Discussion :: RC Circuits - General Questions (Q.No.23)
|Jeyanand Nec said: (Mar 9, 2012)|
|Towqeer said: (Sep 8, 2012)|
|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 !!
|Preplexed said: (Oct 2, 2012)|
|Is it a convention to solve such questions through Admittances? Can I have the reasoning behind it please?|
|Dilip Verma said: (Sep 7, 2013)|
|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
|Rakesh Devaraj said: (Sep 26, 2014)|
WE NOW THAT Z = sqrt(R^2+XC^2) .
FOR PARALLEL Z = 1/(sqrt(1/R^2+XC^2)).
|Beder B said: (Jan 28, 2015)|
|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.
Zeq = 12-j36.
|Zeq| = sqrt((12^2)+(36^2) = 37.9.
|Parneet Kaur said: (Jul 11, 2015)|
|If in case of parallel RC circuit all the parameters are like 1/z and 1/r then why 1 /xc is not done.|
|Sudhir said: (Apr 8, 2016)|
Solve by substituting the value.
|R!Tesh said: (Jun 8, 2016)|
|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.
|Raza said: (Nov 2, 2018)|
|Impedence in perallel is = z = RXc /√ R^2+Xc^2.|
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