Question

(a) For circuits used for transporting electric power, a low power factor implies larger power loss in transmission. Explain.
(b) power factor can oftern be improved by the use of a capacitor of appropraite capacitance in the circuit, Explain.

Answer 1

(a) We know that `P = E_(v) I_(v) cos phi`, where `cos phi` is the power factor.
To supply/transmit a given power P at a given voltage `E_(v)`, if `cos phi` is small, `I_(v)` has to be increased accourdingly. Therefore, power loss
`= I_(v)^(2)` R in transimission would increases.
(b) Power factor,
`cos phi = (R)/(Z) = (R )/(sqrt(R^(2) + (X_(L) - X_(C ))^(2)))`
To improve power factor, `cos phi rarr 1`, for which `Z rarr R`. In a circuit with given value of L, we can use a capacitor of appropriate capacitance so that `X_(C ) rarr X_(L)` and hence `Z rarr R`.