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Kirchhoffova pravila

Kirchhoffova pravila. Prvo Kirchhoffovo pravilo. 5. I 5. Q 1 + Q 4 = Q 2 + Q 3 + Q 5. I 1. I 4. 1. 4. I 1 t + I 4 t = I 2 t + I 3 t + I 5 t. I 2. I 3. 2. I 1 + I 4 = I 2 + I 3 + I 5. 3. R 1. R 2. R 3. B. A.  1 ,R u1.  2 ,R u2. : Q.

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Kirchhoffova pravila

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  1. Kirchhoffova pravila Prvo Kirchhoffovo pravilo 5 I5 Q1 + Q4 = Q2 + Q3 + Q5 I1 I4 1 4 I1t + I4t = I2t + I3t + I5t I2 I3 2 I1 + I4 = I2 + I3 + I5 3

  2. R1 R2 R3 B A 1,Ru1 2,Ru2 : Q Drugo Kirchhoffovo pravilo EepB - EepA = Wne1 – Weu1 – Wev1 – Wev2 + Wne2 – Weu2 – Wev3 Weu = IRuQ , Wev = IRQ EepB - EepA = Wne1 – IRu1Q – IR1Q – IR2Q + Wne2 – IRu2Q – IR3Q B- A= U = 1 – IRu1 – IR1 – IR2 + 2–IRu2 – IR3

  3. R1 R2 R3 1,Ru1 2,Ru2 A, B B- A=0 0 = 1 – IRu1 – IR1 – IR2 + 2–IRu2 – IR3 1 – 2 = IRu1+IRu2 + IR1+ IR2 + IR3 Drugo Kirchhoffovo pravilo

  4. 1,Ru1 R1 F E 2,Ru2 R2 D C A B 3,Ru3 Složeni strujni krugovi Primjer: 1 = 1,5V, 2 = 21, 3 = 6V Ru1 = 3,2, Ru2 = 2Ru1 I1 Ru3 = 0,12 I2 R1 = 8, R2 = 15 I3 I2 + I3 = I1 D: 2 – 3 = - I3Ru3+I2Ru3 + I2Ru2+ I2R2 ABCDA: -2+1 = - I2R2-I2Ru2 – I1Ru1- I1R1 DCEFD:

  5. -6 + 3 = - 0,12 I3 + 6,4 I2 + 15 I2 -3 + 1,5 = -15 I2 – 6,4 I2 – 3,2 I1 – 8 I1 -3 = - 0,12 I3 + 21,4 I2 -1,5 = -21,4 I2 – 11,2 I1 I1 = I2 + I3 -1,5 = -21,4 I2 – 11,2 (I2 + I3) -1,5 = -32,6 I2– 11,2 I3 I2 = - 0,137 A -3 = - 0,12 I3 + 21,4 I2 I1 = 0, 396 A I3 = 0,533 A

  6. Zadatak: Na slici je shema složenog strujnog kruga. Odredite jakost struje kroz otpornike ako je: 1 = 2,1 V D C F I1 I2 2 = 6,3 V R1 R2 Ru1 = 0,01  I3 Ru2 = 0,03  R4 2,Ru2 R1 = 1  1,Ru1 R2 = 2  R3 R3 = 1  A B E R4 = 10  I2 + I3 = I1 D: 1= I1R1+I1Ru1 + I3R4 ABCDA: 2= I2R3+I2Ru2 + I2R2 – I3R4 BEFCB:

  7. 2,1= I1+0,01I1 + 10I3 6,3= I2 +0,03I2 + 2I2 – 10I3 2,1= 1,01I1 + 10I3 I1 = I2 + I3 6,3= 3,03I2 – 10I3 2,1= 1,01(I2 + I3 ) + 10I3 2,1= 1,01I2 + 1,01I3 + 10I3 2,1= 1,01I2 + 11,01I3 I1 = 2,08 A 6,3= 3,03I2 – 10I3 I2 = 0 A I3 = 2,08 A

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