or" E/(I x R); to use the transformation correctly would be to say E/(I x R)=1 (by dividing both sides by (I x R)).
Ah yes, a little brain flatulence excluding those parenthesis.

Thank You
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  #7  
Old 02-03-2006, 06:44 PM
MrCjames's Avatar
California Dreaming
 
Join Date: Jun 2003
Location: SF Bay Area
Posts: 599
Quote:
Originally Posted by Arthur Dalton
Correct..

Remember the old Ohms Circle tool for conversions ????

The circle area is divided into a top half and 2 bottom quarters.
The top is E , the bottom/left is I and the bottom/right is R.
By covering the unknown value you are looking for , you then know to divide, or multipy the others to get your answer..


I=E/R
E=IxR
R=E/I

Hey Alex , let's try Ohms for $600.... ding--ding...
Name:  Ohms_law1.bmp
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  #8  
Old 02-03-2006, 09:57 PM
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Location: Florida / N.H.
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Cool..........that's it,

I had no way to upload it , so a discription was the best I could do..

I,000 words............

I have an old Simpson 260 that has that 'Ohms Peace Sign" scratched right into the rear of the case..... .. [ good meter, too..see them on ebay for short $$$ now ]


Last edited by Arthur Dalton; 02-03-2006 at 10:05 PM.
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