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.
July 1932 Radio News
Wax nostalgic about and learn from the history of early electronics.
See articles from Radio &
Television News, published 1919 - 1959. All copyrights hereby acknowledged.
Student's Radio Physics Course - Series
and Parallel Circuits
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
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:
Figure 1. The
same current flows through every part of the circuit.
R = R1
+ R2 + R3
+ R4 + etc.
The total resistance is R = 380 + 20 + 20 + 20 = 440
The current I flowing in the circuit is:
= 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.
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
R4 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.
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
Figure 2. The
current divides and part flows through each branch.
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.
Any 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
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.
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 r2,
r3, etc., stand for
the individual resistance of the parts of the parallel circuit,
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
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
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
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.
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.
1. Four vacuum-tube
filaments having the following resistances are all connected in
series: 20, 4, 5, 10.
a. Draw the circuit diagram showing
b. What is the total resistance of the combination?
c. How much current will flow if the entire group is connected
to a source of e.m.i. of 50 volts?
2. The resistances
in question (1) are all connected in parallel.
Draw the circuit diagram for this connection.
b. What is the
joint resistance of the combination?
c. What 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?
on This Issue
1. Why do most regenerative detectors
howl when feed-back is adjusted close to the point of oscillation?
2. How may this condition be corrected?
arrangement is used in one interference locator to visibly measure
4. How can the Stenode be used for
5. How is the "wall of light" used
as protection against burglary? What is one main factor which is
retarding the popularization of television?
6. What is the
principle of true "Class A" amplification?
Posted September 9, 2013