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(D) From the given voltage equation $V = 20 \sin(10t + 30^\circ)$,the peak voltage is $V_0 = 20 	ext{ V}$ and the angular frequency is $\omega = 10 \

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$10$

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(D) From the given voltage equation $V = 20 \sin(10t + 30^\circ)$,the peak voltage is $V_0 = 20 	ext{ V}$ and the angular frequency is $\omega = 10 	ext{ rad/s}$.The inductive reactance is $X_L = \omega L = 10 	imes 1 = 10 \, \Omega$.The power dissipated in an $RL$ circuit is given by $P = I_{rms}^2 R = \frac{V_{rms}^2 R}{R^2 + X_L^2} = \frac{V_0^2 R}{2(R^2 + X_L^2)}$.To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero,which gives the condition $R = X_L = 10 \, \Omega$.Substituting $R = 10 \, \Omega$ and $X_L = 10 \, \Omega$ into the power formula:$P_{\max} = \frac{V_0^2 R}{2(R^2 + X_L^2)} = \frac{20^2 	imes 10}{2(10^2 + 10^2)} = \frac{400 	imes 10}{2(100 + 100)} = \frac{4000}{400} = 10 	ext{ W}$.

For the given $RL$ circuit,the maximum power loss across the circuit can be $W$.

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