# short circuits and rf/emf

Discussion in 'Electronic Basics' started by erehwon, Dec 28, 2003.

1. ### erehwonGuest

I have a basic question... )
How is it that when creating an electromagnet using a piece of wire (coiled)
attached to +ve and -ve it doesn't - I assume - create a short circuit. It
seems that somehow the circuit knows that the power is to be used to create
the electro magnetic field rather than just sending it back to the battery
creating a short circuit.

The same question goes for an rf circuit... if I understand correctly, you
generate a large amplitude ac wave (probably modulated with some interesting
signal), and then hang a piece of wire off it to radiate the rf energy...
but how come the wire isn't just ignored, and you just end up with a big
short circuit?

apologies for my naivete.

thanks,
JJ

2. ### John PopelishGuest

The coil puts all the incoming energy into the magnetic field only at
the first moment the battery is connected (when the current is zero).
As the current rises on an L/R time constant curve, more and more of
the energy being delivered by the battery goes into resistive heating
(both the wire resistance and the battery resistance) and less and
less goes into the magnetic field. Once the current approaches a
stable value, all the energy is going into resistive heating, since it
takes no continuous delivery of energy to maintain a magnetic field.
There is also a small component of the magnetic field that escapes as
waves into space anytime the magnetic field is changing.
THe inductance of the coil produces a voltage that cancels that being
applied, as long as the current is changing. And, for a sine wave
signal, the voltage is always changing. The formula that relates
applied (and also internally produced) voltage and the rate of change
of the current passing through is V=L*(di/dt) where di/dt means the
time rate of change of current. Even a straight piece of wire has
some inductance from end ot end and some capacitance to the space
around it.

3. ### John LarkinGuest

Not really. After some initial settling (microseconds to milliseconds
in typical situations) the wire acts just like wire. A typical
electromagnet or solenoid has enough wire (proper diameter and proper
length) to have a resistance that draws the designed correct amount of
current in steady-state operation, without getting too hot. The DC
steady-state current is the same whether that wire is coiled up or
not.
RF is different, because the available voltage is oscillating,
changing rapidly all the time. In this case the inductance of the wire
restricts current flow; just how much depends on the frequency of the
generator and the length of the wire. If a wire is short ehough
relative to the signal wavelength, it does act like a "short circuit."

John

4. ### erehwonGuest

thanks for answering - that makes sense now. )

5. ### erehwonGuest

The same question goes for an rf circuit... if I understand correctly,
you
hmm... thinking about it some more though, the antennas I've seen (on
circuit diagrams) aren't actually 'in' the circuit (i.e. both ends) as far
as I can see, but tend to just be a piece of wire/metal with *one end*
attached to the circuit at the 'appropriate point'.
example:
http://www.uoguelph.ca/~antoon/circ/tx15trak.htm

is the radiation created when the emf (around the inductance coil)
collapses - creating current which somehow gets pushed up the antenna as
well as back into the circuit?

thanks again all,
JJ

7. ### John PopelishGuest

There is more. When the frequency gets high enough that the
conductive paths get to be a significant fraction of a wavelength, you
can no longer assume that the current (and voltage) is the same all
along the path. In fact, at distances 1/2 wavelength apart, the
currents are going the opposite direction and the voltages are of
opposite sign. This applies to wire that is straight and wire wound
into coils (with some more complications). You really need a good
course in fields and waves to get the whole picture. But you now have
some rudimentary mental models to get you started.

8. ### erehwonGuest

isn't it something to do with the capacitance of the antenna? or using the
medium between transmitter/receiver as a dielectric?  