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.
Siva3cool@gmail.com said:
1 decade ago
Please work out.
Vid said:
1 decade ago
R and C are parallel
(R/1+RjWC)=R/sqrt(1+(RwC)2)=4.16
(R/1+RjWC)=R/sqrt(1+(RwC)2)=4.16
Keerthi said:
1 decade ago
Impedance z = sqrt(r^2+xc^2)
Then its value is 13 but how it is 4.46 tell me please?
Then its value is 13 but how it is 4.46 tell me please?
Subbu said:
1 decade ago
Here, capacitor reactance is given so 12*5/(12+5)=3.5 ohms
Plz, say why is it wrong?
Plz, say why is it wrong?
Rathi said:
1 decade ago
Impedance of RC Parallel circuit is RXc/(sqrt(Xc^2+R^2))
= 12x5/sqrt(25+144)
= 4.6154 ohms
= 12x5/sqrt(25+144)
= 4.6154 ohms
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.
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).
Glenn said:
1 decade ago
Rathi is correct the formula is RXc/sqrt(R^2 +Xc^2).
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.
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.
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