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We are dealing with passive and linear circuits.
The simplest circuit consists of an "ideal" battery and a resistor.
In terms of the energy transfer the battery is called the source,
and the resistor is the load. An ideal source
provides the same voltage V, independent of the load, R.
The current is, of course, I=V/R. As you see from this result,
in reality something must happen if the resistor is decreased to low values,
since the current would go up unlimited.
Let us now consider a more complicated load, made of several resistors.
How much current is drawn from the battery? To answer these type of questions
we can always go back to the basic laws of electric circuits - we can solve
the equations we obtain by using the Kirchoff's laws. In some cases we
can also use simple equivalent resistors, replacing two parallel resistors
by using R=R1R2/(R1+R2)
and two resistors in series by R=R1+R2.
Here is an important theorem:
No matter how complicated is the resistor network between two leads,
it can always be replaced by a single resistor of appropriately selected
value.
Naturally, to determine the "appropriate" value could be quite a job for a complicated network. But once we know this resistance R, the current is again simply I=V/R.
What happens if the load is simple, but the source is complex? For example,
here is a potential divider, connected to
an ideal battery. If no load is connected to the output, the voltage is
Vout=VR2/(R1+R2).
If a load is connected, the voltage drops. The current does not increase
to infinity if the load is a short circuit
R=0; instead it becomes I=V/R1. 
Imagine that the voltage source and the potential divider is in a "black box" and only the two leads are sticking out. We can connect all kinds of resistors to the leads, and measure the voltage and the current. Can we tell for sure what is in the box? The answer is NO, for there is another important theorem:
Any source, even if more than one batteries and several resistors
are inside, can be replaced by a single voltage source of appropriate value,
and a single resistor of appropriate value. 
While we may be disappointed by the fact that we can not tell exactly what is in the black box, this theorem is extremely helpful in simplifying complex circuits. Again, finding the "appropriate" replacement values Vsource and r may be challenging, but once that is done, the circuit is really simple. By the way, for the potential divider Vsource=VR2/(R1+R2), and r=R1R2/(R1+R2). (You can check out that this choice gives the right values, as discussed in the previous paragraph, for the open circuit voltage and short circuit current.)
The voltage Vsource is called the source
voltage , the resistance r is the output resistance
or the internal resistance of the source.
Note: A real power source is often nonlinear, and can not be fully described
by a simple circuit. For a NiCd battery used in a computer the concept
of the internal resistance is a good, but only approximate, representation
of the fact that the voltage drops as more current is drawn. A regulated
power supply may behave as a battery with VERY SMALL internal resistance
for modest current, but for higher currents it may stop maintaining the
voltage. See illustrations 
To measure voltages we would like to use a device that does not disturb
the current flow in the circuit. Unfortunately, all measurement devices
draw some current. A "real" device can be represented by an "ideal"
meter and a parallel resistor, Rin. This resistance is
often called input resistance of the meter.
For a good voltmeter Rin can be in the gigaW
range; for an oscilloscope it is often as low as 1MW.
When we measure a real source with a real meter, we have to worry about
the output and input resistances; the best is to have a low resistance
source and a high resistance meter. 
In conclusion we can see that even the most complex passive linear circuit can be reduced to a simple source and load circuit. Let us now turn to the next question: how should we choose a load so that the maximum power is drawn from the source? Your first instinct may be to say that the best choice is to make the load resistance low. In that case the current would be large, but the voltage drops, and therefore the power is low. A little calculation shows that the best choice is a load that is exactly equal to the output resistance of the source.
What happens if the power is transmitted over a "lossy" cable?
The conductors certainly have a finite resistance, and the insulating material
"leaks" some current between the leads. A 1 m segment of our
cable has a resistance of r1, measured between the leads
on one end, as the other end is shorted. We would measure a resistance
of r2 if the other end is open; this is due to the leakage
conductance of the insulator. The resistance r1 and the
conductance 1/r2 scales with the length of the cable.
For a given cable, can we select the output resistance and the load so
that the power transmission is good, and the load is independent of the
length of the cable?
To address this question, we will first represent the cable by a resistor
network. Let us divide the cable to pieces of 1/N meter length.
Each piece is equivalent to a resistor in series, R1=r1/N
, and a resistor in parallel, R2=r2N.
What is the equivalent resistance of an infinite network of R1
and R2? There is a simple way to answer this. Assume
we KNOW that the resistance is R*. Than we can add a
new pair of resistors, and the input resistance of the circuit will not
change - adding one more pair to an infinite network does not make any
difference. This way we obtain a simple, quadratic equation, to be solved
for R*. A further simplification is that, for large N,
R1 is much less than R2. The solution
is R*=(R1R2)1/2.
According to our result, the resistance on one end of an infinite long cable is the geometric mean of the two resistances characterizing the cable. If, for example, r1 = 0.25 x 10-3 W/m and r2 = 10 M W m (watch the units!) the apparent resistance of the cable is 50W. The best conditions for transmitting power are achieved by using a 50W output resistance source, and a 50W load.
What happens if the source voltage is not constant in time? What if we connect and disconnect components in the circuit? When the currents and voltages are time dependent we have to worry about the time variations of the energy stored in the electric and magnetic fields around the conductors. For example, an inductor stores energy as long as current flows; a capacitor has energy if the voltage is not zero. Life gets rather complicated for a general time dependent signal. It turns simple again if the signal is a simple harmonic oscillation, like V=V0 sin ( w t + j), where V0 is the amplitude and j is the phase of the signal. We will discuss that in the next chapter.
The best way to treat an arbitrary time dependent signal is to assume that it is in fact part of a periodic sequence. For periodic signals we have a theorem stating that
As long as the circuits are linear, all periodic signals can be treated as a superposition of sine waves with proper phases and amplitudes.
The mathematical process of determining the phases and amplitudes is called a Fourier transformation. The amplitude as a function of the frequency is called the spectrum of the signal. The transformation also works backwards: if the spectrum and the phases are known, the time dependent signal can be reconstructed. Therefore, knowing the response of an electric circuit to sine waves of all possible frequencies is the same as knowing the response to an arbitrary time dependent signal. This brings us to the next chapter:
Treating AC circuits is a piece of cake if you know complex numbers. Resistance is replaced by the complex impedance Z; the currents and the voltages are also complex numbers. By convention, the real amplitude of a voltage or a current is equal to the absolute value of the complex number; the time delay of the wave is represented by the phase shift, calculated from the ratio of the imaginary and real parts.
The principal elements of the circuit are the resistors (R), capacitors (C) and inductors (L). The impedances are
Here we used j for the imaginary unit, square root of -1. (In math this quantity is denoted by i, but in electrical circuits the symbol i is often used for the current.) Every theorem and formula we discussed above remains valid, except the calculations are more demanding, since we use complex numbers. Let us apply this new found knowledge to the problem of coaxial cables.
For AC signals of high enough frequency the transmission cable is best
characterized by the inductance per unit length and capacitance per unit
length. For example, the coaxial cable used extensively in the lab has
a capacitance of approximately Cft=30pF per foot, and
an inductance of Lft=75nH per foot. Similar to the DC
cable, a one foot segment can be divided up to N pieces, and represented
by a network of elements. In this case the elements are inductors (L=75nH/ft
1/N) and capacitors (C=30pF/ft 1/N). Calculating the impedance of
an infinite long cable leads to results similar to the resistive transmission
line: the input impedance of the cable is the geometric mean of Z1=jwL,
and Z2=1/jwC;
Z*=(L/C)1/2=50W.
Once you start thinking about this, the result is amazing. The circuit we looked at had no resistors, and yet the impedance proved to be entirely ohmic (no "j" in the result)! Inductors and capacitors can store energy, but can not dissipate power, and yet, when connected to a source, the cable sucks up power exactly like a 50W resistor. Where is the energy going? The key to the answer is in the fact that for the calculation to be valid, the cable must be infinite long. The energy from the source just keep flowing along the cable, out to infinity.
With a cable like this the best conditions for signal transmission
are achieved by using a source of 50W output
impedance, and a load of 50W. As long as the
(dissipative) resistivity of the wires is neglected, the signal propagates
without a change in intensity, but it suffers a time delay. For a cable
of length x, the time delay is t=x(LC)1/2.
We created 50ns wide pulses by a 50W output
impedance source, and fed them to a short and long coaxial cable. The copy
of an oscilloscope printout illustrates how the signal (a pulse of about
50ns wide) gets delayed by the longer cable.
In this test we were always careful to use 50W
loads (terminators) at the other end of the cables. Why is it so important?
Terminating the cable with a proper resistor ensures that cable behaves
like an infinite long cable; it exhibits a 50W
impedance to the source. Let us see what happens if the end of the long
cable is left open. 
At the beginning of the pulse the source has no way to "know"
how is the cable terminated (remember, it takes a significant time for
the pulse to reach the end of the cable). So the cable behaves just OK,
and the signal starts its journey to the other end. But when the pulse reaches
the open end it becomes clear that in this circuit there are no dissipative
elements and the power has to go somewhere. Thus the pulse if reflected
back, and it appears at the source again! (The reflected pulse has a slightly
smaller amplitude, due to the losses in the cable). Similarly, there is
no way to dissipate power if the end is shorted. This time the pulse is
reflected back with an opposite sign. The similarity of this phenomena
to waves propagating on a rope with fixed and free end is striking - but
it is of no surprise, since the differential equations describing the two
systems are quite similar. 
The example illustrates that having incorrect termination can really make a mess of signal transmission. In the lab many measurements rely on fast pulses. The cable TV in your home runs from 75W coaxial cables - if you do not use the right connectors and wiring, you will see "ghosts" on your screen - and the reflected signal may also show up in other TV sets on the same line. The Ethernet cable used for some of the local "intranet" are also very sensitive to this problem - it happens to be the same 50W BNC cable used in the lab.
Sometimes we have no choice, but we have to connect cables of different
impedance. Matching of impedances can be still achieved if we use a transformer.
Here is how it works. Assume we have a transformer that changes the voltage
by a ratio a=Vout/Vin.
A resistance R is connected to the output. The power dissipation is Vout2/R.
On the input side, where the voltage is Vin=Vout/a,
the power dissipation is Vin2/R'=Vout2/a2R',
where R' is the impedance seen by the source. But power is not lost
in a (ideal) transformer, so R'=R/a2.
This concludes our "primer" to linear networks. For further reading the introductory chapters of "Elementary Electronics" by D.H. White (several copies are available in the Library, catalog number: TK7816.W45) are recommended.
Created by Laszlo Mihaly; last updated 10/2/97. Here is the Copyright Notice