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Power delivered by Battery to Capacitor in Time Constants

  • #1
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Power delivered by Battery to Capacitor in "Time Constants"

Homework Statement


< Pic 1 >


The Attempt at a Solution



< Pic 2 >

As you see, I'm getting answer as (ln 2)/2
but the answer is just (ln 2)

Please tell me where i'm wrong.
 

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Answers and Replies

  • #2
Delphi51
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The calcs look good.
I'm quite uncomfortable with that P = I²R, which is the formula for the power dissipated by the resistor. But the battery also delivers power to the capacitor. Better to use P = IV. The V is constant, so all you have to worry about is the I. At what time does I decay to half its initial value?
 
  • #3
SammyS
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The calcs look good.
I'm quite uncomfortable with that P = I²R, which is the formula for the power dissipated by the resistor. But the battery also delivers power to the capacitor. Better to use P = IV. The V is constant, so all you have to worry about is the I. At what time does I decay to half its initial value?
I know we usually consider the time constant for an RC circuit to be τ=RC, because that's the time for the current drop to 1/e of its max value. In this case the author may be considering the time constant to be the time it takes for the power to drop to 1/e of its max value, in which case τ=RC/2.

Just a thought.
 
  • #4
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I'm quite uncomfortable with that P = I²R, which is the formula for the power dissipated by the resistor. But the battery also delivers power to the capacitor. Better to use P = IV. The V is constant, so all you have to worry about is the I. At what time does I decay to half its initial value?
Well thanks ... it worked !!!

But i don't get it ... i2R is for power dissipated in resistor but Vi is for battery ... How?

We just used ohm's maw to substitute for iR !!!
 
  • #5
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i know we usually consider the time constant for an rc circuit to be τ=rc, because that's the time for the current drop to 1/e of its max value. In this case the author may be considering the time constant to be the time it takes for the power to drop to 1/e of its max value, in which case τ=rc/2.

Just a thought.
No i'm pretty sure that τ=RC.

I've done many questions of same type and they all used τ=RC
 
  • #6
ehild
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Part of the power delivered by the battery transforms to heat in the resistor, but an other part is used to build up the electric field between the capacitor plates. You know that a capacitor has energy of
Ec=0.5 Q2/C.
The total power delivered by the battery is P=ε*I.

ehild
 
  • #7
Delphi51
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The question asks about "the power delivered by the battery", not the power dissipated in the resistor; hence P = IV.

It is interesting to work it out again using P = i²R + 0.5 Q²/C.
A bit complicated but actually the hardest part was getting the formula for Q that you started with - I had to resort to
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html
for the idea.

Thank you both for an interesting time.
 
  • #8
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It is interesting to work it out again using P = i²R + 0.5 Q²/C.
But 0.5 Q²/C is not power but energy delivered by battery,

P = (Q/C)(dQ/dt)

but i dont know if we can substitute dQ/dt as i or not ...
 
  • #9
Delphi51
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Sure, i = dQ/dt. Oops, I forgot to say "the derivative of" 0.5 Q²/C
P = i²R + d/dt[0.5 Q²/C]
= i²R + Q/C*dQ/dt
= i²R + Q*i/C
Put in your expression for i as a function of time and you will soon get
P = V²/R*e^(-t/RC)
which checks with the other approach, P = iV.
 
  • #10
ehild
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You are right, 0.5 Q²/C is the energy of the capacitor, and P = (Q/C)(dQ/dt).
And i=dQ/dt, the charge flowing through the circuit. The same current charges the capacitor.
The definition of the electric power is voltage times current P=V*i.
In case of a resistor, V=R*i so the power can be written as P=R*i2. In case of a capacitor, the rate of energy delivery is P = (Q/C)(dQ/dt). As V=Q/C and i=dQ/dt, it can also be written as P=V*I.

ehild
 
  • #11
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Ahh!!!! now i get it.

this battery has always been a problem for me!!

Thanks a lot guys for help!!!!!!
 

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