Tag: phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram

Questions Related to phase relations between alternating voltage and alternating current in different types of alternating current circuits and phasor diagram

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:

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