Tag: lr circuit

Questions Related to lr circuit

A $200km$ long telegraph wire has capacity of $0.014\mu F/km$. If it carries an alternating current of $50KHz$, what should be the value of an inductance required to be connected in series so that impedance is minimum?

  1. $0.703H$

  2. $0.303H$

  3. $0.503H$

  4. $0.603H$


Correct Option: A

A $0.19H$ inductor and a $80\Omega$ resistance connected in series to a $220V, 50Hz$ ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

  1. $2.2A, tan^{-1} ({3 \over 4})$

  2. $3A, tan^{-1} ({2 \over 5})$

  3. $5A, tan^{-1} ({8 \over 9})$ 

  4. $6A, tan^{-1} ({7 \over 5})$


Correct Option: A

A metal rod of resistance $20\ \Omega$ is fixed along a diameter of a conducing ring of radius $0.1\ m$ and lies on $x-y$ plane. There is a magnetic field $\vec { B } =\left( 50\ T \right) \vec { k }$. The ring rotates with an angular velocity $\omega=20\ rad\ {s}^{-1}$ about its axis. An external resistance of $10\ \Omega$ is connected across the center of the ring and rim. The current through external resistance is:

  1. $\dfrac { 1 }{ 4 }$

  2. $\dfrac { 1 }{ 2 }$

  3. $\dfrac { 1 }{ 3 }$

  4. $0$


Correct Option: C

A circuit containing an inductance and a resistance connected in series, has an AC source of $200V$, $50Hz$ connected across it. An AC current of $10A$ rms flows through the circuit and the power loss is measured to be $1kW$. Find
(a) the inductance in the circuit
(b) the frequency of the AC when the phase difference between the current and emf becomes $\pi /4$. with the above components.

  1. (a) $\cfrac { \sqrt { 3 }  }{ 70\pi  } H$  (b) $\cfrac { 50 }{ \sqrt { 3 }  } Hz$

  2. (a) $\cfrac { \sqrt { 3 }  }{ 10\pi  } H$  (b) $\cfrac { 50 }{ \sqrt { 3 }  } Hz$

  3. (a) $\cfrac { \sqrt { 3 }  }{ 20\pi  } H$  (b) $\cfrac { 50 }{ \sqrt { 4 }  } Hz$

  4. (a) $\cfrac { \sqrt { 3 }  }{ 60\pi  } H$  (b) $\cfrac { 50 }{ \sqrt { 3 }  } Hz$


Correct Option: B

A solenoid of 10 Henry inductance and 2 ohm resistance, is connected to a 10 volt battery. In how much time the magnetic energy will be reaches to 1/4th of the maximum value?

  1. 3.5 sec

  2. 2.5 sec

  3. 5.5 sec

  4. 7.5 sec


Correct Option: A
Explanation:

Given that,

L = 10 H

R = 2 ohm

V = 10 volt

Now, the maximum current is

  $ {{i} _{0}}=\dfrac{V}{R} $

 $ {{i} _{0}}=\dfrac{10}{2} $

 $ {{i} _{0}}=5A $

The maximum energy is

  $ {{E} _{0}}=\dfrac{1}{2}Li _{0}^{2} $

 $ {{E} _{0}}=\dfrac{1}{2}\times 10\times 5\times 5 $

 $ {{E} _{0}}=125\,J $

Now, the magnetic energy

  $ E=\dfrac{{{E} _{0}}}{4} $

 $ E=\dfrac{125}{4}\,J $

Now, 

  $ E=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{4}=\dfrac{1}{2}L{{i}^{2}} $

 $ \dfrac{125}{2\times 10}={{i}^{2}} $

 $ {{i}^{2}}=\dfrac{25}{2} $

 $ i=\dfrac{5}{2} $

 $ i=2.5\,A $

Now, the time taken to rise current from 0 - 2.5A.

We know that, the instantaneous current during its growth in an L-R circuit.

 $ i={{i} _{0}}\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ 2.5=5\left( 1-{{e}^{-\frac{Rt}{L}}} \right) $

 $ {{e}^{\frac{-Rt}{L}}}=0.5 $

 $ \dfrac{-Rt}{L}=\ln (0.5) $

 $ \dfrac{Rt}{L}=0.693 $

 $ t=\dfrac{6.93}{2} $

 $ t=3.46 $

 $ t=3.5\sec  $

Hence, the magnetic energy will be increases to $\dfrac{1}{4}$ of the maximum value at $3.5$ sec.

 

A coil of inductance $8.4\ mil$ and resistance $6W$ is connected to a $12V$ battery. The current in the coil is $1.0\ A$ at approximately the time

  1. $500\ s$

  2. $25\ $s

  3. $35\ s$

  4. $1\ ms$


Correct Option: D

An inductor coil,a capacitor and an alternating source of virtual value 36 V are connected in series.When the frequency of the source is varied, a maximum virtual current 4  A is observed. If this inductor coil is connected to a battery of emf 18V and internal resistance$ 9\Omega$, the current in the circuit will be:

  1. 1 A

  2. 2 A

  3. 3 A

  4. none of these


Correct Option: A
Explanation:

Given that,

e. m. f = $18\ V$

${{E} _{rms}}=36V$

Internal resistance $r=9\Omega $

Current ${{I} _{rms}}=4\,A$

Now, the external resistance is

We know that,

  $ R=\dfrac{E _{rms}}{I _{rms}} $

 $ R=\dfrac{36}{4} $

 $ R=9\,\Omega  $


When the inductor coil is connected to a 18 V battery with 9 $\Omega$ internal resistance: 

Now, net resistance is

  $ {{R} _{net}}=R+r $

 $ {{R} _{net}}=9+9 $

 $ {{R} _{net}}=18\,\Omega  $


Now, the current is

  $ I=\dfrac{e.m.f}{{{R} _{net}}} $

 $ I=\dfrac{18}{18} $

 $ I=1\,A $

 Hence, the current is $1\ A$ in the circuit

A closed circuit consists of a source of emf $E$ and an inductor coil of inductance $L$, connected in series. The active resistance of whole circuit is $R$. At the moment $t=0$. inductance of coil abruptly decreased to $L/n$. Then current in the circuit immediately after, is:

  1. $zero$

  2. $E/R$

  3. $\dfrac{nE}{R}$

  4. $\dfrac{E}{nR}$


Correct Option: C