Download - rangk_Capasitor
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Kombinasi KapasitorKombinasi Seri
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Kombinasi KapasitorKombinasi Seri
11 C
QV
22 C
QV
33 C
QV
321321321
111
CCCQ
C
Q
C
Q
C
QVVVVab
321
111
CCCQ
Vab
321
1111
CCCCT
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Kombinasi KapasitorKombinasi Parallel
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Kombinasi KapasitorKombinasi Parallel
Q1 = C1V , Q2 = C2V dan Q3 = C3V
321 QQQQ
VCVCVCQ 321
321 CCCCP 321 CCCC
V
QP
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Contoh Soal
Tentukan kapasitas pengganti antara titik a dan b pada gambar dibawah ini, jika C1 = 5.0 F and C2 = 10.0 F and C3 = 2.00 F.
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Contoh Soal
Cs1 Cs2
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Contoh Soal1/Cs1 = 1/C1 + 1/C2 1/Cs1 = 1/(5.0 F) + 1/(10.0 F) 1/Cs1 = (0.20 + 0.10 )/ F 1/Cs1 = 0.30/ F Cs1 = (1/0.30) F Cs1 = 3.33 F
Cs2 = Cs1 = 3.33 F
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Contoh Soal
Cp1 = 3.33 F + 2.0 F + 3.33 F Cp1 = 8.66 F
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Contoh Soal
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Contoh Soal
a b
12 μF 6 μF
6 μF
10 μF
C1 C2
C3
C4
Cab
?
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Contoh Soal
a b
4 μF
6 μF
10 μF
C3
C4
Cab
?
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a b
10 μF 10 μFC4
Cab
?
Jawab : 5 μF
Contoh Soal
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a b5 μF
Cab
?
Contoh Soal
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Contoh Soala)Hitung kapasitas pengganti untuk gambar
berikut
b)Jika rangkaian dihubungkan dengan baterai 12 V, hitung beda potensial, muatan dan
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Contoh Soal
a b
2 μF
2 μF
a b4 μF
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Contoh SoalJawab :
a b
4 μF
Caba).
b)
12 V
Q
Q1
Q2
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b)
12 V
Q
Q1
Q2
Contoh SoalJawab :
Q = CT.V
Q = (4.0 x 10 - 6 F) (12 V) = 4.8 x 10 - 5 C
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Contoh SoalJawab :
Rangkaian parallel maka : V1 = V2 = V = 12 Volt
Q1 = C1.V1 = (2.0 x 10 - 6 F)(12 V) = 2,4 x 10 - 5 C
Q2 = Cs.V2 = (2.0 x 10 - 6 F)(12 V) = 2,4 x 10 - 5 C
3 μF 6 μF=
Cs = 2 μF
Muatan yang tersimpan pada kapasitor 3 μF dan 6 μF adalah Q2 = 2,4 x 10 - 5 C