Currents - RC circuit

  • #1
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Homework Statement


attachment.php?attachmentid=33576&stc=1&d=1301168110.jpg


The Attempt at a Solution



As at t=0 the capacitor behaves as a wire ... so the circuit at t=0 will be like:

attachment.php?attachmentid=33577&stc=1&d=1301168676.png


q at any time for this circuit will be,

[tex]q \ = \ CE(1-e^{-t/\tau}) + Q_oe^{-t/\tau}[/tex]

differentiating it will give current,

[tex]i \ = \ \frac{Ee^{-t/\tau}}{R} \ - \ \frac{Q_oe^{-t/\tau}}{\tau}[/tex]

putting values give 6

but answer is 7
 

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Answers and Replies

  • #2
SammyS
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Is the 1Ω shorting wire temporary? -- that is to say: is it removed before the switch is closed?
 
  • #3
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No ... but the question asks use for t=0 ... i.e. when capacitor behaves as a wire ... so it will short circuit the resistor ... :)
 
  • #4
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SOme help please ????????????
 
  • #5
SammyS
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Yup, at t=0, capacitor A has zero potential across it.

At time t=0, what's the potential across capacitor B ?
 
  • #6
SammyS
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Sorry for the delay. The problem is written in an unfamiliar way (at least for me).

I also came up with 6 amps @ t=0.
 
  • #7
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so that means that the answer in book is wrong?
 
  • #8
SammyS
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I think it must be wrong.
 

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