Funktechnische Aufgaben und Zahlenbeispiele by Dr.-Ing. Karl Mühlbrett (auth.)

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A) Die Stromstarke wird E. I=E· Ri + R Ri+B =E '(R;+R)2=(5+R)2' e) Der Wirkungsgrad N2 R Nl Ri+R 1 1]=-=---=--= 1+ Ri R 1 1+~ R 58 Losung 10. R, If R ° 1 2 3 4 5 6 7 8 10 12 14 00 5,00 2,50 1,67 1,25 1,00 0,83 0,71 0,63 0,50 0,42 0,36 00 0,699 0,398 0,~23 0,097 0,000 -1 +0,919 - I +0,851 -1 +0,799 -1 +0,700 -1 +0,623 -1 +0,556 ° Beim Zeiehnen 00 I u A v w 2,00 1,67 1,43 1,25 1,11 1,00 0,91 0,83 0,77 0,67 0,59 0,53 ° 20,0 16,7 14,3 12,5 11,1 10,0 9,1 8,3 7,7 6,7 5,9 5,3 ° -00 1,67 2,86 3,75 4,44 5,00 5,46 5,83 6,16 6,67 7,05 7,37 10 w ° ° 2,78 4,09 4,69 4,94 5,00 4,97 4,86 4,73 4,44 4,15 3,87 ° ° 0,167 0,286 0,375 0,444 0,500 0,546 0,583 0,616 0,667 0,705 0,737 1 der Kurven wahlt man gewohnlieh R als Abszisse.

A) Die uberschussige Spannung betragt bei der: Normalrohre ex) U = Sparrohre U= 5 - 2,7 = 2,3 V, fJ) U = 3,6 - 3 = 0,6 V, Normalrohre = 2 V, 5- 3 U = 3,6 - = 0,9 V, Normalrohre r) U = 110 - 3 = 107 V, Sparrohre Sparrohre Normalrohre ~) 2,7 U=110-2,7=107,3 V, U = 220 - 3 = 217 V, Sparrohre U = 220 - 2,7 = 217,3 V. Demnach ist vorzuschalten: N ormalrohre I 2 ex) R= =-=4D, U 0,5 R= 2,3 =38 D 0,6 D fJ) R= -0,5 =1,2 , R= 0,9 =15 D D, r) R=10~=214 0,5 R= 107,3 = 1789 D ~) 217 0,5 R=-=434D, Sparrohre 0,06 0,06 ' ' 0,06 R= 217,3 =3622 D 0,06 ' .

D) Es wird insgesamt wahrend dieser 16,7 Stunden eine Arbeit verbraucht A=N·t= 110·2,4·16,7 = 4400 Wh, =4,4 kWh. Daraus errechnet man die Kosten K = 4,4 ·0,20 = 0,88 M. Q=4·30= 120 Who Nach d kostet diese K=0,88 M. Daraus folgt fUr 1 kWh 1000 Kl = 0,88 120 = 7,33 M. f) Jede Zelle braucht bis zu 2,7 V; an 110 V kann man also 110 z=-=40 Zellen 2,7 laden. 54 L6sung 4. g) Die gesamten Ladekosten sind dieselben wie vorher, namlich 0,88 M, da aIle 40 Zellen in Reihe geschaltet und von demselben Strom geladen werden.

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