An $LCR$ circuit is at resonance for a capacitor $C$,inductance $L$,and resistance $R$. If the value of the resistance is halved while keeping all other parameters the same,the current amplitude at resonance will be:

For a series $LCR$ circuit,the $I$ vs $\omega$ curve is shown. Consider the following statements:
$(A)$ To the left of $\omega_{r}$,the circuit is mainly capacitive.
$(B)$ To the left of $\omega_{r}$,the circuit is mainly inductive.
$(C)$ At $\omega_{r}$,the impedance of the circuit is equal to the resistance of the circuit.
$(D)$ At $\omega_{r}$,the impedance of the circuit is $0$.
Choose the most appropriate answer from the options given below:

$A$ series $LCR$ circuit of $R=5 \, \Omega, L=20 \, \text{mH}$ and $C=0.5 \, \mu \text{F}$ is connected across an $AC$ supply of $250 \, \text{V}$,having variable frequency. The power dissipated at resonance condition is $..... \times 10^{2} \, \text{W}$.

$A$ 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

For the series $LCR$ circuit shown in the figure,what is the resonance angular frequency and the amplitude of the current at the resonating frequency? The source voltage is given as $220 \, V$ ($RMS$ value).

The readings of the ammeter and voltmeter in the following circuit are respectively: