What is resistance of a DC circuit has resistor n capacitor connect parallel

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Homework Statement



the title says everything, any formula that to find resistance in a DC circuit of one resistor and one capacitor connect parallel, as well as the current through each of R and C (the known values are R C and V)

Homework Equations



seem to me i = (V/R)e^ (-e/RC), but it only say for series

The Attempt at a Solution



couldn't find answer in text and on any website
 

Answers and Replies

  • #3
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does that mean the current won't go through capcitor branch? and that means if i want to find resistance of the circuit, it's simply R?(nothin to do with C)
and wat about time constant? T = RC? will it be the same like when capacitor n resistor connect series?
 
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  • #4
Hootenanny
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does that mean the current won't go through capcitor branch? and that means if i want to find resistance of the circuit, it's simply R?(nothin to do with C)
It means that after a very short time interval, no current will pass through the capacitor branch, all the current will flow through the resistor.
 
  • #5
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what about time constant T = RC, will it be zero also?
 
  • #6
Hootenanny
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what about time constant T = RC, will it be zero also?
There is no time-constant in this case since after the initial current spike through the capacitor branch, the circuit simply behaves like a a resistor connected across a potential difference.
 
  • #7
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thanx a lot, that quite a lot info for me
 
  • #8
Hootenanny
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thanx a lot, that quite a lot info for me
No problem. If you need to we can treat the network more rigorously, so if you have any more questions please feel free to return.
 
  • #9
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What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get [itex]q/C = 0[/itex] so q = 0 (i.e., it discharges immediately). Hmm.
 
  • #10
Hootenanny
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What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get [itex]q/C = 0[/itex] so q = 0 (i.e., it discharges immediately). Hmm.
That is indeed the case.
 
  • #11
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What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get [itex]q/C = 0[/itex] so q = 0 (i.e., it discharges immediately). Hmm.

Which is why you should be careful while shorting capacitors in the lab :smile:
 

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