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

$A$ resistor of resistance $R$, an inductor of inductive reactance $2R$, and a capacitor of capacitive reactance $X_C$ are connected in series to an $A.C.$ source. If the series $LCR$ circuit is in resonance, then the power factor of the circuit and the value $X_C$ are respectively:

$A$ series $LCR$ circuit driven by $300 \, V$ at a frequency of $50 \, Hz$ contains a resistance $R = 3 \, k\Omega$, an inductor of inductive reactance $X_{L} = 250 \pi \, \Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be: (Take $\pi^{2} = 10$) (in $\mu F$)

In an a.c. $LCR$ circuit,a capacitor and an inductor in series are connected with an a.c. voltage source of $90\,V$. An ammeter and a voltmeter are connected in the circuit to measure the current in the circuit and the voltage across the capacitor plus the inductor combination. If $X_L = X_C = 4\,\Omega$ and $R = 45\,\Omega$,the readings of the ammeter $(I)$ and the voltmeter $(V)$ are:

In the circuit shown,the power factor of the circuit is $1$ and the power factor of the box is $\frac{3}{5}$. Find the reading of the ammeter in $A$.

In a series resonant $LCR$ circuit,the voltage across $R$ is $100 \ V$ and $R = 1 \ k\Omega$ with $C = 2 \ \mu F$. The resonant frequency $\omega$ is $200 \ rad/s$. At resonance,the voltage across $L$ is: