Use a voltmeter and an ammeter to measure the supply voltage and load current; the product of these two readings will give you the apparent power in volt amperes. Use a wattmeter to measure the true power of the load, in watts. Divide the true power by the apparent power, and this will give you the power factor.

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Try using an osciloscope, connect the voltage to one channel and the other to the current (CAUTION: provide proper shunting) You'll see the two wave forms, the distance between them would give you the angle between the voltage phasor and the current phasor, the cosine of this angle in degrees is the power factor.

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You can calculate the power factor if you have a meter that measures voltage in volts and current in amperes and have access to a wattmeter that measures power. Power factor can then be calculated by this formula:

Pf=P/S , S=V*I where S is apparent Power.

Power Factor, simply put is the relationship between real power (Watts) and reactive power (VARS). It isn't related to efficiency, at least not in terms of the ratio of output power/energy to input power. A motor might have a power factor of 0.87 (30 degree phase angle), but an electrical efficiency either *more* than 87%, or *less* than 87%.

Since the input power to a motor should properly be measured as *real* power, the power factor is not considered in calculations. One reason is that the motor's power factor (which is most likely inductive) can readily be corrected back to unity (1.0) by either adding a parallel capacitor (as is done for the inductive ballast coils in fluorescent lights), or by installing a synchronous motor in the circuit and adjusting the amount of excitation.

Strangely enough, a synchronous motor can be made to appear either inductive (like other motors) or *capacitive* according to its excitation power. Such motors are often used for constant-service applications such as airconditioning.

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THIS DOESN'T MATTER IN HOUSEHOLD ELECTRICITY. If a factory has a Power Factor of 95 %, then it will draw 105 % of the current it would draw if it were at 100%, or a Power factor of 1 (also called unity).

The electric company charges a customer more for ineffecient systems, i.e. Power Factor lower than 1 (100 % efficient).

The "Power Factor" is the ratio of volt-amps to watts. To get volt-amps, you also multiply volts times amps. With a resistive load, such as an incandescent lamp, volts times amps equals watts. All of the power gets dissipated heating up the lamp filament to make it glow. In this case, volt-amps is equal to watts, giving a ratio of 1:1, or 100 %. With inductive loads like Transformers, electric motors, fluorescent lamps, etc., there is very little resistance. Something called "reactance" limits current flow. Larger currents flow with little power being dissipated. With a power factor of 50 %, double the current would flow. For example, a 40 watt incandescent lamp draws 0.33 amps. (40 watts / 120 volts = 0.33 amps) This bulb, being a resistive load, has a power factor of 100 %. A single tube fluorescent lamp rated at 40 watts may draw double the current of the 40 watt incandescent, but still only use 40 watts of power. This fixture has a power factor of 50 %.

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Home meters

Electrical meters for homes measure only resistive (real, apparent, or actual) power. They do not measure reactive power.

In the study of alternating current, it will be observed that there are alternating waves of both voltage and current. In a circuit with purely resistance load, the waves of current and voltage are in exact phase relationship to each other. This means that when the voltage is at its peak, the current flow is at its peak as well. An inductive load (that is, a coil) causes the current wave to lag or fall behind the voltage wave, so that the peak current flow is some time after the voltage wave is at its peak level. A capacitive load (that is, a capacitor) causes the current wave to lead or advance ahead of the voltage wave, so that the peak current flow is some time in advance of the peak of the voltage wave.

The consequence of this is that the AVAILABLE REAL POWER is the relationship between the current and voltage waves.

Resistive circuits have a power factor of 1.0, or unity, because the waves are in phase.

The more out of phase the relationship between voltage and current, the less efficient the use of available power, the more "waste" energy.

The less efficient the use of energy, the larger the size of transmission and generating equipment required to provide for energy needs and the more costly the operation of utilization equipment.

Scroll down to related links and look at "What is reactive power?"

No improvement to these good answers. I would just add that single phase power will rarely veer from unity. If you are dealing with household electrical service, you are likely to be at or near unity.

"... likely to be at or near unity..."

Umm... no.

Motors use magnetic windings, and are therefore inductive. Inductive components reduce the power factor below the ideal value of 1.0. Old style fluorescent lights use ballast chokes, which are also inductive. A correction capacitor in the housing will correct the PF back to 1.0, but if the capacitor is faulty/has been removed, the fluoro will also give an inductive power factor. In fact, *most* loads, other than heating elements, are inductive.

So you start up the washing machine, dishwasher, benchtop mixer and electric drill. It is unlikely that any of these appliances is power-factor-corrected. Your PF will drop to less than 1.0. Our electricity supplier does all of their house load calculations based on a PF of 0.8.

"...waste energy..." Yes and no. The extra current does cause extra resistive loss in transformers, cables and switches, but the real problem is back at the generator, where more current must be generated than would be needed by a system with a perfect PF of 1.0.

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Q: How can the power factor be calculated without using a power factor meter?

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Your question is unusual. The point of using a pwer factor meter is to get an actual indication of the power factor *without* needing to do any other measurments or do any calculations. Do a Google search for Power Factor meter and check out the illustrations/photos you get. You will see that the meters read directly in power factor, so no calulation is needed. Rephrase your question and maybe what you actually want to know will be clearer.

there are 3 types of power factor meter: 1). electrodynamic power factor meter, 2).moving-iron power factor meter and 3). nalder-lipman moving-iron power factor meter.

Power Factor Meter

They can measure a lun power is called single phase power factor meter.

power meter,power factor meter, frequency meter, volt meter ,watt meter ,voltage ampere meter,

power factor meters are connected across the supply

Megawatts has the power factor calculated into its formula. KVA doesn't have power factor calculated into its formula.

A wattmeter reads the true power of a load, regardless of its power factor.

Power factor is simply the ratio of a load's true power to its apparent power. True power is expressed in watts, whereas apparent power is expressed in volt amperes.Another way of defining power factor is to say that it is the cosine of the phase angle -i.e. the angle by which a load current lags or leads the supply voltage.To describe the operating principle of a power-factor meter is difficult without referring to a diagram, so I suggest you do a search on the internet for this answer.

Wattmeters are not used for loads which are purely capacitive or inductive, because no watts are consumed and no energy is consumed by the customer. But amps must still be supplied by the power company to supply the customer's capacitive or inductive load and the result is measured in volt-amps reactive (VAR), which can be registered on a meter called a reactive power meter. The ratio of watts to volt-amps is called the power factor (a capacitive load has a power factor close to zero). Industrial customers with a poor power factor are penalised with higher tariffs and encouraged to improve their power factor.

If you are asking whether power-factor improvement has any effect on a wattmeter reading, then the answer is no, it doesn't. Improving the power factor of a load has absolutely no effect on the power of the load, but it can act to reduce the value of the load current.

Because a dynamometer is used to measure the mechanical power output of a motor or engine. There are ways of measuring the power factor directly, for example use a plug-in power and energy monitor.

A poor power factor causes the meter to rotate more slowly than it should, so a poor power factor would reduce your bill. Electric utilities compensate for this in commercial services by billing based on power factor, or they install a meter that actually measures power factor.AnswerEnergy meters 'read' the in-phase component of load current (therefore the load's 'true power' multiplied by time) and, so, are completely unaffected by the power factor of a load. So the power factor of a residential load will have absolutely no effect whatsoever on that residence's 'energy' (not 'power') bill.Industrial and commercial consumers are billed for 'demand' (their rate of consumption of energy -i.e. the power) as well as energy supplied'. In addition, these consumers are usually penalised if the power factor of their load falls below an agreed value. So power factor does affect the overall bill (but not the energy bill) of industrial consumers.

it might have problem with particular capacitor bank that why its showing negative power factor-0.97AnswerYour meter might be reacting to the fact that the direction of power is from the load to the supply, in which case the direction of power 'flow' is considered to be negative. In this case, the negative sign indicates the direction of the power, not a negative power factor. Reversal of power (load to supply) can occur when, for example, a motor overruns and temporarily acts as a generator, or when PV panels are supplying power to the grid.

To measure electrical parameters like voltage, current, power factor etc.

If this is a three phase VAR meter then you have a leading power factor. There is too much capacitive reactance in the distributions load. This could be from a power factor correction bank of capacitors that did not drop off line when the motor that they were correcting dropped out.

The rating of any electrical appliance or piece of electrical equipment is calculated in watts because the watt is the international unit for power. The electrical power of a purely resistive load is calculated using the Power Equation: Power (watts) = Voltage (volts) x Current (amps) For a reactive load - such as an ac electric motor - the Power Equation has to be modified to allow for its Power Factor: Power (watts) = Voltage (volts) x Current (amps) x Power Factor

Lead. Definitely, without a doubt.

When you have a lagging power factor, measuring instruments (i.e. AC energy meters etc..) will read high. For example, if you've actually used 12W, when a lagging power factor is present, the meter might read 13~14W.

K=(voltmeter range*ammeter range*power factor)/wattmeter range

Depending on meter model and type it may. Smart meters have ability to operate without supply to communicate with grid.

Power factor is real power divided by total power. Power factor can be written as: pf = R / sqrt(R^2 + X^2), where X is the reactive resistance of the active elements (in this case, L): pf = R / sqrt (R^2 + (wL)^2) w = frequency in radians of the AC frequency for which the power factor is to be calculated.

When power factor is at unity, the voltage and current waves are aligned or in phase with one another. Since power is the product of voltage and current, power transfer is maximized at unity power factor. When power is transmitted at a lower power factor, greater current is required to deliver the same amount of power. When current is increased, the size of the transmission, distribution and generation systems, all have to be increased accordingly, along with the price of the killowatt-hour at the meter.

no applications just measures the energy consumesd and not able to measure the power factor.... :-)

wattmeter is constructed such that it read only power on a single scale. Only single scale will avilable to read the measured power. In order to take the acuurate reading just we measure the reading and we will multiple along with factor called multificaion factor. That's depends on on which voltage knob we connected and also the current. Multification factor can be dtermind by: Multification factor =(voltage range*current range*pf)/Max scale deflection...