# Electrical Engineering - RC Circuits - Discussion

### Discussion :: 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?

 [A]. 37.9 [B]. 3.7 [C]. 14,400 [D]. 4,800 Explanation:

No answer description available for this question.

 Jeyanand Nec said: (Mar 9, 2012) 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

 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) @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.

 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. Simplify, 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) Znet=(R*jXc)/R-jXc 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.