# Electrical Engineering - RC Circuits - Discussion

### Discussion :: RC Circuits - General Questions (Q.No.3)

3.

A 470 resistor and a 0.2 F capacitor are in parallel across a 2.5 kHz ac source. The admittance, Y, in rectangular form, is

 [A]. 212 [B]. 2.12 mS + j3.14 mS [C]. 3.14 mS + j2.12 mS [D]. 318.3 Explanation:

No answer description available for this question.

 Dilip.Sharan said: (Jan 25, 2012) Wrong answer. Right one is 3.14mS + j318.30mS.

 Abhilash said: (Jan 27, 2012) Given, R=470ohm and c=0.2uf which are in parallel. So, Xc=1/2*pi*f*c Xc = 1/2*3.14*2.5*10^3*0.2*10^-6 = 318.30 We know that, z=R+jXc So z = 470 + j318.30 & z=1/y.

 Anonymous said: (Jan 28, 2012) Y=1/R+jωC So magnitude, |Y|= sqrt[ (1/R)^2 + (ωC)^2 ] Correct answer is B.

 Eslam Wazeer said: (Nov 12, 2012) R=470ohm ,c=0.2uf which are in parallel. Xc=1/2*pi*f*c Xc = 1/2*3.14*2.5*10^3*0.2*10^-6 = 318.30 z=R + jXc z = 470 + j318.30 y = 1/R + 1/Xc y = 2.12 mS + j3.14 mS.

 Mahitha said: (Sep 1, 2013) Xr = 470ohm, Xc = -j/(2*pi*2.5*1000*0.2*10^6) = -318.31j. Now impedance Z = (Xr*Xc)/(Xr+Xc). Admittance Y = 1/Z = 2.12ms+j3.14ms.

 Khougali said: (Aug 26, 2014) In parallel it 1/zt=1/r+1/-jxc then y=1/zt answer correct.

 Minhaj said: (May 25, 2015) Y = 1/R + jW. Y = 1/470 + j*2*3.14*2.5*10^3*0.2*10^-6. Y = 2.12 ms + j 3.14 ms.

 Dileep said: (Nov 8, 2015) Can you explain clearly I can't understand?

 Prasanta said: (Dec 27, 2016) Please explain it briefly.

 Paul said: (Nov 10, 2017) Z = R - jXc = 470 - 318.31j. Y = 1/Z = 1 / (470 - 318.31j). Y = 1.458x10^-3 + 9.878x10^4j S.

 Kuldeep said: (Mar 26, 2018) Here, how to calculate the value of (F)?

 Binny Bansal said: (Aug 20, 2018) F is the frequency which is given in the question in KHz.So it is 2.5KHz.

 Haris said: (Jan 10, 2020) They ask for only Admittance not the magnitude of Admittance so just apply: Y=(1/R+J1/XC). R= 470. XC=1/2*π*F*C. For Magnitude we used to apply; lYl= √(1/R^2+(1/wL-wC)^2.