02. Analisis Rangkaian
Transcript of 02. Analisis Rangkaian
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HUKUMRANGKAIAN
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Hukum Hukum Rangkaian
Hukum OhmHukum Kirchhoff I
Hukum Kirchhoff II
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Hukum Kirchhoff I Untuk suatu lingkungan tertutup : jumlah arus = 0
0
0
654321 =+++++
=∑ I I I I I I
I
I1 I2
I3
I4I5
I6
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Hukum Kirchhoff II Untuk sebuah loop tertutup : jumlah tegangan = 0
0
0
0
87654321
=+++++++
=+++++++
=
∑
HAGH FG EF DE CD BC ABV V V V V V V V
V V V V V V V V
V
A B C
D
EFG
H
V1 V2
V5V6
V3
V4
V8
V7
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Pembagi Arus
N
N
N
S
N
SSS
N N S
N S
R R R R I I I I
R
V I
R
V I
R
V I
R
V I
R I R I R I R I V
I I I I I
I
1...............:
1:
1:
1.............:::
...........;.........;;
.................
...............
0
321
321
3
3
2
2
1
1
332211
321
=
====
=====
++++=
=∑
Vs
+
-
Is I1 I2 I3 IN
R1 R2 R3 RN
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Pembagi Tegangan
N N
N S N SSS
N
N
S
N S
R R R RV V V V
R I V R I V R I V R I V
R
V
R
V
R
V
R
V I
V V V V V
V
..................:::....................:::
.........;.........;;
.....................
...............
0
321321
332211
3
3
2
2
1
1
321
=
====
====
+++=
=∑
Vs+-
IsR1 R2
RN
R2
V2V1
V3
VN
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Drop Tegangan
R
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ANALISIS
RANGKAIAN
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Analisis Rangkaian
Analisis Titik ( Node Analysis )Analisis Loop ( Loop Analysis )
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Analisis Titik
Identifikasi seluruh titik simpul dan arus cabangpada rangkaian
Tuliskan persamaan Kirchhoff I untuk setiap titik
simpul rangkaian Tuliskan persamaan tegangan untuk seluruh
cabang antara 2 titik simpul yang tidak memiliki
sumber arusSelesaikan seluruh persamaan yang terbentuk.
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Lingkungan tertutup
Analisis Titik
i1i3
i2
i4i5
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Analisis TitikPenyelesaian 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 TitikHasil 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
Buat arus loop untuk setiap loop tertutup. Tuliskan persamaan Kirchhoff II untuk setiap loop
yang tidak melewati sumber arus
Buat persamaan khusus untuk loop yang melaluisumber arus
Selesaikan seluruh persamaan yang terbentuk
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Analisis Loop
6 V 8 V
1Ω
2Ω
3Ω1Ω 2Ω
A B3Ω 1Ω
i4i3i1
i2
i5
I1 I2 I3
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Analisis LoopPenyelesaian Soal ??
-6 + 3 I 1 + 3( I 1 –I 2 ) = 0
… … … … … … … … … … … … … . Rapikan !
3 I2 + 1( I 2 - I3 ) + 3( I2 – I 1) = 0
3 I3 + 8 + ( I 3 - I2 ) = 0
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Analisis LoopPenyelesaian Soal ??
6 I1 – 3 I 2 = 6
Selesaikan persamaan yang terbentuk !
-3 I 1 + 7 I 2 - I 3 = 0
-I 2 + 4 I3 = -8
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Analisis LoopHasil 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
i1 i2 i3
i4
i5
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Analisis Titik Penyelesaian Soal ??
Selesaikan persamaan yang terbentuk !
i1 – i2 – 3 = 0
VA – (-2) = -6 i1
i2 – i3 – i4 = 0
VA – VB = 2 i2
i4 – i5 + 3 = 0
VB = 4 i3
VB – VC = 8 i4
VC = 2 i5
VA + 2 = -6 i1 -6 i1 – VA = 2
–2 i2 + VA – VB = 0
–4 i3 +VB = 0
–8 i4 + VB – VC = 0
–2 i5 +VC = 0
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Contoh 1
3 A
I
1
I2
i1 i2 i3
i4
i5
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Analisis LoopPenyelesaian Soal ??
2 + 6 I 1 + 2 (I 1 – 3 ) + 4 (I 1 - I2) = 0
Selesaikan persamaan yang terbentuk !
4 (I 1 - I 2) + 8 (I2 – 3) + 2 I2 = 0
… … … … … … … … Rapikan !
12 I 1 - 4 I2 = 4
-4 I 1 + 14 I 2 = 24substitusi
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Analisis LoopHasil akhir :
I1 = i1 = 1 A
I2 = i5 = 2 A
i4 = I2 – 3 = -1 A
VB = 4 x i3 = -4 V
i3 = I1 – I 2 = -1 A
VA = (2 x i2) +VB = -4 - 4 = -8 V
i2 = I 1 – 3 = -2 A
VC = 2 x i5 = 4 V
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Contoh 2
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Contoh 3
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Contoh 3 : analisis titik
6 V
3 Ω
1 Ω
2 ΩR=2 Ω
1 Ω
2 Ω
1 Ω
3Ω
16 V
2 A
1 A
i1
i2
i3
i4 i
5
i6
A B
CD
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Contoh 3 : analisis titikPenyelesaian Soal ??
(A) i1 + i2 – i3 = 0 VA – 6 = – 3 i1
(B) i3 + 1 + i5 = 0
VA – -16 - VB = 3i3
Selesaikan persamaan yang terbentuk !
(C) -i5 – i4 – i6 = 0
(D) 2 – i2 + i4 = 0
VB – VC = -2i5
VC – VD = 2i4
VC = i6
VA – VD = -i2
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Contoh 3 : analisis loop
6 V
3 Ω
1 Ω
2 ΩR=2 Ω
1 Ω
2 Ω
1 Ω
3Ω
16 V
2 A
1 A
i1
i2
i3
i4 i
5
i6
A B
CD
I 2
I1
I3
-1
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Analisis LoopPenyelesaian Soal ??
Selesaikan persamaan yang terbentuk !
… … … … … … … … Rapikan !
16 + 3 I 1 + 2 (I 1 + 1 ) + 2 (I 1 – I3) + 1 (I 1 – I2) = 0
Loop I1 :
2 = I 3 - I 2
Loop I3 :
Loop I2 : ? Melewati sumber arus ???
Ambil loop yang lebih besar !
-6 + 3 I 2 + 1 (I 2 - I1 ) + 2 (I 3 – I1) + 1 (I3 +1) = 0
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Analisis Loop
Penyelesaian Soal ??
Selesaikan persamaan yang terbentuk !
8 I1 – I 2 - 2 I 3 = -18
I3 – I2 = 2
-3 I 1 + 4 I2 + 3 I3 = -5
I1 = -131/ 47 = -2 37/ 47 = -2.78723 A
I2 = -130/ 47 = -2 36/ 47 = -2.76596 A
I3 = -36/ 47 = -0.76596 A
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Contoh 4
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Contoh 5
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Superposisi
Respon yang disebabkan oleh lebih dari satu buahsumber merupakan penjumlahan linier responyang merupakan kontribusi masing masingsumber
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Superposisi
2R 2R
R
R2R
2R 2R
V1I
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Superposisi
2R 2R
R
R
2R
R
V1
2R 2R
R
2R2R
V1
II
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Superposisi
2R 2R
R
2R2R
V1 I
2R 2R
R
R
V1 I
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Superposisi
2R 2R
R
R
V1 I
2R 2R
2R
V1 I
1)( 2
11
I I v
=
I1
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Superposisi
2RR
2R2R
V2
R
2R 2R
I
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Superposisi
2RR
2R
V2
R
2R
I
I2
2R 2R
22)(4
1.
2
1.
2
12 I I I v ==
½ x I2
½ x ½ x I2
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Superposisi
I3 2R2R 2R
V3
R
2R
I
R
I3
2R
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Superposisi
2R 2R 2R
V3
R
2R
I
R
33)(8
1.
2
1.
2
1.
2
13 I I I v ==
I3
2R
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Superposisi
)()()()(
...321 N
vvvvI I I I I ++++=