June 1944 QST
Wax nostalgic about and learn from the history of early electronics. See articles
QST, published December 1915 - present. All copyrights 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 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.
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.
Fig. 1 - Parallel resistances.
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 tables.
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.
I1 = EG1
I3 = EG3
I2 = EG2
Ip = EGp
The usual formula for finding the parallel resistance of this combination is
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
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,
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 + G3,
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
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 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
since j2 = -1. Now we have
However, we know that
R2 + X2 = Z2.
Fig. 22- Impedance (A) and admittance (B) diagrams for the coil (C).
In the impedance diagram, the current, I, is used as the- Impedance (A) and admittance
(B) diagrams for the coil (C). In the impedance diagram, the current, I, is used as the
reference,,whileethe voltage, E, is used as reference in the diagram of (B).the voltage,
E, is used as reference in the diagram of (B).
Fig. 3 - Series resistor and inductor in parallel with capacitor
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.
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
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
= 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 component.
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
The impedance then is
Fig. 4 - Vector diagram of the currents in the various branches of
the circuit of Fig. 3.
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.
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
Fig. 5 - Resistance-coupled amplifier circuit.
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
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. Therefore,
Zab + Rp = 0.0884 - j0.0123 megohms.
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)
Fig. 6 - Equivalent circuit of the resistance-coupled amplifier in
Fig. 5 for low frequencies.
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
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
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
* ·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 December 11, 2018 (original October 5, 2012)