Tag: electromagnetic induction and alternating currents

Questions Related to electromagnetic induction and alternating currents

If circular coil with $N _{1}$ turns is changed in to a coil of $N _{2}$ turns. What will be the ratio of self inductances in both cases.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

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

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

  3. $\dfrac {N _{1}^{2}}{N _{2}^{2}}$

  4. $\sqrt {\dfrac {N _{1}}{N _{2}}}$

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Correct Option: A
Explanation:

If circular coil with N1 turns is changed in to a coil of N2 turns. 

$L=\dfrac{Nd \phi}{dt} $
$L _1=\dfrac{N _1d \phi}{dt} $
$L _2=\dfrac{N _2d \phi}{dt} $ 
 L is in Henries
        N is the Number of Turns
        Φ is the Magnetic Flux
        Ι  is in Amperes
$\dfrac{L _1}{L _2}=\dfrac{N _1}{N _2} $

If the number of turns per unit length of a coil of a solenoid is doubled the self-inductance of the solenoid will:

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. remain unchanged

  2. be halved

  3. be doubled

  4. become four times

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Correct Option: D

Henry, the SI unit of inductance can be written as :

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. weber ampere$^{-1}$

  2. volt second ampere$^{-1}$

  3. joule ampere$^{-1}$

  4. ohm s$^{-1}$

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Correct Option: B
Explanation:

The SI unit of inductance is Henry,
$\displaystyle L =-\dfrac{e}{\dfrac{di}{dt}}$
SI unit $= \displaystyle \dfrac{volt}{A} \times s$
$=volt \times second \times  ampere^{-1}$

Multiple Correct Answers Type
The SI unit of inductance, henry, can be written as

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. Weber/ampere

  2. Volt-second / ampere

  3. $Joule / (ampere)^2$

  4. Ohm-second

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Correct Option: A,B,C,D
Explanation:

$L=\cfrac{\phi}{i}$

$L=\cfrac{weber}{Ampere}$
$V=L\cfrac{Ldi}{dt}$
$L\to$ volt.second/ampere
$E=\cfrac{1}{2}Li^2$
$L\to $joule/(ampere$)^2$
$\omega L=X _L$
$L\to$ohm.second
Hence all are correct.

A lossless coaxial cable has a capacitance of $7\times { 10 }^{ -11 }$ F and an inductance of $0.39\mu H$. Calculate characteristic impedance of the cable.

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 65

  2. 75

  3. 66

  4. 77

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Correct Option: B
Explanation:

Here,$C=7\times { 10 }^{ -11 }F$,
        $L=0.39\times { 10 }^{ -6 }H$

         ${ Z } _{ o }$ As the cable is lossless,
        $\therefore { Z } _{ o }\sqrt { \dfrac { L }{ C }  } =\sqrt { \dfrac { 0.39\times { 10 }^{ -6 } }{ 7\times { 10 }^{ -11 } }  } =75ohm$

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $\Omega$. Frequency & inductance are 50Hz & 0.7 H then wattless current is 

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 0.5 amp

  2. 0.7 amp

  3. 1.0 amp

  4. None

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Correct Option: A
Explanation:

A source= $220V$

The value of resistance= $220 \Omega$
Frequency= $50 Hz$
Inductance= $0.7H$
Find the wattless current= ?
Wattless component of current is $i=i _v\sin \theta$
                                                            $=\cfrac {Ev}{z}\sin \theta$
where, $z=$ impedance of $L-R$ circuit
                $=\sqrt {R^2+L^2W^2}$ so,
$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,
$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$
$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$
        $=\cfrac {220}{R^2+L^2W^2}LW$
        $=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$
        $=\cfrac {220 \times 220}{(220)^2+(220)^2}$
        $=0.5 A$ .

The time constant of a circuit is 10 sec, When a resistance of $ 100 \Omega $ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $250 H$

  2. $50H$

  3. $150 H$

  4. $25 H$

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Correct Option: A
Explanation:

In LR circuit,

The time constant $\tau=\dfrac{L}{R}$
$10=\dfrac{L}{R}$
$L=10R$. . . . . . .(1)
When Resistance $100\Omega $ is connect in series, than the time constant is
$\tau'=\dfrac{L}{R+100}=2s$
$L=2R+200$. . . . . . .(2)
Equating equation (1 ) and (2), we get
$2R+200=10R$
$8R=200$
$R=25\Omega$
From equation (1),
$L=10R=10\times 25$
$L=250H$

the number of turn of primary and secondary coil of the transformer is 5 and 10 respectively ad the mutual inductance is 25 H. if the number f turns of the primary and secondary is made 10 and 5 , then the mutual inductance of the coils will be

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 6.25 H

  2. 12.5 H

  3. 25 H

  4. 50 H

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Correct Option: D
Explanation:

$M=\mu _o\mu _r \cfrac {N _1N _2}{l}A$

$M \propto N _1N _2$
Since $N _1N _2=10 \times 5= 5 \times 10=50$ in both cases.
Mutual inductance will remain same.

The coefficients of self induction of two inductance coils arc 0.0 1H and 0.03H respectively. When they are connected in series so as to support each other. then the resultant self inductance becomes 0.06 Henry. The value of coefficient of mutual induction will be-

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 0.02 H

  2. 0.05 H

  3. 0.01 H

  4. ZERO

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Correct Option: C
Explanation:

$L1=0.01$

$L2=0.03$
$L eff =0.06$

$Leff = L1+L2+2M$
$0.06=0.01+0.03+2M$
$2M = 0.06-0.03$
$M=0.03/2$
$=0.015$

In an induction coil, the coefficient of mutual induction is 4 henry. If a current of 5 ampere in 1  the primary coil is cut off in $\frac { 1 }{ 1500 } $s, the e.m.f at the terminals of the secondary coil will be:

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. 15 kV

  2. 60 kV

  3. 10 kV

  4. 30 kV

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Correct Option: D
Explanation:

${\phi _{21}} = M.i,$

$ \Rightarrow v = \frac{{d{\phi _{21}}}}{{dt}} = M\frac{{di}}{{dt}}$
$ = 4 \times \frac{5}{{\left( {1/1500} \right)}}$
$ = 20 \times 1500$
$ = 30kv$
Hence,
option $(D)$ is correct answer.