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

Answer: Option B

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.

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