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(C) Given voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}ight) 	ext{ V}$ and current $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}ight) 	ext

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$12$

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(C) Given voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}ight) 	ext{ V}$ and current $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}ight) 	ext{ A}$.Comparing with standard forms $V = V_0 \sin(\omega t + \phi_V)$ and $I = I_0 \sin(\omega t + \phi_I)$,we get $V_0 = 144 	ext{ V}$,$I_0 = 6 	ext{ A}$,and phase difference $\phi = \phi_V - \phi_I = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ$.In an $LR$ series circuit,the phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R}$.$	an(60^\circ) = \frac{X_L}{R} \Rightarrow \sqrt{3} = \frac{X_L}{R} \Rightarrow X_L = \sqrt{3} R$.The impedance $Z$ is given by $Z = \frac{V_0}{I_0} = \frac{144}{6} = 24 \ \Omega$.Also,$Z = \sqrt{R^2 + X_L^2}$.Substituting $X_L = \sqrt{3} R$,we get $24 = \sqrt{R^2 + (\sqrt{3} R)^2} = \sqrt{R^2 + 3R^2} = \sqrt{4R^2} = 2R$.Therefore,$R = \frac{24}{2} = 12 \ \Omega$.

An inductor and a resistor are connected in series to an $AC$ source of voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}\right) \text{ V}$. If the current in the circuit is $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}\right) \text{ A}$,then the resistance of the resistor is: (in $\Omega$)

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