Question

A `25 (mu)F` capacitor, a0.1 H inductor and a `25 Omega` resistorare connected in series with an ac source of emf `E=310 sin 314 t`. Find
(a) the frequency of he emf
(b) The reactance of the circuit
(c ) the impedance of the circuit
(d) the current in the circuit.
(e) the phases angel
(f ) the effective voltages across the capacitor, inductor and resistor.

Answer 1

(a) Given `omega = 314 implies 2pi f=314 implies f=50Hz`
(b) Total reactance of the circuit is combination of capacitive and inductive reactance which is given by
`|X_(C) - X_(L)|=|(1)/(omega C)-omega L| = |(1)/(314xx25xx10^(-6))-314 xx 0.1|`
`=96 Omega`
(c) Impendance of the circuit is
`Z=sqrt(R^(2)+(X_(C)_X_(L))^(2))=sqrt(25^(2)+96^(2))=99.2 Omega`
(d) We have to find rms current in hte circuit in the current . It is given by
`I_(v)=(E_(v))/(Z)=(310//sqrt(2))/(99.2)=2.21 A`
(e) Let `phi` be the angle by which current leads emf, then
`tan (phi)=(X_(C)-X_(L))/(R ) =96/25 implies phi =75.4^(@)`
(f) Effective voltage are given by
`V_(Rv)=I_(v)R=25xx2.21=55.26 V`
`V_(Lv)=I_(v)R_(L)=2.21xx314xx0.1=69.4 V`
`V_(Cv)=I_(v)X_(C)=(2.21xx1)/(314xx25xx10^(-6))=281.6V`.