# A contradiction in Circuits ?

1. Oct 17, 2009

### ╔(σ_σ)╝

I was recently looking at a circuit and I just realized that I didn't understand something.

If we have the circuit ( the one attached below:)

From what I understand the voltage in a capacitor cannot change instantaneously and the current in an inductor cannot also change instantaneously. But the current is a capacitor and the voltage in an inductor can change instantaneously.

So for the below circuit at t = 0 my voltage in my inductor changes instantaneously to 10V but it is in parallel with my capacitor which has to have 0 V across it( no instantaneous changes in voltage) . But my inductor and capacitor are in parallel. Isn't there something wrong here ?

How can my inductor voltage be 10 while my capacitor voltage be zero? They are supposed to be in parallel.

I understand after a long time I would essentially be shorting the 10V DC source but what happens at t=0 ?

PS: This is not my homework or anything, I just stumbled on it while try to learn Pspice.

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2. Oct 17, 2009

### vk6kro

The voltage across the components would immediately drop to zero because the power source has no internal resistance and the inductor has no DC resistance.

However, it would get interesting if you had about 2000 ohms in series with the switch.
The components should resonate at about 8000 Hz so you would probably get some ringing effects (damped oscillation) around that frequency.

If you have it in a simulator already, maybe you would like to try this.

3. Oct 17, 2009

### skeptic2

The instant the switch is closed you may consider the capacitor a short circuit and the inductor an open circuit. What voltage would you expect with a short circuit and an open circuit in parallel? We are assuming of course as vk6kro suggests that the voltage source must have some internal or external resistance.

4. Oct 18, 2009

### ╔(σ_σ)╝

Well I guess due to the internal resistance it the voltages will drop to zero but isn't the internal resistant really small maybe of order ~ 100 ohm.
I would assume that at some point (before the voltages across the inductor drops to zero) the inductor voltage would be close to 10V and it would be in parallel with a 0V capacitor.Thus, a contradiction!

What would happen at that point assuming that there is essentially no internal resistance?

Thanks for your responce skeptic2, but consider the above.

5. Oct 18, 2009

### vk6kro

The internal resistance is whatever the circuit says it is. The circuit says it is zero.

As it stands, you have a voltage source with no internal resistance with zero ohms DC across it. This creates a paradox since such a voltage source will always produce 10 volts regardless of load but how can you develop a voltage across zero ohms?

So, if you would like to change the circuit to include some resistance somewhere, it might be possible to get a result.
2000 ohms is the reactance of those components at resonance, so it seems like an interesting value to put in series.

the inductor voltage would be close to 10V and it would be in parallel with a 0V capacitor.Thus, a contradiction!
Not at all. The capacitor will be low impedance in parallel with the inductor which will be high impedance. That is quite OK but you do need a resistor in series with the switch.

6. Oct 18, 2009

### ╔(σ_σ)╝

I'm sorry but you are not providing me the answer to my question. I don't mean to sound rude but you are gallivanting around the question.

I just want to know what values I would read off a volmeter if I were to hook one across the capacity and another across the inductor at t=0.

I'm not really concerned about if my circuit can produces a ringing sound or if it can make a blackhole. lol

You are talking about high impedance in parallel with low impedance but I don't see the relevance of such abstractions to my question.

Maybe I'm missing out on a key idea that is tied to the answer of my question, because I fail to see an answer.

7. Oct 18, 2009

### ╔(σ_σ)╝

I finally got the simulation to work and I get very disturbing results.

Look at the attachment.

It seems that my inductor voltage actually goes to 10 at t=infinity.

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8. Oct 18, 2009

### vk6kro

Try your simulation with a resistor in series with the power source.
You will get crazy results with the circuit you presented because it is an impossible situation.

If you have a resistor in series with a short circuit and an open circuit in parallel, what would happen?
The short circuit wins, and you get zero volts across the parallel pair.

You can't say there is a contradiction here.

9. Oct 18, 2009

### ╔(σ_σ)╝

I ran the simulation with the circuit below and got the following result which is expected.

I do not know what you mean by impossible but if I can connect and build such a circuit then, that is a ridicules statement!

My whole question is trying to figure out where the results are crazy. Why the inductor voltage goes to 1o instead of going to zero. I just want an explanation. I know such a circuit is useless and should never be built.

If you do not have the answer that is fine; I can always turn to other sources. However, thanks for your help thus far.

My statement contradiction was referring to the situation where I got 10V across the inductor instead of 0V. Surely you would agree that a contradiction exist in such a situation.

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10. Oct 18, 2009

### TurtleMeister

What's the internal resistance of the voltage source and the DC resistance of the inductor? From the looks of the graph they are both zero.

11. Oct 18, 2009

### ╔(σ_σ)╝

Yes.

12. Oct 18, 2009

### arithmetix

I think this is the same kind of problem that philosophy students are asked to deal with when the question is posed: "What is the result when an irresistible force meets an im-movable object".
Impossible situations lie beyond the province of logic.

13. Oct 18, 2009

### TurtleMeister

Then you must give your source voltage an internal resistance (resistance of typical battery) and your inductor a DC resistance. Also, in the real world you would need an oscilloscope to see the results, not a voltmeter.

14. Oct 18, 2009

### vk6kro

I do not know what you mean by impossible but if I can connect and build such a circuit then, that is a ridicules statement!

You CAN'T build such a circuit, though.

A simulator will go to extremes and take your circuit to be perfect. You can't find a voltage source that will give infinite current and you can't find an inductor that will have zero resistance.

What is happening here is that the impedance of the components changes when power is applied. But if you have a power source that can deliver infinite current, you wont see any change in voltage due to the change in impedance.

BUT if you put a resistor in series with the changing impedance, then you will see a change in voltage.

As an example, put a car battery across 100 ohms and 1 ohm. It will still deliver 12 volts.
But put the 100 ohm and 1 ohm (one at a time) in series with 50 ohms and you will see a big difference in the voltage across the 100 ohms and the 1 ohm.

I'm not really concerned about if my circuit can produces a ringing sound or if it can make a blackhole.
Your question asks about the transient effects you get when you apply power. You would get ringing effects.

Looks like the simulation is working OK now.

15. Oct 18, 2009

### ╔(σ_σ)╝

Lmao

I don't want to do that. I like the circuit the way it is. I just thought that the whole things was counter intuitive and a weird turn of events.

I understand what you are saying but my question is not based on the real world. It is based on the idealized world. In the real world you can never get 10 volts from any voltage source, you may get 9.99995 or 10.001 but probably never 10.0 , this is know.

Anyway I guess there is no answer to my ORIGINAL question.

I just wanted to know what would happen in my original circuit assuming idea conditions( no internal resistance etc).

Pspice gave me an answer and I guess I am somewhat satisfied with it.

Thanks for your input and help, though.