Question

A circuit draws a power of 550 watt from a source of 220 volt, `50 Hz`. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it is
A. `75 mu F`
B. `60 mu F`
C. `50 mu F`
D. `65 mu F`

Answer 1

Correct Answer - A
As the current lags behind ghe potentail difference, the circuit contains resistance and inducntace.
Power, `P = V_("rms") xx i_("rms") xx cos phi` ,
Here,
`i_("rms") = (V_("rms"))/(Z)`, where `Z = sqrt([(R^(2) + (omega L)^(2))])`
`P = (V_("rms")^(2) xx cos phi)/(Z)` or `Z = (V_("rms")^(2) xx cos phi)/(P)`
So, `Z = ((220)^(2) xx 0.8)/(550) = 70.8` ohm
Now, power factor `cos phi = (R )/(Z)` or `R = Z cos phi`
`:. R = 70.4 xx 0.8 = 56.32` ohm
Further,
`Z^(2) = R^(2) + (omega L)^(2)` or `(omega L) = sqrt((Z^(2) - R^(2)))` or
`omega L = sqrt((70.4)^(2) - (56.32)^(2)) = 42.2` ohm
When the capacitor is connected in the circuit,
`Z = sqrt([R^(2) + (omega L = (1)/(omega C))^(2)])` and
`cos phi = (R )/(sqrt([R^(2) + (omega L - (1)/(omega C))^(2)]))`
When `cos phi = 1, omega L = (1)/(omega C)`
`:. C = (1)/(omega (omega L)) = (1)/(2 pi f (omega L))`
`= (1)/((2 xx 3.14 xx 50) xx (42.2)) = 75 xx 10^(-6) F = 75 mu F`