Because ordinary ohmmeters are calibrated to measure a range of resistance values that are significantly lower than the resistance of insulation. It should also be realised that we usually test insulation resistance while subjecting the insulation to a high electric field, which a normal ohmmeter is incapable of producing.
since it cant measure the weak current passing through the insulator.
No, you cannot use a multimeter in place of a megger. The megger is used to measure insulation resistance, which could be hundreds of megohms or so. In order to do that, you need a high voltage so as to develop the current needed to sense the resistance. Using an ordinary multimeter would not work as the applied voltage is not high enough, causing the sensed current to be too small to detect.
It can be calculated by a multimeter. [aka. avometer (ampere voltage ohm meter)]. A single voltmeter or ammeter could not measure resistance because this process need calculations. That means the device must be able to measure the voltage and current at the same time.
Put a known small resistance (maybe 1 ohm) in parallel to it. Then use the equation I=V/R (R being the known resistance, and V being the voltage readout) So when using a 1 ohm resistor, if the voltmeter reads 2.35V, it means I=2.35/1 = 2.35 A
You can check if it is shorted, but without coordinating the gate and anode voltages with the cathode, you cannot test a thyrister (SCR) with a multimeter. Check it out of circuit for shorts, then check the circuit, then test it in operation. Anything more advanced will require an oscilloscope.
You cannot use an ordinary multimeter to assess the difference between a normal power transformer winding and a shorted power transformer winding. You need a Q meter and/or to power it up at a reduced voltage with limited current to see what happens. Note that a shorted power transformer is capable of exploding, so you need to take appropriate safety precautions.
No, you cannot use a multimeter in place of a megger. The megger is used to measure insulation resistance, which could be hundreds of megohms or so. In order to do that, you need a high voltage so as to develop the current needed to sense the resistance. Using an ordinary multimeter would not work as the applied voltage is not high enough, causing the sensed current to be too small to detect.
Polarization Index is an indicator which gives an idea of the cleanliness of the windings. It is a ratio of the Insulation Resistance Measured for 10 minutes to the insulation resistance value measured after 1 minute. Since it is a ratio; it does not have any units.Polarization Ratio = Insulation Resistance after 10 minutes/ Insulation Resistance after 1 minuteThe Polarization Index should be above 2.0 to be permissible. Machines having PI below 2.0 cannot be operated. The Polarization Index does not have any significant relation with temperature upto 50 deg. C. However, the Polarization Index test should not be conducted at a temperature beyond 50 deg. C
a BJT may be tested as two diodes using the ohms scale. a FET cannot be tested with just a multimeter.
It can be calculated by a multimeter. [aka. avometer (ampere voltage ohm meter)]. A single voltmeter or ammeter could not measure resistance because this process need calculations. That means the device must be able to measure the voltage and current at the same time.
Put a known small resistance (maybe 1 ohm) in parallel to it. Then use the equation I=V/R (R being the known resistance, and V being the voltage readout) So when using a 1 ohm resistor, if the voltmeter reads 2.35V, it means I=2.35/1 = 2.35 A
You cannot use an ordinary multimeter to assess the difference between a normal power transformer winding and a shorted power transformer winding. You need a Q meter and/or to power it up at a reduced voltage with limited current to see what happens. Note that a shorted power transformer is capable of exploding, so you need to take appropriate safety precautions.
A qualitative measurement is something that cannot be measured in numbers
I cannot answer this question
Loft insulation stops conduction and convection as the insulation which is inserted into the wall cavity prevents the particles of heat energy from traveling through the wall cavity, this stops conduction from taking place. Loft insulation also prevents convection as the cavity insulation stops the air particles from moving, therefore they cannot transfer heat and convection cannot take place.
The foot is a linear measurement. You cannot convert a linear measurement to an area measurement.
You cannot test it with a diagram. You will need a multimeter to test the coil. Click the link.
ml is a liquid measurement and lbs is a weight measurement they cannot be determined