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Cp₂ = 4 + 2 = 6 μF
1/Cs = 1/4 + 1/12 + 1/6
1/Cs = (3 + 1 + 2) / 12
1/Cs = 6/12
Cs = 12/6
Cs = 2 μF
C = Cs
C = 2 μF
C = 2 x 10⁻⁶ F
W = ½ • C • V²
W = ½ • 2 x 10⁻⁶ • 24²
W = 10⁻⁶ • 576
W = 5,76 x 10⁻⁴ Joule
Jadi, besar energi listrik pada kapasitor gabungan adalah 5,76 x 10⁻⁴ Joule (Jawaban C)
Ok?