JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 24)
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8$$\Omega$$, L = 24 mH and C = 60 $$\mu$$F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.
Отвечать
4
Объяснение
At resonance power (P)
$$P = {{{{({V_{rms}})}^2}} \over R}$$
$$ \therefore $$ $$P = {{{{(250/\sqrt 2 )}^2}} \over 8}$$
$$ \Rightarrow $$ P = 3906.25 w
$$ \Rightarrow $$ P $$ \cong $$ 4 Kw
$$P = {{{{({V_{rms}})}^2}} \over R}$$
$$ \therefore $$ $$P = {{{{(250/\sqrt 2 )}^2}} \over 8}$$
$$ \Rightarrow $$ P = 3906.25 w
$$ \Rightarrow $$ P $$ \cong $$ 4 Kw