Ԑ = 6 Volt
R total = Rp + r
→ 1/Rp = 1/R1 + 1/R2
1/Rp = 1/6 + 1/4
1/Rp = 2/12 + 3/12
1/Rp = 5/12
Rp = 12/5 = 2,4 Ω
→ R total = 2,4 + 0,6
R total = 3 Ω
I = Ԑ / R total
I = 6 / 3
I = 2 Ampere
Jadi, besar kuat arus (I) pada rangkaian tersebut adalah
= D. 2,0 A
Jangan lupa jadikan jawaban terbaik, semoga membantu.
Semangat Belajar!!
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Ԑ = 6 Volt
R total = Rp + r
→ 1/Rp = 1/R1 + 1/R2
1/Rp = 1/6 + 1/4
1/Rp = 2/12 + 3/12
1/Rp = 5/12
Rp = 12/5 = 2,4 Ω
→ R total = 2,4 + 0,6
R total = 3 Ω
I = Ԑ / R total
I = 6 / 3
I = 2 Ampere
Jadi, besar kuat arus (I) pada rangkaian tersebut adalah
= D. 2,0 A
Jangan lupa jadikan jawaban terbaik, semoga membantu.
Semangat Belajar!!