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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