Using kirchoffs laws to find current

[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 kirchhoffs 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 ?

 

[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 ..

 

[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

 

[victoriafello]

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 ..

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

You made a mistake from here

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

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

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

 

[ehild]

equation 2 - 4

20i1-9 - 5i2 = 0 (2)
-10i1 + 1 - 15i2=0 (4)

10i1 + 10 -20i2 = 0 Wrong

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

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

go ahead.

ehild

 

[victoriafello]

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

go ahead.

ehild

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,

 

[ehild]

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,

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

 

[rl.bhat]

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

 

[victoriafello]

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

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

Add 10

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

 

[ehild]

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 wrong

i have checked this though so i hope no more errors

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

 

[victoriafello]

How old are you?





ehild

why does that matter - I'm 14 ?

so i make mistakes isn't that why i asked for help, if i knew the answer i wouldn't be here would i ?

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

 

[ehild]

why does that matter - I'm 14 ?

so i make mistakes isn't that why i asked for help, if i knew the answer i wouldn't be here would i ?

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

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

 

[victoriafello]

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

Thats ok,
Thanks for all the help

 

[sand]

using kirchhoffs laws -

1) i3 = i1+i2

FOR THE LOOP AFEBCD

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

THEN FOR LOOP ABCD

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

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.

 

[tomwilliam]

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.

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?

 

[sand]

Thanks