of Contents]These articles are scanned and OCRed from old editions of the Radio & Television News magazine.
Here is a list of the Radio & Television News articles
I have already posted. As time permits, I will be glad to scan articles for you. All copyrights (if any) are hereby
everyone who visits websites like RF Cafe are seasoned electronics veterans. While you and I can do
series and parallel circuit analysis (and series/parallel for that matter, possibly using Fourier or
La Place transforms for reactive AC circuits) in our sleep, many are recently getting into the wonderful
world of electronics who are just coming of age or have suddenly at a later point in life developed a
passion for the craft. Accordingly, this article from Radio News provides yet another tutorial on the
fundamentals of series and parallel circuit analysis. Only resistors and basic Ohms law are covered.
See all available
vintage Radio News articles
Student's Radio Physics Course - Series and Parallel
This series deals with the study of the physical aspects of radio phenomena. It
contains information of particular value to physics teachers and students in high schools and colleges. The Question
Box aids teachers in laying out current class assignments
Lesson Eleven - Series and Parallel Circuits
By Alfred A. Ghirardi*
Figure 1. The same current flows through every part of the circuit.
In order to have current flowing in any conductor the circuit must form a complete conducting path from the positive
terminal to the source of e.m.f. around to the negative terminal (except in the case of a circuit with a condenser).
In actual electrical circuits, electrical devices are connected in either of two ways - or a combination of the
two. When they are connected one after the other in such a way that all of the current flows through each of them,
they are said to be in series. Thus, in Figure 1 the filaments of all three of the vacuum tubes shown are connected
in series with each other and with the resistor R1
across the 110-volt
electric light circuit whose e.m.f. is maintained by the electric dynamo G. In such a circuit the total resistance
of the entire circuit is equal to the sum of the separate resistances. Thus in Figure 1 if the resistances of the
individual parts are as marked, the total resistance is:
R = R1
+ etc. (1)
The total resistance is R = 380 + 20 + 20 + 20 = 440 ohms.
The current I flowing in the circuit is:
I = E/R = 110/440 = 0.25 amperes
Another important fact
regarding the series circuit is that the current is the same through every part of the circuit, since there can
be no accumulation of current at any point along the circuit. If five ammeters were connected at the points marked
I in Figure 1, they would all indicate the same current I, of 0.25 amperes. Also if a series circuit is opened or
broken at any point the current stops flowing.
A voltage drop occurs across each of the various resistances
in a series circuit, depending on its resistance. If a voltmeter were connected across the filament of tube A, it
would indicate E = I X R1
= 0.25 X 20 = 5 volts. This is the voltage
drop or fall of potential across the resistance. Similarly, the voltmeter would read 5 volts if connected across
the filaments of tubes B and C, since they both have resistances of 20 ohms. If it were connected across resistance
it would indicate E = I X R = 0.25 X 380 = 95 volts. The sum of all
these voltage drops around the circuit is equal to 5 + 5 + 5 + 95 = 110 volts. This of course is equal to the voltage
of the source of e.m.f. (G) which is causing the flow of current through the resistances. This illustrates another
law of the series circuit: "The total voltage applied to the circuit is equal to the sum of the voltage drops across
the individual resistances in the circuit." If any unit in a series circuit should become "short circuited," the
current will increase because the total resistance of the circuit would be decreased.
Notice from Figure
1 that the voltage drop across any resistance in the circuit depends upon its resistance. Thus even though the
same current flows through all parts, the voltage drop across the 380-ohm resistance is 95 volts, whereas that across
each 20-ohm resistance is only 5 volts.
In radio receivers series circuits are very common in the plate
circuits of vacuum tubes, as we shall see later. The adding of resistances in series is equivalent to increasing
the length of the conductor, so that the total resistance is equal to the sum of the separate resistances.
Figure 2. The current divides and part flows through each branch.
When parts of a circuit are connected in such a way that they form separate paths through which the current can
divide, they are said to be connected in parallel, multiple, or shunt. Only a portion of the total current flowing
from the source of e.m.f. flows through each path.
Figure 2 shows a parallel circuit consisting of the filaments
of three dissimilar vacuum tubes supplied with current forced through the circuits by the e.m.f. of the storage
battery, E. Only a portion of the total current circulating through the battery passes through each of the circuits,
but of course the sum of the number of amperes of current flowing in the three circuits is equal to the number of
amperes of current circulating through the battery, since all the currents combine again. The actual current in
each wire of the circuit is indicated on the diagram. Notice how the current coming out of the positive terminal
of the battery divides to go through the tube filaments and then combines again at the negative line.
number of electrical devices or circuits may be connected in parallel. The current returning to the negative side
of the source of e.m.f. is exactly equal to the current leaving the positive side. The current is merely circulating
through the circuits. The electrical devices connected in parallel may all have the same resistance or they may
all have unequal resistances. If the resistances are equal, then it is evident that the total current will divide
equally among the various paths, and the combined resistance of all the paths considered together is equal to one
of the resistances divided by the number of resistances. Thus, if five resistances of 100 ohms each are connected
in parallel, the combined resistance will be
100/5 = 20 ohms, since five paths are being presented to the
flow of current instead of only one.
When the parallel resistances are not equal, the combined resistance
must be found by another method, in which the conductances of the various paths are considered. When the resistances
are arranged in parallel, since several paths are being offered for the passage of the current, the effect produced
is the same as if we were to increase the cross-sectional area of the original conductor. The current passing through
the separate resistances is proportional to the conductivity of each path.
It was earlier stated that the
conductance of a circuit is equal to 1/R.
That is, the less the resistance of a wire, the greater is its
conductance or ability to conduct current. Conductance is expressed in mhos. Thus if the resistance of a conductor
is 5 ohms, its conductance is 1/5 = 0.2 mho.
The conductance of the entire parallel circuit is equal to
the sum of the conductances of its individual branches. Thus if R stands for the combined resistance of the parallel
circuit, and r1
etc., stand for the individual resistance of the parts of the parallel circuit, then
from which the combined resistance R may be calculated if the resistances of the individual branches are known.
Thus in Figure 2 the combined resistance of the three filaments in parallel is:
from which R = 2.9 ohms. Ans.
Notice that the combined resistance is less than the resistance of any
of the paths. This should be expected, of course, since even the path of the lowest resistance is having several
additional conducting paths connected in parallel with it so that the resistance must be less. Additional paths
increase the current-carrying ability of the circuit; that is, they decrease the resistance.
We see that
two or more equal resistances in parallel is merely a special case of parallel circuits. Equation (2) can be used
for any condition of equal or unequal resistances.
In a parallel circuit the voltage across each branch
is the same as that across every other branch and is equal to that supplied by the source of e.m.f. The current
which flows through each branch is simply equal to this voltage divided by the resistance of the branch. Thus in
Figure 2, if the battery supplies an e.m.f. of 6 volts, the currents in the various branches are:
Therefore 1=0.3 + 0.6 + 1.2 = 2.1 amps. (This is the total current supplied by the battery.) As a check on this
calculation we may calculate the total current directly from the value of the combined resistance of 2.9 ohms obtained
above for the circuit. Thus
amps (which checks with the value just calculated)
In a parallel circuit, if anyone of the branches is opened,
current will continue to flow through the others. The conditions existing in parallel circuits are as follows:
1. The voltage is equal across all branches.
2.The combined resistance is less than the resistance
of any branch of the circuit.
3. The total current is equal to the sum of the currents through all the branches.
Parallel circuits are very common in radio receivers. In battery-operated receivers the filaments of the
various tubes are usually connected in parallel across the source of e.m.f. (battery). In a.c. electric receivers
the filaments of the tubes are connected in parallel across the filament winding of the transformer. The plate circuits
are connected in parallel across the B supply unit. Question Box
Physics and science
instructors will find these review questions and the "quiz" questions below useful as reading assignments for their
classes. For other readers the questions provide an interesting pastime and permit a check on the reader's grasp
of the material presented in the various articles in this issue.
The "Review Questions" cover material in
this month's installment of the Radio Physics Course. The "General Quiz" questions are based on other articles in
this issue as follows: Eliminating Fringe Howl in Regenerative Detectors; Design and Operation of an Interference
Meter; Operating and Servicing the Stenode Quartz-Crystal Receiver; Radio Guards the Baby; What Television Needs;
Class "B" Tubes-Their Significance in Future Audio Amplifier Design. Review Questions
1. Four vacuum-tube filaments having the following resistances are all connected in series: 20, 4,
a. Draw the circuit diagram showing the connection.
b. What is the total resistance of the
c. How much current will flow if the entire group is connected to a source of e.m.i. of 50
2. The resistances in question (1) are all connected in parallel.
the circuit diagram for this connection.
b. What is the joint resistance of the combination?
current will flow through each filament if the source of e.m.i, is 6 volts?
d. What is the total current
taken from the battery?
3. The filaments of two 201A vacuum tubes having a resistance of 20 ohms
each are connected in parallel. In series with this group is another filament having a resistance of 10 ohms. The
entire group is supplied with current from a 6-volt storage battery. What is the combined resistance of all the
tube filaments, and the total current flowing? General Quiz on This Issue
Why do most regenerative detectors howl when feed-back is adjusted close to the point of oscillation?
How may this condition be corrected?
3. What arrangement is used in one interference locator to visibly
measure interference intensity?
4. How can the Stenode be used for short-wave reception?
is the "wall of light" used as protection against burglary? What is one main factor which is retarding the popularization
6. What is the principle of true "Class A" amplification?
Posted September 9, 2013