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
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 is1,616
161
3,387
1,771
Discussion:
28 comments Page 2 of 3.
Teju said:
1 decade ago
XL=WL=(2*PI*f)L
XC=1/WC=1/(2*PI*f)C
where,L=90mh,C=.015microF, f=1200hz and R=10ohm
Z=sqrt of(R^2+(XL-XC)^2)
XC=1/WC=1/(2*PI*f)C
where,L=90mh,C=.015microF, f=1200hz and R=10ohm
Z=sqrt of(R^2+(XL-XC)^2)
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.
fr = 4331.648.
f = fr-1200Hz=3131.64.
Z = sqrt of(R^2+(XL-XC)^2).
Z = 1617 ohm.
Prashant said:
1 decade ago
I think result is not coming from above all formula I want real answer from you friends please reply.
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.
Resonance frequency:
Fr = 1/2*pi*sqrt(L*C).
Xl = Wl = 2*3.14*f*l.
Xc = 1/wc.
Dinesh said:
1 decade ago
fr = 1/ 2*Pi*sqrt(L*C)
(or)
XL=2*Pi*F*L
Z=Sqrt of(R^2+(XL-XC)^2)
(or)
XL=2*Pi*F*L
Z=Sqrt of(R^2+(XL-XC)^2)
Nosheen Memon said:
5 years ago
If Xc is greater than Xl what will be the formula for impedance.
(1)
Rahul mishra said:
1 decade ago
IN RLC SERIES CIRCUITS WE KNOW:
Z = (jw)^2+(jw)R/L+1/LC.
Z = (jw)^2+(jw)R/L+1/LC.
Sushma said:
6 years ago
1/z = √ [(1/R)^2+(1/Xl-1/Xc)^2] for RLC circuit.
Saikiran said:
5 years ago
I am not getting this, Please the solution clearly.
(2)
Shwetha said:
1 decade ago
Its (xl-xc). Because xl is greater than xc.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers