Impedance of a parallel rc circuit

Thread Starter

ryanmoser

Joined Nov 5, 2015
3
  1. If a circuit consisted of a resistor R parallel with a capacitor C is given an external current excitation with frequency of ω, let the current equals to I0*exp(jωt), what is the response of voltage E? Would the impedance equals to R(1-ωCRj)/(1+ω^2*C^2*R^2), the same as the case when the circuit is excited by AC voltage E0*exp(jωt)?
 

crutschow

Joined Mar 14, 2008
23,349
The impedance of a parallel RC circuit is inversely proportional to frequency.
 

alfacliff

Joined Dec 13, 2013
2,449
there is no resonant point for rc circuits like there is on lc circuits. the impedance, actually reactance, changes with frequency in a fairly linear manner.
 

BillB3857

Joined Feb 28, 2009
2,497
I seem to remember something about phase angle and ELI the ICE man coming into play. Something about a vector???
 

alfacliff

Joined Dec 13, 2013
2,449
eli the ice man is for lcr circuits. phase lead and lag with capacitive and inductive circuits. for rc, its just a ratio of capaccitive reactance and resistance gives phase shift.
 

BillB3857

Joined Feb 28, 2009
2,497
My post (#4) was intended to be a hint to the OP. Please correct me if I'm wrong, but when calculating effective impedance, isn't the answer a vector value of Xc and R? It has been a long time since exposure to those details.
 

Thread Starter

ryanmoser

Joined Nov 5, 2015
3
The impedance of a parallel RC circuit is inversely proportional to frequency.
20151106184554.jpg
The attached figure shows how I derive the response of voltage with regard to a AC current signal. However, additional term (RED) is found in the expression of E? Where is wrong?
 

jjw

Joined Dec 24, 2013
501
Do you really need to derive these equations or can you just use known impedances of R and C and then the parallel impedance of these ?

The term I0/(s-jw) must be wrong.
 

BillB3857

Joined Feb 28, 2009
2,497
It took some digging, but I found support to what I seemed to have remembered from so long ago....

You probably don't have a navigation computer, but a calculator will work.
 

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