Article 37
Alternating Currents.

[Note.¹⁹]

In the general problem treated the cable or wire is regarded as having resistance, distributed capacity, self-induction, and leakage; although some of these may be zero in special cases. The line will also be considered to feed into a receiver circuit of any description; and the general solution will include the particular cases in which the receiving end is either grounded or insulated. The electromotive force may, without loss of generality, be taken as a simple harmonic function of the time, because any periodic function can be expressed in a Fourier series of simple harmonics.²⁰ The E.M.F. and the current, which may differ in phase by any angle, will be supposed to have given values at the terminals of the receiver circuit; and the problem then is to determine the E.M.F. and current that must be kept up at the generator terminals; and also to express the values of these quantities at any intermediate point, distant $x$ from the receiving end; the four line-constants being supposed known, viz.:

in which the maximum values $e_1$, $i_1$, and the phase-angles $\theta$, $\theta \prime$, are all functions of $x$. These simple harmonics will be represented by the vectors $e_1\underset{¯}{/\theta }$, $i_1\underset{¯}{/\theta \prime }$; whose numerical measures are the complexes ²³, , which will be denoted by $ē$, $\bar{ı}$. The relations between $ē$ and $\bar{ı}$ may be obtained from the ordinary equations²⁴

for, since , then $\frac{de}{dt}$ will be represented by the vector $\omega e_1\underset{¯}{/\theta +\frac{1}{2}\pi }$; and $\frac{di}{dx}$ by the sum of the two vectors $G{e}_1\underset{¯}{/\theta },C\omega e_1\underset{¯}{/\theta +\frac{1}{2}\pi }$; whose numerical measures are the complexes $Gē$, $j\omega Cē$; and similarly for $\frac{de}{dx}$ in the second equation; thus the relations between the complexes $ē$, $\bar{ı}$ are

Differentiating and substituting give

and thus $ē,\bar{ı}$ are similar functions of $x$, to be distinguished only by their terminal values.

It is now convenient to define two constants $m$, $m_1$ by the equations²⁷

and the differential equations may then be written

the solutions of which are²⁸

wherein only two of the four constants are arbitrary; for substituting in either of the equations (66), and equating coefficients, give

whence

Next let the assigned terminal values of $ē$, $\bar{ı}$, at the receiver, be denoted by $Ē,Ī$; then putting $x=0$ gives $Ē=A,Ī=A\prime$, whence $B=m_1Ī,B\prime =\frac{Ē}{m_1}$; and thus the general solution is

If desired, these expressions could be thrown into the ordinary complex form $X+jY,X\prime +jY\prime$, by putting for the letters their complex values, and applying the addition-theorems for the hyperbolic sine and cosine. The quantities $X,Y,X\prime ,Y\prime$ would then be expressed as functions of $x$; and the representative vectors of $e,i$, would be $e_1\underset{¯}{/\theta },i_1\underset{¯}{/\theta \prime }$, where ${e_1}^2=X^2+Y^2,i^2={X\prime }^2+{Y\prime }^2,tan\theta =\frac{Y}{X},tan\theta \prime =\frac{Y\prime }{X\prime }$.

For purposes of numerical computation, however, the formulas ($70$) are the most convenient, when either a chart,²⁹ or a table,³⁰ of $coshu$, $sinhu$, is available, for complex values of $u$.

Prob. 106.³¹

Given the four line-constants: $R$ = 2 ohms per mile, $L$ = 20 millihenrys per mile, $C$ = $\frac{1}{2}$ microfarad per mile, $G$ = 0; and given $\omega$, the angular velocity of E.M.F. to be 2000 radians per second; then

Next let the assigned terminal conditions at the receiver be: $I=0$ (line insulated); and $E=1000$ volts, whose phase may be taken as the standard (or zero) phase; then at any distance $x$, by (70),

$$\begin{array}{llllllll}\hfill ē& =Ecoshmx,& \hfill \bar{ı}& =\frac{E}{m_1}sinhmx,& \hfill & & \hfill & \end{array}$$

in which $mx$ is an abstract complex.

Suppose it is required to find the E.M.F. and current that must be kept up at a generator $100$ miles away; then

but, by Prob. 89,

obtained from Table II, by interpolation between and ; hence

where $logtan\theta =log475-log460=.0139$, $\theta =45^{\circ }55\prime$, and $e_1=460sec\theta =661.2$ volts, the required E.M.F.

Similarly , and hence

where $logtan\theta \prime =10.7427$, $\theta \prime =79^{\circ }45\prime$, $i_1=1495sec\frac{\theta \prime }{1601}=5.25$ amperes, the phase and magnitude of required current.

Next let it be required to find $e$ at $x=8$; then

by subtracting $\frac{1}{2}\pi j$, and applying page §. Interpolation between and gives

Hence , where $logtan\theta =.12530$, $\theta =126^{\circ }51\prime$, $e_1=-29.99sec126^{\circ }51\prime =50.01$ volts.

Again, let it be required to find $e$ at $x=16$; here

but

hence

and

where $\theta =180^{\circ }17\prime$, $e_1=1001$ volts. Thus at a distance of about 16 miles the E.M.F. is the same as at the receiver, but in opposite phase. Since $ē$ is proportional to , the value of $x$ for which the phase is exactly $180^{\circ }$is $\frac{\pi }{.2}=15.7$. Similarly the phase of the E.M.F. at $x=7.85$ is $90^{\circ }$. There is agreement in phase at any two points whose distance apart is $31.4$ miles.

In conclusion take the more general terminal conditions in which the line feeds into a receiver circuit, and suppose the current is to be kept at $50$ amperes, in a phase $40^{\circ }$in advance of the electromotive force; then , and substituting the constants in (70) gives

where $\theta =113^{\circ }33\prime$, $e_1=10730$ volts, the E.M.F. at sending end. This is 17 times what was required when the other end was insulated.

Prob. 107.

If $L=0$, $G=0$, $I=0$; then , where $n^2=\frac{\omega RC}{2}$, $n_1^2=\frac{R}{2\omega C}$; and the solution is $$\begin{array}{llllllll}\hfill e_1& =\frac{1}{\sqrt{2}}E\sqrt{cosh2nx+cos2nx},& \hfill tan\theta & =tannxtanhnx,& \hfill & & \hfill & \\ \hfill i_1& =\frac{1}{2n_1}E\sqrt{cosh2nx-cos2nx},& \hfill tan\theta \prime & =tannxcothnx.& \hfill & & \hfill & \end{array}$$

Prob. 108.

If self-induction and capacity be zero, and the receiving end be insulated, show that the graph of the electromotive force is a catenary if $G\ne 0$, a line if $G=0$.

Prob. 109.

Neglecting leakage and capacity, prove that the solution of equations (66) is $\bar{ı}=Ī$, .

Prob. 110.

If $x$ be measured from the sending end, show how equations (65), (66) are to be modified; and prove that

$$ē={Ē}_{0}coshmx-m_1{Ī}_{0}sinhmx,\bar{ı}={Ī}_{0}coshmx-\frac{1}{m_1}{Ē}_{0}sinhmx,$$

where ${Ē}_0$, ${Ī}_0$ refer to the sending end.