# Homework Help: Using kirchoffs laws to find current

1. Feb 8, 2010

### victoriafello

Calculate the current in each of the resistors and indictate the direction

of current flow.

what is the potential difference between A & B

I have attached the circuit & labelled the the circuit directions

using kirchoffs laws -

1) i3 = i1+i2

FOR THE LOOP AFEBCD

2) 20i1 - 10 + 1 - 5i2 = 0

THEN FOR LOOP ABCD

3)-10i3 + 1 - 5i2 = 0

SUB FOR i3

-10(i1+i2) + 1 - 5i2 = 0

-10i1 -10i2 + 1 - 5i2 = 0

4)-10i1 + 1 - 5i2 = 0

SUBTRACT equation 2 - 4

20i1-9 - 5i2 = 0
-10i1 + 1 - 5i2=0

10i1 + 10 = 0

i1 = 1

THEN SUBSTITUTE FOR i1 in equation 4

-10 + 1 - 5i2 = 0

9 - 5i2 = 0

SO

i2 = 1.8

Then from E1 i3 = i1 + i2

i3 = 2.8

hopefuly this is correct, i'm not sure how i go about finding the potential difference between the two points ?

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2. Feb 8, 2010

### thebigstar25

can you please indicate what is the direction of each loop in your solution (clockwise or counterclockwise) .. so I can follow your solution and check ..

3. Feb 8, 2010

### ehild

You made a mistake from here

-10i1 -10i2 + 1 - 5i2 = 0--->

4)-10i1 + 1 - 15i2 = 0 : this is the correct equation.

The potential increases from B to A as the current i3 flows from A to B. The change of the potential across the 10 ohm resistor is 10*i3: U(A)=10*i3+U(B). The potential difference between A and B is U(A)-U(B)=10*i3.

ehild

4. Feb 9, 2010

### victoriafello

The left hand loop is clockwise running EFAB
the right hand loop is counterclockwise running DABC

ok if i use the above equation then i get stuck trying to find a value for i1

4)-10i1 + 1 - 15i2 = 0

equation 2 - 4

20i1-9 - 5i2 = 0
-10i1 + 1 - 15i2=0

10i1 + 10 -20i2 = 0

how do i proceed as I can’t get cancel out the i1 or i2 to get a result

Last edited: Feb 9, 2010
5. Feb 9, 2010

### ehild

There are several ways to proceed. For example, you can multiply eq.(4) by 2 an then add it to eq (2):

-20i1 + 2 - 30i2=0 (4a)
20i1-9 - 5i2 = 0 (2) (2)
_____________________

-7-35i2=0

ehild

6. Feb 9, 2010

### victoriafello

Ok so rearrange

-35i2 = -7

i2 = 5

then substitue to find i1

10i1 + 1 - 15*5 = 0

10i1 -59 = 0

i1 = 5.9 ?

i3 = 10.9

i am really confusing myself now,

7. Feb 9, 2010

### ehild

All wrong. Amazing!

You have the equation:

-7-35*i2=0

add 35i2 to both sides (the equation stays valid if you add, subtract, multiply or divide both sides with the same quantity.)

-7=35*i2

Divide both sides by 35:

-7/35 = i2 --->i2=-0.2 A

Could you follow?

ehild

8. Feb 9, 2010

### rl.bhat

-35i2 = -7
This is wrong. It should be
35i2 = -7

9. Feb 9, 2010

### victoriafello

ok i follow that, sorry for all the mistakes but i am only learning !!!

Now I need to get i1 so if I use E2

20i1 – 9 – (5*-0.2) = 0
20i1 – 9 – 1

20i1 – 10 = 0

20i1 = 10
Divide by 20

So i1 = 0.5A

And therefore i3 = i1 + i2

0.5A + -0.2A = 0.3A

i have checked this though so i hope no more errors

10. Feb 9, 2010

### ehild

How old are you?

20i1 – 9 – (5*-0.2) = --->

5*(-0.2)=-1

-(-1)=+1

20*i1-9+1=0

20*i1 - 8 =0

20*i1 = 8

i1 = 8/20

i1 = 0.4 A

ehild

11. Feb 9, 2010

### victoriafello

why does that matter - i'm 14 ?

so i make mistakes isnt that why i asked for help, if i knew the answer i wouldnt be here would i ?

thanks for the help (and making me feel bad) anyway

12. Feb 9, 2010

### ehild

Sorry, I did not want to make you feel bad. I asked your age because you know a lot of Physics for your age, so I wondered why do you make so silly mistakes in Maths. Forgive, me please...

ehild

13. Feb 9, 2010

### victoriafello

Thats ok,
Thanks for all the help

14. Feb 20, 2010

### sand

Hi. I'd like to ask a question about this. In setting up the equations 2 and 3, I don't really get why the signs of the terms are what they are. For example, in equation2), 20i1 was positive but 5i2 was negative. And then in equation 3 they are both negative. When I had a go at this from scratch, I had the signs in equation 2 the other way around and needless to say i got different answers. I had the direction of the current just as victoriafello did too.
Any clarification is appreciated.

15. Feb 21, 2010

### tomwilliam

If you assign directions to the current, then going round them you have to respect the directions you've allocated. If you go round the loop through a resistor following the same direction as the current, the value will be negative (-i x R) because that's what a resistor does. If you go the other way round when calculating the loop, you have to make the sign positive. When you go through a cell or battery you have to respect the potential difference. When you go from the negative terminal to the positive terminal you use the positive V and negative V for the other direction.
Is that clear?

16. Feb 21, 2010

Thanks