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.
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.
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.
Y = 1/470 + j*2*3.14*2.5*10^3*0.2*10^-6.
Y = 2.12 ms + j 3.14 ms.
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