This is more an Electrical Question

[Eric Nelson]

The voltage across the resistor is equal to the voltage across the resistor.

It had better be!...



-Eric

 

[Terry Woods]

Again...

...the assumption is that the current through the Inductor, before switching, is maximum and that the current is known...

...the current is NOT known! Even if it is maximum!

 

[rsdoran]

I stayed out of this because something didnt seem right and I wasnt sure what it was exactly but off top of my head I think there could be different answer(s) depending on 2 things:
1. Frequency XL=2pifl
2. Parallel or series

I think I get where Terry is going. E = L times di/dt, since current isnt known because frequency isnt known how can you calculate time? I havent used this stuff in far too long so not sure about the details but the question seemed lacking.

I have a page that covers the basics of this. http://www.patchn.com/AC/AC_3.html

 

[Gerry]

Again...

...the assumption is that the current through the Inductor, before switching, is maximum and that the current is known...

...the current is NOT known! Even if it is maximum!

Why should it matter? The voltage across the inductor has reached 810 V. at the point of switching from power supply to shunt resistor. From that time the voltage across the inductor and the resistor are the same and I=E/R.

I suppose you could complain that the problem wasn't stated completely and clearly, but that would be rather pedantic. If you look at the links in Mickey's post which mimic the problem, I think it's reasonable to accept Rick's assumption

I am guessing there is a 810V source connected across the inductor. Then the pos B is when the source is disconnected and the resistor connected across the coil.

as correct.

 

[rsdoran]

Why should it matter? The voltage across the inductor has reached 810 V. at the point of switching from power supply to shunt resistor. From that time the voltage across the inductor and the resistor are the same and I=E/R.

How do you know the voltage across the inductor has reached 810v? Wouldnt that require it to be a parallel circuit? If that was the case then the resistor would have no bearing on the time constant.

If it is a series circuit and E=810v then your calculation may be correct, I am not positive.

The question left out too many factors in my opinion.

 

[Gerry]

How do you know the voltage across the inductor has reached 810v?

I read post #1

Wouldnt that require it to be a parallel circuit? If that was the case then the resistor would have no bearing on the time constant.

If it is a series circuit and E=810v then your calculation may be correct, I am not positive.

Look at the links in post #5 for some indication of the circuitry. Or, simply imagine an ideal SPDT device that can switch in zero time switching one leg of the inductor from the power supply to the resistor (other legs of L&R to P/S common).

 

[rsdoran]

I am game

I read post #1

Where does it say voltage across the inductor is 810v?

Post #1

R=11 Ohms, L=6200 H and E=810 V.

The switch is intially in position A and the inductor is fully charged. Calculate the time after the switch is set to position B for the inductor Voltage to drop 230 V.

I read that E=810v, and inductor fully charged, I dont see any statement that voltage across the inductor is 810v. So what is the voltage across the inductor? If the voltage across the inductor is not known how can you calculate the time till it drops to any value?

I dont see any statement concerning parallel or series, just like DC the impedance, current, voltage etc is different between them. If parallel then 810v is across the inductor, if series...dont know because:
I saw no indication whether AC or DC which also can have a bearing on impedance and current.

I havent done calculations like this in a long time but I still think that the question was insufficient to calculate an accurate answer. I think you "assumed" alot.

Terry you can get back in any time now, you can explain better than I can.

 

[Gerry]

Where does it say voltage across the inductor is 810v?

Post #1

I read that E=810v, and inductor fully charged, I dont see any statement that voltage across the inductor is 810v. So what is the voltage across the inductor? If the voltage across the inductor is not known how can you calculate the time till it drops to any value?

I dont see any statement concerning parallel or series, just like DC the impedance, current, voltage etc is different between them. If parallel then 810v is across the inductor, if series...dont know because:
I saw no indication whether AC or DC which also can have a bearing on impedance and current.

And to quote myself:

I suppose you could complain that the problem wasn't stated completely and clearly, but that would be rather pedantic.

Bear in mind that this is a question from a textbook, paraphrased by a high school student trying to learn some basics. There is probably a circuit diagram in the textbook. I don't think it is the author's intention that the answer to the question should be "indeterminate". Rather it is to emphasise:

1. how to calculate the time constant of an L R circuit
2. the equation for the decaying voltage
3. how to manipulate the equation to find time, given the other variables

I do find "the inductor is fully charged" to be a novel concept, however, I take that to mean that the current through the coil, and the voltage across it, have stabilised and the voltage across the coil is the figure given for E. There really is nothing else in the question for E to apply to, and in the context of the question it makes no sense to talk about AC.

Relax, Ron - I'm not looking for a fight

 

[SimonGoldsworthy]

This particular type of problem has always fascinated me when one considers the realities of the situation. You cannot operate a switch in zero time, so how could you predict:

1. The spectrum of RF emitted ?
2. The amount of switch contact material that evaporates ?

 

[rsdoran]

Gerry, I am not looking for a fight either. I just think answers are being offered to a question that cannot be answered as stated.

Lets say it is DC. E usually implies Source voltage, still does in most books. Inductor voltage is usually stated as EL. I offer this page as a reference: http://www.patchn.com/DC/DC_15.html

If E is source voltage then 810v across the inductor would mean its a parallel circuit and the time constant would be different because the resistor would not be part of the equation.

Textbooks questions do trick people like that.

I was hoping the high school student would come back with more on the subject.

 

[Lancie1]

Guys,

Wolf has been corresponding with me for several weeks with questions. Here is the circuit for his problem:

 

[rsdoran]

Now it can be solved, its DC and a parallel ckt. The time constant will be different you know?

 

[Vic]

Parallel Circuit?

Now it can be solved, its DC and a parallel ckt.

Ron
The question is concerned with what happens after the switch opens. Since there is only one path for the current to flow, it is a series circuit. Its true that it is a parallel circuit before the switch opens (provided you assume it is a perfect inductor with no resistance) but the resistor does not effect the rate of increase of the current in the inductor when the switch is closed.

The time constant will be different you know?

Please explain what you mean by this.

 

[rsdoran]

Explain it any way you want...time constant is R/L so what is R in a parallel ckt? It aint the resistor mentioned.

 

[Vic]

The time constant will be different you know?

You still haven't explained what you meant by this.

Just to set the record straight, the time constant is L/R, not R/L and the only resistance which effects the time constant is resistance in series with the inductor. Resistance in parallel with the inductor does not effect the time constant.