The power factor depends on the phase angle between the voltage and current
on a conductor. The amplitude of the current has no effect on it.
For a fixed resistance (ohms) current increases as voltage increases. Since Watts equals Volts x Amps x Power Factor then Watts would increase as voltage increases. The resistance would usually be fixed, but if you had a variable load resistance as the resistance decreased and the voltage remained constant, the current would increase and watts would therefore increase. Watts = Volts x Amps x Power Factor Volts = Amps x Ohms Power Factor is 1 for a resistive load.
Power factor can be unity. If the load is purely resistive, then the load current and supply voltage are in phase, and the load will have unity power factor.
The power factor of a load is the cosine of the angle by which the load current lags or leads the supply voltage. So if they are in phase (phase angle is zero), then the power factor must be unity (1).
Power is voltage times current. If you want to maintain constant voltage and yet increase power, then current must increase. Its simple math.
In an AC system power is equal to Voltage x Current x Power factor. Power factor is not constant and depends on the type of the load. Ideal value of the Power factor is 1, where as practically remains less then 1.
Not necessarily. If a load has a low power factor, it will be drawing more current than necessary, but its energy consumption will be no different from it having a high power factor.
At least to a certain extent, by increasing the field current. Or In Real Power Plant they decrease the power factor to increase the voltage.
If power factor is increased, current will be reduced for a specific real power (kWs) relative to before. Total power is real power plus the vector of reactive power (you have to do polar math). So total current = (current from real power) + (current from reactive power). By changing the power factor, you decrease the second quantity. Go to Wikipedia.com and search for Power Factor if you need a more extensive description.
Power = voltage * current * cos (power factor); So if voltage increases, and current stays unchanged, power usage will also increase in proportion. A: Power is a factor of voltage and current therefore the power will increase if one or both are increased
P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.
Power factor is the cosine of angle between voltage and current that we all know. And the power factor should be unity or close to unity. Unless if we have the power factor not close to unity or far away from unity is called poor power factor. This termed as poor because it will take large amount of current for the given power. If the large amount of current is drawn from the substation or anything else then the line loss will increase. Line loss is (I^2)R loss. So the current increase line loss will also be increased. So as to avoid such losses every industry should maintain their power factor(Normally in every Industry they maintain .9 and above). Those who are not maintaining power factor will be fined.
The decibel (dB) scale is logarithmic. An increase of power by a factor of 10 is an increase of +10 dB. If power increases by a factor of 100, that is equivalent to +20 dB.The decibel (dB) scale is logarithmic. An increase of power by a factor of 10 is an increase of +10 dB. If power increases by a factor of 100, that is equivalent to +20 dB.The decibel (dB) scale is logarithmic. An increase of power by a factor of 10 is an increase of +10 dB. If power increases by a factor of 100, that is equivalent to +20 dB.The decibel (dB) scale is logarithmic. An increase of power by a factor of 10 is an increase of +10 dB. If power increases by a factor of 100, that is equivalent to +20 dB.
Power factor is the cosine of the angle between current and voltage. Most of the loads are inductive.In inductive loads current always lag the supply voltage.In unity power factor condition the current and the supply are in the same phase. Since, P=V*I*cos(angle between V & I) Now for a constant power as power factor changes i.e: decreases, to keep power demand constant current has to increase as your voltage is constant.In this way the current increases due to increase in copper losses (I*I*R),voltage drops (I*R) and to keep this voltage constant more reactive power is required. KW = KVA * cos(angle between V & I).....so as the power factor is getting lower,the actual output we obtained from the transformer will be lower than the desired output for the particular value of voltage and current in that circuit.That is why in industries have given much more importance for power factor improvement.
It's easier to answer you question the other way around, that is "Why does the load current fall with an increase in power factor?"Before power-factor improvement, the load current is the phasor (or vector) sum of the load's resistive (IR) and inductive currents (IL).Power-factor improvement is achieved by adding a capacitor in parallel to the load so, after power-factor improvement, the load current becomes the phasor sum of the resistive current (IR), inductive current (IL), and the capacitive current (IC).Since the inductive current and capacitive current are displaced from each other by 180 degrees (i.e. are in antiphase), the the phasor sum of IR + IL +IC will be smaller than the phasor sum of just IR and IL.Hence, the supply current reduces as the power factor improves.
Two factors reduce the power used by a piece of equipment compared to the volt-amps drawn from the supply: power factor and harmonic factor. Both factors increase the power transmission losses incurred in supplying a given amount of power.The power factor is less than 1 when voltage and current are out of phase with each other.When they are in phase the power equals the volt-amps except for a nonlinear load with a current that is not proportional to voltage. This generates harmonics in the current and the effect is that the power is less than the volt-amps, by an amount equal to the harmonic factor.
If current increases, then voltage also has to increase, assuming that resistance stay relatively the same. Power will also increase. Since power is the product of voltage and current, then the power increase would be the square of the voltage or current change.
The voltage across a purely inductive load leads the current by 90 degrees.Because of this 90 degree offset the power transfer is inefficient because the current and voltage peaks are never aligned.P = I * ETherefore a capacitor with an reactance equivalent in magnitude but opposite in polarity to that of the inductive load at the operating frequency can be placed in parallel with the load to maximize power transfer.CommentPower-factor improvement does not improve, or increase the efficiency of, power transfer. The power of a load is exactly the same after power-factor improvement as it was before power-factor improvement. All that power-factor improvement does is to reduce the amount of load current.