Saved Bookmarks
| 1. |
A variable frequency 230 V alternating voltage source is connected across a series combination of `L = 5.0 H, C = 80 mu F` and `R = 40 Omega` (i) Calculate the angurla frequency fo the source which drives the circuitin resonance. (ii) Obtain r.m.s. the impedance of the circuit and amplitude of current at resonance frequency (iii) Obtain r.m.s. potential drop across the three elements of the circuit at resonating frequency. (iv) How do you explain the observation that the algerbraic sum of voltages across the three elements obtained is greater than the supplied voltage. |
|
Answer» (i) `omega_(r) = (1)/(sqrt(LC)) = (1)/(sqrt(5.0 xx 80 xx 10^(-6))) = 50 rad s^(-1)` (ii) At resonance, `Z = R = 40 ohm` `I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (1.414 xx 230)/(40) = 8.13 A` (iii) `I_(v) = (E_(v))/(Z) = (230)/(40) = (23)/(4) A` `V_(R ) = I_(v) xx R = 23//4 xx 40 = 230` volt `V_(L) = I_(v) X_(L) = I_(v) (omega L) = 23//4 xx 50 xx 5.0 = 1437.5 V` `V_(C ) = I_(v) X_(C ) = (1_(v))/(omega C) = (23//4)/(50 xx 80 xx 40^(-6)) = 1437 .5 V` The three voltage, `vec V_(R)`, `vec V_(L)` and `vec V_(C )` are phasors/vector. Out of these, `vec V_(L)` and `vec V_(C )` are in opposite phase. Therefore, they cancel out. The vector sum of the three voltage `= vec V_(R ) = 230 V`, which is the source voltage. |
|