# Currents and resistors and series and parallel you get the idea

1. Feb 28, 2005

### thisisfudd

So this problem asks you to determine the magnitudes and irections of the currents in the resistors. I definitely understand how to determine directions, using Kirchhoff's Rules. But here is where I get stuck: the batteries have emfs of E1 = 9.0 V and E2 = 12.0 V and the resistors have vlaues of R1 = 25, R2 = 18 and R3 = 35.

So it looks like this (ignore the DOTS i was trying to separate the lines)

______R1______
|.....................|
E1...................|
|_____R2______|
|.....................|
E2...................R3
|_____________|

Sorry I don't know how else I would go about drawing that

I figured I could use I = E/(R +r) but I get answers that are quite different from the corrected version.

Last edited: Feb 28, 2005
2. Feb 28, 2005

### StatusX

Since this is a linear circuit, you can find the contribution from each voltage source and then add them. Replace one of the batteries by a wire and calculate the current and voltage where ever you need it. Then put that one back and short out the second one, find its contribution, and add it to the first to get your answer.

3. Feb 28, 2005

### Davorak

I can not tell what you have done as of yet, so the general approach is to use Kirchhoff's voltage law in both loops. This is called mesh analysis.

Or you can use superposition where you let one voltage source act as an short circuit, figure out the voltage/currents. Then let the other voltage source act as a short circuit find the voltage/currents. Then add the two answers together. This is what StatusX recommended.

Or you could use node voltage analysis.

I recommend one of the first two if you have not heard of nod voltage analysis.

The resistors are neither in series or parallel that is why you can not reduce the circuit further.

4. Mar 1, 2005

### thisisfudd

OK I still do not understand AT ALL. I don't understand how I am supposed to add volts and ohms together. I tried setting up two equations whose sums equal zero and then substituting, but it didn't work?

9 V + (I1 x 25 ohms) + (I1 + 18 ohms) = 0
9 V + 12 V + (I1 x 25 ohms) + (I2 + 35 ohms) = 0

Umm, it doesn't work though. Could someone give me a shove in the right direction?

5. Mar 1, 2005

### StatusX

amp-ohms is the same unit as volts. Thats why V=IR can be true.

6. Mar 1, 2005

### thisisfudd

Yeah, I get that. But could you tell me if I am going in the right direction? Because I tried doing that and my answers are far larger than those I expect.

7. Mar 1, 2005

### StatusX

the current over R2 isn't the same as the current over R1, it splits at the node between them. use the bottom loop to get a third equation.

8. Mar 1, 2005

### thisisfudd

Ugh, I am totally stressed out by this. Here is what I've done:

9V + 25I1 + 18I2 = 0
12V + 9V +25I1 + 35I3 = 0
12V + 35I3 + 18I2 = 0

I2 = (-9 - 25I1) / 18 = -.5 - 1.39I1
I3 = (-21 - 25I1) / 35 = -.6 - .71I1

12 + 35 (-.6 - .71I1) + 18 (-.5 - 1.39I1) = 0
-21 - 25.56I1 - 9 - 25.02I1 = 0
I1 = .36

And yet, this is incorrect.

9. Mar 1, 2005

### StatusX

keep in mind that the voltage drop in one direction is negative the drop in the opposite direction. specifically, pay attention to the sign of I2, because your loops go over R2 in different directions.

10. Mar 1, 2005

### Gamma

This last of the three equations, I think that's a minus sign at 18I2

12V + 35I3 - 18I2 = 0

11. Mar 1, 2005

### thisisfudd

OK, I think I get what you're saying. So I did this:

9V + 25I1 + 18I2 = 0
21V + 25I1 + 35I3 = 0
12V + 35I3 - 18I2 = 0

I1 = (-9 - 18I2) / 25 = -.36 - .72I2
I3 = (-12 + 18I1) / 35 = -.34 + .51I2

21 + 25 -.36 - .72I2) + 35 (-.34 + .51I2) = 0

I2 = .67

Does this look better? I know it's still not the right answer ...

12. Mar 1, 2005

### Davorak

I can not tell from what you have written exactly what went wrong. Which way do the batteries orientated. Both negative terminal up, or both negative terminal down. For your top and bottom equations to be true the negative terminals must be point toward or way from each other. This is because you have I2 from each loop pointing in the same direction and you have both batteries are adding to the equation. This is assuming you defined the current going out of the positive terminal, that is you did not assume the battery was charging.

It looks like you went for mesh current analysis. Each loop is considered to have its own currents until you add them together in the end. So your I2 in the top equation should be considered different from I2 in your bottom equation. Here is a link on the basic circuit analysis methods I mentioned before, including mesh.

http://www.eas.asu.edu/~holbert/ece201/recipes.html

Last edited by a moderator: Apr 21, 2017
13. Mar 1, 2005

### thisisfudd

Hi, ok so if I2 in the top and bottom equations are different, how can I solve for, for instance, I2? Sorry, I should have mentioned that the positive terminals point outward on each battery. So given that, I'm really not sure how I keep messing this up.

14. Mar 1, 2005

### Davorak

The whole idea of mesh analysis is that one current is going around one loop and another current is going around the other. In the middle you use both currents together.
$$9V = I_1 R_1 +R2(I_1 + I_2)$$
$$12V = I_2 R_3 + R2(I_1 + I_2)$$

Edit:
Sorry what I meant by added the currents together at the end may have been confusing hopefully these equations clear things up.

15. Mar 1, 2005

### StatusX

I'm sorry, I should have said your third equation is I1=I2+I3. (this is what davorak is talking about.) Actually, the big loop should have given you a redundant equation (it is just the equation for the top loop added to the one for the bottom loop, since the R2 term cancels), so I'm not sure how you got an answer out of it.

16. Mar 1, 2005

### thisisfudd

So is the current "in the resistor" the same as the current going around the loop?

17. Mar 1, 2005

### Davorak

This is true for R1 and R3, but R2 as current from both loops going through it.

18. Mar 1, 2005

### thisisfudd

So using these formulas I get the right answer for the first resistor. But when I plug back in to get I2, I instead get the current for resistor number 3. Any ideas?

19. Mar 1, 2005

### Davorak

My I2 and I1 are not your I2 and I1. I only have I2 and I1 in the whole circuit while you have I1, i2, I3.

My I1 is your I1, my I2 is your I3, my I1+I2 is you I2. Take a good look at my equations and the site that I gave a few posts back when you get the chance.