Saved Bookmarks
| 1. |
An inductor 200 mH, capacitor 500 uF, resistor 10 Omegaare connected in series with a 100 V variable frequency a.c. source. Calculate : (i) frequency at which power factor in the circuit is unity, (ii) current amplitude at this frequency, (iii) Q-factor. |
|
Answer» Solution :It is GIVEN that L = 200 mH `=0.2 H, C = 500 muF = 5 xx 10^(-4) F, R = 10 Omega` and `V_(rms) = 100 V` (i) Power factor of the circuit is unity when Z = R or `X_(L) = X_(C)` and for this angular frequency `omega = 1/sqrt(LC) = 1/ sqrt(0.2 xx 5 xx 10^(-4)) = 100 s^(-1)` (ii) Current amplitude at this frequency `I_(m) = sqrt(2) I_(rms) - sqrt(2) V_(rms)/R = (sqrt(2) xx 100)/10 = 14.1 A` (iii) Q-factor `=X_(L)/R = (Lomega)/R = (0.2 xx 100)/10 =2` |
|