LCR Series Circuit

A coil of resistance $200 \mathrm{ohms}$ and self-inductance $1.0 \mathrm{H}$ has been connected to an A.C. source of frequency $\frac{200}{\pi } \mathrm{Hz}$. The phase difference between voltage and current is 

The equation of alternating current is, $i=50\sqrt{2}\sin \left(400\pi t\right) \mathrm{A}$. Then, the frequency and root-mean-square value of current are, respectively,

An alternating voltage $E$(in $\mathrm{volt}$)$=200\sqrt{2}\sin \left(100t\right)$ is connected to a $1 \mathrm{\mu F}$ capacitor through an AC ammeter. The reading of the ammeter shall be,

A coil having an inductance of $\frac{1}{\pi } \mathrm{henry}$ is connected in series with a resistance of $300 \mathrm{\Omega }$. If $20 \mathrm{volts}$ from a $200$ cycle source is impressed across the combination, the value of the phase angle between the voltage and the current is,

In an LR circuit, the value of $L$ is $\left(\frac{0.4}{\pi }\right) \mathrm{henry}$ and the value of $R$ is $30 \mathrm{ohm}$. If in the circuit, an alternating emf of $200 \mathrm{volts}$ at $50 \mathrm{cycles} \mathrm{per} \mathrm{second}$ is connected, the impedance of the circuit and current will be,

In the alternating current shown in the following figure, the currents through inductor and capacitor ar $1.2 \mathrm{amp}$ and $1.0 \mathrm{amp}$, respectively. The current drawn from the generator is 

A circuit has a self-inductance of $1$ $\mathrm{H}$ and carries a current of $2 \mathrm{A}$. To prevent sparking, when the circuit is switched off, a capacitor which can withstand $400 \mathrm{V}$ is used. The least capacitance of capacitor connected across the switch must be equal to,

In a series $\mathrm{LCR}$ circuit, $C=25 \mathrm{\mu F}, L=0.1 \mathrm{H}$ and $R=25 \mathrm{\Omega }$. The emf of the $\mathrm{A}.\mathrm{C}.$ source is $e=311\sin \left(314t\right) \mathrm{volt}$, then the impedance is,

In the LCR series circuit, the voltmeter and ammeter readings are,

An inductive circuit contains a resistance of $10 \mathrm{\Omega }$ and an inductance of $2.0 \mathrm{H}$. If an AC voltage of $120 \mathrm{V}$ and frequency of $60 \mathrm{Hz}$ is applied to this circuit, the current in the circuit would be nearly,

A sinusoidal voltage $V_0\sin \omega t$ is applied across a series combination of resistance $R$ and capacitance $C$. The amplitude of the current in this circuit is,

Resonance frequency of a circuit is $f$. If the capacitance is made $4$ times the initial value, then the resonance frequency will become:

A $10 \mathrm{\Omega }$ resistance, $5 \mathrm{mH}$ coil and $10 \mu \mathrm{F}$ 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:

An $A.C.$ source is in series with $R$ and $L$. If respective potential drops are $200 \mathrm{V}$ and $150 \mathrm{V}$, what is the applied voltage ?

An inductor $20\times {10}^{-3} \mathrm{H}$ , a capacitor $100 \mathrm{\mu F}$, and a resistor $50 \mathrm{\Omega }$ are connected in series across a source of $\mathrm{emf}$$V=10\sin \left(314t\right)$. If resistance is removed from the circuit and the value of inductance is doubled, then the variation of current with time in the new circuit is

The impedance of the given circuit will be-

An alternating $\mathrm{emf}$ $10\cos \left(100t\right)$ $\mathrm{volt}$ is connected in series with a resistance of $10$ $\mathrm{ohm}$ and inductance of $100 \mathrm{mH}$. What is the phase difference?

In a $LR$ circuit the $A.C.$ source has voltage $220 \mathrm{V}$ and the potential difference across the inductance is $176 \mathrm{V}$. The potential difference across the resistance will be:

The current in an $L-R$ circuit builds up to ${\left(\frac{3}{4}\right)}^{\mathrm{th}}$ of its steady state value in $4$ seconds. The time constant of this circuit is

An $LCR$ circuit contains $R=50 \mathrm{\Omega }\text{,} L=1 \mathrm{mH} \text{and} C=0.1 \mu \mathrm{F}$. The impedance of the circuit will be minimum for a frequency of: