In the figure shown,three $AC$ voltmeters are connected. The circuit is at resonance.

$A$ 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 $mH$)

The resonance frequency of an $L-C-R$ $AC$ circuit is $\nu_{0}$. If the capacitance is made $4$ times its initial value,then the new resonance frequency will become . . . . . . .

In a series $L-C-R$ circuit,the capacitance is changed from $C$ to $2C$. To obtain the same resonance frequency,the inductance should be changed from $L$ to:

$A$ series resonant circuit consists of an inductor $L$ and capacitor $C$ which produces resonant frequency $f$. If $L$ is increased by $2L$ (making the new inductance $L' = L + 2L = 3L$) and $C$ is changed to $9C$, the new resonant frequency will be:

An inductor of $0.5 \ mH$,a capacitor of $20 \ \mu F$ and a resistance of $20 \ \Omega$ are connected in series with a $220 \ V$ a.c. source. If the current is in phase with the e.m.f.,the maximum current in the circuit is $\sqrt{x} \ A$. The value of '$x$' is