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
[tex]cos \phi = \dfrac{R}{Z}[/tex]
[tex]R = Z cos \phi[/tex]
[tex]R = 62 \times 0.720[/tex]
R = 44.64 N
Source voltage lags since it is capacitive element, Capacitive reactant can is given by
[tex]X_c = \sqrt{Z^2-R^2}[/tex]
[tex]X_c = \sqrt{55^2-44.64^2}[/tex]
[tex]X_c =32.12\ ohm[/tex]
Power factor is unity
[tex]\dfrac{R'}{Z'} = 1[/tex]
R' = Z'
impedence
[tex]X_L = \sqrt{R^2 + (X_L-X_C)^2}[/tex]
[tex]X_L-X_C = 0[/tex]
[tex]X_L= X_C[/tex]
Inductive reactance
[tex]X_L = L\omega[/tex]
[tex]L = \dfrac{X_L}{\omega}[/tex]
[tex]L = \dfrac{X_L}{2\pi f}[/tex]
[tex]L = \dfrac{32.12}{2\pi \times 47}[/tex]
L = 0.108 H
