Norton equivalen circuit problem

[terryds]

Homework Statement

Determine the norton equivalent Circuit

Homework Equations

Norton theorem

The Attempt at a Solution

First, I transform 5A and 10 ohm (in parallel) into 50 V and 10 ohm in series

So, there are three meshes,
for Mesh 1 (I assume the current is clockwise)

##30 i_1 - 10 i_2 = -20##

for Mesh 2

##40 i_2 - 20 i_3 = 50##

for Mesh 3

##40 i_3 - 20 i_2 = 20##Solving those three equations, I get ##i_1 = 0 A##, ##i_2 = 2 A##, ##i_3 = 1.5 A##

So, I_norton is i1 + i2 - i3 = 0 + 2 - 1.5 = 0.5 A

But, this is wrong answer. Please tell me where I got this wrong.

 

[scottdave]

I cannot figure out where you got the equations for the Mesh Analysis. Where are your loops? Which is i₁ i₂ and i₃? Can you label your picture? Getting zero for one of the currents is possible in some circuits, but I don't think this is one of them (I haven't actually gone through the analysis, but just from looking at the voltages, it doesn't look likely). You may want to revisit the steps for Mesh Analysis. Here are a couple of sources that may help. https://en.wikipedia.org/wiki/Mesh_analysis
https://ocw.mit.edu/courses/electri...pring-2006/lecture-notes/nodal_mesh_methd.pdf

 

[cnh1995]

40i2−20i3=5040i2−20i3=5040 i_2 - 20 i_3 = 50

You didn't include the voltage drop due to i₁ in this equation. Other two equations look fine.

 

[scottdave]

So I went through your calculations, and yes as @cnh1995 pointed out, you did not include the -10*i1 in Mesh 2. But I worked it out, and I arrived at i1 = zero Amperes (imagine that!).

So you can get the Open circuit voltage at A, and Open Circuit voltage at B. Find the difference between these for Δv_ab. So to get the Norton current, you need to short terminals A and B together, then run the calculations to find the current from A to B. The Norton Resistance is (Δv_ab) / (Short Ckt Current from A to B).

 

[haruspex]

transform 5A and 10 ohm (in parallel) into 50 V and 10 ohm in series

I am no expert in this area, but I always understood that ammeters have negligible resistance. The 5A is the current not flowing through the 10 Ohm resistor, no?

 

[scottdave]

I am no expert in this area, but I always understood that ammeters have negligible resistance. The 5A is the current not flowing through the 10 Ohm resistor, no?

It is a 5Amp ideal current source. What @terryds did was replace a Norton circuit with a Thevenin Equivalent, which is a valid step in analysis. They behave the same.

 

[haruspex]

It is a 5Amp ideal current source. What @terryds did was replace a Norton circuit with a Thevenin Equivalent, which is a valid step in analysis. They behave the same.

Ok, thanks for the explanation.

 

[cnh1995]

But I worked it out, and I arrived at i1 = zero Amperes (imagine that!).

Yes, I got the same result.

So you can get the Open circuit voltage at A, and Open Circuit voltage at B. Find the difference between these for Δvab. So to get the Norton current, you need to short terminals A and B together, then run the calculations to find the current from A to B. The Norton Resistance is (Δvab) / (Short Ckt Current from A to B).

Another way to solve the problem is to find the Thevenin equivalent first (where ΔVab would be the Thevenin voltage), and convert it into Norton equivalent using source transformation.
(The resistors are connected in a special way, such that it is possible to find the Norton/Thevenin equivalent resistance only by visual inspection.)

 

[scottdave]

Thanks @cnh1995 . Source Transformation was the term, which I couldn't think of.