Q 12-07-111JEE MainJEE Main 2021 (27 Jul, Shift 1)Easy
A $0.07$ H inductor and a $12\ \Omega$ resistor are connected in series to a $220$ V, $50$ Hz AC source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as $\dfrac{22}{7}$]
Answer: (A) $8.8$ A and $\tan^{-1}\left(\dfrac{11}{6}\right)$
$X_L = 2\pi fL = 2\times\dfrac{22}{7}\times50\times0.07 = 22\ \Omega$.
$Z = \sqrt{12^2 + 22^2} = \sqrt{628} \approx 25.06\ \Omega$, so $I = \dfrac{220}{25.06} \approx 8.8$ A.
$\tan\phi = \dfrac{X_L}{R} = \dfrac{22}{12} = \dfrac{11}{6}$.
Solution by Sreeraj P, M.Sc Physics