An alternating e.m.f. of $0.2 \, V$ is applied across an $LCR$ series circuit having $R=4 \, \Omega$, $C=80 \, \mu F$, and $L=200 \, mH$. At resonance, the voltage drop across the inductor is (in $V$)

$A$ resistor $R$,an inductor $L$,and a capacitor $C$ are connected in series to an oscillator of frequency $N$. If the resonance frequency is $N_R$,then the current lags behind the voltage when:

An $LCR$ circuit contains $R = 50 \, \Omega$,$L = 1 \, \text{mH}$,and $C = 0.1 \, \mu\text{F}$. The impedance of the circuit will be minimum for a frequency of:

In a series $LCR$ circuit,$C = 2\,\mu F$,$L = 1\,mH$,and $R = 10\,\Omega$. When the current in the circuit is maximum,what is the ratio of the energy stored in the capacitor to the energy stored in the inductor?

Two circuits are shown in the figures $(a)$ and $(b)$. At a frequency of $....\,rad/s$,the average power dissipated in one cycle will be the same in both circuits.

$A$ series $LCR$ circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $f_0$. Keeping capacitance constant,if the inductance $(L)$ is increased by $\sqrt{3}$ times and resistance is increased $(R)$ by $1.4$ times,the resonant frequency now is