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diketahui :
V1 = 6 V
V2 = 8 V
R1 = 2 ohm
R2 = 1 ohm
R3 = 2 ohm
ditanyakan :
I = ?
jawab :
rangkaian depan
-6 + 2I1 +2I2 = 0
2 I1 + 2 I2 = 6
I1 + I2 = 3
rangkaian belakang
8+I3 - 2 I2 = 0
I3 - 2I2 = -8
I1 = I2 + I3
I3 =I1 - I2
sehingga pada rangkaian belakang
I3 -2 I2 = -8
I1 - I2 - 2I2 = - 8
I1 - 3I2 = -8
dari persamaan 1 dan 2 eliminasi
I1 + I2 = 3
I1 - 3I2 = -8
_________-
4 I2 = 11
I2 = 11/4 = 2,75
I1 + I2 = 3
I1 + 2,75 = 3
I1 = 3 - 2,75 = 0,25 A
jawaban option A