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An a.c. voltage represented be e = 310 sin 314 t is connected in series to a 24Omega resistor, 0.1 Hinductor and a 25 muFcapacitor. Find the value of the peak voltage rms voltage frequency rectus of the circuit impedance of the circuit and phase angle of the current. Data : R = 24 Omega, L = 0.1 H, C = 25 xx 10^(-6)F |
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Answer» Solution :E = 310 sin 314 t …(i) and `r = E_(0) sin omega t` …(ii) comparing equations (i) & (ii) `E_(0) = 310 V` `E_(rms) = (E_(0))/(sqrt(2)) = (310)/(sqrt(2)) = 219.2 V` `omegat = 314 t` `2pi v = 314` `v = (314)/(2 xx 3.14) = 50 Hz` Reactance `= X_(L) - X_(C) = Lomega - (1)/(C omega` `= L.2pi v - (1)/(C.2pi v)` `= 0.1 xx 2 pi xx 50 - (1)/(25 xx 10^(-6) xx 2pi xx 50)` `= 3.14 - 127.4` `= -96 Omega` `X_(L) - X_(C) = -96 Omega` `:. X_(C) - X_(L) = 96 Omega` `Z = sqrt(R^(2) + (X_(C) - X_(L))^(2))` `= sqrt(24^(2) + 96^(2))` `= sqrt(576 + 9216)` `= 98.9 Omega` `tan phi = (X_(C) - X_(L))/(R )` `= ((127.4 - 31.4)/(24))` `tan phi = (96)/(24) = 4` `phi = 76^(@)` Predominance of capacitive reactance signify that current leads the EMF by `76^(@)`. |
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