$A$ 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 $L-C-R$ circuit,the potential difference across the inductor is $60\,V$,across the capacitor is $30\,V$,and across the resistor is $40\,V$. The supply voltage will be: (in $,V$)

An $ac$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current registered is $I$. If the frequency of the source is changed to $\omega/3$ (while maintaining the same voltage),the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$.

An $AC$ source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current flowing in the circuit is found to be $I$. Now,the frequency of the source is changed to $\frac{\omega}{3}$ (maintaining the same voltage),and the current in the circuit is found to be halved. What is the ratio of reactance to resistance at the original frequency?

In a series $A.C.$ circuit,the current and voltage are given respectively as:
$I = 10 \sin \omega t$; $V = 10 \sin \left( \omega t + \pi / 4 \right)$
Then,which of the following elements may be present in the circuit?
$(a)$ Only $C$
$(b)$ All $L, C$ and $R$
$(c)$ Both $C$ and $R$
$(d)$ Both $L$ and $R$

If a dielectric slab is inserted in the capacitor of the given $RC$ circuit,then the brightness of the bulb will: