Engineering The current in a circuit using nodal analysis

[Gliese123]

Homework Statement

Hi!

"Write an equation for I₁."

Here is the circuit:

That is the question. I suppose one could do this many ways but I've decided to do it with nodal analysis. The problem here is that we only got an ideal current source. I have actually no clue.

So, I'd try to apply Kirchoff's Current Law to do this.

Homework Equations

KCL:
∑ I_i = 0

The Attempt at a Solution

I know that one is to pick a ground node somewhere. Let's pick the lower one for example.
Then one is to identify all nodes, in this case there are just another one.
Then what? Sorry but my teacher isn't very good at explaining this, at least not so I'd understand.

Am I correct if I'm to assume that an ideal current source and a parallell resistance R could be simplified to just an ideal current source?

If one is to apply KCL then it would maybe be something like:

Am I on the right track? Some general advice to do this smooth is worth it's weight in gold!
Thank you! :D


The right answer for this question is: I₁ = -R₂I₀/R₁+R₂

 

[CWatters]

You seem to be on the right lines. Perhaps remember the voltage will be the same across all three devices as they are in parallel. Should give you some more equations.

Sorry I might not be around to follow up.

 

[Gliese123]

You seem to be on the right lines. Perhaps remember the voltage will be the same across all three devices as they are in parallel. Should give you some more equations.

Sorry I might not be around to follow up.

Ahh, thanks for replying anyway :) The current is so strange. My teacher said that it's good to imagine it like water and water tubes but I don't know how that would help me with the mathematical part. Lol.

 

[donpacino]

Ahh, thanks for replying anyway :) The current is so strange. My teacher said that it's good to imagine it like water and water tubes but I don't know how that would help me with the mathematical part. Lol.

water through tubes is a good analogy. If 1 gallon of water passes through a pump (your current source) in 1 second and splits and goes through 2 pipes (R and R2), you know that the sum of the water that passes through the 2 pipes in that 1 second will be 1 gallon.

 

[Gliese123]

water through tubes is a good analogy. If 1 gallon of water passes through a pump (your current source) in 1 second and splits and goes through 2 pipes (R and R2), you know that the sum of the water that passes through the 2 pipes in that 1 second will be 1 gallon.

Hi donpacino! :)
I've got the idea of it clear however how do the calculus...

 

[donpacino]

Hi donpacino! :)
I've got the idea of it clear however how do the calculus...

haha. okay!

so first I can tell you your answer is not correct :(
------------------------
The solution:

first look at ohms law
V=IRwe know the voltage across all three components are the same. we can find that voltage by taking the current through the current source (Io) and multiplying it by the equivalent resistance of the two resistors.

eq1: Req=____

i'll let you solve that. hint: they are two resistors in parallel

eq2: V=Io*Req

so using ohms law the current through R is

eq3: I1=V/R1

so combining eq2 and eq3 we get

I1=Io*Req/R1

so generally we get an equation for current division. In any system comprised of parallel resistors and a current source, you can find the resistance through any of the paths by using the following equation

Ix=Io*Req/Rx

does that make sense?

edit: in your case due to the direction of current your I1 will be inverted
also if you really wanted to do it through nodal analysis, you would need 2 equations

nodal analysis

Io=V0/R+Vo/R2
Vo/R=I1

 

[Gliese123]

haha. okay!

so first I can tell you your answer is not correct :(
------------------------
The solution:

first look at ohms law
V=IRwe know the voltage across all three components are the same. we can find that voltage by taking the current through the current source (Io) and multiplying it by the equivalent resistance of the two resistors.

eq1: Req=____

i'll let you solve that. hint: they are two resistors in parallel

eq2: V=Io*Req

so using ohms law the current through R is

eq3: I1=V/R1

so combining eq2 and eq3 we get

I1=Io*Req/R1

so generally we get an equation for current division. In any system comprised of parallel resistors and a current source, you can find the resistance through any of the paths by using the following equation

Ix=Io*Req/Rx

does that make sense?

edit: in your case due to the direction of current your I1 will be inverted
also if you really wanted to do it through nodal analysis, you would need 2 equations

nodal analysis

Io=V0/R+Vo/R2
Vo/R=I1

eq1: Req=____

i'll let you solve that. hint: they are two resistors in parallel
Hmm. Let's see. We have a ideal current source and the current is splitted up like the water tubes we mentioned earlier. If I ain't misstaken one should use current division rules, obtain that very current through that resistance - in this case R (which is where I is) and then use ohm's law on that very place?

So if I₀ is the total current and I₁ is the sought current, then:

I₀ = RI/R₂+R which is that very current. Now if I ain't completely lost we could take that whole formula and *bang* - use ohm's law? Or do I even want that? I'm so lost. hahha xD
:D

 

[Gliese123]

Seems like nodal analysis in this particular assignment is as wasteful as putting a V8 into a Smart car?
hahah. It's basicall just one thing to do correct?

 

[Gliese123]

HOWEVER since the current that is asked for is pointed at the opposite direction, it should therefore be negative correct? I just want be certain of what I do ;)

 

[donpacino]

I think you have a fundamental misunderstanding of how this works.

First you would write the Kcl
I1=current through R
I2=current through R2
Io=source current

Io+I1+I2=0

then find an expression for Vo
I1=-Vo/R
I2=-Vo/R2
Io=-Vo/R-Vo/R2

then solve for Vo
Vo=-Io/(1/R+1/R2)
Vo=-Io*R1*R2/(R1+R2)

then solve for I1
I1=Vo/R
I1=-Io*R2/(R1+R2)

if you were to solve it using the equation i gave you
1/Req=1/R+1/R2
1/Req=(R+R2)/(R*R2)

then plug it into the equation
I1=-Io*R2/(R1+R2)

 

[Gliese123]

I think you have a fundamental misunderstanding of how this works.

First you would write the Kcl
I1=current through R
I2=current through R2
Io=source current

Io+I1+I2=0

then find an expression for Vo
I1=-Vo/R
I2=-Vo/R2
Io=-Vo/R-Vo/R2

then solve for Vo
Vo=-Io/(1/R+1/R2)
Vo=-Io*R1*R2/(R1+R2)

then solve for I1
I1=Vo/R
I1=-Io*R2/(R1+R2)

if you were to solve it using the equation i gave you
1/Req=1/R+1/R2
1/Req=(R+R2)/(R*R2)

then plug it into the equation
I1=-Io*R2/(R1+R2)

Thank you! Yes it seems I have misunderstood. I have just started with this so it's a quite complex for me. However I think I know what I need to do to make it work.