A circuit is made of 1 Omega resistance and 2 | vedclass

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(C) The phase difference $\phi$ in an $RL$ series circuit is given by the formula: $	an \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given:Resistance $

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$	an^{-1} \left( \frac{\pi}{4} ight)$

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(C) The phase difference $\phi$ in an $RL$ series circuit is given by the formula: $	an \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given:Resistance $R = 1 \Omega$Inductance $L = 2.5 	ext{ mH} = 2.5 	imes 10^{-3} 	ext{ H}$Frequency $
u = 50 	ext{ Hz}$Angular frequency $\omega = 2 \pi 
u = 2 \pi 	imes 50 = 100 \pi 	ext{ rad/s}$.Substituting the values:$	an \phi = \frac{100 \pi 	imes 2.5 	imes 10^{-3}}{1}$$	an \phi = 100 	imes 2.5 	imes 10^{-3} 	imes \pi$$	an \phi = 0.25 \pi = \frac{\pi}{4}$.Therefore,$\phi = 	an^{-1} \left( \frac{\pi}{4} ight)$.

$A$ circuit is made of $1 \Omega$ resistance and $2.5 \text{ mH}$ inductance in series. If an alternating source of $200 \text{ V}$ and $50 \text{ Hz}$ is connected to the circuit,then the phase difference between current and voltage is . . . . . . .

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