Not
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
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
By Alfred A. Ghirardi*
Series Circuit 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 R
_{1}
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 = R
_{1}
+ R
_{2} + R
_{3}
+ R
_{4} + 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 R
_{1}
= 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 R
_{4}
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.
Parallel Circuits
Parallel Circuit 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.
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 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 r
_{1} r
_{2},
r
_{3}, 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, 5, 10.
a. Draw the circuit diagram showing the connection.
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.
a. 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?
General Quiz 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?
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?
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