Tag: electromagnetic induction and alternating currents

Questions Related to electromagnetic induction and alternating currents

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