Docent R. A. Voronov.

Measurement of \(\cos \varphi\) by the Single-Ammeter Method

1.

The methods used in practice for measuring the phase-shift angle \(\varphi\) and the power factor \(\cos \varphi\) require the connection of various instruments, which are encountered comparatively rarely. Most often measurements are made with wattmeters, a voltmeter, and an ammeter, and the power factor is then computed from the equation

\[ \cos \varphi = \frac{P}{V \cdot J} \]

for single-phase current, and

\[ \cos \varphi = \frac{P}{\sqrt{3}\,VJ} \]

for three-phase current with a symmetrical load.

Less often, the readings of two wattmeters connected according to Aron’s circuit are used, with calculation by the equation

\[ \tg \varphi = \sqrt{3}\,\frac{P_1 - P_2}{P_1 + P_2} \]

followed by conversion to \(\cos \varphi\). Still more rarely—because of the scarcity and high cost of the instruments—direct readings from phase meters are used.

In addition to the need to have accurate instruments, which are often absent where it is necessary to measure \(\cos \varphi\), they must also be connected into the network. In doing so, in order to connect the ammeter windings of the instruments, it is necessary to cut the wires, which in most cases can be done only with the network switched off.

At present, in connection with the surveys of installations being carried out everywhere for the purpose of improving \(\cos \varphi\), measurements of the latter are acquiring extremely great importance. In view of the mass character of such measurements and, often, the impossibility of switching off installations in order to connect instruments, carrying out measurements by the usual methods is greatly hindered. Many enterprises, moreover, do not have the necessary instruments at all and therefore are deprived of the possibility of carrying out work on surveying the power factor.

Bearing in mind that the most widespread instruments are simple electromagnetic ammeters, which are already installed in almost all installations at enterprises, or which can easily be installed in advance wherever this is required, it is necessary to have a method of measuring $\cos \varphi$ that would make use precisely of these ammeters. Moreover, because of the possibility of discrepancies in instrument readings, the number of ammeters required for a single measurement must be minimal.

The present article gives a description of a new method for measuring $\cos \varphi$ in three-phase-current circuits, possessing extreme simplicity and requiring only one very simple ammeter. This ammeter may even be very inaccurate, provided only that its errors are approximately proportional to the reading.

With the simplicity of application of this measurement method and with sufficiently accurate results, it may be expected that it will become widely used in those cases where great precision is not required.

2.

Let us suppose that the three-phase-current network shown in Drawing 1 has symmetrical phase voltages, which corresponds to equality of the line voltages. Placing on the diagram (Drawing 2) the vector of the phase voltage of the first phase vertically, the current $J_1$ of this phase, indicated by the ammeter, may be represented in the form of the complex quantity

\[ j_1 = J_1 \cdot \Cos \varphi - j J_1 \cdot \Sin \varphi, \]

Drawing 1.

where $\varphi$ is the phase-displacement angle of the current with respect to the phase voltage, taken as positive for an inductive load.

If now between the wires of this phase and the adjacent (for example, lagging, 2nd phase) one insert

Drawing 2.

an additional ohmic resistance \(R_0\), then the ammeter reading changes, since both the load current and the rheostat current will pass through it. The latter current will be in phase with the line voltage \(V_{12}\) and will lead the phase voltage by an angle equal to \(30^\circ\). Representing this current in complex form,

\[ \dot{J}'_{0}=J_{0}\operatorname{Cos}30^\circ+jJ_{0}\operatorname{Sin}30^\circ =J_{0}\frac{\sqrt{3}}{2}+jJ_{0}\cdot\frac{1}{2} \]

and adding it to the load current, we obtain the current indicated in this case by the ammeter,

\[ \dot{J}_{2}=\dot{J}_{1}+\dot{J}'_{0} =\left(J_{1}\operatorname{Cos}\varphi+J_{0}\frac{\sqrt{3}}{2}\right) -j\left(J_{1}\cdot\operatorname{Sin}\varphi-J_{0}\frac{1}{2}\right) \]

whence

\[ J_{2}^{2}=J_{1}^{2}+J_{0}^{2}+J_{1}J_{0}\left(\sqrt{3}\operatorname{Cos}\varphi-\operatorname{Sin}\varphi\right), \tag{a} \]

If now the same resistance is switched between the other phases, i.e. connected into the leading phase, then, with equality of the line voltages, the current in this resistance will remain of the same magnitude, but will change its displacement with respect to the phase voltage. Since this current will be in phase with the voltage

\[ \dot{V}_{13}=\dot{V}_{1}-\dot{V}_{3} \]

it will lag behind the voltage \(V_{1}\) by an angle of \(30^\circ\), and will be represented by the complex

\[ \dot{J}''_{0}=J_{0}\operatorname{Cos}30^\circ-jJ_{0}\operatorname{Sin}30^\circ =J_{0}\frac{\sqrt{3}}{2}-jJ_{0}\frac{1}{2} \]

The current passing in this case through the ammeter will be equal to the sum

\[ \dot{J}_{3}=\dot{J}_{1}+\dot{J}''_{0} =\left(J_{1}\operatorname{Cos}\varphi+J_{0}\frac{\sqrt{3}}{2}\right) -j\left(J_{1}\operatorname{Sin}\varphi+J_{0}\frac{1}{2}\right) \]

whence

\[ J_{3}^{2}=J_{1}^{2}+J_{0}^{2}+J_{1}J_{0}\left(\sqrt{3}\operatorname{Cos}\varphi+\operatorname{Sin}\varphi\right) \tag{b} \]

If in the latter case the line voltage is taken to be not \(\dot{V}_{13}\), but

\[ \dot{V}_{31}=\dot{V}_{3}-\dot{V}_{1} \]

i.e. directed in the same direction as the voltage \(V_{12}\) (counterclockwise), then the vector of the current \(J_{0}\) will have the directly opposite direction, but to find the current \(J_{3}\) it is then necessary to take the difference of the currents, and not their sum. The result of this, of course, will not change.

Subtracting equation (a) from equation (b), we obtain

\[ J_3^2 - J_2^2 = 2J_1J_0 \operatorname{Sin}\varphi \]

whence

\[ \operatorname{Sin}\varphi = \frac{J_3^2 - J_2^2}{2J_1J_0} = \frac{(J_3 - J_2)(J_3 + J_2)}{2J_1J_0} \tag{c} \]

Adding the same equations, we obtain

\[ J_3^2 + J_2^2 = 2J_1^2 + 2J_0^2 + 2\sqrt{3}\,J_1J_0 \operatorname{Cos}\varphi \]

whence

\[ \operatorname{Cos}\varphi = \frac{J_3^2 + J_2^2 - 2J_1^2 - 2J_0^2}{2\sqrt{3}\,J_1J_0} \tag{d} \]

The last two equations (c) and (d) may serve for finding the angle of displacement and the power factor. In this case, in addition to the three readings of the main ammeter, it is also necessary to know the current in the resistance \(R_0\), or the magnitude of the latter and the line voltage, calculating the current

\[ J_0 = \frac{V}{R_0} \]

Owing to the necessity of taking readings from two instruments, which may have errors of different signs, the accuracy of the result is not very high.

3.

In equations (a) and (b), or, what is the same thing, in equations (c) and (d), there enter only two unknowns—the required phase-displacement angle and the current \(J_0\), which cannot be read on the main ammeter. Solving the system of equations, each of these quantities can be determined separately. This is most simply obtained by squaring equations (c) and (d) and adding separately the right- and left-hand sides. Since

\[ \operatorname{Sin}^2\varphi + \operatorname{Cos}^2\varphi = 1 \]

the displacement angle cancels, and after simplifications we obtain

\[ J_3^4 + J_2^4 + J_1^4 + J_0^4 = J_3^2J_2^2 + J_3^2J_1^2 + J_3^2J_0^2 + J_2^2J_1^2 + J_2^2J_0^2 + J_1^2J_0^2 \]

or

\[ J_0^4 - J_0^2\left(J_3^2 + J_2^2 + J_1^2\right) + \left[ \left(J_3^4 + J_2^4 + J_1^4\right) - \left(J_3^2J_2^2 + J_3^2J_1^2 + J_2^2J_1^2\right) \right] = 0 \]

Introducing the notation

\[ \left. \begin{aligned} J_3^2 + J_2^2 + J_1^2 &= 2A \\ \left(J_3^4 + J_2^4 + J_1^4\right) - \left(J_3^2J_2^2 + J_3^2J_1^2 + J_2^2J_1^2\right) &= B^2 \end{aligned} \right\} \tag{e} \]

we obtain the biquadratic equation

\[ J_0^4 - 2 A J_0^2 + B^2 = 0, \]

solving which we have

\[ J_0 = \pm \sqrt{A - \sqrt{A^2 - B^2}} \tag{f} \]

from which, knowing three readings of the main ammeter, it is easy to calculate the value of the current in the resistance. It should be noted that the other values for the current, obtained as roots of the fourth-degree equation, correspond to an arrangement of vectors that cannot occur in an actual circuit.

The two values obtained, differing only in sign, correspond to capacitive and inductive displacements with equal power factors. In each case it is easy to determine which of the displacements is present in the circuit; but for obtaining the numerical value this choice is not necessary.

Substituting the obtained values of \(J_0\) into equations (c) and (d), we find

\[ \operatorname{Sin}\varphi = \frac{J_3^2 - J_3^2} {2 J_1 \sqrt{A - \sqrt{A^2 - B^2}}} \]

\[ \operatorname{Cos}\varphi = \frac{2\sqrt{A^2 - B^2} - 3J_1^2} {2\sqrt{3}\,J_1\sqrt{A - \sqrt{A^2 - B^2}}} \tag{g} \]

Using these equations, all measurements can be reduced to three ammeter readings—under normal operation \((J_1)\), when some ohmic resistance is connected into the lagging phase \((J_2)\), and when the same resistance is connected into the leading phase \((J_3)\). From the readings obtained, calculating the values of \(\operatorname{Sin}\varphi\) and \(\operatorname{Cos}\varphi\) presents no difficulty, but when a large number of measurements is being made, it proves too lengthy and tiresome.

Drawing 3.

4.

To simplify the work, instead of calculations one may use a graphical construction arising from the vector diagram. Transferring the current vectors to a separate drawing, it is easy to see (Drawing 3) that their mutual arrangement is obtained quite definitely.

The vectors of the three currents \(J_1, J_2\), and \(J_3\) turn out to be arranged so that their ends lie on an equilateral triangle \(ABC\), whose sides are equal to the current \(J_0\) in the resistance \(R_0\), while the side \(BC\) is situated perpendicular to the voltage vector.

Proceeding from this arrangement, the following method may be proposed for constructing the vector diagram. Choosing an arbitrary direction, we lay off on it (Fig. 4) three segments \(OA_1\), \(OB_1\), and \(OC_1\), equal, on the chosen scale, to the readings obtained on the ammeter \(J_1, J_2\), and \(J_3\). Through two of the points obtained (preferably the one nearest to the point \(O\) and the one farthest from it) we draw auxiliary circles with center at the point \(O\). In Figure 4 such arcs are drawn for the points \(A_1\) and \(C_1\).

The equilateral triangle must be of such a size that, with its vertices, it can fall on the two arcs drawn and on the point \(B_1\). Drawing from the latter one more intersection arc with radius \(B_1C\), somewhat greater than the length of the segment \(A_1C_1\), we obtain the isosceles triangle \(AB_1C\). By varying the length of the radius \(BA\) taken, we choose it so that the distance \(AC\) becomes equal to it in magnitude. Such a relation is easily obtained after two or three trial arcs.

Fig. 4.

The triangle \(AB_1C\) obtained by this selection will be equilateral and will correspond to the triangle \(ABC\) of Figure 3. Drawing from the point \(O\) the straight line \(OA\) and dropping the perpendicular \(OD\) to the side \(BC\) of the constructed triangle (or to its extension), we obtain the required angle of phase displacement as the angle between the straight lines \(OA\) and \(OD\). Dropping also from the point \(A\) the perpendicular \(AE\) to the straight line \(OD\), we also obtain the power factor from the ratio of the segments

\[ \cos \varphi = \frac{OE}{OA} \]

The sides of the constructed triangle \(AB_1C\) determine, on the scale, the current in the resistance \(R_0\). In constructing this triangle, the intersection arc from the point \(B\) may also be made on the other side of the axis, which will affect only the direction of the phase angle (capacitive or inductive), and not its magnitude or the value of \(\cos \varphi\). These two arrangements correspond to the two values of the current \(J_0\) obtained from equation (f) and differing only in sign. Knowing what kind of load is present in the circuit, it is easy to decide which arrangement to choose in each case.

If one does not know what kind of load is present in the circuit, but the phase sequence is known, then one may use the rule that follows from

of Fig. 2 and equations (a) and (b). For an inductive load, the larger current will be obtained when the resistance \(R_0\) is inserted in the leading phase, and for a capacitive load—in the lagging one. The same rule may be used to determine the phase sequence.

The construction of the triangle \(ABC\) from the lengths of the segments \(OA\), \(OB\), and \(OC\), without trial-and-error by intersections, is apparently an insoluble problem.

5.

The power factor can be obtained most quickly if one uses a nomogram constructed specially for this purpose, shown in Fig. 5. Taking three ammeter readings \(J_1\), \(J_2\), and \(J_3\), we compute the ratios

\[ b=\frac{J_2}{J_1}; \qquad c=\frac{J_3}{J_1} \]

and find on the nomogram the point corresponding to these coordinates. The position of this point relative to the family of curves constructed for various values of \(\operatorname{Cos}\varphi\) determines the required power factor.

On the same nomogram another series of curves is drawn for the values

\[ a=\frac{J_0}{J_1} \]

by means of which the value of the current \(J_0\) can be determined. The nomogram in Fig. 5 is constructed only for an inductive load, for which \(J_3>J_2\), corresponding to values \(c>b\). For a capacitive load these relations are reversed, which makes it possible to use this same part of the nomogram, but with the values \(b\) and \(c\) interchanged.

To construct the nomogram, let us divide equations (a) and (b) by \(J_1^2\). Taking into account the notation introduced above, after substitution we obtain

\[ b^2=1+a^2+2a\left(\frac{\sqrt{3}}{2}\operatorname{Cos}\varphi-\frac{1}{2}\operatorname{Sin}\varphi\right) \]

\[ c^2=1+a^2+2a\left(\frac{\sqrt{3}}{2}\operatorname{Cos}\varphi+\frac{1}{2}\operatorname{Sin}\varphi\right) \]

or

\[ \begin{aligned} b^2&=1+a^2+2a\operatorname{Cos}(\varphi+30^\circ)\\ c^2&=1+a^2+2a\operatorname{Cos}(\varphi-30^\circ) \end{aligned} \tag{h} \]

Assigning various values to \(a\) and \(\operatorname{Cos}\varphi\) and computing the corresponding values of \(b\) and \(c\), we obtain the possibility of constructing the entire

nomogram from the currents. The calculated values of \(b\) and \(c\) for various power factors are given in the table at the end of the article.

6.

In making measurements, the accuracy of the result is affected by the accuracy of the ammeter, by asymmetry of the line voltages of the network, and by additional voltage drops caused by the inserted resistance. Let us examine what differences may be expected in these cases between the obtained value of \(\cos\varphi\) and its actual value.

Fig. 5.

The inaccuracy of the ammeter, when all readings are made on the same instrument, will have an effect only when

the error is not proportional to the reading. Since in most cases, even if there is no such proportionality as a whole, nevertheless all errors have the same sign, they will partly smooth one another out and have a comparatively small effect on the final result. A check of a number of AC technical instruments from different firms, and an accounting of the resulting errors in the values of the ratios

\[ b=\frac{J_2}{J_1}\quad \text{and}\quad c=\frac{J_3}{J_1} \]

showed that the greatest discrepancies between them do not exceed 1—1.5%. Considering the nomogram, we see that such an error at \(a=(0.3—0.5)\) corresponds, in the worst case, to deviations in the values of \(\mathrm{Cos}\,\varphi\) of the following magnitudes:

\[ \begin{aligned} 0.1 &\pm 0.03\\ 0.5 &\pm 0.04\\ 0.8 &\pm 0.03\\ 1.0 &\pm 0.00 \end{aligned} \]

When accurate ammeters of laboratory or control types are used, the discrepancy in the values does not exceed 0.5—0.8%, and the deviations in the results are reduced approximately by half.

7.

Let us now clarify the question of how asymmetry in the line voltages affects the result.

If this asymmetry is expressed in the fact that one of the line voltages, present between those phases in which there is no ammeter (i.e., phases 2 and 3), has increased or decreased, while the other two voltages have remained unchanged, then this will be reflected in the diagram of drawing 3 only as a change in the angles between the segments \(AB\) and \(AC\) and the voltage vector \(OV\). With an asymmetry of 5%, these angles, instead of \(30^\circ\), will have the values

\[ \arcsin \frac{1.05\,V}{2V}=\arcsin 0.525=31^\circ40' \]

when the third voltage increases, or

\[ \arcsin \frac{0.95\,V}{2V}=\arcsin 0.475=28^\circ22' \]

when it decreases.

Assuming that during measurement the resistance \(R_0\) was connected, from the calculation

\[ a=\frac{J_0}{J_1}=0.5 \]

and that the load had a phase displacement equal to \(\varphi=60^\circ\), we find, from equations \((h)\), with a change in the angles in them by \(30^\circ\), the values obtained above for \(b\) and \(c\):

\[ \begin{aligned} b^2 &= 1 + 0.5^2 + 2 \cdot 0.5^2 \cdot \cos(60^\circ + 31^\circ 40') = 1.221,\\ c^2 &= 1 + 0.5^2 + 2 \cdot 0.5^2 \cdot \cos(60^\circ - 31^\circ 40') = 2.130, \end{aligned} \]

or

\[ b = 1.105 \qquad c = 1.459. \]

Considering these values as obtained from experiment, we find from the nomogram that they correspond to \(\cos\varphi = 0.47\), instead of the true value \(\cos 60^\circ = 0.5\).

In exactly the same way, for a decrease in voltage, i.e., for an angle of \(28^\circ 22'\) instead of \(30^\circ\), we obtain the value \(\cos\varphi = 0.53\). Thus, the deviation in the obtained value of the power factor may, under the given conditions, reach approximately \(0.03\) in either direction.

The indicated case, with a displacement of \(60^\circ\) and distortion of the voltage between the 2nd and 3rd phases, will have the greatest deviation in the values of \(\cos\varphi\). At other phase-displacement angles and other voltage asymmetries, this deviation will be smaller. It is smallest with a wattless load. In this case, when the voltage between the 2nd and 3rd phases is distorted, there will be no error at all in the result; when the voltage between the 1st phase (in which the ammeter is inserted) and one of the other two is distorted, the error may reach only \(0.01\).

Fig. 6.

Performing similar calculations for all possible cases of voltage asymmetry, we obtain the curve of the greatest deviations of the result for \(5\%\) voltage asymmetry and the ratio \(a=0.5\), plotted in Fig. 6.

Fig. 7.

These greatest possible deviations depend to a large extent on the selected resistance \(R_0\), i.e., on the magnitude of the ratio \(a\). Their changes as a function of the latter quantity are presented in Fig. 7, from which we see that, with an increase in the ratio \(a\), the possible error increases rapidly.

The use as the resistance \(R_0\) of a rheostat made from metallic incandescent lamps reduces the error, in view of the increase of its resistance when the voltage is increased.

8.

To determine the effect of the additional voltage drop in the supply wires, which arises in connection with the insertion of the resistance \(R_0\), let us carry out an approximate calculation for one of the loads.

For this calculation we shall take a circuit with a load phase voltage of \(120\ V\) (line voltage \(208\ V\)) and a load consuming \(12\ A\) with an inductive phase displacement of \(60^\circ\) \((\mathrm{Cos}\,\varphi = 0.5)\). We shall take the resistance of the wires as purely ohmic, with such a calculation that the voltage of the current source (mains) is not more than \(125\ V\) (voltage drop about \(4\%\)).

The resistance of the load phase will be equal to

\[ \dot Z_n=\frac{V}{J}(\mathrm{Cos}\,\varphi+j\mathrm{Sin}\,\varphi) =\frac{120}{12}(0.5+j\,0.866)=5+j\,8.66 \]

and the resistance of the wires

\[ R_{pr}=\sqrt{\left(\frac{125}{12}\right)^2-(8.66)^2}-5=0.78\ \text{ohm} \]

In order that the value of the ratio \(a=\dfrac{J_0}{J_1}\) be close to \(0.5\), it is necessary to have a current

\[ J_0=0.5\cdot J_1=6\ \text{amperes} \]

or a resistance

\[ R_0=\frac{208}{6}=34.7\simeq 35\ \text{ohms}. \]

Fig. 8.

The entire connection diagram will appear in the form shown in Fig. 8.

Replacing the triangle formed by the load resistances and the resistance \(R_0\) by its equivalent star, and adding the obtained resistances to the remaining ones, we obtain a simple star with the resistances of the individual phases.

\[ \dot Z_1=\dot Z_3=6.41+j\,4.52 \]

\[ \dot Z_2=5.46+j\,10.71 \]

or

\[ Z_1=Z_3=7.84\ \text{ohms} \]

\[ Z_2=12.01\ \text{ohms} \]

Having now found the admittances of the individual phases,

\[ \dot{Y}_1=\dot{Y}_3=0.104-j\,0.0735 \]

\[ \dot{Y}_2=0.378-j\,0.0742 \]

we calculate the potential of the neutral point of the equivalent star of the load with respect to the neutral point of the network:

\[ \dot{V}_0=\frac{\sum \dot{V}_n\cdot \dot{Y}_n}{\sum \dot{Y}_n} \]

where \(\dot{V}_n\) are the phase voltages of the network,

\[ \dot{V}_2=125+j\cdot 0 \]

\[ \dot{V}_3=-62.5-j\,108.2 \]

\[ \dot{V}_1=-62.5+j\,108.2 \]

The voltages on the phases of the load star will be equal to

\[ \dot{V}'_1=\dot{V}_1-\dot{V}_0=-44.1+j\,125.1 \]

\[ \dot{V}'_2=\dot{V}_2-\dot{V}_0=143.4+j\,16.9 \]

\[ \dot{V}'_3=\dot{V}_3-\dot{V}_0=-44.1-j\,91.3 \]

Whence their magnitudes are

\[ V'_1=132.5\ \text{V};\quad V'_2=144.8\ \text{V and}\quad V'_3=101.5\ \text{V} \]

Dividing by the phase resistances, we find the current strengths in the conductors:

\[ J'_1=16.9\ A;\quad J'_2=12.1\ A;\quad J'_3=12.93\ A \]

It is easy to see that the current \(J'_1\) corresponds to the current passing through the ammeter when the resistance \(R_0\) is inserted in the leading phase (i.e., to the reading \(J_3\)), while the current \(J'_3\) corresponds to insertion of the resistance in the lagging phase (i.e., to the reading \(J_2\)).

Remembering that in operation without the resistance the load consumed 12 amperes (i.e., \(J_1=12\ A\)), we find the ratios

\[ c=\frac{16.9}{12}=1.41 \qquad b=\frac{12.93}{12}=1.08 \]

From these values, using the nomogram we obtain the value \(\operatorname{Cos}\varphi = 0.46\), which differs from the actual value \((\operatorname{Cos}\varphi = 0.5)\) in the direction of decrease by 0.04.

Carrying out similar calculations for other conditions, it is easy to see that this deviation from the true value will be greater at a smaller power factor or a larger current in the additional resistance. At \(\operatorname{Cos}\varphi = 1\) there will be no error at all; at \(\operatorname{Cos}\varphi = 0\) it will reach almost 0.1. The inductance of the leads somewhat improves the result with an inductive load, but worsens it with a capacitive one. With an increase in the voltage drop in the leads, the error increases almost proportionally to it (somewhat more slowly). The presence of a neutral conductor has almost no effect on the resulting error.

For cases of normal operation of networks and loads, such calculations have shown that the error usually does not exceed the value

\[ \Delta \operatorname{Cos}\varphi < 0.1 \,(1 - \operatorname{Cos}\varphi) \]

in the direction of decreasing the result.

9.

In checking the application of this method to measuring the power factors of asynchronous motors under various conditions of their operation, and comparing the results obtained with the readings of a phase meter, it turned out that the error never exceeded 0.05, usually being equal to 0.02–0.03. In most cases, because of the voltage drop in the leads, this error was in the direction of decreasing the result, and only in a few measurements did it turn out to be in the direction of increase (probably because of voltage asymmetry). Since the measurements were made approximately under the same conditions as are usually found in industrial installations, it may be assumed that no large discrepancies will be obtained there either.

Thus, this method is applicable in those cases where it is not necessary to know the exact value of the power factor, but it is sufficient to have it with a deviation of up to 0.05 in one direction or the other. The necessity of connecting the rheostat \(R_0\) sets a limit on the application of this method to powers not exceeding 30–40 KVA in networks with voltages up to 380 volts.

10.

This method of measurement may be modified by connecting, instead of the ohmic resistance \(R_0\), inductive or capacitive ones. In this case, in the diagram of drawing 4, the angle between the segments \(BC\) and \(OD\) should not be \(90^\circ\), but another one, in accordance with the resistance being connected. Denoting the phase-shift angle for the latter by \(\varphi_0\), we obtain

\[ \angle CDO = 90 \pm \varphi_0 \]

Equations \((c)\), \((d)\), \((f)\), \((g)\), and \((h)\) yield another, still more complicated expression; the nomogram of drawing 5, however, gives the values

\[ \operatorname{Cos}(\varphi \pm \varphi_0) \]

When a single capacitance is connected, equation \((c)\) gives the value of \(\operatorname{Cos}\varphi\), and equation \((d)\)—\(\operatorname{Sin}\varphi\); the nomogram of drawing 5 gives the values of \(\operatorname{Sin}\varphi\), but only within the limits for which it has been constructed, and is valid only for a capacitive load.

The replacement of an ohmic resistance by a capacitance may have some significance in measuring \(\operatorname{Cos}\varphi\) in high-voltage networks.

Instead of an ammeter, one may use Dittz clamps, making measurements with them at various points of the circuit. In this case it is also easy to measure \(I_0\), and consequently to make use of equations \((c)\) and \((d)\).

At present, designs of clamps and ammeters are being developed which would make it possible to carry out such measurements for any powers and voltages; moreover, such measurements could also be extended to single-phase current.

In conclusion, it is necessary to note that, when the described method is applied to three-phase-current circuits with an unsymmetrical phase load, the obtained values of \(\operatorname{Cos}\varphi\) will refer to that phase in which the ammeter was located, and will correspond to the angle of displacement between the current in the conductor and the phase voltage of the network, the latter being considered symmetrical.

Table of values required for constructing the nomogram of drawing 5.