The time constant of the given $LR$ circuit is $\tau$. If the battery is replaced by an $AC$ source having voltage $V = V_0 \cos \omega t$,the power factor of the circuit will be

$A$ circuit contains an inductance of $\frac{1}{6 \pi} \text{ H}$ and a resistance of $15 \text{ } \Omega$ in series. If an $AC$ voltage of $100 \text{ V}$ and $60 \text{ Hz}$ is applied to the circuit,then the current in the circuit and the phase difference between voltage and current are,respectively:

In an $A.C.$ circuit,a resistance $R = 40 \ \Omega$ and an inductance $L$ are connected in series. If the phase angle between voltage and current is $45^{\circ}$,then the value of the inductive reactance is $(\tan 45^{\circ} = 1)$. (in $\Omega$)

When $80 \ V$ $d.c.$ is applied across a solenoid,a current of $0.8 \ A$ flows in it. When $80 \ V$ $a.c.$ is applied across the same solenoid,the current becomes $0.4 \ A$. If the frequency of $a.c.$ source is $50 \ Hz$,the impedance and inductance of the solenoid are nearly:

$A$ resistor of $450 \Omega$ and an inductor are connected in series to an ac source of frequency $\frac{75}{\pi} \text{ Hz}$. If the power factor of the circuit is $0.6$,then the inductance connected in the circuit is:

In an $AC$ circuit,the potential differences across an inductance and a resistance joined in series are $16 \, V$ and $20 \, V$ respectively. The total potential difference of the source is .......... $V$.