Phase angle between I and V > 90?

[DeG]

I know that the phase angle between I and V determines the power factor (cos(angle)) and thus the fraction of power supplied that can do real work. Also that a phase angle of 90 degrees means that none of the delivered power can do real work in the system. What happens when the phase angle goes over 90 degrees though? Will this again give real power to the system?

 

[vk6kro]

When the phase angle goes over 90 degrees leading, then it is less than 90 degrees lagging in the previous cycle.

So, the circuit is (or behaves like) an inductor instead of a capacitor combined with some resistance.

If the angle is anything except 90 degrees, then there is some resistive component and power will be dissipated in this resistance if there is a current flowing.

 

[cabraham]

When the phase angle goes over 90 degrees leading, then it is less than 90 degrees lagging in the previous cycle.

So, the circuit is (or behaves like) an inductor instead of a capacitor combined with some resistance.

If the angle is anything except 90 degrees, then there is some resistive component and power will be dissipated in this resistance if there is a current flowing.

Not correct. If I leads V by 100 degrees, it is not lagging V by 260 degrees. I leading V by more than +90 or less than -90 means that power is being delivered instead of absorbed.

This scenario occurs in emitter follower amp stages with capacitive loading. The capacitive load in the emitter is reflected to the base side as a negative resistance plus a reactance. Instability can result from this. The remedy often employed is the addition of a resistor in series with the base lead. The positive resistance of the resistor cancels the negative resistance reflected from the emitter.

Claude

 

[Ratch]

DeG,

I know that the phase angle between I and V determines the power factor (cos(angle)) and thus the fraction of power supplied that can do real work. Also that a phase angle of 90 degrees means that none of the delivered power can do real work in the system. What happens when the phase angle goes over 90 degrees though? Will this again give real power to the system?

If the AC voltage and current are ±90° out of phase, the source provides energy for part of the cycle, and receives the same amount of energy back during the remainder of the cycle, for a net energy transfer of zero. The net energy supplied will be greater than zero when resistance is present. What is "real work"? What other kind of work is there? A passive network can never exceed ±90° between voltage and current. In a circuit with active components, if the phase is greater than ±90°, the signal source will be receiving energy back from the active component energy supply for part of the cycle.

Ratch

 

[rbj]

if you have a two-terminal device, and let's say it's linear (like resistors, capacitors, inductors), and you experimentally derive volt-amp characteristics and if this device looks like a resistive and reactive element with a phase angle between 180° and 90° or between -180° and -90°, that means that there is negative resistance in this device. all by itself, a negative-resistance resistor does not exist because it actually generates energy (out of nothing) and delivers it to the outside. but you can build a circuit with parts including transistors or op-amps that has, over a limited voltage range, behavior like a negative resistor between a pair of nodes and it doesn't violate energy conservation because you are supplying power to the circuit.

 

[DeG]

This all makes sense. Thanks a lot everybody.

@Ratch - I am specifically referring to power distribution systems and how the utility company has to deliver X amount of power to a customer, but if the customer has an inductive load then the customer only uses X*cos(phase angle) of power (i.e. real work) and the rest is somehow dissipated.

It makes sense that if you add capacitors to the customer side to reduce the phase shift and over-corrected, the capacitors could push power back into the transformer.

 

[Ratch]

Deg,

@Ratch - I am specifically referring to power distribution systems and how the utility company has to deliver X amount of power to a customer, but if the customer has an inductive load then the customer only uses X*cos(phase angle) of power (i.e. real work) and the rest is somehow dissipated.

No, the reactive power is not dissipated. It is stored in the inductors as a magnetic field for part a cycle, and then given back to the utility during the remainder of the cycle. What the utililities don't like is the higher current they have to supply to make up for the diverted energy. The higher current results in higher IR losses which is truly dissipated as heat.

It makes sense that if you add capacitors to the customer side to reduce the phase shift and over-corrected, the capacitors could push power back into the transformer.

The same thing happens if there is too much inductance. If there is too much capacitance, some of the power is stored in the capacitors as a electrostatic field for part a cycle, and then given back to the utility during the remainder of the cycle.

So reactance power is not dissipated as heat or any other way. It is stored in the circuit and given back to the source in a cyclical manner. This reactive power is supplied by the source, but is not applied to the load. It is not lost energy, but the source has to increase the current to make up for the diverted power.

Ratch

 

[DeG]

Awesome, thanks for the clarification