02. analisis rangkaian 2015 ide
TRANSCRIPT
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© 2015, Ali Sadiyoko, UNPAR
DASAR RANGKAIAN LISTRIK
ANALISIS RANGKAIAN
Dr. Ali Sadiyoko S.T.,M.T.
Last update : 25 Mei 2016
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© 2015, Ali Sadiyoko, UNPAR
Hukum-Hukum Rangkaian
1. Hukum Ohm
2. Hukum Kirchhoff I (KCL)
3. Hukum Kirchhoff II (KVL)
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© 2015, Ali Sadiyoko, UNPAR
Hukum Ohm
0
0
654321
IIIIII
I
𝑣 = 𝑖𝑅
Dimana,
𝑣 = Tegangan [Volt] 𝑖 = Arus [Ampere] 𝑅 = Resistansi dengan satuan ohm ()
Beda potensial antara dua ujung elemen
resistor sama dengan besar nilai resistansinya
dikalikan dengan besar arus yang mengalir
pada resistor tersebut V R i v + -
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Hukum Ohm
Node adalah:
titik dari hubungan antara dua atau lebih cabang
Cabang adalah:
Representasi sebuah segment di antara 2 buah node, dimana terdapat sumber tegangan/arus atau komponen elektronik lain.
Loop adalah :
lintasan tertutup dalam sebuah rangkaian. Node = 3 buah (a, b, c)
Cabang = 5 cabang (Sumber tegangan, sumber tegangan, 5Ω, 2Ω, 3Ω ) Loop = 3 loop
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Hukum Kirchhoff I (Kirchhoff’s Current Law)
“Jumlah aljabar dari arus-arus yang memasuki setiap node rangkaian adalah nol.“ Untuk suatu lingkungan tertutup : jumlah arus = 0
Konvensi:
Arah tanda panah menunjukkan arah arus. Arah menuju suatu node diartikan positif dan arah sebaliknya diartikan negatif.
N = jumlah cabang yang terhubung ke node yang diamati.
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Hukum Kirchhoff I (Kirchhoff’s Current Law)
0
0
654321
IIIIII
I
6
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Untuk sebuah loop tertutup jumlah aljabar dari tegangan di loop tersebut sama dengan nol jumlah tegangan = 0; 𝑽 = 𝟎.
0
0
0
87654321
HAGHFGEFDECDBCAB VVVVVVVV
VVVVVVVV
V
Hukum Kirchhoff II (Kirchhoff’s Voltage Law, KVL)
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• Untuk sebuah loop tertutup : jumlah tegangan = 0
0
0
0
87654321
HAGHFGEFDECDBCAB VVVVVVVV
VVVVVVVV
V
A B C
D
E F G
H
V1 V2
V5 V6
V3
V4
V8
V7
Hukum Kirchhoff II (Kirchhoff’s Voltage Law, KVL)
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Prinsip Pembagi Arus
NN
N
SN
SSS
NNS
NS
RRRRIIII
R
VI
R
VI
R
VI
R
VI
RIRIRIRIV
IIIII
I
1...............:
1:
1:
1.............:::
...........;.........;;
.................
...............
0
321321
33
22
11
332211
321
Vs +
-
Is I1 I2 I3 IN
R1 R2 R3 RN
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Prinsip Pembagi Tegangan
NN
NSNSSS
N
NS
NS
RRRRVVVV
RIVRIVRIVRIV
R
V
R
V
R
V
R
VI
VVVVV
V
..................:::....................:::
.........;.........;;
.....................
...............
0
321321
332211
3
3
2
2
1
1
321
Vs +
-
Is R1 R2
RN
R2
V2 V1
V3
VN
10
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Drop Tegangan
VS
+
-
RI
R
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Resume
Hukum Ohm : 𝑣 = 𝑖𝑅
Hukum Kirchhoff I (KCL) : 𝑖𝑖 = 0𝑁𝑖=1
Hukum Kirchhoff II (KVL) 𝑣𝑖 = 0𝑁𝑖=1
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ANALISIS RANGKAIAN
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Metode Analisis Rangkaian
• Analisis Titik ( Node Analysis )
• Analisis Loop ( Loop Analysis )
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Analisis Titik
1. Identifikasi seluruh titik simpul dan arus cabang pada rangkaian
2. Tuliskan persamaan Kirchhoff I untuk setiap titik simpul rangkaian
3. Tuliskan persamaan tegangan untuk seluruh cabang antara 2 titik simpul yang tidak memiliki sumber arus
4. Selesaikan seluruh persamaan yang terbentuk.
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6V 8V
1
2
31 2
A B3 1
Lingkungan tertutup
Analisis Titik
i1 i3
i2
i4 i5
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Analisis Titik Penyelesaian Soal ??
i1 – i2 –i3 = 0
VA – 3 i3 = 0
i2 – i4 –i5 = 0
VA – VB – 3i2 = 0
VA + 3 i1 = 6
VB – i4 = 0
VB – 3i5 = 8
Selesaikan persamaan yang terbentuk !
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Analisis Titik Hasil akhir :
i1 = 23/21 = 1 2/21 = 1.095238 A
i3 = 19/21 = 0.904762 A
i2 = 4/21 = 0.190476 A
i4 = 45/21 = 2 1/7 = 2.142857 A
VB = 45/21 = 2 1/7 = 2.142857 V
i5 = -41/21 = -1 20/21 = -1.952381 A
VA = 57/21 = 2.714286 V
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Analisis Loop
1. Buat arus loop untuk setiap loop tertutup.
2. Tuliskan persamaan Kirchhoff II untuk setiap loop yang tidak melewati sumber arus
3. Buat persamaan khusus untuk loop yang melalui sumber arus
4. Selesaikan seluruh persamaan yang terbentuk
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6V 8V
1
2
31 2
A B3 1
Analisis Loop
i1 i3
i2
i4 i5
I1 I2 I3
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Analisis Loop
Penyelesaian Soal ??
-6 + 3 I1 + 3( I1 –I2 ) = 0
…………………………………. Rapikan !
3 I2 + 1( I2 - I3 ) + 3( I2 – I1) = 0
3 I3 + 8 + ( I3 - I2 ) = 0
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Analisis Loop
Penyelesaian Soal ??
6 I1 – 3 I2 = 6
Selesaikan persamaan yang terbentuk !
-3 I1 + 7 I2 - I3 = 0
-I2 + 4 I3 = -8
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Analisis Loop Hasil akhir :
I1 = 23/21 = 1 2/21 = 1.095238 A (= i1)
I2 = 4/21 = 0.190476 A (= i2)
i4 = I2 – I3 = 45/21 = 2 1/7 = 2.142857 A
VB = 1 x 45/21 = 2 1/7 = 2.142857 V
I3 = -41/21 = -1 20/21 = -1.952381 A (= i5)
VA = 3 x 19/21 = 19/7 = 2 5/7 = 2.714286 V
i3 = I1 – I2 = 19/21 = 0.904762 A
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Contoh 1
2V
1
R=2
46
A8
1
1 13A
BC
i1 i2 i3
i4
i5
24
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Contoh 2
16 V
8 V
24 V
1
2
3
R=1 4
2
20
A B
C
25
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Contoh 3
6 V
3
1
2 R=2
1
2
1
3
16 V
2 A
1 A
Rangkaian mengandung sumber arus ! Hati-hati penyelesaiannya (perhatikan cara penyelesaian di kelas)
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Contoh 4
2
1
2
R=2
1
3
1
2A
1V
3V7V
2A
27
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Contoh 5
2
1
2
R=2
1
3
1
2A
1V3V
7V
2A
28