V-i characteristics for voltage divider with current source

[CoolFool]

This is a problem from Foundations of Analog and Digital Electronic Circuits by Agarwal & Lang, I'm going through it by myself, I'm not in a course, so I'm depending on you friendly forum people for help! I really like this book but it has a lot of errors, perhaps this is another error.

1. Homework Statement

Sketch the i-v characteristics for the network. Label intercepts and slopes.

Homework Equations

The solution PDF for this textbook says that the answer is
i = (v/4) + (v/5 + 2)
i = (9/20)v + 2
So the slope is 9/20 and the i-intercept is 2 Amps.

The Attempt at a Solution

My answer is i = v/4 + 2, which I got from connecting a current source to the terminal and solving the circuit for i. So the slope is 1/4, the i-intercept is 2 Amps.

I tried simulating the circuit on partsim.com with a DC sweep and I think it confirmed my answer, but I don't really know my way around circuit simulators yet so maybe I messed up.

I understand that the solution PDF got its answer by plugging Ohm's law into i = i₁ + i₂, where i₁ is current through the 4 Ohm resistor and i₂ is current through the 5 Ohm resistor.

But didn't it get i₂ wrong? Shouldn't i₂ = 2 Amps, because the current through the 5 Ohm resistor goes directly into the ideal current source which is set at 2 Amps?

So in my view, the 5 Ohm resistor could be any resistance, it doesn't matter, because there will always be 2 Amps going through it. Am I correct? (Obviously for real life elements this wouldn't hold up but I'm talking ideal elements here). I still am not totally comfortable with the intuition behind ideal current sources so confirming/denying this would really help me out.

Thanks!

 

[gneill]

Your answer looks good to me.

 

[CWatters]

I agree. I looked at two cases..

Vin = 0 so no current flows through the 4 Ohm. Means Iin = 2A
Iin = 0 so all of the 2 Amp flows through the 4 Ohm. Means Vin = -8V.

Plot both points. Slope is 2/8 = 1/4.

 

[rude man]

My answer is i = v/9 + 10/9
Solved by summing currents to 0 at the output (top of the 5 ohm), & solving for i (and output voltage).
EDIT I drew the picture wrong.
Post #5 is correct. Interesting that the 5 ohm plays no part in the i-V characteristic.

 

[CWatters]

The 5 Ohm is horizontal. Do you mean the top of the 4 Ohm? I tried applying KCL to the node at top of the 4Ohm and get...

Let into the node be +ve then...

I_in - I_4ohm - 2 = 0......(1)

Apply Ohms law to the 4Ohm..

I_4ohm = V_in/4 ......(2)

substitute (2) into (1)

I_in - (V_in/4) - 2 = 0

Rearrange..

I_in = (V_in/4) + 2

Same result as my previous attempt. Slope = 1/4, intersects at +2A

 

[NascentOxygen]

The solution PDF for this textbook says that the answer is
i = (v/4) + (v/5 + 2)

This is wrong. It should be i = (v/4) + 2

The current through the 5Ω resistor is a fixed 2A. End of story.