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RC Circuits - Engineering

Q1:

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 

ANS:B - 2.12 mS + j3.14 mS

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.