A circuit contains an inductance of frac{1}{6 | vedclass

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(C) Given: Inductance $L = \frac{1}{6 \pi} 	ext{ H}$,Resistance $R = 15 	ext{ } \Omega$,Voltage $V = 100 	ext{ V}$,Frequency $f = 60 	ext{ Hz}$.Fi

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$4 	ext{ A}$ and $	an^{-1}\left(\frac{4}{3}ight)$

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(C) Given: Inductance $L = \frac{1}{6 \pi} 	ext{ H}$,Resistance $R = 15 	ext{ } \Omega$,Voltage $V = 100 	ext{ V}$,Frequency $f = 60 	ext{ Hz}$.First,calculate the inductive reactance $X_L$:$X_L = L \omega = L(2 \pi f) = \left(\frac{1}{6 \pi}ight) 	imes 2 \pi 	imes 60 = 20 	ext{ } \Omega$.Next,calculate the impedance $Z$ of the $LR$ series circuit:$Z = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 	ext{ } \Omega$.The current $I$ in the circuit is:$I = \frac{V}{Z} = \frac{100}{25} = 4 	ext{ A}$.The phase difference $\phi$ between voltage and current is given by:$	an \phi = \frac{X_L}{R} = \frac{20}{15} = \frac{4}{3}$.Therefore,$\phi = 	an^{-1}\left(\frac{4}{3}ight)$.

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