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Ac voltage applied to a series lr circuit - class-XII

Description: ac voltage applied to a series lr circuit
Number of Questions: 22
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Tags: physics electromagnetic induction and alternating currents
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A.c across L-R,L-C and L-C-R series circuits. In an LR circuit, $R=10\Omega$ and $L=2H$, If an alternating voltage of $120V$ and $60Hz$ is connected in this circuit, then the value of current flowing in it will be _____ A (nearly)

  1. $0.32$

  2. $0.16$

  3. $0.48$

  4. $0.8$


Correct Option: B

An L-C-R series circuit with $100\omega$ resistance is connected to an A.C source of 200 V and angular frequency $300 rad\,s^{-1}$. When only the capacitor is removed, the current lags behind the voltage by $60^0$ . When only inductor is removed, the current leads the voltage by $60^0$. If all elements are connected , the current in the circuit is

  1. 0.5 A

  2. 1.5 A

  3. 2 A

  4. 2.5 A


Correct Option: C

A coil of self inductance $2H$ carries a $2A$ current. If direction of current is reversed in $1\ sec$., then induced emf in it:

  1. $-8V$

  2. $8V$

  3. $-4V$

  4. $Zero$


Correct Option: B
Explanation:

We know that 

                           $E = \dfrac{ldi}{dt} = 0$

Since current of 2 amperer reverses in 2 second
E.M.F. developed is 
                                 $E = L \times \dfrac{2- (-2)}{1}$
                                  $ E = 2H \times 4As^{-1}$
                                  $E = 8v$
Hence (B) is correct answer

A coil has self-inductance $L = 0.04\, H$ and resistance $R = 12 \Omega$ , connected to $220 V$, 50 Hz supply, what will be the current flow in the coil ?

  1. 11.7 A

  2. 12.7 A

  3. 10.7 A

  4. 14.7 A


Correct Option: B
Explanation:

Given, $L = 0.04 \, H, R= 12\Omega$
$V= 220$ volt and $f= 50 Hz$
The value of current
$I=\dfrac {V}{Z}$
or or $ I=\dfrac {V}{\sqrt{R^2+(\omega L)^2}}$
or $I=\dfrac {V}{\sqrt{R^2+(2\pi fL)^2}}$
or 
$I=\dfrac {220}{\sqrt{144+(2\pi50\times 0.04)^2}}$
$\Rightarrow I= 12.7\, A$

What is the rms value of an alternating current which when passed through a resistor produces heat which is thrice of that produced by a direct current of $2$ amperes in the same resistor:

  1. $6$ amp

  2. $2$ amp

  3. $3.46$ amp

  4. $0.66$ amp


Correct Option: A

When $100$ volt D.C is applied across a coil, a current of one ampere flows through it, when $100V$ ac of $50Hz$ is applied to the same coil, only $0.5amp$ flows. Calculate the resistance and inductance of the coil.

  1. $300\Omega ,\left( \sqrt { 3 } /\pi  \right) Hz$

  2. $100\Omega ,\left( \sqrt { 3 } /\pi  \right) Hz$

  3. $200\Omega ,\left( \sqrt { 3 } /\pi  \right) Hz$

  4. $400\Omega ,\left( \sqrt { 3 } /\pi  \right) Hz$


Correct Option: B

An alternating current of $1.5mA$ and angular frequency $\omega=300rad/s$ flows through $10k\Omega$ resistor and a $0.50\mu F$ capacitor in series. Find the RMS voltage across the capacitor and impedance of the circuit?

  1. $20V,12\Omega$

  2. $10V,12\Omega$

  3. $10V,13\Omega$

  4. $40V,12\Omega$


Correct Option: B

A sinusoidal voltage ${ V } _{ 0 }\sin { \omega t } $ is applied across a series combination of resistance R and inductor L. The amplitude of the current in the circuit is :

  1. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } } $

  2. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }-{ \omega }^{ 2 }{ L }^{ 2 } } } $

  3. $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }+{ \omega }^{ 2 }{ L }^{ 2 } } } \sin { \omega t } \quad $

  4. ${ V } _{ 0 }/R$


Correct Option: B
Explanation:

Impedance of the circuit $\sqrt { { R }^{ 2 }+{ \omega  }^{ 2 }{ L }^{ 2 } } $
Amplitude of voltage$={V} _{0}$
$\therefore$ Amplitude of current $\cfrac { { V } _{ 0 } }{ \sqrt { { R }^{ 2 }-{ \omega  }^{ 2 }{ L }^{ 2 } }  } $

An ideal choke takes a current of $8A$ when connected to an a.c source of $100volt$ and $50Hz$. A pure resistor under the same conditions takes a current of $10A$. If two are connected in series to an a.c supply of $100V$ and $40Hz$, then the current in the series combination of above resistor and inductor is :

  1. $10A$

  2. $8A$

  3. $5\sqrt{2}$ amp

  4. $10\sqrt {2}$ amp


Correct Option: C
Explanation:

${ X } _{ L }=\cfrac { 100 }{ 8 } ;R=\cfrac { 100 }{ 10 } =10\Omega $
$L\times 100\pi =\cfrac { 100 }{ 8 } $
$L=\cfrac { 1 }{ 8\pi  } H$
$Z=\sqrt { { \left( \cfrac { 1 }{ 8\pi  } \times 2\pi \times 40 \right)  }^{ 2 }+{ 10 }^{ 2 } } =10\sqrt { 2 } $
$I=\cfrac { E }{ Z } =\cfrac { 100 }{ 10\sqrt { 2 }  } =\cfrac { 10 }{ \sqrt { 2 }  } =5\sqrt { 2 } A$

A coil of negligible resistance is connected in series with $90\Omega$ resistor across a $120V-60Hz$ line. A voltmenter reads $36V$ across the resistance. Find the voltage across the coil and inductance of the coil.

  1. $114V,1.76H$

  2. $114.5V,0.76H$

  3. $114V,0.86H$

  4. $144V,0.76H$


Correct Option: B
Explanation:

$V=\sqrt{V _R^2+V _L^2}\ \therefore V _L=\sqrt{V^2-V _R^2}\=114.5V\V _R=IR\ \Rightarrow I+\cfrac{36}{90}=0.4A\ \therefore V _L=IX _L\=I\omega L\ \therefore L=\cfrac{V _L}{I\omega}=\cfrac{114.5}{0.4\times2\pi\times60}\=0.76H$

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

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 :

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

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

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

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


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

  1. $0.8$

  2. $1.0$

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

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


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 $

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