Series LCR Circuit

 In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor 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 $=3.0\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }\mathrm{\pi }=3.14$)

In LCR series circuit $L=4H, C=36 \mu F \mathrm{and} R=16\Omega$, then bandwidth of resonance is 

The bandwidth of resonance in the $L-C-R$  A.C. circuit is proportional to inductance.

What is resonance condition in AC circuit?

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 $=3.0\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1},\mathrm{ }\mathrm{\pi }=3.14$)

When resistance $R$ is connected in series with an element $A$, the electric current is found to be lagging behind the voltage $V$ by angle $30°$. When the same resistance is connected in series with element $B$, current leads by $60°$. When $R\text{,} A\text{, }B$ are connected in series, the current now leads voltage by $\theta$ which is equal to ${\tan }^{-1}\left(K/\sqrt{3}\right),$ then the value of $K$ is (assume same $AC$ source is used in all cases)

$\mathrm{A}$ steady current of $2 \mathrm{A}$ flows when an inductor of inductance $2\mathrm{mH}$ is connected to ac source of emf $10 \mathrm{V}$. Now $\mathrm{a}$ capacitor of capacity $2\mathrm{\mu F}$ is connected in series. If the current in the circuit is along the emf, the rms value of the current is

The voltage and current through a circuit element are $V=100\sin \left(314t+45°\right) \mathrm{V}\text{,} i=10\sin \left(314t+315°\right) \mathrm{A}$. Then, circuit element is,

A circuit has a resistance of $11 \mathrm{\Omega },$ an inductive reactance of $25 \mathrm{\Omega }$ and a capacitative reactance of $18 \mathrm{\Omega }.$. It is connected to an ac source of $260 \mathrm{V}$ and $50 \mathrm{Hz}.$ The current through the circuit (in amperes) is
 

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 average power in a $L-C-R$ circuit depends upon

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:

The $Q$ factor of series $LCR$ circuit with $L=2 \mathrm{H}, C=32 \mathrm{\mu }\mathrm{F}$ and $R=10 \mathrm{\Omega }$ is _____.

Explain the term: Impedance.

A series $L-C-R$ circuit of $L=200 \mathrm{mH}$, $C=500 \mathrm{\mu F}$, $R=100 \mathrm{\Omega }$ is connected to $100 \mathrm{V}$ AC source.

Find-

(i) Frequency at which power factor of circuit is $1$.

(ii) Peak value of current at this frequency.

(iii) Quality factor 

Find the resonance frequency of series $L-C-R$ series circuit containing $L=0.1 \mathrm{H},$$C=20 \mathrm{\mu F}$, $R=10 \mathrm{\Omega }$