Question

Two impedances `Z_1` and `Z_2` when connected separately across a `230 V, 50 Hz` supply consume `100 W` and `60 W` at power factor of `0.5` lagging and `0.6` leading respectively. If these impedances are now connected in series across the same supply, find
(a) total power absorbed and overall power factor
(b) the value of reactance to be added in series so as to raise the overall power factor to unity.

Answer 1

Correct Answer - A::B::D
(a) `0.5=R_1/Z_1`
Further, `P=V_("rms")i_("rms")cos phi`
or `100=230xx230/Z_1xx0.5`
`:.Z_1=264.5Omega`
and `R_1=132.25Omega`
Further `X_L=sqrt(Z_1^2-R_1^2)=sqrt3/2Z_1`
`=229Omega`
In second case `0.6=R_2/Z_2`
and `60=(230xx230)/Z_2xx0.6`
`:. Z_2=529Omega`
and `R_2=317.4Omega`
Further, `X_C=sqrt(Z_2^2-R_2^2)`
`=423.2Omega`
When connected in series,
`R=R_1+R_2=449.65Omega`
`X_C-X_L=194.2`
`:. Z=sqrt((449.65)^2+(194.2)^2)`
`=489.79Omega`
Power factor, `cosphi=R/Z=0.92` (leading)
`P=V_("rms")i_("rms")cosphi`
`=(230)(230/489.79)(0.92)`
`=99W`
(b) Since `X_C-X_L=194.C Omega`
Therefore, if `194.2Omega` inductive reactance is to be added in series, then it will become only `R` circuit and power factor will come unity.