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Rtot = 1/(1/R1 + 1/R2 + 1/R3)
= 1/((2+1+3)/120)
= 120/6
= 20 Ω
I = V/Rtot
= 6 V/20 Ω
= 0,3 A
besar arus I2 mengikuti perbandingan terbalik hambatannya,
I1 : I2 : I3 = 1/R1 : 1/R2 : 1/R3
I1 : I2 : I3 = 1/60 : 1/120 : 1/40
I1 : I2 : I3 = 2 : 1 : 3
maka I2 = (1/6).I
= (1/6).0,3 A
= 0,05 A