There is a $5\,\Omega$ resistance in an $AC$ circuit. An inductance of $0.1\,H$ is connected in series with it. If the equation of the $AC$ $EMF$ is $E = 5\sin(50t)$,then the phase difference between the current and the $EMF$ is:

When an inductor $L$ and a resistor $R$ in series are connected across a $12 \, V, 50 \, Hz$ supply, a current of $0.5 \, A$ flows in the circuit. The current differs in phase from the applied voltage by $\frac{\pi}{3}$ radian. Then the value of $R$ is (in $\Omega$)

In the circuit shown,if the $emf$ of the source at an instant is $5 \, V$,and the potential difference across the capacitor at the same instant is $4 \, V$,then the potential difference across $R$ at that instant may be ..... $V$.

$A$ series $R-C$ combination is connected to an $AC$ voltage of angular frequency $\omega = 500 \ rad/s$. If the impedance of the $R-C$ circuit is $R\sqrt{1.25}$,the time constant (in $ms$) of the circuit is:

$A$ bulb connected in series with an air-core solenoid is lit by an $A.C.$ source. If a soft iron core is inserted into the solenoid,then

An inductor of $\left(\frac{100}{\pi}\right) mH$,a capacitor of capacitance $\left(\frac{10^{-3}}{2 \pi}\right) F$,and a resistor of $10 \Omega$ are connected in series with an $AC$ voltage source of $110 \text{ V}, 50 \text{ Hz}$. The tangent of the phase angle $\phi$ between the voltage and the current is: