Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 23)
23.
A 120 
resistor is in parallel with a capacitor with a capacitive reactance of 40 . Both components are across a 12 V ac source. What is the magnitude of the total impedance?
resistor is in parallel with a capacitor with a capacitive reactance of 40 . Both components are across a 12 V ac source. What is the magnitude of the total impedance?37.9
3.7
14,400
4,800
Discussion:
10 comments Page 1 of 1.
RAZA said:
7 years ago
Impedence in perallel is = z = RXc /√ R^2+Xc^2.
R!tesh said:
9 years ago
sqrt(1/Z) = sqrt(1/r) + sqrt(1/Xc).
1/Z = sqrt((1/120)^2 + (1/40)^2).
1/Z = 0.0263.
Z = 37.94.
1/Z = sqrt((1/120)^2 + (1/40)^2).
1/Z = 0.0263.
Z = 37.94.
Sudhir said:
9 years ago
Znet=(R*jXc)/R-jXc
Solve by substituting the value.
Solve by substituting the value.
Parneet kaur said:
1 decade ago
If in case of parallel RC circuit all the parameters are like 1/z and 1/r then why 1 /xc is not done.
Beder B said:
1 decade ago
Zr = 120.
Zc = j(XL-Xc) and XL = 0 and Xc = 40 therefore Zc = -j40.
1/Zeq = 1/Zr+1/Zc = 1/120+1/-j40 = (120-j40)/(-j4800).
Zeq = (-j4800)/(120-j40).
Multiply by (120+j40)/(120+j40).
Zeq = (192000 - j576000)/16000.
Simplify,
Zeq = 12-j36.
|Zeq| = sqrt((12^2)+(36^2) = 37.9.
Zc = j(XL-Xc) and XL = 0 and Xc = 40 therefore Zc = -j40.
1/Zeq = 1/Zr+1/Zc = 1/120+1/-j40 = (120-j40)/(-j4800).
Zeq = (-j4800)/(120-j40).
Multiply by (120+j40)/(120+j40).
Zeq = (192000 - j576000)/16000.
Simplify,
Zeq = 12-j36.
|Zeq| = sqrt((12^2)+(36^2) = 37.9.
RAKESH DEVARAJ said:
1 decade ago
@RAKESH DEVARAJ.
WE NOW THAT Z = sqrt(R^2+XC^2) .
FOR PARALLEL Z = 1/(sqrt(1/R^2+XC^2)).
So z=1/(sqrt(1/120^2+1/40^2)).
z=37.94 Ohm.
WE NOW THAT Z = sqrt(R^2+XC^2) .
FOR PARALLEL Z = 1/(sqrt(1/R^2+XC^2)).
So z=1/(sqrt(1/120^2+1/40^2)).
z=37.94 Ohm.
Dilip verma said:
1 decade ago
Because it is in parallel rc circuit so Ir = V/R = 12/120 = 0.1 and Ic = 12/40 = 0.3
So I = root(Ir2+Ic2) = root.1 = 0.31622.
Z = V/I = 12/.316 = 37.9
So I = root(Ir2+Ic2) = root.1 = 0.31622.
Z = V/I = 12/.316 = 37.9
(1)
Preplexed said:
1 decade ago
Is it a convention to solve such questions through Admittances? Can I have the reasoning behind it please?
Towqeer said:
1 decade ago
Sorry! i'm not getting this....why u solve through Admittance__as we have to find impedance then why not to solve through: Z=sqrt R^2 + Xc ^2 _even we got through this wrong answer.
and by ur method ..answer is right...but i want to the Logic..plz !!
and by ur method ..answer is right...but i want to the Logic..plz !!
Jeyanand NEC said:
1 decade ago
Z=1/Y
Y=(1/R)+(j/Xc)
=sqr(0.0083^2+0.025^2)
=0.02634 mho
Z=1/0.02634=37.9 ohms
Y=(1/R)+(j/Xc)
=sqr(0.0083^2+0.025^2)
=0.02634 mho
Z=1/0.02634=37.9 ohms
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