# Network Current/Voltage Relationship

1. Jun 16, 2010

### jegues

1. The problem statement, all variables and given/known data
See Figure

2. Relevant equations

3. The attempt at a solution

I honestly don't know how to start this problem, I tried cutting out the current source and I'm thinking I should get something like a voltage divider. Does that help in any way?

A slight push to get things rolling would be great!

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2. Jun 17, 2010

### n.karthick

Why can't you try to use superposition theorem? (one source at a time and finally combine effects of all sources)

3. Jun 17, 2010

### Staff: Mentor

You can start by plotting a couple discrete points on the graph to start to get a feel for what is going to happen. What is Iin if Vin=0? What is Iin if Vin=V (why does the voltage source have two negative "-" terminals?) ? What is Iin if Vin=(-V)?

Then, I'd just write the KCL equation for the input node, and look to see if you can fill out the rest of the plot using that.

4. Jun 17, 2010

### xcvxcvvc

You could also transform the current source into a resistor in series with a voltage source. Then, you'd have a simple set of mesh equations that describe the system.

5. Jun 17, 2010

### jegues

I'm confused as to what you're refering as Vin. Is that the Voltage across the 2 ports? Or the voltage from the voltage source?

I'm going to assume it's across the ports and try to answer the questions you posted.

If Vin = 0 then Iin = 0 as well no? (Ohm's Law)

I'm not sure what Iin would be if you were to say Vin=V, Iin=V/Req?

The picture I have doesn't show the voltage source with two negative terminals?

If Vin=-V then Iin= -V/Req?

I trying to get my scanner working so I can show you guys what I've got so far using superposition.

Any more help would be greatly appreciated.

Thanks for the responses so far guys!

6. Jun 17, 2010

### jegues

Here is a sketch of my attempt at superposition, I'm still not sure how to work out my answers from this.

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7. Jun 17, 2010

### xcvxcvvc

[PLAIN]http://img12.imageshack.us/img12/8938/53238315.jpg [Broken]

Don't you think it's easier to write two mesh loops with this set up and then use the system of two equations to solve for I1 in terms of some arbitrary Vin?

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8. Jun 17, 2010

### jegues

I've never learnt how to do that. I'm trying to learn circuit analysis on my own so I may be missing out of some stuff.

EDIT: Can someone tell me if my current efforts at using superposition are correct?

Last edited by a moderator: May 4, 2017
9. Jun 17, 2010

### Staff: Mentor

Yes, I did mean for Vin and Iin to be at the single port at the left. And no, Iin is not zero when Vin is zero. Why not?

10. Jun 17, 2010

### jegues

I'm not really sure.

By looking at the current source I can see the that current will split between two wires and run across R2 and towards the node where i is entering. I'm guessing if part of the current from the current source is entering that node then there must be some current leaving as well. (That would be Iin?)

I'm still not sure how to figure out Iin. As soon as you told me Vin = 0 here is my thought process:

Okay, Ohms law says V = IR, since V = 0; 0 = IR, we know R won't be zero so I must be 0 so I can obtain 0=0.

What part of my understanding do I need to corrected?

11. Jun 17, 2010

### Staff: Mentor

Just picture the Vin=0 case as if you shorted the input terminals. You are correct that there will be no current through the resistor R1, since there is no voltage across it. But what current flows through your short circuit wire? There's current coming from the voltage source on the left through its resistor, and the current source in the middle is an ideal current source -- it puts out its current no matter what the voltage is across it...

12. Jun 17, 2010

### jegues

Wouldn't the current simply be as follows:

Iin = (V/R2) + I ?

13. Jun 17, 2010

### Staff: Mentor

Yes! Good. So that's one point on your graph.

Now set Vin = V (the same as the voltage source). What does that do to the current through the resistor that is in series with the voltage source? That should give you a 2nd point on the graph.

Then try writing the KCL for the input port. Does that help?

I have to bail for a few hours. Keep it up!

14. Jun 17, 2010

### jegues

So if I'm setting Vin = V, everything should be the same expect I have current across the first resistor R1.

So,

Iin = I + V/R1 + V/R2 ?

I can't imagine how to write a KCL at the input port I think I'm getting confused from the diagram.

If I have i running from my port into that first node I know it will split accordingly depending on the values of R1 and R2 as well as with respect to the current source running into that node as well.

15. Jun 18, 2010

### Staff: Mentor

First, sorry, I missed a sign error in your previous post. When Vin=0, Iin doesn't equal I, it equals -I. Why?

Second, when Vin=V, why would there be current through R2? Remember, when writing the KCL, you are using voltage differences across resistors...

16. Jun 18, 2010

### jegues

Alrighty, I was getting real confused so I decided to start from scratch and do a mesh analysis of this circuit and here's what I got. (See Figure)

Now if that is indeed correct how do I go about plotting the i-v curve from that?

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