# I need some help with a circuitry question

1. Apr 3, 2004

### cytokinesis

I'm gonna feel kinda dumb once the answer to this question is revealed. But I cant seem to get the proper answer.

The question is: What are the values of I1, I2, and I3 in the circuit?

The ciruit looks like this:
20.0V I=positive terminal
|-------I|-------|
| |
A I1 A I3
| |
| |
/ R1 |
\ 6.0? |
/ |
\ |
| |
|----v^v^v----|
| R2 2.0? A I2
| |
|----v^v^v----|
R3 4.0?

A = Ammeter

Okay. I have no difficulty finding the amount of current at points I1 and I3, as they are equal, and they both work out to 2.7A.

The difficulty lies in finding I2. The answer according to the book is 0.90 A and no matter how hard I try and how many routes i take to get there, I always wind up with the wrong answer. I'm beginning to think perhaps it's that the book's answer is wrong because the most logical answer I got was 1.0 A.

Any help would be nice! Thanks!

2. Apr 3, 2004

### cytokinesis

wow it really doesnt like that diagram.

I will draw one up, upload it to my website and link to it from there.

Here it is, please excuse the crude size, i did it in AppleWorks paint and I dont have a proper image cropping program here

http://cytokinesis.ath.cx/circuit.jpg [Broken]

Last edited by a moderator: May 1, 2017
3. Apr 3, 2004

### Chen

Code (Text):
20.0V  I=positive terminal
+-------I|-------+
|                |
A  I1            A  I3
|                |
|                |
/                |
\   R1           |
/  6.0?          |
\                |
|                |
+-----v^v^v------+
|    R2  2.0?    A   I2
|                |
+-----v^v^v------+
R3  4.0?

4. Apr 3, 2004

### Chen

The answer is not 1A. R2 and R3 are connected in parallel, right? So the potential drop on each of them is qual:
$$V_2 = V_3$$
The current through R3 is I2, can you see that? As for the current through R2, it is I1 - I2 (because I2 + the current through R2 equals I1). So we get:
$$R_2(I_1 - I_2) = R_3I_2$$
Solve for I2 and you're done.

5. Apr 3, 2004

### cytokinesis

Wow that helps, thanks a lot