In a series $LCR$ resonant circuit,the capacitance is changed from $C$ to $3C$. To obtain the same resonant frequency,the inductance should be changed from $L$ to

The frequency at resonance for the circuit shown in the figure is:

$A$ resistor of $50 \Omega$,an inductor of self-inductance $\left(\frac{3}{\pi^2}\right) H$,and a capacitor of unknown capacity are connected in series to an a.c. source of $100 \ V$ and $50 \ Hz$. When the voltage and current are in phase,the value of capacitance is (nearly):

In a series $R-L-C$ $AC$ circuit,for a particular value of $R, L$ and $C$,the power supplied by the source is $P$ at resonance. If the value of inductance is halved,then the power from the source again at resonance is $P'$. Then:

An alternating $e.m.f.$ of frequency $v = \frac{1}{2\pi \sqrt{LC}}$ is applied to a series $LCR$ circuit. For this frequency of the applied $e.m.f.$:

An inductance of $1\, mH$,a capacitor of $10\, \mu F$,and a resistance of $50\, \Omega$ are connected in series. The reactances of the inductor and the capacitor are the same. The reactance of either of them will be........$\Omega$.