# branch current method

Discussion in 'Math' started by ukiceman, Jan 12, 2010.

1. ### ukiceman Thread Starter New Member

Nov 22, 2009
9
0
Hi Guys

Im on Volume I - DC » DC NETWORK ANALYSIS and Ive just started the
branch current method

towards the end just before the answers for current

The text says For some methods of solution (especially any method involving a calculator)

Because Im just doing this as a hobby and, Im not young enough for a job in the field, this is always my methods of solution.

Im having a stupid moment today, could somebody tell/show me how I crunch/form this into a calculator to get the answers I 1=5amp I 2=4amps and I 3 = -1amps as shown in the text.

Shaun

File size:
21.4 KB
Views:
38
2. ### Fraser_Integration Member

Nov 28, 2009
142
5
First equation, make I1 the subject, and then substitute that value into the second and third equations wherever you see I1.

When you have done that, make I2 the subject of the second formula, and substitute that into the third formula, and you then have a formula with just I3, then work backwards replacing I3 with the value you worked out.

3. ### ukiceman Thread Starter New Member

Nov 22, 2009
9
0

Im now 9 hours into this equation and Im still stuck.. Im obviously missing something, but I cant see what? Maybe a brain!

You say First equation, make I1 the subject, and then substitute that value into the second and third equations wherever you see I1

From somebody who failed math at school, if I substitute I1 into the second and third equations wherever I see I1, I still end up looking at something that looks exactly as it does now, totally confused.

Yet, if I see something worked calculator fashion Im ok when I come across a similar problem because, I just apply the same logic inputting it into the calculator and come up with the correct answers.

I know its probably frustrating dealing with people like me but, Im happiest using a calculator because at least I can continue turning pages

Last edited: Jan 12, 2010
4. ### Fraser_Integration Member

Nov 28, 2009
142
5
it's very difficult to write it all out, subscript and all, but if you make I1 the subject in equation 1, then I1 = I2 - I3

Sub that into eqn 2, noticing that there is no I3 becuase it has a zero before it.

4(I2 - I3) + 2I2 = 28

this expands to 6I2 - 4I3 = 28

then make I2 the subject, this is (28 + 4I3) all over 6

now sub that into the last equation, there are too many brackets to do it, but just write it as you see it, and use your calculator if you need to!

5. ### thyristor Active Member

Dec 27, 2009
94
0
A slightly easier method:

From Eq.1 we know I2 = I1 + I3 let this be Eq.4

Then if we add Eqs. 2 & 3 we get:

4I1 - I3 = 21

therefore I3 = 4I1 - 21

Substituting this into Eq.4 gives

I2 = 5I1 - 21 call this Eq.5

Subtract Eq.3 from Eq.1 to get

-I1 + 3I2 = 7 which rearranges to give

I2 = (I1 + 7)/3 Eq.6

therefore from Eqs.5 & 6 we get

5I1 - 21 = (I1 + 7)/3

which means that 15I1 - 63 = I1 + 7

which gives I1 = 5

therefore from Eq.5, I2 = 5I1 - 21 = 4

and from Eq.4, I3 = I2 - I1 = 4 - 5 = -1

QED

6. ### AlexR Well-Known Member

Jan 16, 2008
735
54
All fairly straight forward but what I would imagine the text alludes to is the fact that many scientific calculators these days have an inbuilt facility for solving simultaneous equations. You just plug in the coefficients and out pops the solution.

7. ### Fraser_Integration Member

Nov 28, 2009
142
5
true but that is basic Maths. GCSE level (age 16) here in UK.

8. ### ukiceman Thread Starter New Member

Nov 22, 2009
9
0
Sorry for not saying thanks for the posts, not been around for a few days.
After Fraser_Integration made his second post it made a bit more sense but, thyristors reply nailed it. Thanks guys