June 1944 QST
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.
When 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
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
A.C. Calculations for Parallel and Series-Parallel Circuits
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.
1 - Parallel resistances.
I1 = EG1
I3 = EG3
I2 = EG2
Ip = EGp
The usual formula for finding the parallel resistance of this
However, this formula is only a special case of the more general
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
1/R1 + 1/R2 + 1/R3
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.
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 is
Yp = Y1
+ Y2 ... etc.
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
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 thusly,
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, R. Then,
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
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 impedance. 2
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.
To find 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 admittance, or
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 (electrical radians).
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 mhos.
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 component.
The impedance of the inductive branch is
ZL = RL
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
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.
Y1 + Y2 = 10 + j0 + 2.44 + j1.95 + 12.44 +
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
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)
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.
Posted October 5, 2012