# VI Characteristic

1. Jan 28, 2008

### abeltyukov

Hi,

I have the following circuit: http://i137.photobucket.com/albums/q208/infinitbelt/ProblemSet3Circuit1.png" [Broken]

Also, the following is known:
X=60 V
R1= 10 Ω
R2= 20 Ω
R3= 10 Ω
R4=5 Ω
R5=20 Ω
R6 = 5 Ω

How do I go about finding a V-I characteristic of this circuit? I know that the slope of the graph is the resistance but I am confused as to how to start this one. I found the resistance across the source by using circuit reduction. I got that to be 32 ohms. So is the V-I characteristic just V = 32i? or is it just a constant line y = 60? (in this case V is on the y-axis and I is on the x-axis)

Thanks!

Last edited by a moderator: May 3, 2017
2. Jan 28, 2008

### Corneo

What exactly do you mean find the V-I characteristic? Your battery supply is 60V and you already can solve for V and I.

3. Jan 29, 2008

### CEL

If you vary the voltage V, what is the variation of the current I?
The question was also posed at the homework forum, so it should be deleted from this forum.

4. Jan 31, 2008

### TVP45

Is something missing or perhaps drawn incorrectly? The problem looks odd and I can't figure out the purpose of that ground (just something to throw you off?).

5. Jan 31, 2008

### CEL

The ground has no meaning at all, since the input V is taken across R5.

6. Jan 31, 2008

### TVP45

Well, I think I agree that the ground has no meaning, but why is it there? Is this just supposed to be tricky? And, why the very strange configuration of R1, R3, and X? Again, is this just tricky, or is something missing?

7. Jan 31, 2008

### CEL

R1, X and R3 are in series (forget the ground), so you can draw the characteristic of this sub circuit (call it I).
I is in parallel with R3 and you can draw the characteristic of this new sub circuit (call it II).
II is in series with R4 and R6 (call this sub circuit III).
Finally III is in parallel with R5.

8. Feb 1, 2008

### TVP45

I know how to do circuit analysis; that wasn't my question.

9. Feb 1, 2008