14. RECTANGULAR COORDINATES 64. The vector diagram of sine waves gives the best insight into the mutual relations of alternating currents and e.m.fs. For numerical calculation from the vector diagram either the trigonometric method or the method of rectangular components is used. The method of rectangular components, as explained in the above paragraphs, is usually simpler and more convenient than the trigonometric method. In the method of rectangular components it is desirable to distinguish the two components from each other and from the resultant or total value by their notation. To distinguish the components from the resultant, small letters are used for the components, capitals for the resultant. Thus in the transformer diagram of Section 13 the secondary current I\ has the horizontal component ii = — I\ cos 0i, and the vertical component i'\ — + I\ sin 0\. To distinguish horizontal and vertical components from each other, either different types of letters can be used, or indices, or a prefix or coefficient. Different types of letters are inconvenient, indices distinguish- ing the components undesirable, since indices are reserved for distinguishing different e.m.fs., currents, etc., from each other. Thus the most convenient way is the addition of a prefix or coefficient to one of the components, and as such the letter j is commonly used with the vertical component. Thus the secondary current in the transformer diagram, Section 13, can be written i\ + ji* = Ii cos 0i + jli sin 0i. (1) This method offers the further advantage that the two com- ponents can be written side by side, with the plus sign between them, since the addition of the prefix j distinguishes the value jit or jli sin 0i as vertical component from the horizontal com- ponent i\ or 1 1 cos 0i. 1 1 = ii + ji* (2) thus means that I\ consists of a horizontal component i\ and a vertical component iz, and the plus sign signifies that i\ and iz are combined by the parallelogram of sine waves. 78 ELEMENTS OF ELECTRICAL ENGINEERING The secondary e.m.f. of the transformer in Section 13, Fig. 34, is written in this manner, E\ = — ei, that is, it has the hori- zontal component — e\ and no vertical component. The primary generated e.m.f. is E'==' 00 and the e.m.f. consumed thereby E' = + ei- w The secondary current is where ii = Ii cos 0i, iz = Ii sin 0i, (6) and the primary load current corresponding thereto is I' = - aii = aii - jaiz. (7) The primary exciting current, Joo = h - jg, (8) where h = J0o sin a is the hysteresis current, g = I0o cos a the reactive magnetizing current. Thus the total primary current is J0 = I' + J00 = (aii + h) -j (aiz + g). (9) The e.m.f. consumed by primary resistance rQ is r0Jo = TQ (aii + h) - jr0 (aiz + 0). (10) The horizontal component of primary current (aii + h) gives as e.m.f. consumed by reactance XQ a negative vertical com- ponent, denoted by JXQ (aii + h). The vertical component of primary current j (aiz + g) gives as e.m.f. consumed by react- ance XQ a positive horizontal component, denoted by XQ (aiz + (/)• Thus the total e.m.f. consumed by primary reactance XQ is XQ (aiz + g) + jxQ (aii + h), (11) and the total e.m.f. consumed by primary impedance is r0 (aii + A) + x0 (aiz + g) - j[rQ (aiz + g) - XQ (aii + h)]. (12) RECTANGULAR COORDINATES 79 Thus, to get from the current the e.m.f. consumed in react- ance XQ by the horizontal component of current, the coefficient j has to be added; in the vertical component the coefficient — j omitted; or, we can say the reactance is denoted by jxQ for the horizontal and by r- for the vertical component of current. In other words, if 7 = i — ji' is a current, x the reactance of its circuit, the e.m.f. consumed by the reactance is jxi H- xi' = xi' + jxi. 65. If instead of omitting — j in deriving the reactance e.m.f. for the vertical component of current we would add j also (as done when deriving the reactance e.m f. for the horizontal component of current), we get the reactance e.m.f. jxi — fxi', which gives the correct value jxi + xi', if f = - 1; (13) that is, we can say, in deriving the e.m.f. consumed by reactance, x, from the current, we multiply the current by jx, and substitute By defining, and substituting, j2 = — 1, jx can thus be called the reactance in the representation in rectangular coordinates and r -+- jx the impedance. The primary impedance voltage of the transformer in the preceding could thus be derived directly by multiplying the current, /o = (aii + h) - j (aii + g), (9) by the impedance, Z0 = r0 -f jxQ, which gives E'o = Zo/o = (r0 + jx<>) [(aii + h) - j (ai2 + g)] = r0 (aii + h) - jrQ (ai2 + g) + jxQ (aii + h) - j2xQ (ai2 -f g), and substituting j2 = — 1, E'o = [r0 (aii + h) + XQ (ai2 + g)] - j [r0 (ai2 + g) - XQ (aii + h)], (14) 80 ELEMENTS OF ELECTRICAL ENGINEERING and the total primary impressed e.m.f. is thus EQ = E -["• E o = [^ + r0 (ail + h) + x0 ( + i'2> (18) The capital letter I in the symbolic expression / = i + jif thus represents more than the / used in the preceding for total current, etc., and gives not only the intensity but also the phase. It is thus necessary to distinguish by the type of the latter the capital letters denoting the resultant current in symbolic expres- sion (that is, giving intensity and phase) from the capital letters giving merely the intensity regardless of phase; that is, I = denotes a current of intensity / = and phase tan 0 = — . ^ RECTANGULAR COORDINATES 81 In the following, dotted italics wfll be used for the symbolic expressions and plain italics for the absolute values of alternating waves. In the same way z = \/r2 + x2 is denoted in symbolic repre- sentation of its rectangular components by Z = r + jx. (91) When using the symbolic expression of rectangular coordinates it is necessary ultimately to reduce to common expressions. ; Thus in the above discussed transformer the symbolic expres- sion of primary impressed e.m.f. EQ = |j^ + rQ (aii + h) + X0(ai2 +g) J -j [r0(ai2+0) -z0(a*'i+/i)J (15) means that the primary impressed e.m.f. has the intensity (ai'2+flf)J (20) and the phase tan 0o = -1 + r0 (aii + h) + X0 (aiz + flf) This symbolism of rectangular components is the quickest and simplest method of dealing with alternating-current phenom- ena, and is in many more complicated cases the only method which can solve the problem at all, and therefore the reader must become fully familiar with this method. EXAMPLES 67. (1) In a 20-kw. transformer the ratio of turns is 20 : 1, and 100 volts are required at the secondary terminals at full load. What is the primary current, the primary impressed e.m.f., and the primary lag, (a) at non-inductive load, 0i = 0; (6) with 0i = 60 degrees time lag in the external secondary circuit; (c) with 61 = — 60 degrees time lead in the external secondary circuit? 82 ELEMENTS OF ELECTRICAL ENGINEERING i i «!s 32 i • t*» s§ + + I II OS §2 o •* • d § 111 + 3" O O (NO d«o CO CO ^ «° » 1 fe : + : £ ^J- f 11 + « ^ a l' ^ E 8 *• & S ill!! i •D a c S D. g .S : s z » 1 & So. : : S t> ' « £ : tf • 05 n s s 'N o. -S V 3 3 « « & ** 03 T3 S fl 'E i ja l?l> +* ° « + 0 +N S5S » ««• •-! oo 1 d S| H " 84 ELEMENTS OF ELECTRICAL ENGINEERING The exciting current is /'00 = 0.3 — 0.4 j amp. at e = 2000 volts impressed, or rather, primary counter-generated e.m.f. The primary impedance, Z0 = 2 -f 5 j ohms. The secondary impedance, Z\ = 0.004 + 0.01 j ohm. We have, in symbolic expression, choosing the secondary current /i as real axis, the results calculated in tabulated form on page 82. 68. (2) eQ = 2000 volts are impressed upon the primary circuit of a transformer of ratio of turns 20:1. The primary impedance is ZQ = 2 -f 5 j, the secondary impedance, Zi = 0.004 + 0.01 j, and the exciting current at er = 2000 volts counter-generated e.m.f. is 70o = 0.3 — 0.4 j; thus the exciting admittance, Y = ^ = (0.15 - 0.2 j)10~3. 6 What is the secondary current and secondary terminal voltage and the primary current if the total impedance of the secondary circuit (internal impedance plus external load) consists of (a) resistance, Z = r = 0.5 — non-inductive circuit. (6) impedance, Z = r + jx = 0.3 + 0.4 j — inductive circuit. (c) impedance, Z = r -\- jx = 0.3 — 0.4 j — anti-inductive circuit. Let e = secondary e.m.f., assumed as real axis in symbolic expression, and carrying out the calculation in tabulated form, on page 83. 69. (3) A transmission line of impedance Z = r -}- jx = 20 + 50 j ohms feeds a receiving circuit. At the receiving end an apparatus is connected which produces reactive lagging or leading currents at will (synchronous machine) ; 12,000 volts are impressed upon the line. How much lagging and leading currents respectively must be produced at the receiving end of the line to get 10,000 volts (a) at no load, (6) at 50 amp. power current as load, (c) at 100 amp. power current as load? Let e = 10,000 = e.m.f. received at end of line, ii = power current, and i% = reactive lagging current; then total line current. LOAD CHARACTERISTIC OF TRANSMISSION LINE 85 The voltage at the generator end of the line is then E0 = e + ZI = e + (r + jx) (ii — jiz) = (e + rii + xi2) — j (n'2 — xii) = (10,000 + 20 ti + 50 1*2) - j (20 za - 50t'i); or, reduced, •f-ni + zi2)2 + (ri2 - thus, since E0 = 12,000, 12,000 = V(10,000 + 20*i + 50^)2 + (20 iz - 50^i)2. . (a) At no load i\ = 0, and 12,000 = hence, i2 = -j- 39.5 amp., reactive lagging current, I = — 39.5 ./. (6) At half load ii = 50, and 12,000 = VuMXX) + 50i2)2 + (20z2 - 2500)2;- hence, is = + 16 amp., lagging current,/ = 50 — 16 j. (c) At full load ii = 100, and 12,000 = V(12,000 + 50i2)2 + (20 i - 5000)2; hence, i2 = - 27.13 amp., leading current, I = 100 + 27.13 j.