Electrical Engineering - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 14)
14.
If the value of C in a series RLC circuit is decreased, the resonant frequency
Discussion:
6 comments Page 1 of 1.
Manju yadav said:
9 months ago
Fr is inversely proportional to C.
Jadeppa bgm said:
6 years ago
It's simple,
Xl=xc is resonant frequency.
Xl>xc is resonant frequency decrease.
Xl<xc is resonant frequency increase i.e c decrease means xc increase xc=1/2πfc.
Xl=xc is resonant frequency.
Xl>xc is resonant frequency decrease.
Xl<xc is resonant frequency increase i.e c decrease means xc increase xc=1/2πfc.
SUBHAJIT RAY said:
8 years ago
@Srinu wrote resonant fq = 1/r(root (L/C)).
But resonant fq = 1/r(root (LC)). I can't understand.
But resonant fq = 1/r(root (LC)). I can't understand.
Roshni singh said:
9 years ago
Because in resonance wL= 1/wC then, when L is increases then C is decreases and vice-versa.
Saurav kumar said:
1 decade ago
xc=xl
1\wc=wl
1\w2lc=1
w2=1\lc
f=1\2*3.14root(l*c)
1\wc=wl
1\w2lc=1
w2=1\lc
f=1\2*3.14root(l*c)
Srinu said:
1 decade ago
Fr=(1/R(root(L/C))). Hence as C decreases fr increases.
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