Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 11)
11.
What is the resonant frequency in the given circuit?
Discussion:
21 comments Page 1 of 3.
Hussain said:
1 decade ago
Please provide the explanation
Fr = 1/(2*pi*sqrt(L*C));
L = 75mH C = 0.033 UF on substituting these values in above formula Fr = 0.101 KHz.
where i went wrong?
Fr = 1/(2*pi*sqrt(L*C));
L = 75mH C = 0.033 UF on substituting these values in above formula Fr = 0.101 KHz.
where i went wrong?
Akki said:
1 decade ago
Dude you are applying correct formula you once again check your calculation m getting correct ans.
Razif said:
1 decade ago
F = 1/(2*pi*sqrt(L*C)) = 1/(2*pi*4.98*10^-5) = 3.2 kHz
Yogesh said:
1 decade ago
@Hussain
you insert 1000 for inducter and (e-6) for capacitor Milli and you value.
you insert 1000 for inducter and (e-6) for capacitor Milli and you value.
Manikanta said:
1 decade ago
Here what is Rw?
Monica said:
1 decade ago
@Manikanta
Here Rw is nothing resistance but it is simply given its not of use
Resonant Frequency=1/(2*pi*sqrt(L*C))
=1/(2*pi*qrt(75*10^-3*0.033*10^-6))
= 31.99
=3.2khz
Here Rw is nothing resistance but it is simply given its not of use
Resonant Frequency=1/(2*pi*sqrt(L*C))
=1/(2*pi*qrt(75*10^-3*0.033*10^-6))
= 31.99
=3.2khz
Guru said:
1 decade ago
It is parallel resonance circuit then why you are applying series resonance formula please. Explain me.
Weldeabreha beyene said:
1 decade ago
@Guru.
Let me explain, it is because resonance does not bother whether, series or parallel, simply (C&L).
Let me explain, it is because resonance does not bother whether, series or parallel, simply (C&L).
Nilesh said:
1 decade ago
@Hussain said is right because after calculation it comes 0.101khz not 3.2khz
Check out calculation correctly
Fr = 1/(2*pi*sqrt(L*C));
L = 75mH C = 0.033 UF on substituting these values in above formula Fr = 0.101 KHz.
Check out calculation correctly
Fr = 1/(2*pi*sqrt(L*C));
L = 75mH C = 0.033 UF on substituting these values in above formula Fr = 0.101 KHz.
Pushpendra chaurasiya said:
1 decade ago
When i calculated first my ans was 0.101
n now my ans is 3.2KHz
Let me expalin,
Fr = 1/[2*3.14*sqrt{(75*10^-3)(0.033*10^-6)}]
= 1/[2*3.14*sqrt{2.475*10^-9}]
=1/[2*3.14*4.975*10^-5]
=1/[3.124*10^-4]
n now my ans is 3.2KHz
Let me expalin,
Fr = 1/[2*3.14*sqrt{(75*10^-3)(0.033*10^-6)}]
= 1/[2*3.14*sqrt{2.475*10^-9}]
=1/[2*3.14*4.975*10^-5]
=1/[3.124*10^-4]
(1)
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