JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 16)
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 $${{22} \over 7}$$]
8.8 A and $${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$$
88 A and $${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$$
0.88 A and $${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$$
8.8 A and $${\tan ^{ - 1}}\left( {{{6} \over 11}} \right)$$
Explanation
$$\phi = {\tan ^{ - 1}}\left( {{{{X_L}} \over R}} \right)$$
$${X_L} = \omega L$$
$${X_L} = 2 \times {{22} \over 7} \times 50 \times 0.07 = 22\Omega $$
$$\phi = {\tan ^{ - 1}}\left( {{{22} \over {12}}} \right)$$
$$R = 12\Omega $$
$$\phi = {\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$$
$$Z = \sqrt {X_L^2 + {R^2}} = 25.059$$
$$I = {V \over Z} = {{220} \over {25.059}} = 8.77A$$
$${X_L} = \omega L$$
$${X_L} = 2 \times {{22} \over 7} \times 50 \times 0.07 = 22\Omega $$
$$\phi = {\tan ^{ - 1}}\left( {{{22} \over {12}}} \right)$$
$$R = 12\Omega $$
$$\phi = {\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$$
$$Z = \sqrt {X_L^2 + {R^2}} = 25.059$$
$$I = {V \over Z} = {{220} \over {25.059}} = 8.77A$$
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