Two circuits are shown in the figures (a) and | vedclass

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(C) The average power dissipated in a purely resistive circuit (figure $a$) is given by $P_a = \frac{V_{	ext{rms}}^2}{R}$.The average power dissipate

Vedclass · Accepted Answer

$500$

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(C) The average power dissipated in a purely resistive circuit (figure $a$) is given by $P_a = \frac{V_{	ext{rms}}^2}{R}$.The average power dissipated in an $LCR$ series circuit (figure $b$) is given by $P_b = I_{	ext{rms}}^2 R = \left(\frac{V_{	ext{rms}}}{Z}ight)^2 R = \frac{V_{	ext{rms}}^2 R}{Z^2}$,where $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Given that the average power is the same in both circuits,$P_a = P_b$:$\frac{V_{	ext{rms}}^2}{R} = \frac{V_{	ext{rms}}^2 R}{Z^2}$This implies $R^2 = Z^2$,which means $Z = R$.Substituting the expression for $Z$:$R^2 = R^2 + (X_L - X_C)^2$$(X_L - X_C)^2 = 0$$X_L = X_C$This is the condition for resonance,where $\omega L = \frac{1}{\omega C}$.Solving for $\omega$:$\omega^2 = \frac{1}{LC} = \frac{1}{0.1 	imes 40 	imes 10^{-6}}$$\omega^2 = \frac{1}{4 	imes 10^{-6}} = 0.25 	imes 10^6 = 250000$$\omega = \sqrt{250000} = 500\,rad/s$.

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

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