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.
Anonymous said:
1 decade ago
Y=1/R+jωC
So magnitude, |Y|= sqrt[ (1/R)^2 + (ωC)^2 ]
Correct answer is B.
So magnitude, |Y|= sqrt[ (1/R)^2 + (ωC)^2 ]
Correct answer is B.
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.
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.
Dilip.sharan said:
1 decade ago
Wrong answer.
Right one is 3.14mS + j318.30mS.
Right one is 3.14mS + j318.30mS.
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