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

## 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

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?

Last edited:
Hootenanny
Staff Emeritus
Gold Member
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.

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

Hootenanny
Staff Emeritus
Gold Member
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.

thanx a lot, that quite a lot info for me

Hootenanny
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
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.
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.