# Electronics - RL Circuits - Discussion

Discussion Forum : RL Circuits - General Questions (Q.No. 1)
1.
As frequency increases
both series and parallel RL impedance decrease
series RL impedance decreases and parallel RL impedance increases
series RL impedance increases and parallel RL impedance decreases
both series and parallel RL impedance increase
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

SAMurtza said:   6 years ago
Can be seen via MATLAB (Option C is correct)

w=0:1/10:2*π;
% RL Impedance Increases.

R=1; L=1;
Zs=R+j*w*L;
Zp=1./(1./R+1./j*w*L);
% normalized plots.
plot(w,abs(Zs)/max(abs(Zs)),w,abs(Zp)/max(abs(Zp))).

SAMurtza said:   6 years ago
Z(w)_Series=R+jwL.
Z(0)_Series= R & |Z(infinite)_Series|= infinite (w increases , |Z| increases in SERIES) [Open at High frequency].

Z(w)_Parallel=1/(1/R-j/wL) .
|Z(0)_Parallel |=infinite & Z(infinite )_Parallel = R ( w increases , |Z| decreases in PARALLEL) [Short at DC].

Jyothis said:   8 years ago
Find effective impedance of both series and parallel circuits. Analyse what happens when omega increases.

Raheel ahmed said:   1 decade ago
When the frequency of Parallel RL Circuit Increases, XL increases which causes IL (current through inductor) decreases. Decrease in IL causes It (It=Il+Ir) to decrease, which means by relation IT=Vs/Zt, the Zt (Total Impedance) Increases.

JAydeep Gidhvani said:   1 decade ago
Z=impedance
Z=R+XL
XL=2*Pi*F*L.

F Increase
XL Increase
SO Impedance increase.

In parallel the real impedence is in dinominator so optiion c is coorrect.

Sri Harshith Rajam said:   1 decade ago
Inductive Impedance (X)= Omega*Inductance
As omega is directly proportional to the frequency (2*pi*f) , the impedance increases and the inductor circuit tends to enter into the open circuit state !

Situ dash said:   1 decade ago
In both series & parallel

Xl=2*pi*f*L

So as freq increases Xl value increases & hence impedance increases.