A sinusoidal voltage V(t) = 210 sin(3000t) te | vedclass

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(A) Given: $V(t) = 210 \sin(3000t) 	ext{ V}$,$L = 10 	ext{ mH} = 10^{-2} 	ext{ H}$,$C = 25 \mu	ext{F} = 25 	imes 10^{-6} 	ext{ F}$,$R = 100 \Ome

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$	an^{-1}(0.17)$

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(A) Given: $V(t) = 210 \sin(3000t) 	ext{ V}$,$L = 10 	ext{ mH} = 10^{-2} 	ext{ H}$,$C = 25 \mu	ext{F} = 25 	imes 10^{-6} 	ext{ F}$,$R = 100 \Omega$.Angular frequency $\omega = 3000 	ext{ rad/s}$.Inductive reactance $X_L = \omega L = 3000 	imes 10^{-2} = 30 \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{3000 	imes 25 	imes 10^{-6}} = \frac{1}{0.075} = \frac{1000}{75} = \frac{40}{3} \Omega \approx 13.33 \Omega$.The phase difference $\Phi$ is given by $	an \Phi = \frac{X_L - X_C}{R}$.$	an \Phi = \frac{30 - 13.33}{100} = \frac{16.67}{100} = 0.1667 \approx 0.17$.Therefore,$\Phi = 	an^{-1}(0.17)$.

$A$ sinusoidal voltage $V(t) = 210 \sin(3000t) \text{ V}$ is applied to a series $LCR$ circuit in which $L = 10 \text{ mH}$,$C = 25 \mu\text{F}$,and $R = 100 \Omega$. The phase difference $(\Phi)$ between the applied voltage and resultant current will be

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