Question

Current Electricity

In the circuit of figure, find the current in \(20 \ohm\) Resistor by Kirchoff's Law.

Answer 1

@Miracrown @hartnn @BSwan @Kainui @phi

Answer 2

Sorry, I don't know how to write Ohm in LaTeX .

Answer 3

Oh, I missed something.

Answer 4

so first lets use the juntion rule ok?
so imput= output of electrical curret

Answer 5

Well, I'm making arrows currently .. thats quite tough to do :P

Answer 6

choose your favorive juntion

Answer 7

how about F ?
so lets choose this direction

Answer 8

This is what I have now.

Answer 9

Have I done any mistake there in marking that mesh currents?

Answer 10

http://screencast.com/t/hQt5jrSp2geN

ok so 6 different current pathways
so we can already create some junction pathway equations
so at F what equation can we get?

Answer 11

Hmm , first I will have to check those good looking arrows of Miracrown with mine.

Answer 12

@BSwan everyone is different, so don't expect everyone to be the same!
And @mathslover it looks good but very confusing with the currents

Answer 13

I suggest you label each current separately

Answer 14

I wish we had a \circuitikz ... this would have made understanding a lot much easier.

Okay, Mira, as you say! I will recheck mine.

Answer 15

@Miracrown

Answer 16

Sorry,

for taking such a long time.. but the software I used was a little bit nasty

I couldn't label individually as it was taking a lot of time :(

Answer 17

Will you check them please? Are they rightly labelled?

Answer 18

that's alright! - Sure. One sec..

Answer 19

AS I had not marked A,B,C ... there. So, just edited it.

Answer 20

so for the loop A
what equation do we get?

Answer 21

and great job!

Answer 22

For loop : ABFEA

we have :

\(\epsilon - (i_1 + i_2) R_1 - (i_1 - i_3 )R_3 = 0 \)

Is that right?

Answer 23

correct! :)
ok so for loop B what do we have?

Answer 24

Oh.. that was right? Great! I'm not used to make such equations..

Answer 25

you're doing good so far !

Answer 26

Loop B : \(\bf{BFDCB}\)

\(-(i_1 + i_2 ) R_1 - (i_2 + i_3) R_4 - i_2 R_2 - \epsilon _ B = 0\)

\(\epsilon B = 120 V\) and in the first equation for loop A , \(\epsilon = 45 V\)

Is that right?

Answer 27

What I just follow is :

If the direction of current flowing through resistance is same to the direction of current flowing through the loop then I just need to put it like this \( \bf{-} iR\)
and when the direction is opposite then the sign changes to \(\bf{+}\) iR

And for the batteries, if the direction of the loop ends up at - terminal , then it is \(-\epsilon\) and when the direction of the loop ends up at + terminal, then it is \(+ \epsilon\)

Is this concept right?

Answer 28

you were close
it is a positive eB for the second equation

http://screencast.com/t/efItClok2nOl

Answer 29

So, the concept I stated above for the batteries should be reversed?

Answer 30

yes that is correct !
ok so now you have 2 euqations and 3 unknowns
so all you need now is 1 more equaiton
and you can use the junction rule for that

Answer 31

nevermind
you need to do the third loop

Answer 32

so how is the third loop equation?

Answer 33

we hv more than 3 loops here

Answer 34

So, third loop is actually EFDHGE

\((i_1 - i_3) R_3 - (i_2 - i_3 ) R_4 - i_3 R_5 + \(\epsilon _ C\) = 0 \)

Answer 35

o.O sorry ... I meant : \(\epsilon_C = 0 \) ..

Answer 36

@shamim -> Yes, there are more than 3 loops, but I think, getting 3 equations will be enough.

Answer 37

ya

Answer 38

right enough is enough :P

Answer 39

Mira, is the 3rd eqn right?

Answer 40

alright great!
we have 3 equations to solve for 3 unknowns
plug in values and simplify each equation ok?

Answer 41

Oh I did the mistake there... \(i_2 + i_3\) ... not \(i_2 - i_3\)

Answer 42

yes

Answer 43

Okay, so, this will be my test for algebra skills..! right? ;)

Answer 44

haha yeah :P

Answer 45

but easy once you get to matrix form

Answer 46

http://screencast.com/t/tqNHwSWHf

Answer 47

its mess, I know. But am to tired :(

Answer 48

http://screencast.com/t/XclXTX48r

Answer 49

so what do you get?

Answer 50

Its really complicated... just give me 30 sec.

Answer 51

i got it
http://screencast.com/t/bGCMnWPX1T

Answer 52

so what is the current though the 20 ohm resister?

Answer 53

Naah... Is that your answer?

i1 = -2.5
i2 = 4.6
and i3 = -2.8 ?

Answer 54

ok so from the diagram which current flows through the resister?

Answer 55

i1 + i2 will flow through the required 20 ohm resistor.

But, if those are your values, unfortunately, those don't match the books' one.

Answer 56

no... look at the question
what is the question?

Answer 57

hmmm
what does the book give?

Answer 58

The question says to find the current in 20 ohm resistor by Kirchoff's law.

So, the current flowing through it is i1 + i2...

and the book gives

i1 = 3.63 and i2 = 0.767

Answer 59

Oops .. i2 = -0.767

Answer 60

Though, my answer was very close to the books' one ... But I think, we did a mistake somewhere. I will restart the whole process, and check what I get.

Answer 61

I know this, I'm half asleep unfortunately >.<. Sorry.

Answer 62

I think you can take it from here and figure out the issue, I need to go.
Good luck!

Answer 63

No worries! I'll try to solve it. And will post what I get.

Answer 64

Sure. Bbye! Good Night.

Answer 65

Thanks.

Answer 66

Here is the solution I reached:

Answer 67

@Vincent-Lyon.Fr Thanks for such a nice and clear solution, but unfortunately , the answer you got doesn't match with the answer given in the book.

The answer given by the book is 2.866 in the correct direction (\(\downarrow\) )

Answer 68

@Miracrown @vishweshshrimali5

Answer 69

Current division at junction E is wrong.

Answer 70

Oh, why so?

Answer 71

Similarly at F

Answer 72

Okay see:

For junc. E:

Input current = \(i_3\)

Output current = \(i_3 + i_1\)

They are not equal

Answer 73

Similarly at F:

Input = \(i_3 + i_1 + i_2\)

Output = \(i_2 + i_3\)

Answer 74

They are not equal

Answer 75

You're seeing this - http://assets.openstudy.com/updates/attachments/5390610be4b0172a938e56b8-mathslover-1401972700846-problemele.c2.jpg ?

Answer 76

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
This is what I have now.
\(\color{blue}{\text{End of Quote}}\)

Answer 77

http://assets.openstudy.com/updates/attachments/5390610be4b0172a938e56b8-mathslover-1401971695233-problemelec..jpg

Answer 78

Please delete the rest.

Answer 79

Okay, sorry, yeah that was wrong.


The correct and final one is -
http://assets.openstudy.com/updates/attachments/5390610be4b0172a938e56b8-mathslover-1401972700846-problemele.c2.jpg

Answer 80

Okay

Answer 81

Current division is correct.

Answer 82

Now what is the problem?

Answer 83

The whole problem is that we're unable to get correct answer for \(i_1 + i_2\)

Answer 84

And what are your equations or answers till now.

Answer 85

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
For loop : \(\bf{ABFEA}\)

we have :

\(\epsilon - (i_1 + i_2) R_1 - (i_1 - i_3 )R_3 = 0 \)

Is that right?

Loop B : \(\bf{BFDCB}\)

\(-(i_1 + i_2 ) R_1 - (i_2 + i_3) R_4 - i_2 R_2 - \epsilon _ B = 0\)

\(\epsilon B = 120 V\) and in the first equation for loop A , \(\epsilon = 45 V\)

Is that right?

So, third loop is actually \(\bf{EFDHGE }\)

\((i_1 - i_3) R_3 - (i_2 - i_3 ) R_4 - i_3 R_5 + \epsilon _ C = 0 \)
\(\color{blue}{\text{End of Quote}}\)

Answer 86

Those are the equations that we got.

Answer 87

Second equation is wrong

Answer 88

It must be \(+e_b\)

Answer 89

Yeah.. miracrown corrected me there,

it is \(+\epsilon _B\)

Answer 90

Sorry...

Answer 91

Okay

Equation 1 is correct

Answer 92

Eq. 3 is wrong

Answer 93

It must be \(-(i_2+i_3)R_4\)