you read a lot of tutorials about introductory electronics on the Internet,
most are the same format where stoic, scholarly presentations of the
facts are given. Those of you who don't have enough fingers and toes
to count all of the college textbooks like that which you have read
know of what I speak. When hobby articles are written in a similar fashion,
it can quickly discourage the neophyte tinkerer or maybe even a future
Bob Pease. QST
has printed a plethora of articles over the years that are more of a
story than just a presentation of the facts. My guess is the reason
is because often the authors are not university professors who have
forgotten how to speak to beginners. This article on basic calculations
for AC series and parallel circuits is a prime example.
June 1944 QST
of Contents]These articles are scanned and OCRed from old editions of the
ARRL's QST magazine. Here is a list of the
QST articles I have already posted. All copyrights are hereby acknowledged.
A.C. Calculations for Parallel and Series-Parallel
Solving for Current at Voltage by Means of Admittance
By S.E. Spittle, * W4HSG
The problem of finding the resultant impedance of a group of impedances
in parallel is not usually discussed in the more elementary texts on
alternating currents, probably because the solution of such problems
is rather lengthy and difficult without the use of complex algebra.
The term "complex algebra" may have a mysterious and rather terrifying
sound to those who have never heard of it, but actually the process
is comparatively simple for anyone having a knowledge of the elements
of plain algebra.
The recent QST article entitled "Meet
Mr. j"1 provides a good explanation of the application of
complex algebra to alternating-current circuits and should be read as
an introduction to this discussion. In that article one method of computing
the impedance of parallel circuits by means of their phase angles was
described. An alternative method, described here, makes use of the resistive
and reactive components without requiring a knowledge of the phase angles
as such, and therefore can be applied without the use of trigonometric
Since the laws of alternating-current circuits
are merely extensions of the laws governing direct-current circuits,
taking into account the effects produced by the storage of energy in
the electric and magnetic fields, it is logical to explain the solution
of a.c. problems in terms of the familiar operations used with d.c.
problems. For example, take a circuit consisting of two resistances,
R1 and R2 in parallel, as shown in Fig. 1.
Fig. 1 -
I1 = EG1
I3 = EG3
I2 = EG2
Ip = EGp
The usual formula for finding the parallel resistance of this combination
However, this formula is only a special case of the more general formula
1/Rp = 1/R1 + 1/R2
+ 1/R3 ... etc.,
applied to the case of only two resistances. In the special case
of two resistances, the second formula is converted into the first by
means of a few simple transpositions; When there are more than two resistances,
the general formula is more practical, as will readily be seen if we
apply the same process to the case of three resistances. Thus
1/Rp = 1/R1 + 1/R2
when transposed becomes
It may be interesting to note that the process of finding the resulting
resistance is the same as finding the total current in the parallel
circuit, using any assumed value of applied e.m.f. Thus
Ip = E/Rp = E/R1
+ E/R2 + E/R3.
Changing the value of E will change Ip but not Rp.
Therefore we can assume E to be one volt, giving the formula for the
parallel resistance, which is usually expressed as
The same formula can be applied to a number of parallel impedances
in an a.c. circuit, giving
The catch in this is that the currents in the various impedances
are usually not in phase with each other, so that the phase difference
must be taken into account in obtaining the correct value of parallel
impedance. In the d.c. case the quantity 1/R is known as the conductance,
represented by G whose unit of measurement is the mho. Thus, a resistance
of 5 ohms corresponds to a conductance of 1/5 or 0.2 mho. The total
conductance of a parallel circuit is the sum of the conductances of
the individual branches. Thus, making 1/R1 = G1,
etc., we have for three resistances in parallel
Gp = G1 + G2
and the equivalent or parallel resistance, Rp, is equal
to 1/Gp. The conductance, G, is numerically equal to the
current which would flow with one volt applied to the circuit. Hence,
in finding the total conductance of a parallel circuit by adding the
individual conductances we are merely following the same process as
in finding the total current by adding the individual currents. If this
can be done for d.c. circuits we should likewise be able to do the same
thing for a.c. circuits. In the latter case, 1/Z is called the admittance
and is represented by Y. Also, the total admittance of a parallel circuit
Yp = Y1 + Y2
The value of Y is also measured in mhos, just as resistance, reactance
and impedance all are measured in ohms.
Here we remember our
forgotten friend, the phase-angle. We know that if we apply an alternating
voltage to a circuit having several parallel branches, the absolute
values of the branch currents will add up to more than the absolute
value of total current unless all the currents happen to be in phase.
Since the admittance is a measure of the value of current that would
flow in an impedance when one volt is applied, it is necessary to add
admittances in the same manner as we add currents in a parallel circuit.
The only practical ways of adding a number of currents, voltages or
impedances having various phase angles are by drawing scale diagrams
or by splitting each one into two components at right angles to each
other, adding the two sets of components separately and then combining
the two sums by the familiar right-triangle rule, or Pythagorean theorem.
One component represents the condition of current in phase with voltage,
or 100 per cent energy consumption, and is often called the real component
. The other component represents the condition of current and voltage
90 degrees out of phase, or 100 per cent energy storage, and is often
called the imaginary component. This component is usually prefixed by
the letter j to show that it has been rotated 90 degrees with respect
to the real component. The letter j has been given the value of √(-1)
and when it occurs in a computation it is treated as a multiplier having
this value, as has already been explained in the article mentioned previously.
Algebraic numbers having √(-1) as a factor are known as imaginary
numbers, which explains the name given the components of voltage, etc.,
to which the letter j is applied. The method of adding the real and
imaginary components of voltage or current also is covered very well
by Mr. Noll and therefore only its application to parallel circuits
will be discussed here.
In many problems of parallel impedance
the mathematical solutions can be simplified by the use of admittances.
In this way frequent reference to trigonometric tables is unnecessary,
since phase angles no longer are factors in the computations. Practical
application of the method is discussed in this article and illustrated
with typical examples.
The preceding paragraph implies that
we must split up each of our admittances into two components before
we can add them. To obtain the components of the admittance we make
use of the complex expression for impedance, which means the impedance
when split into its components of resistance and reactance. Expressed
Z = R + jX,
the j means that, if a diagram of the components of impedance is
drawn, the reactance, X, will be drawn at right angles to the resistance,
since Y = 1/Z, the corresponding admittance would be
In order to get the complex expression out of the denominator we
multiply the fraction by
since j2 = -1. Now we have
However, we know that
R2 + X2 = Z2.
Therefore the two parts of the admittance, Y, are R/Z2
and - jX/Z2. If Z happened to be a pure resistance, X would
equal zero, and R/Z2 would become equal to R/R2
or 1/R, which is called G, or conductance in d.c. circuits. The term
R/Z2 is therefore called the a.c. conductance and is also
represented by the letter G. The term X/Z2, which is the
reactive or imaginary part of the admittance, is known as the susceptance
and is represented by B. Thus,
Y = G -jB,
where G = R/Z2 and B = X/Z2. Note that the
phase angle of the admittance is of sign opposite to that of the corresponding
Fig. 2 - Impedance (A) and admittance
(B) diagrams for the coil (C).
In the impedance diagram, the current,
I, is used as the reference,
while the voltage, E, is used as reference
in the diagram of (B).
Fig. 2 shows an impedance diagram at A and the corresponding
admittance diagram at B for a coil having 4 ohms resistance and 3 ohms
reactance (C). The diagrams are shown in terms of the components of
voltage and current, and an applied e.m.f. of 25 volts is assumed, so
that both diagrams will be to the same scale.
the resultant impedance of a number of impedances in parallel, we add
the admittances of the individual branches to obtain the total admittance.
The impedance of the combination is then the reciprocal of the total
Z = 1/Yt.
The application of this method will be shown by a couple of examples.
For the first example we shall take the tuned circuit of Fig. 3,
consisting of a condenser of 400-μμfd. capacity with negligible
resistance, and a 100-microhenry inductance coil having a resistance
of 20 ohms. A circuit with these values will be series resonant at 795.58
kilocycles, as can be verified by the formula for resonance,
We shall now calculate the impedance of the circuit at this frequency,
assuming a voltage to be applied between the points A and B, making
it a parallel-resonant circuit. Since the factor ω = 2πf is
used in calculating both inductive and capacitive reactances, we start
by computing its value as follows:
ω = (2) (3.1416) (795,580) = 5,000,000
The impedance of the capacity branch is Rc - jXc
Rc = 0 and
(There are 1,000,000,000,000 micromicrofarads in one farad). Therefore
Z = 0 - j500 ohms. The admittance of this branch is
Yc = Gc - jBc
Gc = Rc/Zc2 = 0
Yc = 0 - (- j0.002) = 0 + j0.002
Since the impedance in this particular case is a pure reactance
we could have found B directly, since 1/Xc = ωC. This
short cut cannot be used, however, if the impedance also has a resistive
The impedance of the inductive branch is
ZL = RL +jXL
RL = 20 ohms and
XL = ωL = (5,000,000)
(0.0001) henries = 500 ohms.
ZL = 20 + j500 ohms.
YL = GL -jBL
The admittance of the parallel circuit is
Yab = Yc + YL
= (0 + j0.002) + (0.00008 - j0.001997),
which is added thusly,
0 + 0.00008 + j0.0002 - j0.001997 = 0.00008
+ j0.000003 mho.
The impedance then is
The result shows that the inductive and capacitive reactances in
a parallel circuit do not cancel out in the same manner as they do in
a series circuit. This results from the fact that the resistance in
the inductive branch shifts the phase of the current slightly, so that
it is not exactly 180 degrees out of phase with the current in the capacity
branch. The current for a given applied voltage is equal to E/Z or EY,
since Y = 1/Z. Therefore a diagram of the relative values and phases
of currents in various parts of the circuit can be drawn by using the
values of Y to represent the currents, since current is proportional
to Y. Fig. 4 is such a diagram, with the conductance and total current
exaggerated to show the effect of resistance in the circuit. Actually
the frequency at which the resultant reactance is zero is so close to
the frequency at which XL = Xc that for all practical
purposes they can be considered identical except in circuits having
lower values of Q than are ordinarily used in radio work. Such low-Q
circuits are, however, found in television amplifiers and are also likely
to be found in audio-frequency work.
Fig. 4 - Vector diagram of the
currents in the various branches of the circuit of Fig. 3.
Fig. 5 - Resistance-coupled amplifier
A second example, illustrating a series-parallel circuit, is the
resistance-coupled amplifier shown in Fig. 5. An exact calculation for
such an amplifier, taking into account all possible current paths, is
quite complicated, so that it is customary to reduce the amplifier to
a simplified circuit which approximately represents the conditions existing
for the frequency of interest. For low frequencies the circuit of Fig.
5 can be represented by Fig. 6, where the applied voltage is the a.c.
voltage developed between plate and cathode by a signal, and is equal
to μ times the a.c. voltage applied between grid and cathode.
The voltage eg2 between grid and cathode of the following
tube is practically equal to the voltage drop across the grid leak,
since the voltage across the cathode bypass condenser is negligible
if the bypass condenser has a fairly large capacity (considering the
drop caused by the applied grid voltage only). This voltage will, therefore,
be a portion of the voltage between A and B and will then depend upon
the frequency as well as the constants of the circuit. Assuming the
frequency of the applied signal to be 53 cycles per second, the reactance
of the coupling condenser will be 200,000 ohms. With an applied signal
of 1 volt and an amplification factor of 20 the amplified a.c. voltage
applied to the network of Fig. 6 will be 20 volts. This voltage will
be divided between the internal resistance of the tube, Rp,
and the impedance, Zab, between points A and B.
The admittance Yab, will be the sum of the admittances of
the two branches, one being the plate-coupling resistance, R1,
and the other consisting of the grid leak, R2, in series
with the coupling condenser. For the first branch
Z1 = R1 = 100,000
+ j0 ohms and
Y1 = 0.00001 - j0 mho.
For the second branch
Z2 = R2 - jX2
= 250,000 - j200,000 ohms and
The calculations for a high impedance such as this can be performed
more conveniently by expressing the impedance in megohms. The corresponding
admittance will then be in micromhos. Thus,
and Y1 = 10 - j0 micromhos.
Yab = Y1 + Y2
= 10 + j0 + 2.44 + j1.95 + 12.44 + j1.95 micromhos.
The impedance between points A and B will be
In order to find the voltage eg2 we must first find the voltage
between the points A and B. The impedance Zab in series with
the plate resistance Rp forms a voltage divider. Therefore,
the voltage across Zab will be equal to
Rp is 10,000 ohms resistance or 0.01 + j0 megohms.
Zab + Rp = 0.0884
- j0.0123 megohms.
Fig. 6 - Equivalent circuit of the resistance-coupled
amplifier in Fig. 5 for low frequencies.
eab=(20) (0.876- j0.0165) =17.52- j0.33 volts. The
voltage eab is the sum of eg2 and the voltage
drop across the coupling condenser since they are in series. Therefore,
eg2 = (eab) (Rg/Z2)
The absolute value of this voltage is
The overall amplification or ratio of voltage at grid No. 2 to that
at grid No. 1 will then be 13.6, since one volt was assumed to be applied
to grid No. 1. The voltage at grid No.2 will lead the voltage at plate
No.1 by an angle whose tangent is
It is not necessary to find any phase angles to determine the voltage
or current anywhere in the circuit. In fact, the relative phase of any
voltage in the circuit can be found by graphical methods with sufficient
accuracy for ordinary purposes. The fact that the voltages are expressed
as complex quantities makes it easy to draw a scale diagram of the various
voltages in the circuit. Such a diagram is useful for checking the accuracy
of the calculations. A diagram of the voltages in this circuit is shown
in Fig. 7, and a diagram of the relative currents is shown in Fig. 8.
The values of alternating current in microamperes can be found by multiplying
the admittance in micromhos by the applied e.m.f. of 20 volts (I = EY).
The voltages and currents considered in this problem are, of course,
only the alternating parts of the pulsating voltages and currents in
the actual amplifier. The d.c. values have no other effect than to establish
the values of Rp and μ and to determine the amplitude
of voltage that can be applied before distortion begins.
Fig. 7 - Vector diagrams useful
in checking calculations.
(A) shows the relative values of voltage
in the circuit of Fig. 6, while
(B) shows relative currents.
In practical work the performance of amplifiers is ordinarily computed
by means of graphs or approximate formulas rather than such detailed
calculations. The examples given above were used instead of the usual
problems concerning miscellaneous collections of inductance and capacity
because they are typical of the circuits actually found in radio equipment,
and thus should indicate that the methods of calculations illustrated
have practical applications.
One desirable feature of using
complex notation is that it eliminates the necessity for extracting
square roots except for one such operation as the last step in the calculations,
when the absolute value of current or voltage is desired, as is usually
the case. A vector diagram of all the values in the problem should be
drawn as a check on the calculations and to make sure that plus signs
have not been slipped in where there should be minus signs or vice versa.
Such errors are usually readily apparent as soon as construction of
the diagram is attempted. The vector diagram need only be drawn roughly
to scale, and may be in terms of voltages, currents, impedances or admittances,
whichever is most convenient.
* ·12 Griggs St.. Allston 34, Mass.
1 Noll, "Meet Mister j,"
QST, October, 1943, p. 21.
22 B is sometimes defined as -
X / Z2, making Y = G + jB, This of course, is equivalent
to the relation given above.