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 2 of 2.
KULDEEP said:
7 years ago
Here, how to calculate the value of (F)?
Binny bansal said:
7 years ago
F is the frequency which is given in the question in KHz.So it is 2.5KHz.
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.
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