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

[Learner123]

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

 

[Hootenanny]

Parallel RC networks aren't really that interesting since the capacitor becomes charged very quick and hence acts like an open branch. See http://en.wikipedia.org/wiki/RC_circuit#Parallel_circuit for more information.

 

[Learner123]

does that mean the current won't go through capacitor 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?

 

[Hootenanny]

does that mean the current won't go through capacitor 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.

 

[Learner123]

what about time constant T = RC, will it be zero also?

 

[Hootenanny]

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.

 

[Learner123]

thanx a lot, that quite a lot info for me

 

[Hootenanny]

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.

 

[jdstokes]

What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get $q/C = 0$ so q = 0 (i.e., it discharges immediately). Hmm.

 

[Hootenanny]

What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get $q/C = 0$ so q = 0 (i.e., it discharges immediately). Hmm.

That is indeed the case.

 

[naresh]

What happens if we remove the resistor so the circuit consists just of a charged capacitor? Going around the loop we get $q/C = 0$ so q = 0 (i.e., it discharges immediately). Hmm.

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