# RL transient circuit

1. Feb 13, 2013

### freshbox

1. The problem statement, all variables and given/known data
A series R-L transient circuit is connected to a voltage source of E= 40V through a switch as shown. Suppose the switch is closed at time t=0μs and it was observed that the current iL flowing through the inductor rises to 4mA at time t=20μs and remains at 4mA after t=20μs.

(i) Find the R-L time constant τ in μs for the storage phase. Resistor given is 10kΩ

3. The attempt at a solution
iL=E/R(1-e-t/τ)
0.004=40/10000(1-e-20x10-6/τ)
0.004=0.004(1-e-20x10-6/τ) -> 0.004-0.004=0
0=-e-20x10-6/τ -> Wanted to take ln for both side, ln 0 =error

(ii) Calculate the amount of energy Wl in nJ, stored in the inductor after 1 minute.
How do we know 1min=60sec>>5time constant?

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Last edited: Feb 13, 2013
2. Feb 13, 2013

### Staff: Mentor

There's a useful rule of thumb for these exponential curves that essentially states that all the interesting action is over after 5 time constants

3. Feb 13, 2013

### freshbox

I know that after 5 time constants, the inductor will become a short circuit.

But how would i know after 1minute, it's 5 time constant?

And if you don't mind can you help me take a look at my working please.

Thanks gneill.

4. Feb 13, 2013

### cepheid

Staff Emeritus
Yeah, the exponential term has to decay to 0 in order for the current to approach the asymptotic value of E/R. This happens when the exponent "blah" is large enough that e^-blah = 0. (To put it loosely: (-20 us)/tau → ln0 ≈ -infinity). So the time of 20 us is much much larger than a time constant, so that the exponent will be much larger than -1, and you have many factors of "e" of decay. Typically in these engineering classes they give you a "rule of thumb" that says that you can consider the exponential to have effectively decayed to zero after n time constants. I'm inferring from you comment in part ii that your class is taking n = 5 as your rule of thumb. If that's true, you can solve for tau by assuming that the 20 us is = 5 time constants. Without such a rule of thumb, you can't really solve for a specific value of tau, all you can say is that tau is much smaller than 20 us.

EDIT: scooped by gneill!

5. Feb 13, 2013

### freshbox

Then how do i determine whether the time given by the question for example 10s,20s or 30s etc is after 5 tau? It's not clear at all. For this instance he just say that after 1min. But actually 1min >> 5 time constants

Last edited: Feb 13, 2013
6. Feb 13, 2013

### Staff: Mentor

Part (ii) of the question follows part (i) In part (i) you determined the time constant, and it must be less than 20 μs. One minute is 60 million microseconds, which is much more than 5 times 20 μs. In fact, anything over a tenth of a millisecond is guaranteed to be more than $5\tau$.

7. Feb 13, 2013

### freshbox

oic, btw where have i gone wrong for my working?

8. Feb 13, 2013

### cepheid

Staff Emeritus
Agreed, except for this part. It should say, "which is much more than 20 μs", since it is 5τ that equals 20 μs, not τ. Right?

9. Feb 13, 2013

### Staff: Mentor

I used 5 times 20 μs since 20 μs is a given upper bound for τ (although not a least upper bound) and 5τ is the settling time. But sure, your phrasing is arguably more accurate.

10. Feb 13, 2013

### Staff: Mentor

You're trying to find $\tau$ such that e-20x10-6 = 0. But the function $e^{-x}$ only approaches zero asymptotically as x goes to infinity. You can't find τ exactly by this method. Use the "excitement's over after 5τ" rule of thumb instead.

11. Feb 13, 2013

### freshbox

You mean part (i) is actually after 5 time constant already?

12. Feb 13, 2013

### Staff: Mentor

"(i) Find the R-L time constant τ in μs for the storage phase. Resistor given is 10kΩ"

You're told that the current reaches essentially steady state after 20 μs. So τ must be less than 20 μs. In fact, it's about 5 times smaller than 20 μs.

13. Feb 13, 2013

### freshbox

which part of the question says that the circuit is at steady state?

14. Feb 13, 2013

### Staff: Mentor

"Suppose the switch is closed at time t=0μs and it was observed that the current iL flowing through the inductor rises to 4mA at time t=20μs and remains at 4mA after t=20μs."

15. Feb 13, 2013

### freshbox

Ok here's another question. A series RC charging circuit is connected to a voltage source of E = 80V through a switch as sown. Assume there was no charge initially stored in the capactior. Suppose the switch is closed at time =0s and it was observed that the voltage vc across the capacitor, rises to 80v after time t=25millisec.

For this question, it did not say that remains at xxx after t=xxxx. So how would i know it is after 5 time constant (steady state) ?

***Ah ok i understand already, is it because initially there was no charge and since battery is 80v and after 25millisec, it's fully charged hence steady state. ?

16. Feb 13, 2013

### Staff: Mentor

Because E = 80V and it reached 80V after 25 ms. 80V is as high as it can go; the steady state.

17. Feb 13, 2013

thanks :)