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(B) Given,peak voltage $V_0 = 150 	ext{ V}$ and peak current $I_0 = 5 	ext{ A}$.Phase difference $\phi = (100 \pi t + \pi) - (100 \pi t + \frac{2 \p

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$187.5 	ext{ W}, 15 \Omega$

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(B) Given,peak voltage $V_0 = 150 	ext{ V}$ and peak current $I_0 = 5 	ext{ A}$.Phase difference $\phi = (100 \pi t + \pi) - (100 \pi t + \frac{2 \pi}{3}) = \frac{\pi}{3} = 60^{\circ}$.Average power dissipated $P_{av} = V_{rms} I_{rms} \cos \phi = \frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ} = \frac{150 	imes 5}{2} 	imes \frac{1}{2} = 187.5 	ext{ W}$.Impedance $Z = \frac{V_0}{I_0} = \frac{150}{5} = 30 \Omega$.Since $\cos \phi = \frac{R}{Z}$,we have $R = Z \cos 60^{\circ} = 30 	imes 0.5 = 15 \Omega$.

$A$ resistor and an inductor are connected in series to an $AC$ source of voltage $V = 150 \sin (100 \pi t + \pi) \text{ V}$. If the current in the circuit is $I = 5 \sin (100 \pi t + \frac{2 \pi}{3}) \text{ A}$,then the average power dissipated and the resistance of the resistor are respectively:

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