In the given circuit,the potential difference across the resistance is $54 \, V$ and the power consumed by it is $16 \, W$. If the $AC$ frequency is $60 \, Hz$,find the value of $L$ in $H$.

An $AC$ circuit having supply voltage $E = 10 \sin \omega t$ consists of a resistor of resistance $R = 3 \Omega$ and an inductor of reactance $X_L = 4 \Omega$ as shown in the figure. The voltage across the inductor at $t = \pi / \omega$ is: (in $\text{ V}$)

When alternating current is passed through an $L-R$ series circuit,the power factor is $\frac{\sqrt{3}}{2}$ and $R=50 \ \Omega$. If the frequency of the source is $50 \ Hz$,then the value of $L$ is (Assume $\pi \approx 3.14$):
$\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

If $i$ is the current in an $L-C-R$ series $AC$ circuit,what are the formulas for:
$(i)$ Voltage across resistance.
$(ii)$ Voltage across inductor.
$(iii)$ Voltage across capacitor.

$A$ resistor and a capacitor are connected in series with an $AC$ source. If the potential drop across the capacitor is $5 \,V$ and that across the resistor is $12 \,V$, then the applied voltage is (in $\,V$)

$A$ long solenoid connected to a $12\,V\, DC$ source passes a steady current of $2\,A.$ When the solenoid is connected to a source of $12\,V\, RMS$ at $50\,Hz,$ the current flowing is $1\,A\, RMS.$ Then the inductance of the solenoid is ....$mH$.