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.
(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.

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.

Mahitha said:   1 decade ago
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:   1 decade ago
In parallel it 1/zt=1/r+1/-jxc then y=1/zt answer correct.

Minhaj said:   1 decade ago
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:   10 years ago
Can you explain clearly I can't understand?

Prasanta said:   9 years ago
Please explain it briefly.


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