In a series $LCR$ circuit,the voltages across the capacitor,resistor,and inductor are in the ratio $2:3:6$. If the voltage of the ac source in the circuit is $240 \ V$,then the voltage across the inductor is (in $V$)

In the following circuit,an $AC$ input $V_i(t) = (20 \text{ mV}) \sin(10^5 t)$ is applied at the left end. The amplitude of the output voltage $V_0$ at the right end across the capacitor will be (in $mV$)

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.

What will be the reading in the voltmeter and ammeter of the circuit shown?

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 $R = 5 \ \Omega$.

$A$ sinusoidal voltage of peak value $283 \, V$ and angular frequency $320 \, rad/s$ is applied to a series $LCR$ circuit. Given that $R = 5 \, \Omega$,$L = 25 \, mH$,and $C = 1000 \, \mu F$. The total impedance and the phase difference between the voltage across the source and the current will respectively be: