Jun/Jul 1958 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. 
Many people end on RF cafe
as a result of a Google (or other) search about electronics, so even though regular visitors
might find this primer on Ohm's law to be redundant review (is that phrase redundant?),
it will be valuable to the aforementioned people. Electronics technology has moved forward
at lightning speed in the last century, but the fundamentals of Ohm's law remain unchanged.
Indeed, we would be in trouble if voltage no longer equaled the product of current and
voltage (E = I x R). National RadioTV News magazine was
published monthly (1928  1980) by
National
Radio Institute, a correspondence school that did business from 1914 through 2002.
A bonus electronicsthemed comic is included.
How Ohms Law is Used in Service Work
By Dale Stafford, NRI Consultant
Ohm's Law defines the relationship between voltage, current, and resistance (in a
dc circuit) or impedance (in an ac circuit). Since resistance and impedance both refer
to the opposition offered to the flow of current, no effort will be made to differentiate
between them in this article. The term resistance will be used to refer to any opposition
to current flow.
The current flowing in a circuit or through any part in a circuit depends on two things:
(1) the voltage applied across the circuit or part and (2) the resistance of the circuit
or part.
The current is directly proportional to the applied voltage. It will vary in direct
proportion to any changes in the voltage. If the voltage is increased without changing
the resistance, the current will increase. If the voltage is reduced without changing
the resistance, the current will decrease. If the voltage is doubled, the current will
double. If the voltage is cut in half, the current will be cut in half, also.
The current is inversely proportional to the resistance and will vary inversely with
any change in the resistance. If the resistance is increased without changing the applied
voltage, the current will decrease. If the resistance is reduced without changing the
applied voltage, the current will increase. If the resistance is doubled, the current
will be cut in half. If the resistance is cut in half, the current will be doubled.
These relationships make it easy for us to find one of these values if the other two
are known. Three formulas are used for these computations. They are:
I = E ÷ R
R = E ÷ I
E = I x R
Here is my certificate for completing a National Radio Institute (NRI)
electronics technician course back in the 1980's.
In these formulas E stands for the applied voltage, R stands for resistance, and I
stands for current. In calculations involving ac circuits, Z, which stands for impedance,
is substituted for R in the formulas. Otherwise the formulas are the same.
Just why is Ohm's Law so important to the serviceman? Only on rare occasions is it
necessary for him to use it in finding the value of a part needed as a replacement. Ordinarily,
diagrams are available giving the value of the parts used in a radio or TV receiver.
He needs only to glance at the diagram to obtain the information he needs.
The answer is that he uses Ohm's Law (perhaps unconsciously) every time he attempts
to locate a defect in a receiver by measuring the operating voltages or taking resistance
measurements. For example, suppose he finds that the voltage across a cathode bias resistor
is too high. Because E equals I x R, he immediately knows that the resistor has increased
in value or that the tube is drawing too much current.
Or suppose he finds the voltage too high on a screen grid that has a dropping resistor
between the grid and B plus. At once, he suspects that the resistor has decreased in
value or that, for some reason, the tube is not drawing normal screen current.
The complaint may be a severe case of distortion. When our technician connects his
VTVM across the grid resistor in the output stage, he finds a positive dc voltage on
the grid. Normally, the grid does not draw current, and with no current through the grid
resistor, there should be no voltage drop across it. Pulling the output tube (if it is
an ac set) or disconnecting the coupling capacitor (if it is an acdc set) will enable
the serviceman to determine if a gassy output tube or a leaky coupling capacitor is responsible
for the trouble.
Another condition often encountered is that of an excessive voltage drop across a
decoupling resistor in the plate circuit of a tube. This might be caused by a change
in the value of the resistor. It could also be caused by some defect that caused the
current through the resistor to increase, such as a leaky bypass capacitor on the plate
side of the resistor, a shorted cathode bias resistor or bypass capacitor, or a defective
tube.
It does not take a very large change in resistance or current to cause a considerable
increase in the voltage drop across the resistor with a corresponding decrease in the
voltage available at the plate of the tube.
"You don't have to fix it young man  just find it!"
Even if the serviceman does not know what the plate voltage or the voltage drop across
the decoupling resistor should be, a tube manual will usually give him this information.
These manuals list operating voltages and currents for typical operating conditions and
a comparison of these voltages with those available in the receiver will usually make
it possible to determine the approximate correct value.
These and many, many others are all practical applications of Ohm's Law. Automatically,
almost unconsciously, the serviceman uses Ohm's Law to analyze circuit conditions even
though he doesn't sit down with paper and pencil to calculate exact values.
However, there are certain occasions when he may find it necessary to make such computations.
One such occasion might be in computing the value of a series voltagedropping resistor
in the filament circuit of a radio receiver. In acdc receivers, the tube filaments are
connected in series and connected across the power line. If the voltage required for
the tubes is less than the line voltage, it is necessary to insert a voltagedropping
resistor in the filament circuit to use up the excess voltage. This resistor may burn
out, leaving no clue to the proper value.
To find the value of this resistor, it is first necessary to find out how much voltage
must be dropped across it. Adding the voltage drops across all the tubes, and subtracting
this figure from the line voltage gives us this value. Of course, tubes in these sets
are designed to operate with the same value of filament current, and this current must
also flow through the resistor. Dividing the voltage drop across the resistor by the
filament current gives us the value of the resistor.
If the resistor is a chassismounted resistor, it is necessary to compute the required
wattage rating of the replacement to be sure that the necessary amount of heat can be
safely dissipated. This can be found by multiplying the voltage drop across the resistor
by the resistor current. To allow a margin of safety, a higher value should be used (at
least 1½ times the computed value).
If a linecord resistor is used, there is no need to figure the wattage as these resistors
can dissipate more heat than they will ever be required to.
A serviceman may find it necessary to figure the value of a cathode bias resistor
to be used in a circuit he is wiring or to replace a burnedout resistor whose value
is unknown. A cathode bias resistor carries the plate and screengrid currents of the
tube. The current through the resistor can be found by adding the plate and screengrid
currents which will flow at the value of plate voltage used. Again a tube manual can
be used to determine the approximate value.
The bias voltage developed across the resistor will be equal to the current through
the resistor multiplied by the value of the resistor (E equals I x R). The proper resistor
value can be found by dividing the bias voltage needed by the current through the resistor
(sum of plate and screen grid currents).
Pilot lamps in many receivers are connected in series with the tube filaments. When
this is done, it is necessary to protect the lamp against excessive current flow. This
can be done either by shunting the lamp across a portion of the rectifier tube filament
or by connecting a small resistor in series with the filament circuit and connecting
the lamp across the resistor.
The resistor is chosen so that the voltage drop across the resistor and lamp will
be 4.25 volts when the normal filament current is flowing. The resistor must carry a
current equal to the difference between the current rating of the lamp and the current
rating of the tubes.
The proper resistance value can be found by subtracting the current rating of the
lamp from the current rating of the tubes and then dividing 4.25 by this figure.
Many other applications of Ohm's Law will be encountered in practical servicing work.
The busy technician uses his knowledge of this subject every hour of every day. Whether
he uses it to figure the exact value of a new or replacement part or merely to interpret
the meaning of measurements he has taken, he finds it one of the most valuable "tools
of the trade."
There are other reasons why the student of electronics should know the subject thoroughly.
Not every student starts his own shop when he completes his training. Many go to work
for others. Before a company hires a technician, he will be required to take a written
or oral examination to prove that his training is adequate. Quite likely several of the
questions will be on practical applications of Ohm's Law.
Those who go to work for someone else may find themselves helping to design and construct
circuits of many kinds. Perhaps one may be helping an engineer with design work while
another is doing maintenance work on old circuits that must be modified or rewired.
If one of these men is asked by his superior to compile certain information, he must
know how to obtain that information as quickly and accurately as possible. For example,
if he is asked for the current flow through a resistor, he should know that measuring
the voltage drop across it and dividing this figure by the value of the resister is a
quicker and easier method of finding the current than actually measuring the current
with a meter. To measure the current, the circuit must be broken so the meter can be
inserted in series with the current, the measurement must be taken, the meter removed,
and the circuit reconnected. Unless the meter is connected as a permanent part of the
circuit, the first method is, naturally, preferable.
Fig. 1  Ohm's Law in use.
Fig. 2  Ohm's Law in dc series circuits.
The following are some examples of Ohm's Law in use:
In Fig. 1(A), we have a simple dc circuit consisting of a resistor in series with
a battery.
If we know the battery voltage and the resistance, we can find the current. Suppose
the battery voltage is 90 volts and the resistance is 30 ohms. Since I = E ÷ R,
the current through the resistor is 90 divided by 30, which equals 3 amperes.
If the battery voltage and the current are known, as shown in Fig. 1(B), the resistance
can be found. Here the battery voltage is 90 volts and the current is 2 amperes. R = E ÷ I,
so the resistance equals 90 divided by 2, or 45 ohms.
If both the resistance and current are known, the battery voltage can be found. (Fig.
1(C). Here the resistance is 60 ohms and the current is 1.5 amperes. E = R x
I, so the battery voltage equals 60 x 1.5, or 90 volts.
The same procedure can be used in simple dc series circuits containing several resistance
values in series, as shown in Fig. 2.
In Fig. 2(A), we have three resistors of known value in series with a 90volt battery.
The first step in finding the current is to add the value of all the resistors to find
the total circuit resistance. (Rt = R1 + R2 + R3, 5 + 10 + 30).
We find the total resistance to be 45 ohms. Then, using the formula E+R=I; we have 90
divided by 45, which equals 2 amperes, the circuit current.
In Fig. 2(B), we have three resistors of known value in series with a battery. If
the circuit current is known, we can use the formula E = I x R to
find the voltage across each resistor, since in a series de circuit the same current
flows through each part in the circuit. In this case, 5 x 2 equals 10 volts
(the voltage across R1), 10 x 2 equals 20 volts (across R2), and 30 x 2
equals 60 volts (across R3).
In Fig. 2(C), we have the same circuit except that we do not know the value of the
resistors. Since R = E ÷ I, it is easy to find the total resistance of the circuit
by dividing the battery voltage by the circuit current. However, we need to know the
voltage drop across each resistor to find the value of each. Measuring these voltages,
suppose we find 10 volts across R1, 20 volts across R2, and 60 volts across R3 as shown
by the dotted lines in the figure. Then 10 divided by 2 equals 5 ohms (value of R2, 20
divided by 2 equals 10 ohms (value of R2), and 60 divided by 2 equals 30 ohms (value
of R3).
Fig. 3  Ohm's Law in dc parallel circuits.
Fig. 4  Circuit problems (answers at bottom of page).
We can find unknown values In parallel dc circuits in the same manner. In Fig. 3(A),
a 90volt battery is connected in series with two parallelconnected 20ohm resistors.
Since I = E รท R, the circuit current will be equal to 90 volts divided
by the equivalent resistance of the parallelconnected resistors. Let's see how to find
the equivalent resistance.
If resistors of equal value are connected in parallel, the equivalent resistance of
the combination is equal to the value of one of the resistors divided by the number of
resistors used. In this case, 20 divided by 2 equals 10 ohms. Had the resistors been
of unequal value, the formula could have been used to find
the equivalent resistance.
The circuit current will be 90 divided by 10, or 9 amperes. In a parallel circuit,
the total current divides among the various branches in proportion to the resistance
of each branch. Since the resistors are equal in value, each branch will draw half the
total current, or 4.5 amperes.
In Fig. 3(B), we know the battery voltage and the circuit current, but, to find the
value of each resistor, we need to know the current flowing in each branch. Suppose we
measure these currents and find 4.5 amperes flowing through each branch.
When a voltage is applied across a parallel circuit, this voltage appears across each
of the branches, so, using the formula R = E ÷ I, the value
of each resistor equals 90 divided by 4.5, which equals 20.
In Fig. 3(C), two 20ohm resistors are connected in parallel with 4.5 amperes flowing
in each branch. Since E = R x I, we can find the source voltage by
multiplying 20 by 4.5. This gives us 90 volts, the battery voltage.
Ohm's Law could be applied to an ac circuit with equal ease if it were possible to
have such a circuit which was purely resistive in nature. However, when capacity and
inductance are added to a circuit, the procedure becomes somewhat more involved.
Special formulas must be used to figure the reactance of the parts and the impedance
of the circuit. Once this is done, Ohm's Law can be used as easily as in dc circuits.
In Fig. 4 are some problems which can be used for practice. In Figs. 4(A) and 4(C)
you are to find the current. In Fig. 4(B) you are to find the voltage drop across each
resistor. The answers will be found printed upside down below Fig. 4.
Answers: Fig. 4(A), 0.25 amp.; Fig. 4(B), R1  3.2 volts, R2  4
volts, R3  2.4 volts; Fig. 4(C), 0.5 amp.
Posted May 14, 2018
