In an $L-C-R$ circuit,the capacitance is changed from $C$ to $2C$. For the resonant frequency to remain unchanged,the inductance should be changed from $L$ to:

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:

$A$ series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit,the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit,the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is:

In a series $LCR$ circuit,the voltage leads the current when (Given that $\omega_0$ is the resonant angular frequency):

An inductor of inductance $2\,\mu\text{H}$ is connected in series with a resistance,a variable capacitor,and an $AC$ source of frequency $7\,\text{kHz}$. The value of capacitance for which maximum current is drawn into the circuit is $\frac{1}{x}\text{ F}$,where the value of $x$ is $.........$. (Take $\pi = \frac{22}{7}$)

$A$ 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).