$A$ coil of $200 \, \Omega$ resistance and $1.0 \, H$ inductance is connected to an $ac$ source of frequency $200/2\pi \, Hz.$ The phase angle between potential and current will be.....$^o$

$A$ coil has $L = 0.04 \, H$ and $R = 12 \, \Omega$. When it is connected to a $220 \, V$,$50 \, Hz$ supply,the current flowing through the coil in amperes is: (in $.7$)

An a.c. voltage source $V=V_0 \sin \omega t$ is connected across resistance $R$ and capacitance $C$ in series. It is given that $R=\frac{1}{\omega C}$ and the peak current is $I_0$. If the angular frequency of the voltage source is changed to $\frac{\omega}{\sqrt{3}}$,then the new peak current in the circuit is

$A$ light bulb connected in series with a capacitor and an $a.c.$ source is glowing with certain brightness. On reducing the value of capacitance and frequency respectively, the brightness of the bulb

When a coil is connected to an $AC$ supply of frequency $50 \, Hz$, a current of $4 \, A$ flows in it and it consumes $240 \, W$ power. If the potential difference across the coil is $100 \, V$, then the inductance value of the coil is

For an $LCR$ circuit driven at frequency $\omega $,the equation reads $L\frac{di}{dt} + Ri + \frac{q}{C} = V_i = V_m \sin \omega t$.
$(a)$ Multiply the equation by $i$ and simplify where possible.
$(b)$ Interpret each term physically.
$(c)$ Cast the equation in the form of a conservation of energy statement.
$(d)$ Integrate the equation over one cycle to find that the phase difference between $V$ and $i$ must be acute.