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RL transient circuit

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Homework Statement


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



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

unable to continue, please help, thanks.


(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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  • #2
gneill
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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 :wink:
 
  • #3
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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
cepheid
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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
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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
 
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  • #6
gneill
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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
Part (ii) of the question follows part (i) :smile: 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
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oic, btw where have i gone wrong for my working?
 
  • #8
cepheid
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Part (ii) of the question follows part (i) :smile: 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##.
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
gneill
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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?
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
gneill
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oic, btw where have i gone wrong for my working?
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
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You mean part (i) is actually after 5 time constant already?
 
  • #12
gneill
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You mean part (i) is actually after 5 time constant already?
"(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
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which part of the question says that the circuit is at steady state?
 
  • #14
gneill
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which part of the question says that the circuit is at steady state?
"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
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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
gneill
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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) ?
Because E = 80V and it reached 80V after 25 ms. 80V is as high as it can go; the steady state.
 
  • #17
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thanks :)
 

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