Linear Equations with Kirchoffs Law

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meano

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Hi, is anybody familiar with Kirchoffs Law? It is basically electric circuit theory which will enable you to calculate circuit current in different parts of the circuit.

I = Current measured in Amps
i1 = Current 1
i2 = Current 2
i3 = Current 3

I have a formula where yet again I am struggling with the substitution method. There are 3 formulas comprising of 3 different values. The only examples I have available to work through with explanations have 3 formulas but only one has 3 different values the others have only 2. Thus making it easier to solve.

The formula I am struggling with is

i1 + 8i2 + 3i3 = -31

3i1 - 2i2 + i3 = -5

2i1 - 3i2 + 2i3 = 6

Remember, 8i2 is 8 x Current 2 not 8 x i x 2. 3i3 is 3 x Current 3 not 3 x i x 3 etc etc.

I need to solve the formula to get values for i1, i2 and i3.
 
I can't figure it out, I think I am starting the equation all wrong.

I began by subtracting equation 1 from 2 giving 2a - 10b - 2c = 26

I don't think this is right though, anyway, I worked through this and got the value of b at 4.57.

Can you give me a head start?
 
Well I tried multiplying equation 1 by 3. Deducting this from equation 2 to get

-26b - 8c = 88 lets call this eq5

Then multiplied equation 2 by 2 then deducted equation 3 from this result to get

4a - 3b = -16 lets call this eq6

Then multiplied equation 1 by 4 and deducted eq6 from this to give

35b + 12c = -108 lets call this eq7

Multiplied eq5 by 1.5 to get

-39b - 12c =132

Deduct eq7 from the above

-39b - 12c = 132
35b + 12c = -108

Equals -74b = 240

b = 240/74 = 3.24

Where have I gone wrong?
 
Denis said:
You simply need to solve (a,b,c easier to use):
a + 8b + 3c = -31
3a - 2b + c = -5
2a - 3b + 2c = 6
Click to expand...
meano said:
...35b + 12c = -108 lets call this eq7

Multiplied eq5 by 1.5 to get: -39b - 12c =132

Deduct eq7 from the above

-39b - 12c = 132
35b + 12c = -108

Equals -74b = 240
Click to expand...
If you're adding -12c and +12c, then why are you not adding -39b and +35b?

Maybe try to clear columns out in an orderly fashion.

-3R1: -3a - 24b - 9c = 93
-2R1: -2a - 16b - 6c = 62

Add -3R1 and R2:

+3a - 2b + c = -5
-3a - 24b - 9c = 93
0a - 26b - 8c = 88 <== new R2

...and add -2R1 and R3:

+2a - 3b + 2c = 6
-2a - 16b - 6c = 62
0a - 19b - 4c = 68 <== new R3

Divide new R2 by -2 and new R3 by -1:

13b + 4c = 44 <== new R2
19b + 4c = -68 <== new R3

So now the system is:

a + 8b + 3c = -31
. ..13b + 4c = +44
. ..19b + 4c = -68

Now only one row has an entry in the first, "a", column. Normally, I'd then move on to the second, "b", column, but the third, "c", column seems set up already for cancellation, so why not? Let's multiply R2 by -1 and add to R3:

-13b - 4c = -44
+19b + 4c = -68
6b + 0c = -112 <== new R3

Divide through by 2 to get:

3b = -56 <== new R3

Now the system is:

a + 8b + 3c = -31
. ..13b + 4c = +44
. . .3b . . . . = -56

I hate dealing with fractions or decimals before I absolutely have to, so I'll multiply R3 by -4 and add to R2, and multiply by -3 and add to R1:

+13b + 4c = +44
-12b . . . . = 224
1b + 4c = 268 <== new R2

a + 8b + 3c = -31
. . .-9b . . . . = 168
a - 1b + 3c = 137 <== new R1

Now the system is:

a - 1b + 3c = 137
. . .1b + 4c = 268
. . .3b . . .. = -56

Can you take it from there (after checking all of my work!!)? ;)
 
This is what I have just done for b. Not sure its right though. Hope you can follow.

6a - 4b + 2c = -10 (Equation 2x2)(4)
2a - 3b + 2c = 6 (3)

(2) - (3)

4a - b = -4 (5)

a + 8b + 3c = -31 (1)
9a - 6b + 3c = -15 (Eq 2 x 3)

(2) - (1) =

8a - 14b = -16 (6) Multiply by 4 32a - 56b = -64 (7)
4a - b = -4 (5) Multiply by 8 32a - 8b = -32 (8)

(7) - (8)

-48b = -32

b = -32/-48
b = 0.66
 
meano said:
This is what I have just done for b. Not sure its right though. Hope you can follow.

6a - 4b + 2c = -10 (Equation 2x2)(4)
2a - 3b + 2c = 6 (3)

(2) - (3)

4a - b = -4 (5)
Click to expand...
No, again you are confusing operations. To "eliminate" those "2c" terms, you need to subtract one equation from another: 2c- 2c= 0. Then 6a- 2a= 4a and -4b- (-3b)= -4b+ 3b= -b but -10- 6= -16, not -4. That would be -10+ 6.

a + 8b + 3c = -31 (1)
9a - 6b + 3c = -15 (Eq 2 x 3)

(2) - (1) =

8a - 14b = -16 (6) Multiply by 4 32a - 56b = -64 (7)
4a - b = -4 (5) Multiply by 8 32a - 8b = -32 (8)

(7) - (8)

-48b = -32

b = -32/-48
b = 0.66
Click to expand...
 
meano said:
Hi, is anybody familiar with Kirchoffs Law? It is basically electric circuit theory which will enable you to calculate circuit current in different parts of the circuit.

I = Current measured in Amps
i1 = Current 1
i2 = Current 2
i3 = Current 3

I have a formula where yet again I am struggling with the substitution method. There are 3 formulas comprising of 3 different values. The only examples I have available to work through with explanations have 3 formulas but only one has 3 different values the others have only 2. Thus making it easier to solve.

The formula I am struggling with is

i1 + 8i2 + 3i3 = -31

3i1 - 2i2 + i3 = -5

2i1 - 3i2 + 2i3 = 6

Remember, 8i2 is 8 x Current 2 not 8 x i x 2. 3i3 is 3 x Current 3 not 3 x i x 3 etc etc.

I need to solve the formula to get values for i1, i2 and i3.
Click to expand...

What method/s have you been taught to solve simultaneous linear equations - Gauss elimination, matrix inversion or something else?
 
Hi, thanks for the replies. I don't know how I got -4 instead of -16 so thanks for pointing that out.

I have been taught the elimination and substitution methods in solving linear equations.

I haven't got time to look over any work tonight so it'll probably be the weekend before I get chance. I'm in the UK by the way so on a different time zone to most of you. This is a great forum by the way, it is helping me out no end.
 
meano said:
Hi, thanks for the replies. I don't know how I got -4 instead of -16 so thanks for pointing that out.

I have been taught the elimination and substitution methods in solving linear equations.

I haven't got time to look over any work tonight so it'll probably be the weekend before I get chance. I'm in the UK by the way so on a different time zone to most of you. This is a great forum by the way, it is helping me out no end.
Click to expand...
Multiple rows by 6, 2, and 3
Subtract row 1 from row 2 and subtract row 1 from row 3
Multiple rows by 1, 27, and 22
Subtract row 2 from row 3
c = -264 / 168 ~ -1.57142857142857
b = -(4752 + 432 c) / 594 ~ -6.85714285714286
a = (-185 - 18 c - 18 b) / 6 ~ -5.71428571428572

_______a_______b_______c
133-31
3-21-5
2-326
661818-186
26-42-10
36-9618
=61818-186
=0-22-16176
=0-27-12204
161818-186
270-594-4324752
220-594-2644488
=61818-186
=0-594-4324752
=00168-264
 
Hey guys, right I have finally solved this equation, solved it correctly first time and did it without referring to anything that could help me. I think a fresh mind and plenty of coffee has helped.

To make the original equation look clearer, I changed the I values to a, b and c as someone had already mentioned to do. Then I worked out as follows.

a + 8b + 3c = -31 (EQ1)
3a - 2b + c = -5 (EQ2)
2a - 3b + 2c = 6 (EQ3)

Multiply EQ1 by -3 which gives -3a - 24b - 9c = 93 then add to EQ2 which cancels out a. We now have

a + 8b + 3c = -31 (EQ1)
-26b - 8c = 88 (EQ2)
2a - 3b + 2c = 6 (EQ3)

Multiply EQ1 by -2 which gives -2a - 16b - 6c = 62 then add to EQ3 which cancels out a. We now have

a + 8b + 3c = -31 (EQ1)
-26b - 8c = 88 (EQ2)
-19b - 4c = 68 (EQ3)

Multiply EQ3 by -2 which gives 38b + 8c = -136 then add this to EQ2 which cancels out c and leaves us with

12b = -48
b = -48/12
b = -4

We can now solve for c

-26b - 8c = 88
(-26 x -4) - 8c = 88
104 - 8c =88
-8c = 88 - 104
-8c = -16
c = -16/-8
c = 2

We can now solve for a

a + 8b + 3c = -31
a + (8 x -4) + (3 x 2) = -31
a -32 + 6 = -31
a - 26 = -31
a = -31 + 26
a = -5

Solutions are

a = -5
b = -4
c = 2

So referring to my original post

I1 = -5
I2 = -4
I3 = 2

I think I have finally got it!!!

But no doubt I'll be back again soon with another problem!
 
Subhotosh Khan said:
If I1, I2 and I3 are currents in a circuit,

what does "negative I1 (-5)" mean in this case?
Click to expand...
Entering is positive, leaving is negative or vice versa I can't remember the convention. Never worked much with circuit analysis.
 
Ishuda said:
Entering is positive, leaving is negative or vice versa I can't remember the convention. Never worked much with circuit analysis.
Click to expand...

I know the answer - but I wondered if the OP asked that question after the solution!
 
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