I have heard that you can add an external resistor to the positive in line, after estimating what the amps might be, to the meter and than use a formula which incorporates the ohms of the resistor and the amount of amps that the meter displays. Basically what is the formula for this and how do I go about it?
2006-07-06 09:48:41 · 8 answers · asked by ↓ImWithStupid ░░▒▒▓▓ 4 in Science & Mathematics ➔ Engineering
Yes: The resistor is called a "Shunt". The shunt is connected in series with the load (the device or circuit being measured) and the resistance of the shunt developes a voltage across it directly proportional to the current passing through it. For example a 1-Ohm (one Ohm) shunt with 1 Amp flowing will develope 1 volt across its terminals. Almost any low resistance can be used as a shunt, a long coil of wire can be used as a shunt if you measure it's dc resistance accurately. (wire coils do not work well for AC circuits due to inductance but are fine for DC circuits)Keep in mind the accuracy of your Amp calculation is only as good as your measurement of the shunt value.
Therefore set your meter to measure "VOLTAGE"
Place the Shunt resistor in series with the load,
turn on the device or load.
Read the Voltage across the shunt.
You Know the Voltage now,
You Know the resistance .
to determine the current:
Divide the Voltage reading on your meter by the Olm value of the shunt" That number is the value of the current in amps.
Amps= voltage/resistance (olms Law) 101
example: shunt value = .5 Ohm (1/2 Ohm)
Voltage reading across the shunt = 2 volts
Divide .5 into 2 = 4 Amps
2006-07-06 10:44:18 · answer #1 · answered by Grumpy 6 · 5⤊ 0⤋
By measuring voltage drop across a resistor inserted into a circuit in order to measure current, you have to consider the effect of that voltage on the cicuit, as was mentioned above. This depends on what voltages are present in the circuit. If the circuit voltage at the point of measurement is 100v, you can tolerate a 1-2v drop across a shunt; it all depends on the accuracy you require . It is better to use a shunt across the meter in its ammeter mode, but to do this you need to know the internal impedance of the meter (this should be part of the meter's specs). This is usually quite low in the high current ranges, but special very low value (fractional ohm) precision resistors are made for this purpose. Say the internal meter resistance is Ri and the shunt resistor is Rs. If the meter reads a current i(m) then the circuit current is i(m)*(1+Ri/Rs). For example if Ri = Rs, then the circuit current is 2x the meter reading.
2006-07-06 13:43:01 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
When measuring voltage the internal resistance of the meter is very high, almost like an open circuit. When measuring current, do so in series and the internal resistance of the meter is very low as not to affect the circuit. You will want to measure voltage by adding a shunt resistor. You could possibly use the meter to measure current only if you're dealing with 5VDC digital stuff.
2006-07-06 12:58:05 · answer #3 · answered by Dr. Answer 2 · 0⤊ 0⤋
NO. In measuring amperes, the meter is connected in series, but a shunt (low value) resistor must be used to bypass most of the current or the meter will be Kaput. The calculated value of the shunt depends on the ratio of current through the meter and the amount "shunted" around the meter. The "in-line" resistor will probably disturb your circuit.
The formula ? I can't help you without a pencil and paper.
2006-07-06 10:06:26 · answer #4 · answered by Puzzleman 5 · 0⤊ 0⤋
Suppose you place a 1k-ohm resistor, then measure the voltage across the resistor. Take that voltage and divide my 1000 and thats your amperage.
But I'm not completely sure why you want to add an extra resistor. I would need to see your circuit setup to understand.
2006-07-06 10:04:56 · answer #5 · answered by cw 3 · 0⤊ 0⤋
Yes, add a high precision shunt resistor and read voltage, then you convert back into milliamps using E=IR.
Example using a 20 ohm shunt resistor, you read 10.0000 volt DC.it means your current was 500 mA
2006-07-06 10:15:35 · answer #6 · answered by Anonymous · 0⤊ 0⤋
How about measuring the resistance of a given weld, and then the voltage across it? Provided you're only measuring across the weld (and a known amount of pipe on either side, I suppose) and not the welding terminals, then you should be able to apply Ohm's law to determine the current flowing. Even cheapo multi-meters tend to have pretty substantial voltage ranges.
2016-03-27 06:53:54 · answer #7 · answered by Regenna 4 · 0⤊ 0⤋
Make sure you connect the multimeter in series!!!!
2006-07-06 16:02:40 · answer #8 · answered by Mr.answers 2 · 0⤊ 0⤋