Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 19)
19.
What is the impedance of the circuit in the given circuit?
1.05 k
1.33 k
2.19 k
3.39 k
Discussion:
10 comments Page 1 of 1.
Pritam said:
5 years ago
Thanks @Ehtesham.
(1)
Louie Bacalla said:
6 years ago
r=1.2K ohms.
Xl=2*π*f*L = 423π ohms.
So, Xc = 1/(2*π*f*C) = 3386.275385 ohms.
For parallel circuit:
Y = G+i(Bc-Bl).
Where:
G = 1/r.
Bc = 1/Xc,
Bl = 1/Xl.
Finally, solving for Impedance
Z = 1/Y
Z = 1052.064861 ohms or 1.052k ohms.
Xl=2*π*f*L = 423π ohms.
So, Xc = 1/(2*π*f*C) = 3386.275385 ohms.
For parallel circuit:
Y = G+i(Bc-Bl).
Where:
G = 1/r.
Bc = 1/Xc,
Bl = 1/Xl.
Finally, solving for Impedance
Z = 1/Y
Z = 1052.064861 ohms or 1.052k ohms.
Winny said:
6 years ago
Please, can anyone explain it in brief?
Murugan said:
7 years ago
Z= √ (1/1 ÷ r^2+(1 ÷ xl-1 ÷ xc)^2).
Ehtesham said:
1 decade ago
This is the best way to calculate impedance.
First calculate total current into the circuit.
R = 10mA.
xc = 3.54mA.
xl = 9.03mA.
I = sqrt(10m*10m+(9.03m*3.03m-3.54m3.54m)).
Now I = 11.40mA.
Now,
Total impedance z = vs/total current.
12/11.40m.
z = 1.05km.
First calculate total current into the circuit.
R = 10mA.
xc = 3.54mA.
xl = 9.03mA.
I = sqrt(10m*10m+(9.03m*3.03m-3.54m3.54m)).
Now I = 11.40mA.
Now,
Total impedance z = vs/total current.
12/11.40m.
z = 1.05km.
Sachin said:
1 decade ago
And w=295.3*10^3.
G=8.33*10^-4.
Bc=2.953*10^-4.
Bl=7.525*10^-4.
Bc-Bl=-4.572*10^-4.
-Ve Sign means Ic<IL.
Y = 9.5023*10^-4.
So Z = 1/Y, = 1052.3 or 1.05 kohms.
BL = Inductive Susceptance.
Bc = Capacitive Susceptance.
G=8.33*10^-4.
Bc=2.953*10^-4.
Bl=7.525*10^-4.
Bc-Bl=-4.572*10^-4.
-Ve Sign means Ic<IL.
Y = 9.5023*10^-4.
So Z = 1/Y, = 1052.3 or 1.05 kohms.
BL = Inductive Susceptance.
Bc = Capacitive Susceptance.
Sachin said:
1 decade ago
Z = 1/Y.
And in II circuit,
Y = G+j(Bc-Bl).
Where G = 1/r, Bc = wC, Bl = 1/w And w = 2*pi*f.
And take square root of complex value of Y.
And then take inverse.
And in II circuit,
Y = G+j(Bc-Bl).
Where G = 1/r, Bc = wC, Bl = 1/w And w = 2*pi*f.
And take square root of complex value of Y.
And then take inverse.
Vasu said:
1 decade ago
The correct formula for a parallel RLC circuit is given below.
1/Z=1/R + 1/Xc + 1/Xl.
as 1/r= 8.33*10^-4
1/Xl= -j7.5*10^-4
1/Xc=j2.95*10^-4
which means , 1/Z= 8.33*10^-4 -j7.5*10^-4+j2.95*10^-4
= 8.33*10^-4 -j4.55*10^-4
So Z= 1/sq.root (square (8.33*10^-4) +square(4.55*10^-4))
=0.105*10^4
Z= 1.05 Kohms
1/Z=1/R + 1/Xc + 1/Xl.
as 1/r= 8.33*10^-4
1/Xl= -j7.5*10^-4
1/Xc=j2.95*10^-4
which means , 1/Z= 8.33*10^-4 -j7.5*10^-4+j2.95*10^-4
= 8.33*10^-4 -j4.55*10^-4
So Z= 1/sq.root (square (8.33*10^-4) +square(4.55*10^-4))
=0.105*10^4
Z= 1.05 Kohms
Mahi said:
1 decade ago
Its got 2 b wc-1/wl
Elkaa said:
1 decade ago
I calculate xl=1328 ohm, xc=3388 ohm
xc-xl = 2060 ohm
z= 1037 ohm ( 1/ sqr root of (1/R2 + 1/xt2)
Please advise where this assumption is wrong ?
Thanks.
xc-xl = 2060 ohm
z= 1037 ohm ( 1/ sqr root of (1/R2 + 1/xt2)
Please advise where this assumption is wrong ?
Thanks.
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