Re: AC electricity

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding:

> I know that AC delivers energy - that is obvious.
> But I still have trouble understanding how this is possible
> since half the time the potential is positve and the other
> have it is negative.  It seems they should cancel out.

Consider two long parallel wires going out horizontally away from you in
the direction in which you are facing.  Assume one wire is situated to
the right of the other one.  Suppose these two wires each carry a current
such that the current in each wire is the negative of the current in
the other wire.  Maybe there is an energy source connected to your end of
the wires and there is some resistive load connected across the wires at
the distant end.  Let's analyze the direction of energy flow along the
wires.

Case 1. The right hand wire (from your point of view) carries current
(using the usual positive convention for current) down the wire away
from you, and the left hand wire carries a return current back up the
wire toward you.  Also, the right hand wire is assumed to have a positive
voltage relative to the left hand wire.

In this situation the magnetic field is relatively strong in the region
between the wires.  The direction of the magnetic field there is
predominantly upward as the magnetic field lines converge below the
wires, go up between the wires, and split above them with half of them
looping back (rightward and down) around the right wire, and half of them
looping back (leftward and down) around the left wire.  This magnetic
field pattern is determined by the assumed direction of the currents in
the wires via application of Ampere's law using the right hand rule for
each wire and superposing the results.  Since the right side wire is
positive w.r.t. the left wire there is an electric field which is also
relatively strong in the region between the wires and this field points
to the left from the positive right side wire to the negative left side
wire in that intermediate region.

The energy current vector (EM energy transported per unit time per unit
area perpendicular to the direction of the flow) for this situation is
given by the Poynting vector which is proportional to the cross product
of the electric and magnetic fields (E x H).  Since in the region where
the fields are relatively strong H points up and E points to the left, we
see (via the right hand rule) that the cross product points down along
the wires away from you.  Thus, we understand that the wires guide the
transport of EM energy down along the wires away from you.

Case 2.  *Both* the direction of the current flow in the wires and the
electric polarity of the wires is reversed from that of case 1.  Since
the currents are reversed the magnetic field is also reversed.  And since
the electric polarity is reversed the electric field is reversed as well.
But (-E) x (-H) = E x H so we see that the Poynting vector, i.e. the
energy current *still* points down along the wire away from you, and EM
energy is still being transported down the wire to the supposed load at
the far end.

Case 3.  Now suppose that the load is highly reactive and there is a low
power factor it.  In this case the current is not in phase with the
electric voltage polarity of the wires and for part of the AC cycle there
is a situation where the current is still flowing in the direction
indicated in case 1, but the electric voltage of the wires is reversed
relative to that case.  Since (-E) x H = - E x H we see that the Poynting
vector of the energy current is now reversed from case 1 and energy is
being transported back up along the wires from the "load" to you and the
supposed energy "source".

Case 4.  If the load is highly reactive there will also be a time
during some portion of the AC cycle where the current in the wires is
reversed relative to case 1, but the electric voltage polarity of the
wires is the same as that case.  In this case the magnetic field is
reversed but the electric field is not (relative to case 1.).  Since
E x (-H) = - E x H we again see that energy is being transported back up
the wire from the "load".

If the phase shift between the current and the voltage of the wires is
less than 90 degrees in magnitude, then more energy will be sent down to
the load during the cases 1 & 2 situations than will be sent back
during the cases 3 & 4 situations because of the relative time durations
during the cycle for which these conditions obtain.  If the phase shift
is precisely +/- 90 degrees, then the system spends 1/2 the time sending
energy down the wire, and 1/2 the time receiving energy up from the load
and the load consumes no net energy over an integer number of AC cycles.

If the phase shift between the current and the voltage of the wires is
greater that 90 degrees in magnitude, then more time is spent with
energy being recieved from the supposed 'load' than is spent sending
energy down to it.  In this case the 'load' is really the energy source,
and the supposed 'source' is really the load, and a net transfer of
energy takes place in a reversed sense up the wire toward you.

BTW, in order to characterize the temporal mismatch between the current
and the voltage in AC cycle in terms of an angle, I have implicitly
assumed that the AC waveform was sinusoidal.  If this is not the case
then it is not a good idea to try to use of a phase shift angle to
describe the process.

Note that the above analysis does not depend in any way on how the
"load" uses the energy delivered to it.  It could be stored by maybe
charging a rechargable battery (via suitable rectification), radiated
into space from an antenna, or dissipated by heating an ordinary
resistance that obeys Ohm's law (or something else).  The issue of
the direction of energy flow is separate from the issue of energy
dissipation.

David_Bowman@georgetowncollege.