Translation:
After being closed for a while the circuit opens at t = 0.
What is the max value for the voltage across the coil when the current cuts off.


I actually got a solution for this problem already seen here.

I don't understand what he's doing at this part:
di / i = -10.1*10^3 * (50*10^-3) * dt

So I guess 50*10^-3 comes from the value of L.

But why don't we divide the right side with L if we want to remove it from the left side of the equation ? (Here he multiplied the right side with L and it magically disappears from the left? what is this sorcery

 

Try read this
https://electronics.stackexchange.c...r-calculates-a-value-i-do-not-u/392663#392663

 

Translation:
After being closed for a while the circuit opens at t = 0.
What is the max value for the voltage across the coil when the current cuts off.


I actually got a solution for this problem already seen here.

I don't understand what he's doing at this part:
di / i = -10.1*10^3 * (50*10^-3) * dt

So I guess 50*10^-3 comes from the value of L.

But why don't we divide the right side with L if we want to remove it from the left side of the equation ? (Here he multiplied the right side with L and it magically disappears from the left? what is this sorcery

Not sorcery, just really poor work combined with pure luck.

I certainly hope that whoever did this solution isn't the teacher or grader, because they have no business doing either, as far as I'm concerned.

Since I don't know what the students have seen at this point, I don't know if it was reasonable to set up the differential equation and solve it completely just to find the max voltage, which we know happens at t = 0 and which we can determine by inspection. The steady-state current before the switch opens is 10 V / 100 Ω which is 0.1 A. When the switch opens this same current must now flow through the 10 kΩ resistor, resulting in (0.1 A)·(10 kΩ) = 1 kV across it.

Notice that the size of the inductor has absolutely no impact on the max voltage across it when the switch opens. That's why the person doing the solution could be so sloppy and still get the right value.

But their complete answer is very wrong, and had they done either of the two things that I am always harping about they would have known it.

First, always, always, ALWAYS track the units through the work -- don't just tack the units that you would like the answer to have onto the end. Had they don't this they would have realized that they did something wrong because the units would not have worked out. Specifically, their time constant would not have had units of time.

Second, always, always, ALWAYS ask if the answer makes sense. The person doing the solution should know that the time constant of an L-R circuit is L/R, which in this case is 50 mH / 10.1 kΩ. This is going to be a number that is approximately 5 μs. Since this is in the denominator, the variable t is multiplied by the reciprocal of this, which is about 200,000 s^-1, and not 500.

 

Not sorcery, just really poor work combined with pure luck.

Wow thank you, good explanation.

The guy who became up with this solution was a teacher on tutorme.com he got some money to help me with this. Still there are some very good teachers on that site but not everyone I realize now.

 

Here's a simulation to show what the inductor voltage and current look like.

 

Hi,

What i find interesting about this question although just a side note is that it does not look like we can find the maximum using the usual route of using calculus to find the max because the derivative at t=0 is not continuous. Perhaps someone else has an idea for this.

All we can do is assume we know how the voltage will show up across the resistor knowing how the initial current in an inductor behaves in a first order system. Adding a cap across the resistor helps, but perhaps someone else has a better idea here.

 

Hi,

What i find interesting about this question although just a side note is that it does not look like we can find the maximum using the usual route of using calculus to find the max because the derivative at t=0 is not continuous. Perhaps someone else has an idea for this.

All we can do is assume we know how the voltage will show up across the resistor knowing how the initial current in an inductor behaves in a first order system. Adding a cap across the resistor helps, but perhaps someone else has a better idea here.

In an "ideal" circuit the value of dI/dt is undefined at a corner. In the actual simulation involving "real" components there is a transition that limits the maximum value of the fast edge in the voltage waveform. In fact the series resistance of the inductor should be modeled explicitly.

 

In an "ideal" circuit the value of dI/dt is undefined at a corner. In the actual simulation involving "real" components there is a transition that limits the maximum value of the fast edge in the voltage waveform. In fact the series resistance of the inductor should be modeled explicitly.

Hi,

Yes the series resistance is included already so that wont help. In fact, no added resistance will help.
The di/dt could be called infinite at t=0 so maybe we can just call this an impulse response with impulse E/RL.
In any case, yes it is a 'corner' so the derivative doesnt exist, and so we cant use the usual method of finding the max. That's what was interesting. Adding a cap across R however changes all that and tames the derivative. We thus come out with a workable problem where the solution comes out very very close to t=0 with a very small cap value. Yeah, a little harder to solve though.

 

The di/dt could be called infinite at t=0

Is it not just V/L where V is I*R?
At t=0 the inductor current instantly starts flowing through R, which gives -1000V, so the initial di/dt would be 1000V/50mH = 2V/μs.

 

Hi,

What i find interesting about this question although just a side note is that it does not look like we can find the maximum using the usual route of using calculus to find the max because the derivative at t=0 is not continuous. Perhaps someone else has an idea for this.

Why can't we use the usual route of using calculus to find the max voltage here?

What is the first thing that is usually presented when they teach about using calculus to find a max/min over a range? That it occurs at one of three places. Where the first derivative is zero, at points of discontinuity, or at the endpoints of the range. They usually show several examples each situation specifically to try to disabuse people of the notion that simply setting the derivative equal to zero and solving for it will yield the min/max.

All we can do is assume we know how the voltage will show up across the resistor knowing how the initial current in an inductor behaves in a first order system. Adding a cap across the resistor helps, but perhaps someone else has a better idea here.

How does adding the capacitor help? It makes the problem much harder because now we DO have to develop the entire equation for v(t) over T and then solve for the max, keeping in mind that if the system is underdamped that will occur at many points in time and we have to be sure to find the one that is the actual max. Since it seems you don't want to assume we know how the voltage will show up across the resistor, we can't assume that the first peak will be the actual peak.

 

Is it not just V/L where V is I*R?
At t=0 the inductor current instantly starts flowing through R, which gives -1000V, so the initial di/dt would be 1000V/50mH = 2V/μs.

Hi,

Oh wait i meant the voltage not the current. My mistake.

 

Why can't we use the usual route of using calculus to find the max voltage here?

What is the first thing that is usually presented when they teach about using calculus to find a max/min over a range? That it occurs at one of three places. Where the first derivative is zero, at points of discontinuity, or at the endpoints of the range. They usually show several examples each situation specifically to try to disabuse people of the notion that simply setting the derivative equal to zero and solving for it will yield the min/max.



How does adding the capacitor help? It makes the problem much harder because now we DO have to develop the entire equation for v(t) over T and then solve for the max, keeping in mind that if the system is underdamped that will occur at many points in time and we have to be sure to find the one that is the actual max. Since it seems you don't want to assume we know how the voltage will show up across the resistor, we can't assume that the first peak will be the actual peak.

What i meant was that we cant just set the derivative to zero and solve.

Adding the cap tames the derivative, i said that before. We get a smooth function then which behaves ALMOST like the true response. It's the response we might see with some parasitic capacitance. With the smooth function we can actually set the derivative to zero and solve. The result we get approaches t=0 as C approaches zero.