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I'm trying to figure out the reason why in the following circuit, at steady state, the current ix is zero. My best guess is that because there is a parallel branch which by KCL should equal 100ix and would be zero because of the open circuit provided by the capacitor.

However, this seems counter intuitive because wouldn't the electricity want to go around the outer loop. And in this case, how do you deal with a loop that has a dependent current source dependent on its own current? Is that even possible?

Thank you!

Here is the circuit:

circuit

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    \$\begingroup\$ You say steady state, but yet you have this unknown function 15u(t) as part of the voltage. These two statements are inconsistant unless u(t) is always zero, in which case it makes no sense to show it. \$\endgroup\$ – Olin Lathrop Nov 22 '13 at 14:02
  • \$\begingroup\$ You'll need to provide some more detail: What do you mean by steady state? Your input voltage v_i is specified in the circuit diagram to be time-varying. Assuming you meant it to be constant instead, your argument is entirely valid (assuming I may pretend that the element labelled 99 i_x, which I do not know, was absent from the circuit). \$\endgroup\$ – pyramids Nov 22 '13 at 14:04
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    \$\begingroup\$ @OlinLathrop, this is a valid question since \$u(t)\$ is the unit step function so the independent voltage source produces 5V for \$t<0\$ and 20V for \$t\ge 0\$. To find the steady state solution as \$t \rightarrow \infty\$, simply replace the voltage source with a 20VDC source. \$\endgroup\$ – Alfred Centauri Nov 22 '13 at 14:47
  • \$\begingroup\$ To everybody that has no idea what steady state means: Steady state means after all transients have died down, effectively after a very long time (mathematically, infinity, like Alfred said). Transients will occur when a switch opens/closes, after an impulse, step input transition, etc. \$\endgroup\$ – apalopohapa Nov 22 '13 at 16:27
  • \$\begingroup\$ (...) In a linear circuit, mathematically speaking, after all exponentials with negative real powers have killed the corresponding sinusoidals they multiply, and only the sinusoidals without those exponentials remain. Those, and DC, are the steady state response. \$\endgroup\$ – apalopohapa Nov 22 '13 at 16:36
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Despite the comments to the contrary, this circuit does have a steady state solution since the voltage source produces 20V for \$t \ge 0\$.

My best guess is that because there is a parallel branch which by KCL should equal 100ix and would be zero because of the open circuit provided by the capacitor.

That's correct. The steady state KCL at the node in question is:

$$i_x + 99i_x = i_C(\infty) = 0 \rightarrow i_x = 0$$

However, this seems counter intuitive because wouldn't the electricity want to go around the outer loop.

It may seem counter intuitive but that's because your intuition hasn't fully developed yet. Once you come to fully understand the implication of that current source, the result will seem obvious.

What you must fully appreciate is that a current source completely determines the current through its branch. If there is a current source in a branch and you set its value to zero, the branch is open, i.e., there can be no current through for any voltage across.

And in this case, how do you deal with a loop that has a dependent current source dependent on its own current? Is that even possible?

But this isn't the case here*. There are two meshes (loops), one with current \$i_x\$ and the other with current \$99i_x\$. So the controlling variable of the dependent current source is not "its own current".

But, if it were the case, then the only way for the source to produce a non-zero current is for the current gain to be precisely 1:

$$i_x = ki_x \rightarrow i_x = 0$$

unless \$k=1\$ in which case you have

$$i_x = i_x$$

Since any value of \$i_x\$ satisfies the equation, the current is indeterminate. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

In this circuit, the voltage and current are not determined. The only equation one can write is:

$$V_{CCCS1} = I_{CCCS1} \cdot 100\Omega$$

But, we cannot determine what the current or voltage actually is since we have two unknowns and just one equation.

*Yes, in steady state, one might argue that it is the case here thus the remainder of the answer.


The equivalent circuit to the right of the resistor

It is straightforward to show that the equivalent circuit looking to the right of the resistor is:

schematic

simulate this circuit

In other words, for the purposes of calculating \$i_x(t)\$, one can replace the circuit to the right of the resistor with the above equivalent. Now, one can see by inspection that \$i_x(\infty) = 0\$


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  • \$\begingroup\$ @RedGrittyBrick, I greatly appreciate the reduction of the schematic size and would also greatly appreciate understanding how you did that. I use Ubuntu and Firefox and CircuitLab always complains about my zoom setting or something but I don't have the browser window zoomed. Any insight would be appreciated. \$\endgroup\$ – Alfred Centauri Nov 22 '13 at 17:08
  • \$\begingroup\$ I just add an inconspicuous text object (. or *) on the far right of the canvas. You have to scroll down to near the bottom of the object pallette to find the text tool. It's an ugly workaround because, so far as I know, CircuitLab lacks any more sensible way to specify a scale. \$\endgroup\$ – RedGrittyBrick Nov 22 '13 at 17:23
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Steady state current is zero because, given that excitation is dc:

  1. The cap is an open circuit, so the current through that branch is zero.
  2. The input current to that branch come from two branches, one has \$i_x\$ and the other has \$99i_x\$.

So you have that \$100i_x\$ is zero, therefore \$i_x\$ is zero.

Regarding your comment of the current wanting to go through the outer loop, like some sort of feedback loop (since one current is a factor of the other, and they are connected to each other):

First you need to realize that the inductor on the right has zero effect, no matter what. This is because all an inductor does is generate a voltage across its terminals that oppose current changes. But the nature of a current source (dependent or independent) is that it will vary its voltage as much as required to make its defined current flow. So it will counter any voltage that may develop through the inductor. So you can put resistors, voltage sources, inductors, etc. in series with the current source, and they won't matter.

Once you see that, you realize that that right branch is just proportionally adding to whatever flows through the left branch. It becomes a current amplifier, and the output factor is 100 (=99+1, since the original current also flows through). The actual current value ix will depend on how the middle branch reacts to the voltage source + series resistor excitation, knowing that the current through them is being amplified by the 99ix "injection".

Alfred gave you an equivalent circuit with the cap 100 times smaller and an inductor 100 times larger.

This is because with a capacitor, the voltage effect of up-scaling the current is equivalent to down-scaling the capacitance:

$$ dV_C = \frac{(100i)}{C} dt = \frac{i}{(C/100)} dt $$

And with an inductor, the voltage effect of up-scaling the current is equivalent to up-scaling the inductance:

$$ V_L = L\frac{d(100i)}{dt} = (100L)\frac{di}{dt} $$

So in this circuit, the effect of scaling the current is equivalent to scaling the components it feeds accordingly, in terms of the voltage developed across the 2 nodes that contain them.

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If u(t) really is the unit step function, then first you should clearly say so, and second the current is clearly 0 in steady state from inspection.

In steady state, capacitors are open circuits and inductors shorts. The middle vertical leg of the circuit is therefore effectively not there. Now you have two current sources that must be equal and opposite, but the only possible converged solution is that they are both zero.

Put mathematically, Ix = -99 Ix, with Ix = 0 being the obvious answer. Lots of fancy analisys isn't necessary here.

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