For the $L-R$ circuit shown in the figure,calculate the phase angle in degrees if the frequency is $f = 100/\pi \ Hz$. The inductance is $L = 0.025 \ H$ and the resistance is $R = 5 \ \Omega$.

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

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $i$ will be:

In a $L-R$ circuit of $4 \text{ mH}$ inductance and $3 \text{ } \Omega$ resistance, e.m.f. $E = \cos(1000t) \text{ V}$ is applied. The amplitude of current is (in $\text{ A}$)

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.

$A$ 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: