Kirchhoffova pravila Prvo Kirchhoffovo pravilo 5 1 2
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Kirchhoffova pravila Prvo Kirchhoffovo pravilo 5 1 2 I 5 I 1 I 2 I 4 I 3 3 Q 1 + Q 4 = Q 2 + Q 3 + Q 5 4 I 1 t + I 4 t = I 2 t + I 3 t + I 5 t I 1 + I 4 = I 2 + I 3 + I 5
Drugo Kirchhoffovo pravilo R 1 A 1, Ru 1 R 2 R 3 2, Ru 2 B Eep. B - Eep. A = Wne 1 – Weu 1 – Wev 2 + Wne 2 – Weu 2 – Wev 3 Weu = IRu. Q , Wev = IRQ Eep. B - Eep. A = Wne 1 – IRu 1 Q – IR 2 Q + Wne 2 – IRu 2 Q – IR 3 Q : Q B - A = U = 1 – IRu 1 – IR 2 + 2 – IRu 2 – IR 3
R 1 R 2 1, Ru 1 R 3 2, Ru 2 A, B B - A = 0 0 = 1 – IRu 1 – IR 2 + 2 – IRu 2 – IR 3 1 – 2 = IRu 1 + IRu 2 + IR 1 + IR 2 + IR 3 Drugo Kirchhoffovo pravilo
Složeni strujni krugovi Primjer: F D A 1 = 1, 5 V, 2 = 2 1, 3 = 6 V E Ru 1 = 3, 2 , Ru 2 = 2 Ru 1 2, Ru 2 R 2 I 1 I 3 1, Ru 1 R 1 C I 2 3, Ru 3 = 0, 12 R 1 = 8 , R 2 = 15 B D: I 2 + I 3 = I 1 ABCDA: 2 – 3 = - I 3 Ru 3 + I 2 Ru 2 + I 2 R 2 DCEFD: - 2 + 1 = - I 2 R 2 - I 2 Ru 2 – I 1 Ru 1 - I 1 R 1
- 6 + 3 = - 0, 12 I 3 + 6, 4 I 2 + 15 I 2 -3 + 1, 5 = -15 I 2 – 6, 4 I 2 – 3, 2 I 1 – 8 I 1 -3 = - 0, 12 I 3 + 21, 4 I 2 -1, 5 = -21, 4 I 2 – 11, 2 I 1 = I 2 + I 3 -1, 5 = -21, 4 I 2 – 11, 2 (I 2 + I 3) -1, 5 = -32, 6 I 2 – 11, 2 I 3 I 2 = - 0, 137 A -3 = - 0, 12 I 3 + 21, 4 I 2 I 1 = 0, 396 A I 3 = 0, 533 A
Zadatak: Na slici je shema složenog strujnog kruga. Odredite jakost struje kroz otpornike ako je: 1 = 2, 1 V 2 = 6, 3 V Ru 1 = 0, 01 Ru 2 = 0, 03 R 1 = 1 R 2 = 2 R 3 = 1 R 4 = 10 D R 1 1, Ru 1 A I 1 C I 2 I 3 R 4 B R 2 F 2, Ru 2 R 3 E D: I 2 + I 3 = I 1 ABCDA: 1 = I 1 R 1 + I 1 Ru 1 + I 3 R 4 BEFCB: 2 = I 2 R 3 + I 2 Ru 2 + I 2 R 2 – I 3 R 4
2, 1 = I 1 + 0, 01 I 1 + 10 I 3 6, 3 = I 2 + 0, 03 I 2 + 2 I 2 – 10 I 3 2, 1 = 1, 01 I 1 + 10 I 3 6, 3 = 3, 03 I 2 – 10 I 3 I 1 = I 2 + I 3 2, 1 = 1, 01(I 2 + I 3 ) + 10 I 3 2, 1 = 1, 01 I 2 + 1, 01 I 3 + 10 I 3 2, 1 = 1, 01 I 2 + 11, 01 I 3 I 1 = 2, 08 A 6, 3 = 3, 03 I 2 – 10 I 3 I 2 = 0 A I 3 = 2, 08 A
