ep19z05f

# ep19z05f - 199 Zadania a b 1 I 12 V 2 b I 2 3A 2 6V 12 V I...

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Zadania 199 a”) b’) b”) Rozwi ą zania : 9 3 1 12 0 = - = U V ; 3 5 , 1 2 0 = = U V ; 3 18 5 , 10 2 0 = - = U V ; 1 = w R ; 1 = w R ; 1 = w R ; 3 2 1 9 = + = I A . 5 2 1 12 3 = + + = I A . 1 2 1 3 = + = I A . c) Rozwi ą zanie : 5 , 13 5 , 4 3 5 , 4 6 0 = - = U V ; 5 , 4 = w R ; 3 2 5 , 4 6 5 , 13 = + + = I A . d) Rozwi ą zanie : 3 28 14 3 2 0 = = U V (dzielnik napi ę cia); 3 2 = w R ; 5 4 3 2 14 3 28 = + + = I A . e’) e”) Odpowied ź : I = 3 A . Odpowied ź : I = 1 A . ( do samodzielnego rozwi ą zania ) 3 A 3 A 1 12 V U 0 3 A 1 12 V 2 I 10,5 A U 0 2 18 V 24 V 2 I 2 18 V 24 V 2 2 2 12 V 6 V 2 2 I 1,5 A U 0 6 V 2 2 3 9 A 6 V I 6 6 2 3 U 0 3 9 A 4,5 A 6 6 3 4,5 A U 0 1 2 14 V 1 7 A 14 V 2 4 I 7 V U 0 1 7 A 2 7 V 4 A 2 3 1 3 A 12 V I 4 A I 2 3 1 3 A 12 V

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Elektrotechnika podstawowa 200 Zad. 5-13. Oblicz warto ść pr ą du
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## This note was uploaded on 11/20/2011 for the course ECON 00502 taught by Professor Khazadi during the Fall '07 term at Oxford Brookes.

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ep19z05f - 199 Zadania a b 1 I 12 V 2 b I 2 3A 2 6V 12 V I...

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