Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 1)
1.
A 10 resistor, a 90 mH coil, and a 0.015 F capacitor are in series across an ac source. The impedance magnitude at 1,200 Hz below fr is
1,616
161
3,387
1,771
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 2 of 3.

Shailu said:   9 years ago
Is it correct? Above explained examples?

Rajat Dey said:   9 years ago
I know above formula but this is very long process and heavy time span. So how can do it quickly? Please help me.

Shwetha said:   9 years ago
Its (xl-xc). Because xl is greater than xc.

Prashant said:   9 years ago
I think result is not coming from above all formula I want real answer from you friends please reply.

ROBERT said:   10 years ago
XL=2*3.14*F*L THIS IS THE FORMAL.

I WAS ASK TO MAKE SO I CAN FIND UNKNOWN F AND L.

MOVE THING AROUND I CAME UP WITH SO ARE THESE RIGHT.

F=2*PI*L*XL.
L=2*PI*F*XL.

AGGREY KERE said:   10 years ago
From R-L-C circuit above, we know that; Z=root of R^2+(XL-XC)^2.
but XL=2pi*fr*L.
XC=1/2Pi*fr*C.

Manindra said:   10 years ago
Fr = 1/2*pi*sqrt(L*c)=4331.648.
f = fr-1200Hz = 3131.64.

Xl = Wl = 2*3.14*3131.64*l.

Xc = 1/wc.
Z = sqrt of(R^2+(XL-XC)^2) = 1616 ohm.

DURAI said:   1 decade ago
fr = 1/2*pi*sqrt (L*c).

Babu said:   1 decade ago
We know that,

Resonance frequency:

Fr = 1/2*pi*sqrt(L*C).

Xl = Wl = 2*3.14*f*l.

Xc = 1/wc.

Mrakovic said:   1 decade ago
fr = 1/2*pi*sqrt of LC.
fr = 4331.648.

f = fr-1200Hz=3131.64.
Z = sqrt of(R^2+(XL-XC)^2).

Z = 1617 ohm.


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