e = 2 7r/n$ sin r the instantaneous generated e.m.f.Symbolic AC And Complex Quantities
Why This Family Matters
Section titled “Why This Family Matters”Routes equations where Steinmetz turns alternating quantities into symbolic algebra.
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candidate records in this family.
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reviewable relation candidates.
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sources represented.
Candidate Formula Cards
Section titled “Candidate Formula Cards”B = 6’ + jh” = 6(cos 13 + j sin /3)is r - j (x -f x0} = r = .6, x + x0 = 0, and tan S>0 = 0 ;is r - j {x + Xo) = r = 0.6, x -{- Xo = 0, and tan do = 0; that= - J = (tan a - j) (7)If an alternating current i = I0 sin 6 passes through a resist-1.) Ohm’s law : i = e j r, where r, the resistance, is a1.) Ohm’s law : i = e j r, where r, the resistance, is aor, if E = e -\-je’ is the impressed E.M.F., and 7 = i ’ -\- ji’/, upon the e.m.f., or by IE cos d, where 9 = angle of time-but E = E^y I^E^j z. If x^ > - 2,t-, it raises, if ;r < - 2 jr,a maximum for dP^j db = 0, that is, according to § 61, if -e.m.f., e = EQ sin 6.at A and is tan C02 = Vijv^ (thus being greater) when theR = j/VK. (15)copper of No. 5, or j of ;j = ^: Cu. = ^FH = DH sin a, and DL = DH sin av (1)cubic hyperbolas: e^i = kz2; or, el =- £j and since we find forfc1 = 2 TT / sin <t>dfa (3)cos r = - - - - = 0.4iron is /z4 = 280 at B4 = 2850. Thus the field intensity/ H = - je = 2 irfn$ sin 2 IT} (t - ti)or, e = 2-jrfn& sin (6 - 0i),i*r = 702r sin2 0 = ^r C1 ~ cos 2 0),e = E0 sin (0 - 0i) = 273 sin (0 - 0i) ;e = 273 sin 210 (t - h).e = - -j-. L 108 absolute units34. An alternating current i = IQ sin 2irft or i - I0 sin 0i = /0 sin 2 IT/ (t - t’),i = IQ sin 6 passes through a circuit of resistance r and induc-2 irft = /o sin 0) of effective valuee’2 = - xI0 cos 2 irft = - xIQ cos 6,e’z = - xIQ cos 0,ty : i = 7i sin 2 TT N(t - A) + 7, sin 6 TT N (t - /3)i = 1 sin 2 -wjl,/ = / sin =^ (/ - A) = /sin 2 tt A” (/ - A) ;and i = I cos 2 tt/^.i = /1 sin 2-KJ{t - ti) + h sin 4 7r/(i - ^2)i = Ai sin 2 tt/^ + A2 sin 4 x// + A3 sin 6 tt/^ + …i = 7] sin 2 Trf(t - ti) + 1 3 sin 6 Trf(t - ts)i = A\ sin 2Tvft + A3 sin 6x/f + ^5 sin lOvr/^ + …i = Ai sin 2 vNt + A, sin 4 ttNI + A, sin C irNl + …E0 = V(^ cos w + Ir)2 -f- (E sin w + Ix)z.Thus, for instance, at the amplitude AOB^ == </>j = 2ir/j/ T’i = I cos (?> - ??2),i = / cos (?? - /3)e = E cos (i? 4- a’) -i = I cos {t} + /3’)i = I cos (?? - /3)ffo = Vfr2 + S^2 + 20^ sin Wi,since y’2 = - 1, j = V- 1 ;Ea = V{E COS 6 4- /r)2 + (E sin 6 + Ixy.100 volts, and /j = 75 amperes, for a non-inductive secon-- /V) , tan w0 =$Fo = V^H^i’ + 2 IFSFi sin Wi ,h = 7 sin ^ is the vertical component of the sine wave.j’ = - 1,system of ordinary numbers, this definition of j = V - 1 doesI = {a + a’) +j(6 + 6’).or e.o = \/(e + ir)- + {ix^, tan ^o = - 77^-0 = 0’+ jc” = c(cos T + j sin 7)^ = -[cos (a + |3 - 7) + j sin (a + |8 - 7)],c = OC cos (0 - 0o).Z -jx0 = r-j(x +#e).i = I cos’ (?9 - 0)circuit •- z= 1 Qj r = 1>0> x= 0 (Curve j)&Q = ro JXoi ZQ = V f0 -j- Xo ,is r - y (.r + ;r^) = r = .6, x + x^ = 0, and tan w^ = ;X = .9, X = Oy X = - .9, and Z^ = .3 -j .4.this case, a = ::^j 2 r,,, while in a continuous-current circuitZo = r„+ j”j:0 =0.1 +0.3j;Assuming tan a = 0.6, which is a fair value for a closed mag-