Question

A series circuit has an impedance of 62.0 Ω and a power factor of 0.720 at a frequency of 47.0 Hz . The source voltage lags the current.Part A What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? a. inductor b. capacitorPart B What size element will raise the power factor to unity? L = ___.

Answer 1

Answer:

L = 0.108 H

Explanation:

given,

impedance in the series circuit(Z) = 62 Ω

Power factor = 0.720

frequency = 47 Hz

A) Inductor will be placed in series so, that circuit raises to its power factor.

B)  Using expression of power factor

    $cos \phi = \dfrac{R}{Z}$

    $R = Z cos \phi$

    $R = 62 \times 0.720$

          R = 44.64 N

Source voltage lags since it is capacitive element, Capacitive reactant can is given by

  $X_c = \sqrt{Z^2-R^2}$

  $X_c = \sqrt{55^2-44.64^2}$

  $X_c =32.12\ ohm$

Power factor is unity

    $\dfrac{R'}{Z'} = 1$

           R' = Z'

impedence

$X_L = \sqrt{R^2 + (X_L-X_C)^2}$

$X_L-X_C = 0$

$X_L= X_C$

Inductive reactance

$X_L = L\omega$

$L = \dfrac{X_L}{\omega}$

$L = \dfrac{X_L}{2\pi f}$

$L = \dfrac{32.12}{2\pi \times 47}$

L = 0.108 H