A 0 . 07H inductor and a 12Ω resistor are connected in series to a...

A $0 . 07H$ inductor and a $12Ω$ resistor are connected in series to a $220V, 50Hz$ AC source. The approximate current in the circuit and the phase angle between current and source voltage are respectively.

[Take $\pi$ as \left\dfrac{22}{7}\right]

Options

A8 . 8A $and$ \tan^{- 1}\left(\dfrac{11}{6}\right)
B88A $and$ \tan^{- 1}\left(\dfrac{11}{6}\right)
C0 . 88A $and$ \tan^{- 1}\left(\dfrac{11}{6}\right)
D8 . 8A $and$ \tan^{- 1}\left(\dfrac{6}{11}\right)

Correct answer

A. 8 . 8A $and$ \tan^{- 1}\left(\dfrac{11}{6}\right)

Step-by-step solution

$ϕ = \tan^{- 1}\left(\dfrac{{X}_{L}}{R}\right)$

${X}_{L} = \omegaL$

${X}_{L} = 2 \times \dfrac{22}{7} \times 50 \times 0 . 07 = 22Ω$

$ϕ = \tan^{- 1}\left(\dfrac{22}{12}\right)R = 12Ω$

$ϕ = \tan^{- 1}\left(\dfrac{11}{6}\right)$
$Z = \sqrt{{X}_{L}^{2} + {R}^{2}} = 25 . 059$
$I = \dfrac{V}{Z} = \dfrac{220}{25 . 059} = 8 . 77A$

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