Because 4-20ma is an industry standard.
300 / 25, or 12/1.
It depends on the shunt feedback resistor on the op-amp, for example with a 10k feedback resistor connecting the output to the inverting input, 1 mA input current gives 10 volts signal output. The input terminal stays near zero voltage because of the high open-loop gain of the op-amp, so the inverting input is termed a 'virtual earth'.
It depends on how the diode is damaged. There are generally two cases. One, the diode is shorted, and conducts with a low impedance in both directions. The other, the diode is open, and does not conduct, having a high impedance, in both directions. The effect depends on the particular circuit. In a power supply, a shorted diode will often blow the fuse, while an open diode will result in no output, or in high ripple voltage output. Is it possible that diode has normal voltage output but wrong current,meaning low mA?
It depends on the particular zener diode. Typically, they will pull 75 ma of current.
No, a Diac cannot trigger an SCR because when the Diac turns ON, the current through the Diac is around 9 mA. The gate threshold current of an SCR is typically 5 mA, which is less. So the SCR can get damaged due to this high gate current.
A: Assuming 100% efficiency 320 ma
There are .42 amps in 420 mA. Equation 420/1000 = .42 amps
mA stands for mili-Amp, or one-thousandth of an Amp (or Ampere; being the unit of measure of current, or charge carriers; ie electrons in a circuit) 1 mA = 0.001 A
By having a minimum current in your current loop it is possible to detect when there is a fault in the line or the device at the other end has been disconnected. If these errors conditions occur, the current falls to zero, which should never happen in normal operation.
No, Your original adaptor has an output of 3 amps or 3000 ma. As you can see, the one you want to use for a replacement adaptor only has 1000 ma output, one third of the current capacity that you need.
The MA would be 6cm because the formula to find the MA is input arm divided by output arm
Yes. (For any pairing of power supply and device, as long as the voltages are a match (in your case: 9v), and the output (in amps or milliamps (A or mA) of the power supply IS EQUAL TO OR GREATER THAN the current required by the device (in your case 1300mA or higher) then you will be fine. Yes it is suitable: The OUTPUT VOLTAGE (5v, 9v, 12v, etc) of a power supply MUST BE EQUIVALENT to the required voltage of the device to which it is to be connected, whereas the output CURRENT (500mA, 1A, 1500mA, 2A... etc) offered by the power supply MUST BE AT LEAST EQUAL TO OR GREATER THAN the current required by the device to which it is to be connected. (in your case, for example, as long as the power supply is rated at 9v, you could use one that has a rating of 1300mA, 1400mA, 1500mA, 1A...and so-on, without any damage to either device)
yes.
Supply voltage (VCC) 4.5 to 15 V Supply current (VCC = +5 V) 3 to 6 mA Supply current (VCC = +15 V) 10 to 15 mA Output current (maximum) 200 mA Maximum Power dissipation 600mW Power consumption (minimum operating) 30mW@5V, 225mW@15V Operating temperature 0 to 70 °C
MA = Revolutions of input shaft / Revolutions of output shaft. (Input torque * MA) * efficiency = Output torque Note 100% efficiency = 1.0
Pronunciation: pray ve ma ANSWER: ρεύμα (REV-ma) - if you are referring to the current in a stream
If the resistance is 1.2k and the current is 0.024 ma, then the voltage is 0.0288 volts. (Voltage = resistance times current) If the voltage is 0.0288V and the current is 0.024 ma, then the power is 0.6912 microwatts. (Power = voltage times current)