Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.

Explanation

Suppose in a given series ac circuit, current I lags behind the voltage by an angle `phi`. Here power factor is `cos phi = ( R )/( | Z| )`
Here resultant current phasor can be represented as,
`vec( I ) = vec( I_(p)) + vec( I_(q))`
where `I_(p) `= power component of `vec( I)`
`I_(q) ` = wattless component of `vec( I)`
`( :.` Angle between `vec( I_(q))` and `vec( V)` is `90^(@)` and so it is wattless )
Now , if we cannot a capacitor of an appropriate capacitance in parallel to above circuit then we can have `vec( I_(q) ^(.))` equal and opposite to `vec( I_(q))`. Here` vec(I_(q)^(.))` is leading and `vec( I_(q))` is lagging by same amount and so they cancel out each other and then `phi` becomes zero and so `cos phi ` =1 =` max and so,
`I_(rms) =( P)/( V_(rms) cos phi ) `= minimum
`rArr` Power loss would be minimum