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Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having a relationship between V or I..

Can some one clarify this.

Regards, Sridhar

What do you mean Cos(Pi)?

affects affects power, then frequency affects power.

For example the frequency of the AC mains supply is, in North America = 60 hertz) and elsewhere is often = 50 hertz.

Maybe what is being referred to is the affect of 'reactance' in an AC circuit? Reactance (as compared to pure DC resistance usually designated 'R') is the affect on current/voltage by the presence of inductors or capacitors in a circuit.

In an inductive circuit the current flowing will 'lag' behind the applied voltage due inductive reactance often designated Xl. In a capacitive circuit the current flowing will 'lead' the applied voltage due to capacitive reactance often designated Xc.

Barring anything else frequency will not 'affect' induction. Current and voltage will be 'affected' (if we must use that affected word) by the frequency of the voltage applied; this is how, for example, low pass and high pass filters work!

Also in this discussion thread suppose the frequency is zero? i.e. DC (Direct Current). Inductive reactance would be nil. Capacitive reactance would be infinite (after initial charging). The affect of 'inductive reactance' on the current flowing would be nil.

I have been wondering why this thread did not get moved to:

alt.engineering.electrical

snipped-for-privacy@optonline.net wrote:

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

snipped-for-privacy@optonline.net wrote:

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

snipped-for-privacy@optonline.net wrote:

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

I believe you will find that's the square root of 3.

Nick

#### Site Timeline

- posted on October 21, 2005, 10:10 am

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having a relationship between V or I..

Can some one clarify this.

Regards, Sridhar

- posted on October 21, 2005, 10:31 am

- posted on October 21, 2005, 10:41 am

Frequency just describes the cycles per second, or Hertz, that the
Voltage is. Here in the US its 60 Hertz. It is constant. The Pi in the
formula is also a constant.

- posted on October 21, 2005, 11:34 am

"Frequency just describes the cycles per second, or Hertz, that the
Voltage is. Here in the US its 60 Hertz. It is constant. The Pi in the
formula is also a constant. "

The argument of the cosine function is the angle between voltage and current in an AC load. It varies depending on the inductance and/or capacitance of a particular load. The symbol used for the angle is normally theta, not Pi as the OP stated.

The argument of the cosine function is the angle between voltage and current in an AC load. It varies depending on the inductance and/or capacitance of a particular load. The symbol used for the angle is normally theta, not Pi as the OP stated.

- posted on October 21, 2005, 10:53 am

"Hi,
Can some one explain the relationship between Voltage, Frequency. Is
frequency is affecting the electrical consumption.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having a relationship between V or I..

Can some one clarify this. "

The power formula you have is for a three phase AC load. For a pure resistance load, eg a simple load like a heater, the voltage and current are always in line with each other. If you drew graphs of the two, and placed one over the other, they would line up perfectly.

That is not true for a load that has inductance or capacitance, eg a motor. In that case, the inductance of the motor will cause a phase shift between the voltage and the current. If you place the graphs together, you will see that while the frequency is exactly the same, one curve is shifted slightly relative to the other.

Since power is the product of voltage and current, the instantaneous power is still V*I, but the average power is affected by the amount of shift between the voltage and current curves. That's where the Cos() function comes in. The angle used in the cosine function is the angle between the voltage and current in the load. For a pure resistive load, the voltage and current would be in perfect alignment and the angle would be zero, giving cos(0)=1. As you add inductance or capacitance, the angle will become non-zero, resulting in a reduction in power. Another way of looking at this intuitively is that as the the voltage and current go out of alignment, when multiplying instantaneous voltage and power along the two curves, since they no longe line up, the power will obviously be reduced.

According to the formula P= Root (3) V . I. Cos (Pi).

Frequency is not causing any affect on Power until unless it is having a relationship between V or I..

Can some one clarify this. "

The power formula you have is for a three phase AC load. For a pure resistance load, eg a simple load like a heater, the voltage and current are always in line with each other. If you drew graphs of the two, and placed one over the other, they would line up perfectly.

That is not true for a load that has inductance or capacitance, eg a motor. In that case, the inductance of the motor will cause a phase shift between the voltage and the current. If you place the graphs together, you will see that while the frequency is exactly the same, one curve is shifted slightly relative to the other.

Since power is the product of voltage and current, the instantaneous power is still V*I, but the average power is affected by the amount of shift between the voltage and current curves. That's where the Cos() function comes in. The angle used in the cosine function is the angle between the voltage and current in the load. For a pure resistive load, the voltage and current would be in perfect alignment and the angle would be zero, giving cos(0)=1. As you add inductance or capacitance, the angle will become non-zero, resulting in a reduction in power. Another way of looking at this intuitively is that as the the voltage and current go out of alignment, when multiplying instantaneous voltage and power along the two curves, since they no longe line up, the power will obviously be reduced.

- posted on October 21, 2005, 11:12 am

Just to further clarify, regarding frequency, the OP is correct.
Frequency of an AC load does not affect power

- posted on October 21, 2005, 6:53 pm

affects affects power, then frequency affects power.

- posted on October 21, 2005, 12:16 pm

"Are you sure about that? Frequency affects induction, and if
induction
affects affects power, then frequency affects power"

Well, you've got me there. That was an incorrect statement. I had the equation for 3 Phase power that the OP gave in mind and the fact that freq is not part of it. But of course you are right, the frequency has a big effect on any load with inductance or capacitance. And that effect gets into the OP's power equation by virture of the fact that the phase angle between voltage and current contained in the equation is itself a function of the frequency.

Thanks for correcting that!

Well, you've got me there. That was an incorrect statement. I had the equation for 3 Phase power that the OP gave in mind and the fact that freq is not part of it. But of course you are right, the frequency has a big effect on any load with inductance or capacitance. And that effect gets into the OP's power equation by virture of the fact that the phase angle between voltage and current contained in the equation is itself a function of the frequency.

Thanks for correcting that!

- posted on October 22, 2005, 5:21 pm

For example the frequency of the AC mains supply is, in North America = 60 hertz) and elsewhere is often = 50 hertz.

Maybe what is being referred to is the affect of 'reactance' in an AC circuit? Reactance (as compared to pure DC resistance usually designated 'R') is the affect on current/voltage by the presence of inductors or capacitors in a circuit.

In an inductive circuit the current flowing will 'lag' behind the applied voltage due inductive reactance often designated Xl. In a capacitive circuit the current flowing will 'lead' the applied voltage due to capacitive reactance often designated Xc.

Barring anything else frequency will not 'affect' induction. Current and voltage will be 'affected' (if we must use that affected word) by the frequency of the voltage applied; this is how, for example, low pass and high pass filters work!

Also in this discussion thread suppose the frequency is zero? i.e. DC (Direct Current). Inductive reactance would be nil. Capacitive reactance would be infinite (after initial charging). The affect of 'inductive reactance' on the current flowing would be nil.

- posted on October 22, 2005, 2:53 pm

"Are you sure about that? Frequency affects induction, and if

"Induction is NOT 'affected' by frequency! "

"Induction" most certainly is affected by frequency. The basic definition of induction is the inducing of a voltage in a conductor by a changing magnetic field. With zero frequncy, you have zero induced current. Start increasing the frequency of the magnetic field, eg by moving a magnet back and forth, or allowing a nearby coil's magnetic field to vary, and there will be an increasing induced current in the conductor.

I think what your referring to is that the "inductance" of say a particular coil is fixed and determined strictly from the physical properties of the coil and it's measured in henrys and not dependent on frequency. I think what Toller meant was fairly clear from the context of the discussion concerning power. And that is that in the general case, the amount of power consumed by a given piece of equipment depends on the frequency of the source it is being driven with. That's because induction is frequency dependent and the impedance of a load containing inductance and/or capacitance will change as a function of frequency.

"Induction is NOT 'affected' by frequency! "

"Induction" most certainly is affected by frequency. The basic definition of induction is the inducing of a voltage in a conductor by a changing magnetic field. With zero frequncy, you have zero induced current. Start increasing the frequency of the magnetic field, eg by moving a magnet back and forth, or allowing a nearby coil's magnetic field to vary, and there will be an increasing induced current in the conductor.

I think what your referring to is that the "inductance" of say a particular coil is fixed and determined strictly from the physical properties of the coil and it's measured in henrys and not dependent on frequency. I think what Toller meant was fairly clear from the context of the discussion concerning power. And that is that in the general case, the amount of power consumed by a given piece of equipment depends on the frequency of the source it is being driven with. That's because induction is frequency dependent and the impedance of a load containing inductance and/or capacitance will change as a function of frequency.

- posted on October 22, 2005, 7:20 pm

Toller ( snipped-for-privacy@yahoo.com) said...

Frequency does not effect power, as a purely inductive or capacitive load does not draw any power (not counting any losses due to the tiny resistance of the conductors themselves, which means that you can never really have a purely inductive or capacitive load!).

Since the impedance will be affected by frequency, the total current draw will be changed as a result of a frequency change.

The cosine of the angle between current and voltage is known as the "power factor". In a purely resistive load, it is 1. In a purely inductive or capacitive load, it is zero. Typical loads lie inbetween, but it is best to keep it as close to 1 as possible.

Industrial customers of electric utilities must maintain as high a power factor as possible. They usually have banks of capacitors that can be automatically switched on and off as needed as they will have heavy inductive loads from motors. If they maintain too low a power factor, they will be charged for kVA-hours instead of kW-hours.

For instance, if you maintained a 0.5 power factor, then a 100 A load at 120 V would be 100 x 120 x 0.5 = 6000 watts. An hour of this would be 6 kWh, but it is 12 kVAh.

Why be charged for more energy than you actually used? Simply because you are a burden on the system. Even though your load was only 6000 watts, you drew double the current than was really necessary for that amount of power. Therefore, the infrastructure needed to deliver that power had to have twice the capacity than was really necessary.

Low power factor loads tend to be inductive, so a bank of capacitors can cancel it out. Inductors cause current to lag behind the voltage, while capacitors cause current to lead voltage. The two cancel each other out. In the example above, since capacitors store and release current, they supply the "extra" current needed for the inductive load, so the only current draw on the supply is for the current actually needed to provide power. If perfectly matched, the 6000 watt load would draw 50 amps from the supply while the other 50 amps would be current between the inductive load and the capacitors.

As an interesting side note: your home probably has a leading power factor most of the time. The wiring in your home actually acts as a capacitor. When I was a student, I worked on weekends as a watchman at a factory. There was a power factor meter where the bank of capacitors was. When the factory was shut down and the only load was lighting, the power factor was usually 0.7-0.8 leading. As machines were started up and the inductance of the load increased, the PF would rise to 1 then start dropping on the lagging side. I believe a bank of capacitors would be switched in when it dropped to 0.7. In addition to the kWh meter, there was a kVAh meter.

Frequency does not effect power, as a purely inductive or capacitive load does not draw any power (not counting any losses due to the tiny resistance of the conductors themselves, which means that you can never really have a purely inductive or capacitive load!).

Since the impedance will be affected by frequency, the total current draw will be changed as a result of a frequency change.

The cosine of the angle between current and voltage is known as the "power factor". In a purely resistive load, it is 1. In a purely inductive or capacitive load, it is zero. Typical loads lie inbetween, but it is best to keep it as close to 1 as possible.

Industrial customers of electric utilities must maintain as high a power factor as possible. They usually have banks of capacitors that can be automatically switched on and off as needed as they will have heavy inductive loads from motors. If they maintain too low a power factor, they will be charged for kVA-hours instead of kW-hours.

For instance, if you maintained a 0.5 power factor, then a 100 A load at 120 V would be 100 x 120 x 0.5 = 6000 watts. An hour of this would be 6 kWh, but it is 12 kVAh.

Why be charged for more energy than you actually used? Simply because you are a burden on the system. Even though your load was only 6000 watts, you drew double the current than was really necessary for that amount of power. Therefore, the infrastructure needed to deliver that power had to have twice the capacity than was really necessary.

Low power factor loads tend to be inductive, so a bank of capacitors can cancel it out. Inductors cause current to lag behind the voltage, while capacitors cause current to lead voltage. The two cancel each other out. In the example above, since capacitors store and release current, they supply the "extra" current needed for the inductive load, so the only current draw on the supply is for the current actually needed to provide power. If perfectly matched, the 6000 watt load would draw 50 amps from the supply while the other 50 amps would be current between the inductive load and the capacitors.

As an interesting side note: your home probably has a leading power factor most of the time. The wiring in your home actually acts as a capacitor. When I was a student, I worked on weekends as a watchman at a factory. There was a power factor meter where the bank of capacitors was. When the factory was shut down and the only load was lighting, the power factor was usually 0.7-0.8 leading. As machines were started up and the inductance of the load increased, the PF would rise to 1 then start dropping on the lagging side. I believe a bank of capacitors would be switched in when it dropped to 0.7. In addition to the kWh meter, there was a kVAh meter.

--

Calvin Henry-Cotnam

"Never ascribe to malice what can equally be explained by incompetence."

Calvin Henry-Cotnam

"Never ascribe to malice what can equally be explained by incompetence."

Click to see the full signature.

- posted on October 22, 2005, 12:27 pm

"Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never

really have a purely inductive or capacitive load!). "

That's true for a purely inductive or capacitve load, but isn't true in the general case. Consider a load consisting of a resistor in series with an inductor. Connect a voltage source with a frequency of zero (DC) and the inductor offers no impedance and the load is purely resistive. It looks just like a resistor and the power dissipated is V******2/R.

Now change the frequency to the other extreme, making it virtually infinite. Now the inductor effectively blocks current and the power dissipated (in the resistor) is close to zero.

really have a purely inductive or capacitive load!). "

That's true for a purely inductive or capacitve load, but isn't true in the general case. Consider a load consisting of a resistor in series with an inductor. Connect a voltage source with a frequency of zero (DC) and the inductor offers no impedance and the load is purely resistive. It looks just like a resistor and the power dissipated is V

Now change the frequency to the other extreme, making it virtually infinite. Now the inductor effectively blocks current and the power dissipated (in the resistor) is close to zero.

- posted on October 24, 2005, 4:33 pm

"Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never

really have a purely inductive or capacitive load!). "

Frequency does not affect power ONLY in a load that is purely inductive or capacitive, because there is no power dissipation there. However, in the general case, frequency does affect power consumption. Consider a load consisting of a 5 ohm resistor and an inductor in series driven by a 10volt source of variable frequency. At 0 freq, ie DC, the series load looks just like a resistor and the power dissipated is 20watts. At inifinite frequency, zero current flows through the resistor, and the power dissipated is 0 watts.

really have a purely inductive or capacitive load!). "

Frequency does not affect power ONLY in a load that is purely inductive or capacitive, because there is no power dissipation there. However, in the general case, frequency does affect power consumption. Consider a load consisting of a 5 ohm resistor and an inductor in series driven by a 10volt source of variable frequency. At 0 freq, ie DC, the series load looks just like a resistor and the power dissipated is 20watts. At inifinite frequency, zero current flows through the resistor, and the power dissipated is 0 watts.

- posted on October 24, 2005, 6:25 pm

I have been wondering why this thread did not get moved to:

alt.engineering.electrical

- posted on October 22, 2005, 12:56 pm

snipped-for-privacy@optonline.net wrote:

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

- posted on October 22, 2005, 12:56 pm

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

- posted on October 22, 2005, 12:56 pm

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

- posted on October 22, 2005, 12:57 pm

The formula works just as well for single phase power. Your description is quite correct for any number of phases.

Bill Gill

- posted on October 22, 2005, 12:07 pm

"The formula works just as well for single phase power.
Your description is quite correct for any number of phases. "

P= Root (3) V . I. Cos (Pi).

It does? The cube root arises because it's a three phase circuit.

P= Root (3) V . I. Cos (Pi).

It does? The cube root arises because it's a three phase circuit.

- posted on October 22, 2005, 10:03 pm

I believe you will find that's the square root of 3.

Nick

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