# Energy in AC-Driven Purely Inductive/Capacitive Circuit

1. Aug 10, 2012

### modulus

The power losses in an AC-driven circuit depend not only upon the rms voltages and currents in the circuit, but also the power factor, cos$\phi$, where $\phi$ is the phase difference between the current and the applied alternating voltage.
In purely capacitive and purely inductive circuits, the phase difference is 90°, and so, the power factor is:
cos(90°) = 0​
.
Which means there is no power transferred from the alternating voltage source to any part of the circuit.
But how is that possible? The alternating voltage produced by a dynamo (or nay other AC source), during every cycle involve new energy inputs (the rotating armature) every cycle. So, there should be an energy buildup somewhere....
The energy seen in the form of voltage in one direction can't go back to the ac source during the time the voltage decays, because then there would be a buildup of energy in the source - the energy that is returning to it during voltage decay, and the new energy that produces the voltage of the next half of the cycle.
There seems to be a contradiction....

2. Aug 10, 2012

### RankineCycle

Energy will just slosh back and forth between the generator and the capacitor or inductor. Once 1/2CV2 or 1/2LI2 has been generated, the amount of mechanical power required to turn the generator will only be whatever is required to overcome parasitic loss. It will be like picking a bicycle wheel off the ground and giving it a spin: once the wheel is up to speed it takes almost zero power to keep it spinning.

When the ideal generator voltage drops during the AC cycle, the ideal C/L will feed all of the energy back to the generator. The generator will become an ideal motor, giving back the mechanical energy which was previously supplied. It will be like running the steam turbine or engine or whatever mechanical energy source with no load.

And of course in the real world the circulating current has to plow through the resistance of the circuit, which is why poor power factor is undesirable...it's current that's clogging up the wires, transformers and such without doing any useful work.

Last edited: Aug 10, 2012
3. Aug 11, 2012

### sophiecentaur

@modulus
There is little point in considering such 'ideal' situations because they lead to conclusions that have no relevance to real-life. You really have to include some resistance in your circuit or the scenario makes no sense. Remember that, in any case, if there are changing currents and voltages in any circuit, there will be a finite amount of energy radiated as EM waves. That, in itself, is equivalent to having a resistor in the circuit.
I think you just need to bite the bullet and use the equations in their complete form, including the resistive components. You can then investigate the limiting behaviour as the resistance approaches zero.

4. Aug 11, 2012

### modulus

@RankineCycle
Ok...I get it! The bicycle wheel analogy really helped. From what I have understood, the AC generator will not input extra energy into the circuit periodically, it will do so as and when it is needed by the circuit. This is similar to how we would apply a force to the spinning wheel only when it begins to slow down. Please correct me if I'm wrong.
And I noticed this is your first post....so, welcome to Physics Forums! I really appreciate that you took the time to create an account to answer my question.
Thanks a bunch!

@sophiecentuar
Yes, of course. That is what I strive to do.
But I think that sort of analysis is more relevant and effective when you're trying to make something practically. This past month, I was working on a coilgun.....and I realized how much different real-world situations can be. The inductor coil I was making had considerable resistance, and I had to carefully assess the effect of its internal resistance on the entire system.
But this post was meant for a theoretical discussion.
Though I completely agree with you, and understand you. :)