Dec '40/Jan '41 National Radio News
[Table
of Contents] These articles are scanned and OCRed from old editions of the
National Radio News magazine. Here is a list of the
National Radio News articles I have already posted. All copyrights are hereby acknowledged. 
As mentioned in an earlier article, National Radio Institute (NRI)
was one of the first companies to exploit the burgeoning field of
electronics in the early part of the last century. They invested
heavily in facilities and resources for producing educational material
for both classroom and correspondence courses. Up until sometime
in the early 1990s when throwaway electronics and almost total
offshore manufacturing became the rule rather than the exception,
there were always large advertisements in magazines offering to
rescue floundering career seekers with promises of untold opportunities
and riches from servicing radios, televisions, home appliances,
and more. That is not to say the courses were not valuable
 they were. Regular newsletters and other publications were mailed
to participants, usually containing good advice on troubleshooting,
circuit theory, component selection, how to handle customers who
try to skip out on paying you, etc. This particular edition of National
Radio New discuses current carrying capabilities of various size
wires  a topic that never goes out of style.
The Laboratory Page: CurrentCarrying Capacity of Wires
By George J. Rohrich
George J. Rohrich, Engineer in Charge N. R. I. Laboratory

The purpose of this department is to furnish supplemental experiments
to students who have completed their Home Laboratory course, but
who wish additional laboratory experience. You are not required
to perform these experiments, but you will gain increased knowledge
by doing so.
Most of the material required will be that received as part of
the Laboratory Course. Any other material necessary can be purchased
very reasonably and will constitute an investment rather than an
expense, as it will serve as replacements in service work or be
useful in your shop later.
The CurrentCarrying Ability of Resistors
A resistor can be constructed to have a given value of resistance
by employing different sizes of wire. For instance, we can construct
a 500 ohm resistance of very fine wire and it will be necessary
to use only a few feet of such wire in order to obtain 500 ohms.
This particular wire can carry only a given amount of current
without becoming excessively heated.
If we find that the 500 ohm resistor becomes excessively hot,
then it will be necessary to obtain another resistor which is constructed
of larger wire. In order to maintain the same value of resistance,
it will be necessary to use a greater length of the larger sized
wire. The same amount of heat will be produced in the larger wire
but its larger surface will radiate heat quickly, keeping it cool.
If we find that the resistor has to carry considerable current,
then it may be necessary to use very large wire.
From the above explanation, we see a resistor has to be constructed
so as to carry a given ·amount of current. Of course. a resistor
which is capable of carrying a large current can also safely carry
a smaller current.. The usual practice is to construct a resistor
which can safely handle the maximum current which will be flowing
through it at any time. There is no economy in constructing a larger
resistance when a smaller one will do, unless the smallest allowable
size is mechanically fragile and difficult to handle.
Instead of rating a resistor in its current carrying capacity,
it is customary to rate the resistor as "capable of dissipating
a given amount of heat." This rating is given in watts. This rating
in watts is obtained by multiplying the value of current itself,
and finally multiplying this result by the value of resistance.
This is expressed in a formula; Watt = I^{2}R.
For example, let us say that we find a 500 ohm resistance carries
2 amperes. Multiplying 2 amperes by itself, we obtain 4. Multiplying
4 by 500 ohms we obtain 2000 watts. This is the rating of the resistor
and we say that it is capable of carrying 2 amperes.
The rating in watts can be obtained by a second method. This
requires you to multiply the voltage across the resistor by the
current. This is expressed by the formula: Watts = E x I.
A 500 ohm resistance which has a lower rating in watts is intended
to carry a smaller current. This current carrying capacity can be
obtained by dividing the value in watts by the value of resistance,
and then taking the square root of the result. Thus, a 1/2watt,
500ohm resistor is intended to carry currents which are not greater
than 0.031 ampere or only 31 milliamperes.
You
will notice that the relation between watts, voltage and current
is expressed by an equation which is similar to the equation of
Ohm's Law which shows the relation between voltage, current and
resistance. Remember these two equations:
Voltage = current x resistance.
Power = voltage x current.
In order to help you remember them keep in mind these four numbers:
2, 3, 6 and 12 and associate the 2 with current, the 3 with resistance,
the 6 with voltage and the 12 with power.
We know that 6 = 2 x 3.
Therefore voltage = current x resistance.
We also know that 2= 6 ÷ 3.
Therefore current = voltage ÷ resistance.
We know that 3 = 6 ÷ 2.
Therefore, resistance = voltage ÷ current.
We know that 12 = 6 x 2.
Therefore, power = voltage x current.
We know that 12 = 2 x 2 x 3.
Therefore, power = current x current x resistance. This represents
the equation W = I^{2}R.
We know that 12 = (6 x 6) ÷ 3. Thus, power = voltage x
voltage, divided by the resistance. This represents the equation
W = E^{2}+R.
We know that 3 = (6 x 6) ÷ 12. Thus, resistance = voltage
x voltage divided by the power. This represents the equation R =
E^{2} ÷ W.
We know that 2 = √ (12 ÷ 3). Thus, current =the
square root of the result which is obtained when the power is divided
by the resistance. This is represented by the equation: I = √
(W ÷ R).
We know that 6 = √ (12 x 3). Therefore, voltage equals
the square root of the result which is obtained when the power is
multiplied by the resistance. This represents the equation E = √
(W x R).
We know that 3 = 12 ÷ (2 x 2). Therefore, resistance equals
power divided by the square of the current. This is represented
by the equation: W ÷ I^{2}.
We know that 6 = 12 ÷ 2. Therefore voltage = power ÷
current. The symbol equation is represented by E=W ÷ I.
We know that 2 = 12 ÷ 6. Therefore, current = power ÷
voltage. The symbol equation is I = W ÷ E.
With these twelve equations you can find the value of anyone of
the four quantities, provided you know the value of two other quantities.
You can see how easy it will be for you to work out anyone of the
twelve if you will only remember the two equations to which I called
your attention above.
The current carryingcapacity of a wire used in a circuit or
electrical instrument is dependent on the size of the wire. It is
customary to allow 1000 circular mills of crosssectional area for
each ampere where copper wire is exposed to good ventilation. Where
the wire is used so that air cannot reach it freely for carrying
off the generated heat, it is customary to allow 1500 circular mills
for each ampere.
The crosssection area of various wires is shown in the accompanying
table.
For instance a No. 30 wire has a cross section area of approximately
100 circular mills. This wire is used for carrying 0.1 ampere where
the wire is exposed and it is used to carry a maximum of 0.066 ampere
where the wire is confined to close places. A No. 30 wire will burn
out when a current of 10 amperes flows through the wire.
A No. 38 wire has a cross sectional area of 15.72 circular mills
and it is rated to carry 16 milliamperes and 10 ma. under the two
conditions. This wire will burn out with 2.5 amperes.
The milliammeter furnished with the experimental parts uses No.
38 wire.
For ordinary purposes it may be remembered that a wire will not
burn out until a current of 50 times to 150 times the allowable
currentcarrying capacity of the wire flows through the wire.
Posted May 28, 2014
