# Problem with Kirchhoff

1. Jan 21, 2010

### fluidistic

1. The problem statement, all variables and given/known data
See picture for the problem.
1)Calculate the work done to move a charge a= 0.7 x 10 ^(-3)C from A to B, passing by R2.
2)The power dissipated in R3.
3)The current passing by R1.

2. Given equations
R1=R3=100 omhs, R2=200 ohms, R4=300 omhs.
e1=e3= 5 V, e2=10 V and e4=15 V.

3. The attempt at a solution
I realize that the problem is very simple if only I get the current passing through each branch of the circuit. However, applying Kirchhoff's law of voltage, I reach non sense.

More precisely, let i1 be the current passing through e3, let's suppose it in the clockwise direction. Let i3 be the current passing through the e1 branch, suppose its direction as down. Let i2 be the current passing through R3, anticlockwise.
Then I get, according to Kirchhoff's law of voltage:
$$\varepsilon _3 - \varepsilon _2 - \varepsilon _1 - i_1 R_2 -i_3 R_1=0$$ and $$\varepsilon _4 - i_2(R_3+R_4)- \varepsilon _1 - i_3R_1 =0$$.
From it I reach $$i_1= -20V+400 i_2$$. Of course units cannot match so there's at least an error. But I don't know where. I don't see it in the arithmetic so I guess I've used wrongly Kirchhoff's law, but I don't see it...
Thanks for any help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jan 21, 2010

### rl.bhat

The equation becomes
200*i1 = -20V + 400*i2.

3. Jan 22, 2010

### fluidistic

Ok thanks, I'm going to check it out right now. However assuming that you're right, I'm still stuck with the same problem. I have that a current is equal to a a constant times another current, plus a voltage. What am I misunderstanding? I mean a current should have at least one unit, and a current units one.
So I guess I made an error writing down Kirchhoff's law?

Edit: Ok, I reached the equation you provided, I forgot the 200 term in the last step.
But I'm still stuck, as stated above, in the same post. :)

Last edited: Jan 22, 2010
4. Jan 22, 2010

### rl.bhat

Consider i3 as (i1 + i2). The two equations become
-10 + 200i1 + 100(i1 + i2) =0 .....(1)
10 + 400i2 + 100(i1 + i2) = 0.......(2) Solve these two equations to find i1 and i2.

5. Jan 22, 2010

### fluidistic

Ok I will do it. I'm just curious about what I found, that a current is worth something senseless. I'm surprised there is no error in this. So you really confirm there is no error?

6. Jan 22, 2010

### rl.bhat

I don't understand what you mean by senseless.

7. Jan 22, 2010

### fluidistic

Ok. I reached the following equation: $$i_1=2i_2-\frac{1V}{10}$$. Although I realize I've not reached the final answer for $$i_1$$, I do realize that the left side of the equation is measured in amperes and that the right side of the equation is something measured in amperes + something measured in volts. It doesn't make sense to me. Could you please explain to me why it is possible? Or I'm missing something obvious?

I thank you for all your help and time, sir. I'm willing to learn as much as possible.

8. Jan 22, 2010

### rl.bhat

You can write 1V/10 as 1V/(10 ohm). In that case have you got any problem?
Any way how did you get
i1 = 2*i2 - 1V/10 ?

If you add the two equations
-10 + 200i1 + 100(i1 + i2) =0 .....(1)
10 + 400i2 + 100(i1 + i2) = 0.......(2)
you get
2*i1 + 3*i2 = 0

9. Jan 23, 2010

### fluidistic

Ah yes, you're right! I totally missed this part, I should have carried the units.
I reached this from
I was about to do this .

Thanks a lot for all. If I have any further problem I'll let you know.

10. Jan 23, 2010

### fluidistic

I still have some problems.

Are you sure you didn't make any error? In my first post I reached
$$\varepsilon _3 - \varepsilon _2 - \varepsilon _1 - i_1 R_2 -i_3 R_1=0$$, and
$$\varepsilon _4 - i_2(R_3+R_4)- \varepsilon _1 - i_3R_1 =0$$.
When I plug the values, I get these 2 equations:
$$-100(i_1+i_2)-10-200i_1=0$$, and
$$10-100i_2-100(i_1+i_2)-300i_2=0$$, which differ from your equations.
Solving for $$i_1$$, I reached $$i_1=-\frac{5i_2}{4}$$.

While I had reached $$i_1=2i_2-\frac{1V}{10 \Omega}$$.
So I can solve for $$i_2$$. I get $$i_2=\frac{2}{65}A$$. And thus $$i_1=-\frac{1}{26}A$$. Did I do something wrong?

How did you get -10 + 200i1 + 100(i1 + i2) =0 .....(1)
10 + 400i2 + 100(i1 + i2) = 0.......(2) ?

11. Jan 23, 2010

### rl.bhat

How did you get -10 + 200i1 + 100(i1 + i2) =0 .....(1)
10 + 400i2 + 100(i1 + i2) = 0.......(2) ?

Yes. You are right.
It should be
+10 + 200i1 + 100(i1 + i2) =0 .....(1)
-10 + 400i2 + 100(i1 + i2) = 0.......(2) ?

12. Jan 23, 2010

### fluidistic

We are still not in agreement. :uhh: This problem is a headache!

13. Jan 23, 2010

### rl.bhat

I am getting i1 = 3/70 A and i2 = 2/70 A.
I think there is a problem in your sign convention. Will you state your sign conventions for Ε and I*R.
You have taken i1 in the clockwise direction. You have traversing the loop in the direction of i1. You have taken I*R product negative assuming here is a drop of potential. you have taken E3 positive, where as you are crossing it from positive terminal to negative terminal. Is it not a drop in potential?

14. Jan 23, 2010

### fluidistic

Right about the clockwise direction for $$i_1$$. I might be wrong of course, but as you said, I took $$\varepsilon _3$$ as a "climb of potential", i.e. not a drop. Should it be a drop of potential?
If so, then I should take $$\varepsilon _1$$ and $$\varepsilon _2$$ as a climb of potential instead of a drop of potential as I took.

15. Jan 23, 2010

### rl.bhat

According to the convention, current moves from positive terminal to negative terminal in the external circuit, and from negative terminal to positive terminal inside the cell. So we are assuming that +ve terminal is at the higher potential.
So if you cross the cell from +ve to -ve terminal then it is a fall in potential.

16. Jan 24, 2010

### fluidistic

Ok thank you for all. I have to restart the whole problem.

17. Jan 25, 2010

### fluidistic

I get $$i_1=\frac{3}{70}A$$ as you, but I get $$i_2=-\frac{1}{35}A$$ which differs from your value. The most extraneous thing is that I got $$i_1$$ via $$i_2$$.
I have that $$i_1=\frac{1}{10}+2i_2$$.

Anyway, let's assume you are right.
To answer the first question, is the work done $$q(\varepsilon _2 - i_1 R_2)$$?

2)Is the power dissipated in $$R_3$$ worth $$i_2 ^2 R_3$$?

In my result of a negative current, does that mean that it is positive but in the other than assumed direction?

18. Jan 25, 2010

### rl.bhat

i2 = 2/70 A = 1/35 A. I have not considered the sign of the current.
Other results are correct.
The negative sign indicates that the assumed direction of the current not correct.

19. Jan 26, 2010

### fluidistic

Thank you very much for all, once again. Problem solved!