shorting a dependent current source

Out of curiosity, which textbook are you using?

Yours is not necessarily a dumb question, but if this is how your textbook is presenting the information, you need a different book. It is, at best, confusing to see the labels such as V_1 amps.

First, what is the purpose of the lesson? Are you studying basic series dc circuits, independent and dependent current and voltage sources, or KCL and KVL?

From a very basic standpoint, you end up with a series circuit consisting of a voltage source (1 volt?) and a resistor (1 ohm). Since current equals voltage (in volts) divided by resistance (in ohms), your current is 1 amp. You seemed to understand that in the first part of your post when you talk about using KVL.

For the second part, your diagram does not label I_2, so it is a bit more difficult. If I_2 is the current through the short, then 1 amp would still be correct. However, if I_2 is the current through what is being referred to as the dependent current source, it would be zero.

If your drawings are correct and the if the text says that the current through a shorted dependent current source is 1 amp, then it is incorrect. Not an unusual occurance. Nobody is perfect, least of all those who write the answers to textbook problems and me.

Richard

It is David E. Johnson, Johnny R. Johnson, John L. Hilburn and Peter D. Scott, Electric Circuit Analysis, 3rd Edition. The problem is gave above is Excercise 16.2.1, p. 663.

That was my own for illustration here only. That only means the dependent current source is provising current numericall equal to the voltage source. Sorry if it was confusing.

Ah! My error again. I_2 is current through the short; I_2 = I_xy in the figure. I have redrawn the corrected figure below (it was very late at night when I was drawing those figures). I_1 R=1 Ohms x I_2 |--->----\\/\\/\\/\\-------o-------->------ + | | V_1 / \\ | - / | \\ | | \\ v / V_1 amps. | | \\ / | | | | -----------------------o--------------- y

So, given these corrections in labels, I_2 = I_1 and current through dependent sourse = 0?

Thanks, HS.

I assume the current source direction is from x to y, namely that the thing inside the diamond is an arrow pointing down.

Being pedantic for a moment, what was the voltage from x to y before it was shorted?

John

Correct.

It is given by by the symbol V_2 (so, in my figure above V_xy = V_2). Hence, V_2 = V_1 - V_R where V_R is the drop across the resistor.

I guess my question boils to this basic form: by shorting the dependent current source, do we effectively exclude it from the circuit? It is assumed to be ideal.

HS.

But what *is* V2 before you connect the short?

You certainly can't exclude it from affecting the current through the short, which was your original question, I think.

John

It is not given, it is just symbol, a variable. Actually, the given circuit is a two port network and I am trying to find one of its parameters. This particular case, the one I gave above, is what isn't clear to me.

BTW, how does it matter what V_2 was before shorting?

Okay. If that is the case, then when I short the dependent current source and the circuit is then given by:

I_1 R=1 Ohms x I_2 |--->----\\/\\/\\/\\-------o-------->------ + | | V_1 / \\ | - / | \\ | | \\ v / V_1 amps. | | \\ / | | | | -----------------------o--------------- y

then applying KCL at node x gives: I_1 = V_1 + I_2 ........................ eq.1

Now, by applying KVL around the lefthand side mesh, I get: V_1 = I_1 (1) + 0 => V_1 = I_1 ....... eq.2

Using eq. 2 in eq. 1 gives: I_2 = 0,

which means no current in the wire shorting the dependent source. Is this correct?

thanks, HS.

Right. Assuming V1 = 1 volt, there's 1 amp being dumped into the short by the resistor, and 1 amp being sucked out by the current source, net zero.

The reason I asked about V2 before the short was applied is that V2=0, so the short, when applied, conducts no current.

Applying mesh equations is a good way to obscure the simplicity of what's going on. Another problem with relying in an all-eggs-in-one-basket math approach is that a simple mistake, like a sign reversal or something, can produce absurd results that are just accepted. It's better, when you can, to reason through circuits one step at a time, adhering to physical reality at each step. Instructors tend to not like this.

John

Thanks. However, the textbook says that I_2 = 1 in the case above and not 0. I am thinking it might be a mistake in that book.

Ok, I'll take a shot at this. This is my opinion and only an opinion..

I think maybe something is getting miss interpreted here.

The triangle symbol could be nothing more than an inline AMP meter taking measurements. or inline current sense device. the ideal measuring device would have 0 ohms.. there for, you don't subject anything you see there as part of the equation but maybe a place holder to log in the current. Shorting the symbol only means the current sense isn't sensing anything. This could be a hall effect current monitor for example. So, to put it in basic form.

I1=I2= V_1/R = 1

That's my take on it. Just my 2 cents.

That device is a voltage controlled current source and not a measuring device. The convention used in the text book in question is clear on that.

regards, HS

If V1=1 volt, then the triangle-doobie dependant source (as I interpret it) is 1 amp in the down direction. So 1 amp flows through the 1 ohm resistor, so the voltage drop across R is one volt. So, with the short not yet applied, the voltage at X is already zero, with a thevenin impedance of one ohm. So the short does nothing and conducts no current.

The reasoning applies for any value of V1; it all washes.

Is that right?

John

If you put a short across the current source, there will be no voltage across it - so the question becomes how much current does the current source source or sink when there is no voltage across it?

My common-sense technician's view is that all the current will flow through the short, and none through the current source.

It is an idea voltage controlled current source.

In this case, I_2 will then be the same as I_1. This would then agree with the answer given in the text book. However, how do you prove this using basic electric laws then? That is where I am stumped. Yours is the only way I could explain why I_2 = I_1. But if I use KVL and KCL, I_2 comes out to be equal to 0.

->HS

is that anything like how much current could a current source source if a current source could source? :)

Here is another thought that just occurred to me. If we short the dependent current source, nodes x and y become one single node. Now, since ideal nodes (in KCL) neither store nor provide any charge, that means the current going through the current source then must be zero (all elements are ideal in this problem). Hence the current I_1 must pass through the short, therefore I_1 = I_2. Makes sense?

->HS

How can no current flow through a current source?

In fact, 1 amp flows through the current source, and zero amps flow through the short.

John

Nope, no sense whatsoever.

John

hmm .. nodes are just internconnections, so you are right.

->HS

But it is a voltage-controlled current course, not a fixed 1 amp current source - if there is no voltage across it (due to the short) what current does it pass?