jika saklar s ditutup,kuat arus listrik I sebesar? ..tolong pk cara makasih

1/R paralel = 1/20+1/30 = 3/60+2/60 = 5/60 = 1/12
R paralel = 12 ohm
R total = 14,2+12+3,3+0,5 = 30 ohm
kuat arus =12/30 = 0,4 Ampere

[tex]rp = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{60}{5} = 12 \\ rtot = 12 + 14.2 + 3.3 + 0.5 = 30 \\ i = \frac{v}{r} \\ \frac{12}{30} = \frac{2}{5} = 0.4 \: ampere[/tex]

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