Resonating Circuit

For what frequency the voltage across the resistance $R$ will be maximum?

The inductance of the oscillatory circuit of the radio station is $10 \mathrm{mH}$ and its capacitance is $0.25 \mathrm{\mu F}$. Taking the effect of resistance negligible, wavelength of the broadcasted waves will be (velocity of light $=3.0\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }\mathrm{\pi }=3.14$)

The inductance of the oscillatory circuit of the radio station is $10 \mathrm{mH}$ and its capacitance is $0.25 \mathrm{\mu F}$. Taking the effect of resistance negligible, wavelength of the broadcasted waves will be (velocity of light $=3.0\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }\mathrm{\pi }=3.14$)

Which one is the expression of resonant frequency in a series resonant circuit?

A parallel L-C circuit comprises of a 5H inductor and $5\mathrm{\mu F}$ capacitor. The resonant frequency of the circuit in $\mathrm{Hz}$ is 

In the given figure, which point shows the resonant state?

In a series resonant circuit, having $\mathrm{L}, \mathrm{C}$ and $\mathrm{R}$as its elements, the resonant current is $i$. The power dissipated in the circuit at resonance is

Which of the following combinations should be selected for better tuning an L.C.R. circuit used for communication?

Which of the following combination should be selected for better sharpness of $LCR$ tuning circuit?

If value of $R$ is changed, then :-

The resonance frequency of the tank circuit of an oscillator when $L=\frac{10}{{\pi }^2} \mathrm{mH}$ and $C=0.04 \text{μF}$ are connected in parallel is

In a series resonant circuit, the AC voltage across resistance $R$, inductor $L$ and capacitor $C$ are $5 \mathrm{V}, 10 \mathrm{V} \mathrm{and} 10 \mathrm{V}$, respectively. The AC voltage applied to the current will be

In $LCR$ series circuit, the resonance condition in terms of capacitive reactance $\left(X_C\right)$ and inductive reactance $\left(X_L\right)$ is

The inductance of the oscillatory circuit of the radio station is $10 \mathrm{mH}$ and its capacitance is $0.25 \mathrm{\mu F}$. Taking the effect of resistance negligible, wavelength of the broadcasted waves will be (velocity of light $=3.0\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }\mathrm{\pi }=3.14$)

A $1\mathrm{0 }\mathrm{\Omega }$ resistance, $5 \mathrm{mH}$ coil and $1\mathrm{0 }\mathrm{\mu F}$ capacitor are joined in series. When a variable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency:

In the diagram, the point which indicated, resonance is-

In $\mathrm{LCR}$ series circuit the value of $C=1 \mathrm{\mu F}, L=1 \mathrm{\mu H}$ at resonance, the frequency will be-

An ac circuit is resonant at $10 \mathrm{kHz}$. If frequency is raised to $12 \mathrm{kHz}$. The impedance of circuit will-

In the circuit shown in figure, if source resistance is neglected, then what will, respectively, be the voltmeter and ammeter readings?

What happens if the value of $R$ in the adjoining circuit is changed?