Tag: electromagnetic induction and alternating currents

Questions Related to electromagnetic induction and alternating currents

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:

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

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

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

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

  4. $0$

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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.

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  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) $\cfrac { 50 }{ \sqrt { 3 }  } Hz$

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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?

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. 3.5 sec

  2. 2.5 sec

  3. 5.5 sec

  4. 7.5 sec

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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

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. $500\ s$

  2. $25\ $s

  3. $35\ s$

  4. $1\ ms$

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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:

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. 1 A

  2. 2 A

  3. 3 A

  4. none of these

Show answer
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

In series LCR circuit, if $C=9\mu F, V = 0.2, V, L =0.01 H and $R=$4C=9\mu $ then the voluage across inductor is:

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. 5/9

  2. 5/3

  3. 9/5

  4. 3/5

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Correct Option: B

In a choke coil, the reactance $X _L$ and resistance R are such that :-

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. $X _L = R$

  2. $X _L >>> R$

  3. $X _L <<< R$

  4. $X _L = \infty$

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Correct Option: B

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:

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. $zero$

  2. $E/R$

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

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

Show answer
Correct Option: C

A $0.21\space H$ inductor and a $12\Omega$ resistance are connected in series to a $220\space V, 50\space Hz$ ac source. The current in the circuit is :

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. $\displaystyle\frac{220}{\sqrt{4400}}A$

  2. $\displaystyle\frac{22}{3\sqrt5}A$

  3. $\displaystyle\frac{220}{\sqrt{4550}}A$

  4. $\displaystyle\frac{22}{5\sqrt3}A$

Show answer
Correct Option: B
Explanation:

$ X _L = L 2 \pi f = 0.21 \times 314 \Omega $
$ R = 12 \Omega $
$ I = \dfrac{ 220}{ \sqrt{ X _L^2 + R^2 } } = 3.28 = \dfrac{22}{3\sqrt{5}} A  $

An A.C voltage $V=5\cos { \left( 1000t \right) V } $ is applied to a L-R series circuit of inductance $3mH$ and resistance $4\Omega$. The value of maximum current in the circuit is  _______ A

ac voltage applied to a series lr circuit lr circuit phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram electromagnetic induction and alternating currents physics

  1. $0.8$

  2. $1.0$

  3. $\cfrac{5}{7}$

  4. $\cfrac { 5 }{ \sqrt { 7 } } $

Show answer
Correct Option: B
Explanation:

$V=5\cos { \left( 1000t \right) V } $

The standard equation for the voltage is:
$V={ V } _{ 0 }\cos { \omega t } $

So, from the equation, ${ V } _{ 0 }=5volt;\omega =1000rad/s$
$L=3\times { 10 }^{ -3 }H,R=4\Omega $

Maximum current $10=\cfrac { { V } _{ 0 } }{ Z } \quad $
$10=\cfrac { 5 }{ \sqrt { { \omega  }^{ 2 }{ L }^{ 2 }+{ R }^{ 2 } }  } =\cfrac { 5 }{ 5 } =1A\quad \quad $