. Electric traction and transmission engineering . ) Fig. 65. where the resistances of the various branches and thevoltage at the substations are known and the equivalentresistances R of the load and x of the rest of the conductingsystem, out and back from both substations and consideredas connected in parallel, are to be found. The problem issolved by applying Kirchhoffs laws, which result in thefollowing equations, where the resistances THE DISTRIBUTING SYSTEM. i A=a + b + c B=d+f+g ohms. C = b + d + h Ah -bis +IR = E Bh -dh -IR =-E -bh -dh +Ch = o h -h = I\ 147 (7) (8) Solving for R by mean
. Electric traction and transmission engineering . ) Fig. 65. where the resistances of the various branches and thevoltage at the substations are known and the equivalentresistances R of the load and x of the rest of the conductingsystem, out and back from both substations and consideredas connected in parallel, are to be found. The problem issolved by applying Kirchhoffs laws, which result in thefollowing equations, where the resistances THE DISTRIBUTING SYSTEM. i A=a + b + c B=d+f+g ohms. C = b + d + h Ah -bis +IR = E Bh -dh -IR =-E -bh -dh +Ch = o h -h = I\ 147 (7) (8) Solving for R by means of determinantsA o -b E R = 0 B -d- E■b —d C o 1 —I o / A -b (E-AI)/!B -d -Ell-{b-\-d) C b -b I A B -{b+d) -d -I /-I - b -d — I c C 0 I 0 0 0 ohms. (9) A o 0 B-b -d 1 —IWhence the voltage impressed upon the load is E{b+dY-E{A+B)C-{Ad-+Bb-ABC)I . ^^= ib + df -{A+ B)C ^°^- ^°^ The drop e between either substation and the load is e = xl = E- RIvolts, (11) where x is the equivalent resistance in ohms of the con-ducting system between
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