Then, if E0 = impressed E.M.F.,-Inductance, Capacity, And Stored Energy
Why This Family Matters
Section titled “Why This Family Matters”Routes equations involving field storage, charge, discharge, and frequency-dependent opposition.
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reviewable relation candidates.
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sources represented.
Candidate Formula Cards
Section titled “Candidate Formula Cards”e-o.296 1 = 0.5, - 0.296 Mog e = log 0.5, t =for the constant impressed E.M.F., E0 = 100 ; for the con-(a) M strength: i = ~, hence (1 - €-°-29«0 = 0.5.(d) If i = 0 at * = 0.0005, then(e) If t = - I = - 90 at t = 0.001, thencircuits. Hence the inductance is L = $/ i = ;/2/(R.circuits. Hence the inductance is L = ^ / 1 = n^ / iR.but E = E0,I= E0/z. If x0 < - 2 x, it raises, if x0 > - Zv,d.} If x = 0, that is, if the receiver circuit is non-maximum for x0= +1.0, x = - 1.0, and r = 0, whered.) If ;r = 0, that is, if the receiver circuit is non-since x = ~Vz2 - r2, if rr0 -f- x0 ~\/z2 - r2 is a maximum.f = rr0 + *0 Vs2 - r2 = maximum or minimum, ifmaximum for x^ = +10, x = - 1.0, and r = 0, whereei = E € L = i0re £ the generated e.m.f.In this case, at t = 0, e\ = E, that is, the e.m.f. does not rise.after impressing the required e.m.f. E = 230 volts will it takeimpressing the e.m.f. E = 500 volts will it take for the field tothe impressed e.m.f. is E = 230, the final value of currentIMPRESSED E.M.F. CONSTANT, E0=IOOIMPRESSED E.M.F. CONSTANT, E0 = IOOO VOLTS.L2 = inductance of the second circuit, and M = mutual induc-or Li = Si + - M L2 = Sz + - M,or M2 = (Li - Si)(Lz - Sz).H = 0.4 TT/ = - - , and the flux in the zone dlx is-’ «i = 0.ei = - = - 0.368 E.iei = io2 (r + n) c L ;io = - = 6.95 amp. Thus the current at time t ist = * - 6(2) To get io = 6.95 amp., with E = 500 volts, a resist-ance r = ^-f-= = 72 ohms, and thus a rheostat having a resist-i = io (l € 2)(a) i = ^, after t = 1.08 seconds.(b) i = 0.9 i0, after i = 3.6 seconds.(3) Impressing E = 500 volts upon a circuit of r = 33.2,33. (5) A coil of resistance r = 0.002 ohm and inductanceL = 0.005 mh., carrying current / = 90 amp., is short circuited.(6) i = 0.1 7, c-400< = 0.1, after t = 0.00576 second.E = 1 volt is inserted in the circuit of this coil, in opposite direc-Thus, - E + 61 = - E - L jt, the total e.m.f.;At t = 0, i = /, thus c = / + -;(6) i = o, e-400 « = 0.85, after t = 0.000405 second.(c) i = - I = - 90, e-400 « = 0.694, after t = 0.00091 second.E = joTITi = °-81 volt-E = - = 0.91 volt.or i = /osin (6 - 8’),machine is E = 4«4>7V10~8 volts, independent of the num-and *%E = 0 in a closed circuit,Z = r -jx, z = Vr2 + x’2,Z + r0 = r + r0-jx\E = EnJ >* + *2 . Eoa.) r0 = .2 ohm (Curve I.)b.) r0 = .8 ohm (Curve II.)with values of reactance, x = V^2 - r2, for abscissae, fromx = + 1.0 to x = - 1.0 ohm.As shown, / and E are smallest for x = 0, r = 1.0,For r0 = .8, and x = 0, x = + .8, x = - .8, the polarZ = r - jx, z = -\/r2 -|- x’2.c.) E = E0 , or the insertion of a series inductance, x0 ,^z*-\-2xx0 + x02 = 2;or, x0 = - 2 x.II. r=.6 X=H-,8111. r=.e i=-.8E^ = const. = 100 volts, -cr = 1 ohm, and -X = + 1.0 to ;r = + 1.0 ohm.As shown, / and E are smallest for ;r = 0, r = 1.0,For r^ = .8 and x = 0,x = + .8, and x = - .8, the polar2=1.0, r= .6,^= .8(CurveII.)2= 1.0, r= .6, AT= - .8 (Curve III.)This rise is a maximum for x0 = i .8, or, x0 = - x (the