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(C) Given: Peak voltage $V_0 = 150 	ext{ V}$, Resistance $R = 25 \ \Omega$, Impedance $Z = 75 \ \Omega$. The average power dissipated in an $AC$ circ

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$50$

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(C) Given: Peak voltage $V_0 = 150 	ext{ V}$, Resistance $R = 25 \ \Omega$, Impedance $Z = 75 \ \Omega$. The average power dissipated in an $AC$ circuit is given by the formula: $P_{	ext{avg}} = I_{	ext{rms}} V_{	ext{rms}} \cos \phi$ Since $\cos \phi = \frac{R}{Z}$, $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{Z}$, and $V_{	ext{rms}} = \frac{V_0}{\sqrt{2}}$, we can write: $P_{	ext{avg}} = \left( \frac{V_{	ext{rms}}}{Z} ight) V_{	ext{rms}} \left( \frac{R}{Z} ight) = \frac{V_{	ext{rms}}^2 R}{Z^2} = \frac{(V_0 / \sqrt{2})^2 R}{Z^2} = \frac{V_0^2 R}{2 Z^2}$ Substituting the given values: $P_{	ext{avg}} = \frac{150^2 	imes 25}{2 	imes 75^2} = \frac{22500 	imes 25}{2 	imes 5625} = \frac{562500}{11250} = 50 	ext{ W}$

$A$ series $LCR$ circuit is connected to an $AC$ source of voltage $V = 150 \sin(80 \pi t) \text{ V}$. If the resistance of the resistor in the circuit is $25 \ \Omega$ and the impedance in the circuit is $75 \ \Omega$, the average power dissipated per cycle in the circuit is: (in $\text{ W}$)

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