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:

$A$ telephone wire of length $200\, km$ has a capacitance of $0.014\, \mu F$ per km. If it carries an $AC$ of frequency $5\, kHz$,what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum? (in $mH$)

$A$ parallel combination of a pure inductor and a capacitor is connected across a source of alternating e.m.f. '$e$'. The currents flowing through the inductor and the capacitor are $i_{L}$ and $i_{C}$ respectively. In this parallel resonant circuit,the condition for the currents $i$,$i_{L}$,and $i_{C}$ is ($i =$ net r.m.s. current in the circuit).

In a series resonant $R-L-C$ circuit, the voltage across $R$ is $100 \, V$ and the value of $R = 1000 \, \Omega$. The capacitance of the capacitor is $2 \times 10^{-6} \, F$; the angular frequency of the $AC$ source is $200 \, rad \, s^{-1}$. Then the potential difference across the inductance coil is: (in $V$)

In an $LCR$ circuit,the resonance frequency of the circuit increases to two times its initial value by changing the capacitance from $C$ to $C^{\prime}$ and the resistance from $100 \ \Omega$ to $400 \ \Omega$,while the inductance $L$ is kept constant. The ratio $C / C^{\prime}$ is:

In an $LCR$ circuit, at a particular angular frequency $\omega_0$, the capacitive reactance and inductive reactance are the same. If the angular frequency is doubled to $2\omega_0$, what will be the ratio of the reactance of the capacitor to that of the inductor?