An a.c. voltage represented be e = 310 sin 314 t is connected in series to a `24 Omega` resistor, 0.1 H inductor and a `25 muF` capacitor. 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`

Explanation

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^(@)`.