Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 70)
70.
What is the impedance of an ac RC parallel circuit if the resistance is 12 ohms and the capacitive reactance equals 5 ohms?
Discussion:
24 comments Page 1 of 3.
Chayan said:
1 decade ago
Hello @Subbu,
The formula you have used is for parallely connected DC Resistance. Here we are given the reactance (resistance for ac analysis, the resistance a capacitor and an inductor introduces in a circuit is called reactance, which is given by Xc and Xl) and resistance. And as they are connected in parallel so the net resistance i.e. impedance will be R/sqrt (1+ (RWC) ^2).
The formula you have used is for parallely connected DC Resistance. Here we are given the reactance (resistance for ac analysis, the resistance a capacitor and an inductor introduces in a circuit is called reactance, which is given by Xc and Xl) and resistance. And as they are connected in parallel so the net resistance i.e. impedance will be R/sqrt (1+ (RWC) ^2).
Chayan said:
1 decade ago
Hello Vid,
Your concept is right but the calculation you have done, I think is not correct the result is 0.198 or 0.2ohm.
I can't understand how the other people gets the answer 4.7 ohm.
Please make it clear.
Your concept is right but the calculation you have done, I think is not correct the result is 0.198 or 0.2ohm.
I can't understand how the other people gets the answer 4.7 ohm.
Please make it clear.
Meow said:
6 years ago
In parallel RC/RLC/RL Ckts, get the admittance (Y) first. Y= G + jBc - jBL;
G =1/R; Bc =2pi*f*C; BL= 1/(2pi*f*L);
Y= 1/Z >>> Z is impedance.
Y = 1/12 + j1/5 >>> Z=1/Y=4.61 angle -67.38.
G =1/R; Bc =2pi*f*C; BL= 1/(2pi*f*L);
Y= 1/Z >>> Z is impedance.
Y = 1/12 + j1/5 >>> Z=1/Y=4.61 angle -67.38.
Vikas choudhary said:
1 decade ago
It is given in parallel so net impedance is given:
Z = (R*-jXc)/R-jXc= -j60/5-j12 by rationalisation we get:
720-j300/168 now taking magnitude.
Square root(720^2+300^2)/169 == 4.6153.
Z = (R*-jXc)/R-jXc= -j60/5-j12 by rationalisation we get:
720-j300/168 now taking magnitude.
Square root(720^2+300^2)/169 == 4.6153.
Samir said:
9 years ago
It's parallel circuit connection that's why the equation for it,
Z = RXC / SQRT OF (R^2 + XC^2).
And for series connection as you calculated above method:
z = sqrt of r^2 + xc^2.
Thanks.
Z = RXC / SQRT OF (R^2 + XC^2).
And for series connection as you calculated above method:
z = sqrt of r^2 + xc^2.
Thanks.
Almansur said:
6 years ago
For parallel circuit finding the total impedance is you need to find 1st the total admittance Y.
Y = R+jXc.
Y = 13 ohms,
Z = 1/Y,
Z = 0.076 ohms.
Y = R+jXc.
Y = 13 ohms,
Z = 1/Y,
Z = 0.076 ohms.
Akshar said:
8 years ago
Impedance Z in parallel RC circuit.
1/Z = √[(1/R^2)+(1/Xc)^2].
=√[(1/144)+(1/25)]=0 .21656.
Z = 1/0.21656= 4.617ohm.
1/Z = √[(1/R^2)+(1/Xc)^2].
=√[(1/144)+(1/25)]=0 .21656.
Z = 1/0.21656= 4.617ohm.
Prp said:
8 years ago
Impedance Z in parallel RC circuit.
1/Z= √[(1/R^2)+(1/Xc)^2].
=°[(1/144)+(1/25)]=0 .21656.
Z=1/0.21656= 4.617ohm.
1/Z= √[(1/R^2)+(1/Xc)^2].
=°[(1/144)+(1/25)]=0 .21656.
Z=1/0.21656= 4.617ohm.
BHADMANATHAN.C said:
1 decade ago
Impedance values for, parallel circuit is,
Z = V/I.
=> IR/I = R. So R = Xc = 5ohm.
For series circuit, Z = R+jXc.
Z = V/I.
=> IR/I = R. So R = Xc = 5ohm.
For series circuit, Z = R+jXc.
Kate said:
8 years ago
Impedance of parallel RC circuit is:
Z = R*Xc/(sqrt(R^2+Xc^2).
Plug in all values then you'll get 4.615384615 ~ 4.61
Z = R*Xc/(sqrt(R^2+Xc^2).
Plug in all values then you'll get 4.615384615 ~ 4.61
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