A 0.07 H inductor and a 12$\Omega$ resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as ${{22} \over 7}$]
Solution
$\phi = {\tan ^{ - 1}}\left( {{{{X_L}} \over R}} \right)$<br><br>${X_L} = \omega L$<br><br>${X_L} = 2 \times {{22} \over 7} \times 50 \times 0.07 = 22\Omega$<br><br>$\phi = {\tan ^{ - 1}}\left( {{{22} \over {12}}} \right)$<br><br>$R = 12\Omega$<br><br>$\phi = {\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$<br><br>$Z = \sqrt {X_L^2 + {R^2}} = 25.059$<br><br>$I = {V \over Z} = {{220} \over {25.059}} = 8.77A$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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