JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 21)
Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.
_25th_July_Evening_Shift_en_21_1.png)
Answer
500
Explanation
For figure (a)
$${P_{avg}} = {{v_{rms}^2} \over R}$$
$${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$$
$${R^2} = {Z^2}$$
$$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$$
$$ = 25{({x_C} - {x_L})^2} + 25$$
$${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$$
$${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$$
$$\omega = 500$$
$${P_{avg}} = {{v_{rms}^2} \over R}$$
$${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$$
$${R^2} = {Z^2}$$
$$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$$
$$ = 25{({x_C} - {x_L})^2} + 25$$
$${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$$
$${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$$
$$\omega = 500$$
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