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7.Alternating Current
hard
Two circuit are shown in the figure $(a)$ and $(b).$ At a frequency of $....\,rad/s$ the average power dissipated in one cycle will be same in both the circuits.

A
$1000$
B
$200$
C
$500$
D
$5$
(JEE MAIN-2021)
Solution
$P_{\text {avg }}=\frac{v_{\text {rms }}^{2}}{R}$
$\frac{v_{ rms }^{2}}{Z^{2}} \times R=\frac{v_{r m s}^{2}}{R} \times 1$
$R^{2}=Z^{2}$
$25=\left(\sqrt{\left(x_{C}-x_{L}\right)^{2}+5^{2}}\right)^{2}$
$25=\left(x_{c}-x_{L}\right)^{2}+25$
$x_{c}=x_{L} \Rightarrow \frac{1}{\omega C}=\omega L$
$\omega^{2}=\frac{1}{L C}=\frac{10^{6}}{0.1 \times 40}$
$\omega=500$
Standard 12
Physics
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