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KIRCHOFF LAW

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Siti Aissah

on 18 December 2013

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Transcript of KIRCHOFF LAW

KIRCHOFF LAW

MESH ANALYSIS WITH VOLTAGE SOURCE
MESH ANALYSIS WITH CURRENT SOURCE
2 loops
MESH ANALYSIS
mesh is a loop which does not contain any other loops within it.
mesh analysis applies KVL to find unknown current.
it is not quite as general as nodal analysis because it is only applicable to a circuit that is planar
-7 + 8i1 + 6i2 = 0
8i1 + 6i2 = 7 eq 1
I2 = i1 + 6 eq 2
2 into 1
8i1 + 6(I1 + 6) = 7
8i1 + 6i1 + 36 = 7
14i1 = -29
i1 = -2.07 A

subs i1 = -2.07 A
i2 = i1 + 6
i2 = (-2.07) + 6
i2 = 3.93 A

Given I2 = -4 A

Equation for mesh 1:
10I1 + (I1-I2)5 = 10
15I1 – 5I2 = 10

Equations for mesh 2:
2I3 + (I3-I2)20 = 20
- 20I2 + 22I3 = 20

I1 = -0.667 A
I2 = - 4 A
I3 = - 2.73 A

i4 = 2A
i3 = -5A

Mesh 1
-4 + 10i1 + 10(i1 – i2) = 0
20i1 – 10i2 = 4

Mesh 2
-10(i2 – i1) + 5(i2 – i3) + 2(i2 – i4) =0
-3i2 + 10i1 -5i3 -2i4 = 0
1 & 2 into 4
-3i2 + 10i1 – 5i3 – 2i4 = 0
-3i2 + 10i1 – 5(-5) – 2(2) = 0
10i1 – 3i2 = -21

From 3
20i1 – 10i2 = 4
I2 = (4-20i1)/(-10)
6 into 5
10i1 – 3i2 = -21
10i1 – 3( (4-20I1)/(-10) ) = - 21

i1 = -1.3875A
i2 = (4-20i1)/(-10)
i2 = -3.175A
4 loops
2 loops

4I1 + 6(I1 – I2) = 10 – 2 eq 1
6(I2 – I1) + 2I2 + 7I2 = 2 + 20 eq 2
10I1 – 6I2 = 8 eq 3
-6I1 + 15I2 = 22 eq 4

From equation 3
I1 = (8 + 6 I2)/10 eq 5

From eq 5 insert in eq 4
-6 ( (8 + 6 I2)/10 ) + 15 I2 = 22
I1 = 2.2105
I2 = 2.3509
3 loops
Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10
Mesh 2: 4(I2 – I1) + 11(I2 – I3) + 3I2 = - 10 - 8
Mesh 3: 9I3 + 11(I3 – I2) + 10(I3 – I1) = 12 + 8

20I1 – 4I2 – 10I3 = 30 eq 1
-4I1 + 18I2 – 11I3 = -18 eq 2
-10I1 – 11I2 + 30I3 = 20 eq 3

I1 = 2.4731
I2 = 0.5938
I3 = 1.7088

4 loops
Mesh 1
-124 + 4(i1-i2) + 8(i1-i4) = 0
4i1 – 4i2 + 8i1 -8i4 = 124
12i1 – 4i2 - 8i4 = 124

Mesh 2
-16i2 + 4(i2 - i1) + 80(I2 - i3) = 0
-162 + 4i2 – 4i1 + 80i2 – 80i3 = 0
-4i1 + 68i2 – 80i3 = 0

Mesh3
-1 + 80(i3 – i2) = 0
80i3 – 80i2 = 1

Mesh 4
1 + 80(i4 – i1) + 12i4 = 0
8i4 – 8i1 + 12i4 = -1
-8i1 + 20i4 = -1

From 3
80i3 – 80i2 = 1
i2 = (1-80i3)/(-80)

From 4
-8i1 – 20i4 = -1
i4 = (-1+8i1)/20 eq 6

Subs 5 & 6 into 1
12i1 – 4i2 – 8i4 = 124
12i1 + 4((1-80i3)/(-80)) - 8((-1+8i1)/20) = 124
i1 = (2471/20-4i3)/(44/5)

7 into 2
-4i1 + 68i2 – 80i2 – 80 i3 = 0
-4((2471/20-4i3)/(44/5)) +68 ((1-80i3)/(-80)) – 80i3 = 0
i3 = -0.39A

i2 = (1-80i3)/(-80)
i2 = -0.4025A

i1 = (2471/20-4i3)/(44/5)
i1 = 14.22

i4 = (-1+8i1)/20
i4 = 5.638A
3 LOOPS
io= a+b
loop 1
i1= 4A
loop 2
-5a + 1(i2-i3) + 3(i2-i1) + 2i2 = 0
-5a + 6i2 – 3i1 – i3 = 0 1
loop 3
5a + 4i3 + 5(i3-i1) + (i3-i2) = 0
5a + 10i3 – 5i1 – i2 = 0 2
At node 0
i3 + a = i1
i3 = i1 - a 3
i1 = 4A into 3
i3 = 4 - a 4
4 & 3 into 1
-5a + 6i2 – 3(4) – 1(4 – a) = 0
3i2 – 2a = 8 5
4 & 3 into 2
5a + 10i3 – 5i1 – i2 = 0
-5a + 20 – i2 = 0
i2 = 20 – 5a 6
6 into 5
3i2 – 2a = 8
60 – 15a – 2a = 8
a = 52/(17 ) A
loop 4
-5b + 1 (i4 – i5) + 3i4 + 2i4 = 0
-5b + 6i4 – i5 = 0 7

loop 5
5b + 4i5 – 20 + 5i5 + 1(i5 – i4) = 0
5b + 10i5 – i4 – 20 = 0 8
i5 = - b into 7 & 8
7
-5b + 6i4 – i5 = 0
-4b + 6i4 = 0 9
8
5b + 10i5 – i4 – 20 = 0
5b + i4 = - 20
i4 = -20 – 5b 10
10 into 9
-4b + 6i4 = 0
-4b – 120 – 30b = 0
b = (-60)/( 17) A
i0 = a + b
= 52/17 + ((-60)/( 17) )
= - 0.4706 A

application of mesh analysis
superposition
RTh = ( (1 )/4 + 1/12 ) + 1
RTh = 3 + 1 = 4Ω
Find VTh
loop 1
-32 + 4i1 + 12(i1 – i2) = 0
16i1 – 12i2 = 32 1
loop 2
i2 = - 2 A 2
2 into 1
16i1 – 12(- 2) = 32
i1 = 0.5 A
VTh = 12 ( i1 – i2 )
= 12 (0.5 + 2 )
= 30 V
thevenin's theorem
RN = (( 8+4+8 ))/5 = 4Ω
loop 1
i1 = 2 1
loop 2
-12 + 4 ( i2 – i1 ) + 8i2 + 8i2 = 0
20i2 – 4i1 = 12 2
1 into 2
20i2 – 4i1 = 12
20i2 – 4 (2) = 12
i2 = 1 A

i2 = ISC = IN
IN = 1 A

norton's theorem
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