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
resistor and a 0.2 F capacitor are in parallel across a 2.5 kHz ac source. The admittance, Y, in rectangular form, is212
2.12 mS + j3.14 mS
3.14 mS + j2.12 mS
318.3
Discussion:
13 comments Page 1 of 2.
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.
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.
Binny bansal said:
7 years ago
F is the frequency which is given in the question in KHz.So it is 2.5KHz.
KULDEEP said:
7 years ago
Here, how to calculate the value of (F)?
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.
Y = 1/Z = 1 / (470 - 318.31j).
Y = 1.458x10^-3 + 9.878x10^4j S.
(1)
Prasanta said:
9 years ago
Please explain it briefly.
Dileep said:
10 years ago
Can you explain clearly I can't understand?
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.
Y = 1/470 + j*2*3.14*2.5*10^3*0.2*10^-6.
Y = 2.12 ms + j 3.14 ms.
Khougali said:
1 decade ago
In parallel it 1/zt=1/r+1/-jxc then y=1/zt answer correct.
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.
Now impedance Z = (Xr*Xc)/(Xr+Xc).
Admittance Y = 1/Z = 2.12ms+j3.14ms.
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.
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.
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