Electrical Engineering - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 21)
21.
A certain series RLC circuit with a 200 Hz, 15 V ac source has the following values: R = 12 
, C = 80 
F, and L = 10 mH. The total impedance, expressed in polar form, is
, C = 80 F, and L = 10 mH. The total impedance, expressed in polar form, is12.28
12.34°
12.34° 12.5712.34°
12.34° 9.9512.34°
12.34° 12.6212.34°
12.34° Discussion:
6 comments Page 1 of 1.
SANJAY said:
9 years ago
Z = R+j(Xl-Xc).
Xl = 2*3.14*f*l.
Xc = 1/(2*3.14*f*c).
Option A.
Xl = 2*3.14*f*l.
Xc = 1/(2*3.14*f*c).
Option A.
Arun said:
8 years ago
How is this calculation please explain?
GAYATRI said:
8 years ago
Z = sq.root of [r2+(xl-xc)2] where,
Xl = 2*3.14*f*l = 2*3.14*200*10x10-3 = 12.56637.
Xc = 1/(2*3.14*f*c) = 1/(2*3.14*200*80*10-6) = 9.947.
Z = Sq.root of [12*12+(12.566 - 9.947)*(12.566 - 9.947)].
= Sq.root of [144+(2.619*2.619).
= Sq.root of [144+6.86].
= Sq.root of 150.860.
= 12.28.
Xl = 2*3.14*f*l = 2*3.14*200*10x10-3 = 12.56637.
Xc = 1/(2*3.14*f*c) = 1/(2*3.14*200*80*10-6) = 9.947.
Z = Sq.root of [12*12+(12.566 - 9.947)*(12.566 - 9.947)].
= Sq.root of [144+(2.619*2.619).
= Sq.root of [144+6.86].
= Sq.root of 150.860.
= 12.28.
Quest77 said:
5 years ago
The angle should be 29.99 degrees. Inverse tangent of (xl-xc)\r.
HAris said:
3 years ago
Z=√R^2+(XL-XC)^2
Z=√12^2+((2*PIE*F*L)-(1/2*PIE*F*C))^2
Z=√144+12.566-9.947.
Z=12.28 ohm.
FOR ANGLE θ due to Polar Form as Required,
θ = tangent inverse*(XL-XC/R).
= tan.inv*(12.566-9.947/12).
= 12.31.
So,
= 12.28<12.31* ohm , Option A.
Z=√12^2+((2*PIE*F*L)-(1/2*PIE*F*C))^2
Z=√144+12.566-9.947.
Z=12.28 ohm.
FOR ANGLE θ due to Polar Form as Required,
θ = tangent inverse*(XL-XC/R).
= tan.inv*(12.566-9.947/12).
= 12.31.
So,
= 12.28<12.31* ohm , Option A.
SONAL said:
1 year ago
@Haris.
How to calculate the below step? Please explain.
θ = tangent inverse*(XL-XC/R).
= tan.inv*(12.566-9.947/12).
= 12.31.
How to calculate the below step? Please explain.
θ = tangent inverse*(XL-XC/R).
= tan.inv*(12.566-9.947/12).
= 12.31.
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