Find the resistance and inductance ensuing a short-circuit

[Parad0x88]

Homework Statement

To measure the inductance and resistance of a real inductor, a physicist first connects the inductor to a 3V battery. At these conditions, the final, steady, current is 24 A. Then the physicist suddenly short-circuits the inductor with a thick (resistance-less) wire place across its terminals. The current then decreases from 24 A to 12 A in 0.22 s. Find the resistance and the inductance.

Homework Equations

ε = 3V
I₁ = 24A
I₂ = 12A
t = 0.22s

L = (ε_L X dt) / DI
R = L / time constant

The Attempt at a Solution

For L: (3v X .22s) / 12A = 0.0275H

For R: 0.0275H / 0.22s = 0.125Ω

Would that be right? It seems like what I did is so simple for a physics assignment, I get suspicious of it..

 

[gneill]

Find the resistance first; You've got the steady-state current and the potential drop.

For the inductance, note that the current will follow an exponential function from its initial value, finally decaying down to zero as the energy is bled off by the resistance. What's the equation for the current with respect to time?

 

[Parad0x88]

Find the resistance first; You've got the steady-state current and the potential drop.

For the inductance, note that the current will follow an exponential function from its initial value, finally decaying down to zero as the energy is bled off by the resistance. What's the equation for the current with respect to time?

Should I assume it's an RL circuit then? Because if I do then I would get a completely different set of equation...

For R, would (assuming it's an RL circuit) the equation be:

I = (ε/R) (I being the initial current at t = 0) ==> R = ε/I = 0.125Ω.. which I just noticed is what I previously found. Does that make sense?

 

[gneill]

Should I assume it's an RL circuit then? Because if I do then I would get a completely different set of equation...

Well, it's clearly an RL circuit, since it's an inductance in series with a resistance.

For R, would (assuming it's an RL circuit) the equation be:

I = (ε/R) (I being the initial current at t = 0) ==> R = ε/I = 0.125Ω.. which I just noticed is what I previously found. Does that make sense?

Yes, that's correct.

 

[Parad0x88]

Well, it's clearly an RL circuit, since it's an inductance in series with a resistance.

Yes, that's correct.

Ok, then the next step would be to find the time constant to get from 24A to 12A, here's how I would do it:

I = ε/R X (1 - e^{-t/time constant}, isolate for time constant:
IR / ε = 1 - e^{-t/time constant}
1 - (IR / ε) = e^{-t/time constant}
ln (1 - (IR / ε) = -t / time constant
time constant = -t / ln (1 - (IR / ε)

Once I have the time constant, I would then use:

Time constant = L/R;

L = Time constant X R

I would you 12A as my I since we have the time value for how long it took to reach 24 to 12A

Does that work as well for part B?

 

[gneill]

Ok, then the next step would be to find the time constant to get from 24A to 12A, here's how I would do it:

I = ε/R X (1 - e^{-t/time constant}, isolate for time constant:

The current starts with a maximum value (24A) and then decays to zero over time. What's the shape of the resulting curve? What's the shape of the curve that you've specified in the above equation?

<snip>

Once I have the time constant, I would then use:

Time constant = L/R;

L = Time constant X R

I would you 12A as my I since we have the time value for how long it took to reach 24 to 12A

Does that work as well for part B?

That would work, yes.

 

[Parad0x88]

The current starts with a maximum value (24A) and then decays to zero over time. What's the shape of the resulting curve? What's the shape of the curve that you've specified in the above equation?

<snip>

That would work, yes.

Ok well the current decaying over time, I'm assuming it's non-linear, otherwise it would have been specified, so it's an inverse non-linear graph (I don't know it that is the right term, it's been a while since I studied graphs)

The formula that I have posted, wouldn't also yield a non-linear result?

Oh wait I think I see what you mean... The formula, should it simply be: I = ε/R X (e-^{t/time constant})

 

[gneill]

Oh wait I think I see what you mean... The formula, should it simply be: I = ε/R X (e-^{t/time constant})

Bingo!

 

[Parad0x88]

Bingo!

Awesome thanks, I just want to make sure I understand correctly though.

On one end, we have the current that decay from 24A to 0A with a sudden drop from 24A to 12A, thus giving an inversely proportional graph

On the other end, according to my book (I don't have the reflex to picture the graphs from a formula) the second formula I gave you gives out that graph.

Makes sense. But I don't understand how that formula gives that graph, do you mind giving me a bit of insight?

 

[gneill]

Okay, consider the formula:

$y = e^{-x}$

What value does y take on when x = 0, and when x → +∞ ?

 

[Parad0x88]

Okay, consider the formula:

$y = e^{-x}$

What value does y take on when x = 0, and when x → +∞ ?

Ahh put it like that, it makes sense.. Awesome, thanks!