Electrical Engineering - RC Circuits - Discussion

Discussion Forum : 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
212
2.12 mS + j3.14 mS
3.14 mS + j2.12 mS
318.3
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Paul said:   8 years ago
Z = R - jXc = 470 - 318.31j.
Y = 1/Z = 1 / (470 - 318.31j).
Y = 1.458x10^-3 + 9.878x10^4j S.
(2)

Binny bansal said:   8 years ago
F is the frequency which is given in the question in KHz.So it is 2.5KHz.
(2)

Eslam Wazeer said:   1 decade ago
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.
(1)

Dileep said:   1 decade ago
Can you explain clearly I can't understand?
(1)

Prasanta said:   9 years ago
Please explain it briefly.
(1)

KULDEEP said:   8 years ago
Here, how to calculate the value of (F)?
(1)

HAris said:   6 years ago
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.
(1)

Dilip.sharan said:   1 decade ago
Wrong answer.

Right one is 3.14mS + j318.30mS.

ABHILASH said:   1 decade ago
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:   1 decade ago
Y=1/R+jωC

So magnitude, |Y|= sqrt[ (1/R)^2 + (ωC)^2 ]

Correct answer is B.


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