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Multiple choice 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

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. $0.603H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice 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

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. $5A, tan^{-1} ({8 \over 9})$<span>&nbsp;</span>

  4. $6A, tan^{-1} ({7 \over 5})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Z = sqrt(R^2 + (wL)^2). w = 2*pi*50 = 314 rad/s. wL = 314 * 0.19 approx 60 ohms. Z = sqrt(80^2 + 60^2) = 100 ohms. I = V/Z = 220/100 = 2.2A. Phase angle tan(phi) = wL/R = 60/80 = 3/4. Thus phi = tan^-1(3/4).

Multiple choice 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

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. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice 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

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) $\cfrac { \sqrt { 3 }&nbsp; }{ 70\pi&nbsp; } H$&nbsp; (b) $\cfrac { 50 }{ \sqrt { 3 }&nbsp; } Hz$

  2. (a) $\cfrac { \sqrt { 3 }&nbsp; }{ 10\pi&nbsp; } H$&nbsp; (b) $\cfrac { 50 }{ \sqrt { 3 }&nbsp; } Hz$

  3. (a) $\cfrac { \sqrt { 3 }&nbsp; }{ 20\pi&nbsp; } H$&nbsp; (b) $\cfrac { 50 }{ \sqrt { 4 }&nbsp; } Hz$

  4. (a) $\cfrac { \sqrt { 3 }&nbsp; }{ 60\pi&nbsp; } H$&nbsp; (b) $\cfrac { 50 }{ \sqrt { 3 }&nbsp; } Hz$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = I^2 * R. 1000 = 100 * R, so R = 10 ohms. Z = V/I = 200/10 = 20 ohms. Z^2 = R^2 + (wL)^2, so 400 = 100 + (2*pi*50*L)^2, 300 = (100*pi*L)^2, L = sqrt(3)/(10*pi). For phase angle pi/4, tan(pi/4) = 1 = w'L/R, so w' = R/L = 10 / (sqrt(3)/(10*pi)) = 100*pi/sqrt(3). f' = w'/(2*pi) = 50/sqrt(3).

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

 

Multiple choice 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

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. $1\ ms$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The time constant tau = L/R = 8.4mH / 6 ohms = 1.4ms. The current in an RL circuit is I(t) = (V/R)(1 - e^(-t/tau)). Here V/R = 12/6 = 2A. We want I(t) = 1A, so 1 = 2(1 - e^(-t/tau)), 0.5 = 1 - e^(-t/tau), e^(-t/tau) = 0.5. t = tau * ln(2) = 1.4ms * 0.693 approx 1ms.

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice 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

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

  1. $X _L = R$

  2. $X _L &gt;&gt;&gt; R$

  3. $X _L &lt;&lt;&lt; R$

  4. $X _L = \infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A choke coil is designed to limit current in AC circuits with minimal power dissipation. Since power loss P = I^2 * R, we need R to be as small as possible compared to the inductive reactance X_L.

Multiple choice 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

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. $\dfrac{E}{nR}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The magnetic flux through an inductor cannot change instantaneously. Phi = L * I. If L changes to L/n, the current I must change to I' such that L * I = (L/n) * I'. Therefore, I' = n * I. Since the initial current was I = E/R, the new current is nE/R.