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Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. |
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Answer» <P> Solution : 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 eachother 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 |
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