Why Does Terminal a Show Negative Potential in This Circuit Analysis?

[PainterGuy]

Hi

This is confusing me very much. I would try to ask it now and try my best to convey my confusion. Please have a look on the following link: http://img708.imageshack.us/img708/3581/img0002ri.jpg

The scan also has my comments there which are an attempt to explain what is troubling me.

The Circuit #2 could be found here in full context here: http://img843.imageshack.us/img843/4292/norton1.jpg (Original link)

As you see the author in the "Original link" says that the circuit on left of the terminals a-b is supplying power but I don't get it. From the viewpoint of current source Io the terminal "a" should be +ve because current source is pointing toward it; that means its +ve terminal is connected with "a". But from the viewpoint of the circuit on the left of terminal a-b the terminal "a" is at -4V potential which I don't get. Do you see my confusion?

Thank you very much for all the help.

Cheers

 

[gneill]

Current sources don't care about the potential difference that exists across them. They use (or consume) whatever energy is required to move the current carriers from one terminal to the other. An analogy might be a water pump that delivers a certain flow of water. It doesn't matter if the outlet is below the inlet, it pumps water at the same rate of flow regardless.

 

[MalcolmKee]

I don't understand what does the original text discussing, anyway, I guess that your confusion is that why the right-hand side current source is pointing upwards, but the potential at the upper node is lower than the ground.

I think that the reason is that the current source is not the only power source. There is another dependent current source in the circuit, which makes the scenario possible. It's like having two voltage source in a circuit, and then you cannot predict the potential at certain node by just looking at the polarity of single voltage source.

Is this what you looking for?

 

[PainterGuy]

Thank you, gneill, Malcolm. gneill, I'm very much grateful to you for all your help. You have helped me with many problem in the past.

I'm still having difficulty with understanding the concept. Perhaps this statement of Malcolm could be useful to explain the reason for negative potential at 'a' even though arrow of the current source is pointing toward 'a'.

I drew this diagram: http://img6.imageshack.us/img6/7424/img0003rz.jpg

You can also some of the text there which may help you understand what's going on in my mind.

My other question is what source is responsible for creating -ve potential at 'a'. Is it the current source Io?

Many thanks for all the help.

Cheers

 

[gneill]

Your circuit diagram does not correspond to what appeared in the earlier circuits. Two ideal voltage sources connected in such a fashion are going to cause problems -- infinite currents and other 'bad things'.

What is causing the negative potential at "a" is the dependent current source 2I_x, which produces a current that depends upon conditions at another location in the circuit; it is an "active" component that uses some unspecified source of energy to produce its effects. Keep in mid that ideal current sources will manifest any voltage required in order to inject their specified current.

 

[PainterGuy]

Thank you for letting me know that it's dependent current source which is responsible for negative potential at "a". I think one more question can make it easier to understand for me. What do we deduce from this? I deduce that no matter what value for Io we choose, the voltage appearing across dependent current source would be greater than appearing across Io?

Perhaps, we can also say that dependent current source will always be pumping more coulombs than Io?

 

[gneill]

Thank you for letting me know that it's dependent current source which is responsible for negative potential at "a". I think one more question can make it easier to understand for me. What do we deduce from this? I deduce that no matter what value for Io we choose, the voltage appearing across dependent current source would be greater than appearing across Io?

Perhaps, we can also say that dependent current source will always be pumping more coulombs than Io?

If you look at the circuit, all the components are connected in parallel via the top and bottom "rails". So the dependent current source and I_o are in fact in parallel, and must therefore have the same voltage across them.

What is clear from the circuit, though, is that I_x = 2*I_o.

 

[PainterGuy]

Thank you, gneill.

I understand it could be getting a little frustrating for you to explain the same thing repeatedly. I'm sorry for this. Sometimes, it takes a lot of time to understand simple things, especially for persons like me.

1:
Please have a see on this link: http://img696.imageshack.us/img696/4900/negativepotential.jpg

The linked diagram is redrawing of the Figure 4.35(b) found in the Original Link in my first post above:

The Circuit #2 could be found here in full context here: http://img843.imageshack.us/img843/4292/norton1.jpg (Original link)

Let's hope I have it right this time. Please let me know.

2:

What is clear from the circuit, though, is that Ix = 2*Io.

I'm sorry I don't see how Ix = 2*Io? But I'm almost sure about one thing that 2Ix should be bigger than Io because it should have more voltage appearing across it which means a larger current through than through Io.

Many thanks for all the help.

Best wishes

 

[gneill]

Thank you, gneill.

I understand it could be getting a little frustrating for you to explain the same thing repeatedly. I'm sorry for this. Sometimes, it takes a lot of time to understand simple things, especially for persons like me.

Not at all. Willingness to learn trumps all.

1:
Please have a see on this link: http://img696.imageshack.us/img696/4900/negativepotential.jpg

The linked diagram is redrawing of the Figure 4.35(b) found in the Original Link in my first post above:

Let's hope I have it right this time. Please let me know.

2:


I'm sorry I don't see how Ix = 2*Io? But I'm almost sure about one thing that 2Ix should be bigger than Io because it should have more voltage appearing across it which means a larger current through than through Io.

Many thanks for all the help.

Best wishes

Okay, I think I can tackle both questions in one go.

Note that your labeled points a and c are the same node: they are both connected to the same conductor that runs the width of the circuit along the top. Similarly, the bottom horizontal conductor is all one node.

First, you should be aware that when voltage sources are tied together to the same nodes, the net voltage is not their sum or difference. Usually, "bad things" happen. If the voltage sources are ideal, then both will ABSOLUTELY INSIST that the potential difference be their own potential difference, and will produce ANY AMOUNT of current required to achieve this end. Needless to say, things will get messy. If they are not ideal sources and are afflicted with some internal resistance, then a very high (but not infinite) current will flow that will cause voltage drops on these resistances equal to the difference in potential between the sources. Things may still get rather hot.

It's not a practical approach to assign arbitrary voltages to current sources to see how a circuit behaves, because just as a voltage source will produce any amount of current to achieve its aims, a current source will likewise produce any amount of voltage in order to achieve its aims of producing the specified amount of current. That voltage can be positive or negative, and of any magnitude.

That said, an approach that is practical in this case is to employ superposition to explore the relationship between I_o and I_x.

Both resistors in the circuit are in parallel and in turn are in parallel with the current supplies. When a current source feeds parallel branches you have a current divider. Take first the case of current supply I_o, faced with the branches of 2Ω and 4Ω. The current it will push through the 2Ω resistor will be
$$I_1 = I_0 \frac{4}{4+2} = \frac{2}{3}I_o$$
This current flows from the top of the 2Ω resistor to the bottom, in the opposite direction to the I_x label in your diagram.

Now take the 2I_x source. Its contribution to the current through the 2Ω resistor will be from the bottom to top, in direction of the I_x labeled current.
$$I_2 = 2 I_x \frac{4}{4+2} = \frac{4}{3}I_x$$
But the current I_x is sum of the currents I₁ and I₂. Taking into account their directions with respect to the I_x labeled current,
$$I_x = I_2 - I_1$$
$$I_x = \frac{4}{3}I_x - \frac{2}{3}I_o$$
Solving for I_x,
$$I_x = 2 I_o$$