GR9277 #11



Alternate Solutions 
niux 20091106 07:02:08  My way was:
As Q is proportional to , then you will need a factor of () for Q (that is, a factor of ) in order to obtain as the problem requires. Then just equate this to time dependece for discharge:
and solving for t, you will get answer (E)
  QM320 20081108 06:07:58  You can also remember (or derive) . So . Squaring this, and using , we find . Plug in and solve. Voila!   QM320 20081108 06:07:16  You can also remember (or derive) V=V_0 e^{t/(RC)}U=\frac{1}{2}C V^2U=U_0 e^{2t/(RC)}U=U_0/2$ and solve. Voila!  

Comments 
psychonautQQ 20131007 14:07:40  What am I doing wrong..?
Qf = Qie^(t/rc)
U > Q^2
so Qf^2 = (Qi^2)/(2^2 )> (Qi^2)/4 = (Qi^2)e^(2t/rc)
1/4 = e^(t/rc)
ln(1/4) = t/rc
rc*ln(1/4) = t.
I realize this is wrong, I feel like the given answer skips some steps, it'd be coo if someone could elaborate.
  kstephe6 20121104 17:30:56  ONLY FAST ANSWERS MATTER!
half life (of charger in this case) is ln(2)*(tau)
this gives the time it takes for (1/2) of the charge to remain, which is the same (by the energy equation) as (1/4) of the energy to remain.
Well, you want twice as much energy left as this, so you only want half of this time to pass. Just the divide the half life by 2.
DONE!
kstephe6 20121105 05:09:16 
I meant this as an alternate solution.

  Quark 20111025 13:40:35  What if instead of U= you remembered U=. Then you would think that the energy was proportional to Q instead of and get the wrong answer...
Dr. D.R. Dopetec 20111031 20:44:52 
Realize that the charge is constant in this circuit, but the voltage across the capacitor is not. There is no voltage pump (e.g. battery) in this circuit.

  niux 20091106 07:02:08  My way was:
As Q is proportional to , then you will need a factor of () for Q (that is, a factor of ) in order to obtain as the problem requires. Then just equate this to time dependece for discharge:
and solving for t, you will get answer (E)
  tsharky87 20091030 09:30:42  In the official solution, he uses Ohm's law as
,
but isn't Ohm's law actually
?
kroner 20091101 16:16:31 
Yosun is implicitly using Kirchhoff's voltage law here: .
Then from and , we get .

  QM320 20081108 06:07:58  You can also remember (or derive) . So . Squaring this, and using , we find . Plug in and solve. Voila!
gravity 20101107 23:06:46 
If you're like me and get retarded wondering if it should be divide by two or multiplied by two (debating between Answer D or E), remember that the Energy is proportional to the voltage and so the energy will dissipate faster than the voltage, and thus current. The time constant is the time that it takes for the current to half, so we know it should be shorter than this. D is out, E is in.
And that's how you pass the GRE by being too dumb to figure out the U = U/2 stuff. (I don't know, it takes me longer to think about that than it does to think about the answer I gave).

  QM320 20081108 06:07:50  You can also remember (or derive) . So . Squaring this, and using , we find . Plug in and solve. Voila!   QM320 20081108 06:07:16  You can also remember (or derive) V=V_0 e^{t/(RC)}U=\frac{1}{2}C V^2U=U_0 e^{2t/(RC)}U=U_0/2$ and solve. Voila!
QM320 20081108 06:08:38 
Oops  bad LaTeX. See the version above!

  dcb 20070916 14:18:22  (I'm no expert, so feel free to poke holes in my method).
Mathless solution:
If you can recall these two simple things from your electronics or basic physics class:
1) Time constant for full charge is RC
2) Time relation is logarithmic.
By 1, we can rule out anything greater than RC (eliminate D).
By 2, we can rule out anything without an "ln" (eliminate A, B, C)
That leaves E.
ewhite2 20071027 10:57:35 
RC is the time to dissipate 2/3 of the charge in the capacitor, not the full charge. However even after putting this value into your line of reasoning it still works for a quick way to find a solution under the time constraints.

  boundforthefloor 20061119 17:51:50  The problem asks for the time it takes to lose half of the initial stored energy.
I'd rather think of the equation to solve as
which will multiply U by 1/2 as shown in the first equation because C is constant.   agaliarept 20061118 15:26:46  What do you mean by "dissipate by 2"? I am trying to write out the whole equation and something isnt coming out quite right.
StrangeQuark 20070522 12:46:13 
Try this,
we are looking for the time in which the energy is half of the intial, i.e.
Now just solve for t

 

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