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  4. An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 Ω. The phase difference between current and voltage is

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Q. An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 $\Omega$. The phase difference between current and voltage is

KEAMKEAM 2005Alternating Current

Solution:

$ \tan \, \phi = \frac{ X_L}{ R} = \frac{ \omega L}{ R} = \frac{ 2 \pi f L}{ R } $
$ \therefore \tan \, \phi = \frac{ 2 \times 3.14 \times 50 \times 0.5 }{ 157 } = 1 $
$ \therefore \phi = 45^\circ$