How does a Ground Resistance Clamp Meter work and what are its limitations?
827069
The meter should have some literature with it.\n\nWrite to the manufacturer and ask for it.
438682
sure. go to a science store or look up schematics on the web.
980282
Make sure you have a good, solid connection between your sample and your ohmmeter probes.  Use a scale on your meter that's appropriate for the resistance you are measuring.  If your resistance is 1/2 ohm, then it will read zero on a 0-20Kohm scale, which isn't useful.  It will read half-way on a 0-1ohm scale, where you can get a good reading.\n\nThe resistance of a sample depends not just on the material and temperature, but also the geometry.  The resistance of a uniform wire will be proportional to its length and inversely proportional to its cross-section area.
823030
You could use a resistor in series for the voltmeter and a shunt resistor in parallel with the ammeter.  You can also use a signal amplifier to transform the input signal to one that is within the range of the meter.  If it is a panel type meter, do not forget to change the scale behind the meter.\n\nJeffM
1085999
You can check with your local county engineering office.  They may or may not be willing to loan you this information.
455603
If it's for a simple hobbyist application they need to know only the resistance value and its power rating.\n\nA full specification would also include:\n\ntolerance,  thermal coefficient,  maximum voltage.
521856
Constant current source, diode junction for temperature sensing, and a gain/offset circuit for scaling to 10mV/deg C.\n\nIf you look at the diagram in the product folder, its pretty easy to see whats going on.\nhttp://www.national.com/pf/LM/LM35.html\n\nThey get the accuracy without taking 2 measurements at different current levels, as is commonly done in diode temperature sensors, by wafer trimming.... which is why the LM35 costs so much more, than a simple diode sensor. The current source, and gain/offset circuitry uses minimal real estate on the die, with interconnects and physicial handling issues consuming almost as much as the circuit needs. The cost is a funtion of the extra time needed for trimming, and of course the manufacturing yield.
964185
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
16724
As you are building a computer, ESD sensitive parts can get damaged.  You can buy an ESD wrist strap and hook it to the AC outlets center screw (between the two plugs).  This will let the static electricity discharge through ground and not hurt your components.
1014845
if its a single wire .. find where something is "connecting" live to ground.\n\nif its multiple wires:  find where live & ground are being joined/connected.\n\nhere's how elect. fences work:\nhttp://www.sureguard.com.au/electric_fence_design.html
370125
you are almost right, electrical ground is allways relative to a higer potential source like a live wire and the current flows toward ground from the higer source, when you ar isolated (you have rubber shoes, you have a high resistance in the way of the current), there still a current flow, but it is so small (nano amperes) that you body cant feel it
572940
A voltmeter consists of a galvanometer(to detect current) and a high resistance(usually thousands) connected in series. It measures the voltage across two points. Theorectically a voltmeter should have infinite resistance.
249398
I suppose you are reffering to the small digital scales. They have a device whos electrical resistance changes under pressure.
795766
I work on diesel - electric locomotives .The power system and components ( high voltage and amperage ), such as the traction motors,are protected by a 'ground relay ' that 'trips' and cuts power when a ground fault exists.\n In our terms a 'ground fault' means that the negative or positive cables or components are grounded, meaning the insulation is bad and the circuit touches the body at that bad spot, referred to as ground. (ground is referred to as the general mass of earth ) and the locomotive is basically on earth through the wheels on the rails.\nWe use a 1000 volt meggar meter , one lead coupled to the positive or negative power circuit, and the other lead coupled to ground, and then we send 1000 volts into the system and it indicates/reflects to what extent the circuit is grounded and by isolating different parts of the circuit we can detect exactly where the ground fault is situated by reading the power schematic .(electrical drawing or plan )Hope it explains enough.
58713
Measure its capacitance.  Many multimeters have this capability.  If the body is not a two-terminal device (e.g., a capacitor), measure the capacitance from the body to a ground plane.  Then measure its voltage with a very high-impedance voltmeter.  The act of measuring will drain off some charge, so you have to take a reading quickly.  The charge in Coulombs is V*C, where V is voltage in volts, C is capacitance in Farads.
902547
How to Calculate Voltage Drop For Long Paired Wire Runs\n \nA primary concern when installing lengths of wire is voltage drop. The amount of voltage lost between the originating power supply and the device being powered can be significant. Improper selection of wire gauge can lead to an unacceptable voltage drop at load end. The following chart is designed to help calculate voltage drop per 100 feet of paired wire as a function of wire gauge and load current.\n\nBy matching load current (in AMPs) across the top of the chart with wire gauge (AWG) down the left side of the chart, one can determine voltage drop per 100 feet of paired wire run.\n\nNOTE: A paired wire run represents the feed and return line to the load. Therefore, a 500 foot wire pair is equivalent to 1000 feet of total wire.\n\n\nEXAMPLE ONE:\n\nGiven a load current of 1 AMP, and using 18 AWG wire, how much voltage drop can we expect at the load end for a 350 foot run of paired wire?\n\nUsing the chart, we match the row for 18 AWG and the column for 1 AMP and determine that voltage drop per 100 feet is 1.27 Volts. By dividing the paired wire length by 100, we get the factor by which we need to multiply voltage drop per 100 feet to determine total voltage drop. Therefore, 350 feet divided by 100 equals 3.5. Multiply 3.5 by 1.27 volts drop per 100 feet to get your total voltage drop. Thus the total voltage drop is 3.5 times 1.27, or 4.445 voltage drop for 350 feet. \n\nRead more --> http://www.securityideas.com/howtocalvold.html
940296
gnd for the total system and the single circuit is over all the same. That's why its call ground.\nSo yes you can use any ground in the system as ground point for your probe