Mix Test - 4

AC generator works on the principle of Faraday's Law. The AC Generator's input supply is mechanical energy supplied by steam turbines, gas turbines and combustion engines. The output is alternating electrical power in the form of alternating voltage and current.

AC generators convert mechanical energy into electrical energy. The generated energy is in the form of a sinusoidal waveform (alternating current).

So, all the statements regarding AC generators correct.

Hence, the correct option is (D).

Given: Charge on Capacitor (\(\mathrm{Q}_{0})=2 \times 10^{-5} \mathrm{C}\), and Natural frequency\((\omega_{0})=4 \times 10^{3} \mathrm{rad} / \mathrm{sec}\)

The relation between the maximum charge and the maximum current in the LC oscillation circuit is given as,

\(I_{0}=\omega_{0} Q_{0}\)

\(=4 \times 10^{3} \times 2 \times 10^{-5}\)

\(=8 \times 10^{-2} \mathrm{~A}\)

Hence, the correct option is (A).

By the first experiment of Faraday and Henry, we can say that when a bar magnet is moved towards or away from a coil an emf gets induced in the coil.

If the circuit is closed in which the coil is connected then a current will also get induced in the coil.

Here, it is not clear that the circuit in which the coil is connected is open or closed.

Therefore, in this case we can say that when a magnet is moved axially towards a coil, an emf must be induced but current may or may not be induced in the coil.

Hence, the correct option is (A).

Given that:

\(\mathrm{R}=0 \Omega,\) \(\mathrm{X}_{\mathrm{L}}=8 \Omega \text { and } \mathrm{X}_{\mathrm{C}}=6 \Omega \)

\(tan \theta=\frac{X_{L}-X_{C}}{R}\)

Where, Resistance is (R), Capacitance Reactance is \( (\mathrm{X}_\mathrm{c}) \), Inductance Reactance is \( \mathrm({X}_{\mathrm{L}}) \).

Phase difference, \(\theta=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)=\tan ^{-1}\left(\frac{8-6}{0}\right)=\tan ^{-1}(\infty)=90^{\circ}\)

Hence, the correct option is (B).

The AC circuit containing a resistor, inductor, and a capacitor is called an LCR circuit.

The meanings of inductor, capacitor and resistance are as follows:

• The device which stores magnetic energy in a magnetic field is called an inductor.
• The device that stores electrostatic energy in an electric field is called a capacitor.
• The properties of a conductor which opposes the flow of current are called resistance.

The capacitor and inductor are the storage devices that store the energy in it.

The inductor and capacitor can’t dissipate energy. As the resistance opposes the flow of electric current. So, the resistance of any circuit dissipates the power.

Hence, the correct option is (A).

Given: Initial current \(I_{1}=4 \mathrm{~A}\), Final current \(\mathrm{I}_{2}=0 \mathrm{~A}\), Inductance \(\mathrm{~L}=100 \mathrm{~mH}=0.1 \mathrm{H}\) and Elctromotive Forc\(\mathrm{e}=5 \mathrm{~V}\)

This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.

Self-inductance of a solenoid is given by:

The time required to induce \(5 \mathrm{~V}\) across the inductor is:

\(dt=L \frac{d i}{e}=\frac{0.1 \times(4-0)}{5}=0.08 \mathrm{~sec}\)

Hence, the correct option is (B).

Given:

Voltage across the Resistor R (\(V_{\mathrm{R}})=80 \mathrm{~V}\)

Voltage across the Inductor L (\(\mathrm{V}_{\mathrm{L}})=40 \mathrm{~V}\)

Voltage across the Capacitor C (\(\mathrm{V}_{\mathrm{C}})=100 \mathrm{~V}\)

For a series LCR circuit, the total potential difference of the circuit is given by:

\(V=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)

Putting all the above values in the equation,

\(V=\sqrt{80^{2}+(100-40)^{2}}\)

\(=\sqrt{6400+3600}\)

\(=\sqrt{10000}\)

\(=100 \mathrm{~V}\)

Hence, the correct option is (C).

Given:

Initial magnetic flux \(\left(\phi_{1}\right)=20 \times 10^{-2}\) Weber

Final magnetic flux \(\left(\phi_{2}\right)=12 \times 10^{-2}\) Weber

Change in time \(\Delta \mathrm{t}=0.02\) s

This change can be produced by changing the magnetic field strength, moving the coil into or out of the magnetic field,  moving a magnet toward or away from the coil, rotating the coil relative to the magnet, etc.

The value of induced emf in the coil is given by:

\(e=-\frac{\Delta \phi}{\Delta t}\)

\(=-\frac{(12 \times 10^{-2}-20 \times 10^{-2})}{0.02}\)

\(=4 V\)

Hence, the correct option is (A).

Given: \(\mathrm{M}_{12}=\mathrm{M}_{21}=\mathrm{M}=200 \mathrm{H}, \mathrm{N}_{1}=100, \mathrm{~N}_{2}=120\), and \(\mathrm{I}_{2}=5 \mathrm{~A}\)

We know that when two coils of the number of turns \(N_{1}\) and \(N_{2}\) are placed close to each other, then the mutual inductance of coil 1 with respect to coil 2 is given as,

\(M_{12}=N_{1} \frac{\phi_{1}}{I_{2}} \quad \cdots\) (1)

Where, \(\mathrm{N}_{1}=\) number of turns in coil \(1, \phi_{1}=\) flux linked with coil 1, and \(\mathrm{I}_{2}\) current in coil 2 

By equation (1) the flux linkage with the inner solenoid is given as,

\(\phi_{1}=\frac{M_{12} \times I_{2}}{N_{1}}\)

\(\Rightarrow \phi_{1}=\frac{200 \times 5}{100}\)

\(\Rightarrow \phi_{1}=10 \mathrm{~Wb}\)

Hence, the correct option is (B).

Given: \(N=2 N_{1}\) and \(B=2 B_{1}\)

The magnetic flux through the electric generator when the magnetic field is \(B\), current flowing is A and the number of turns is \(\mathrm{N}\)

\(\phi=N B A \cos \theta \quad \ldots (1)\)

The magnetic flux through the electric generator when the magnetic field and number of turns of the coil of an electric generator is doubled

\(\phi_{1}=N_{1} B_{1} A \cos \theta \quad \ldots (2)\)

\(\Rightarrow \phi_{1}=(2 N)(2 B) A \cos \theta=4 N B A \cos \theta=4 \phi \quad[\because \phi=N B A \cos \theta]\)

Hence, the correct option is (D).

Given: \(l=2 \mathrm{~m}, \mathrm{n}_{1}=\mathrm{n}_{2}=100, \mathrm{r}_{1}=7 \mathrm{~cm}=7 \times 10^{-2} \mathrm{~m}\), and \(\mathrm{r}_{2}=14 \mathrm{~cm}=14 \times 10^{-2} \mathrm{~m}\)

We know that, if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.

The mutual inductance of both the solenoids is given as,

\(\mathrm{M}_{12}=\mathrm{M}_{21}=\mathrm{M}=\mu_{0} \mathrm{n}_{1} \mathrm{n}_{2} \pi \mathrm{r}_{1}^{2} l \quad \cdots\) (1)

Where, \(n_{1}=\) number of turns per unit length of solenoid 1, \(\mathrm{n}_{2}=\) number of turns per unit length of solenoid 2, \(r_{1}=\) radius of the inner solenoid, and \(l=\) length of both the solenoids

By equation (1) the mutual inductance of both solenoids is given as,

\(\mathrm{M}=\mu_{0} \mathrm{n}_{1} \mathrm{n}_{2} \pi \mathrm{r}_{1}^{2}{l}\)

\(\Rightarrow \mathrm{M}=\mu_{0} \mathrm{n}_{1} \mathrm{n}_{2} \pi \mathrm{r}_{1}^{2} \mathrm{l} \times \frac{4 \pi}{4 \pi}\)

\(\Rightarrow \mathrm{M}=\frac{\mu_{0}}{4 \pi} \times 4 \pi^{2} \mathrm{n}_{1} \mathrm{n}_{2} \mathrm{r}_{1}^{2} l \quad \cdots\) (2)

We know that,

\(\frac{\mu_{0}}{4 \pi}=10^{-7} \mathrm{T}\frac{\mathrm{m}}{\mathrm{A}} \quad \cdots\) (3)

By equation (2) and equation (3),

\(\mathrm{M}=10^{-7} \times \frac{22}{7} \times \frac{22}{7} \times 100 \times 100 \times\left(7 \times 10^{-2}\right)^{2} \times 2\)

\(\Rightarrow \mathrm{M}=9.68 \times 10^{-5} \mathrm{H}\)

Hence, the correct option is (C).

Given: 

Resistance(\(\mathrm{R})=8 \Omega\), Capacitance Reactance \((\mathrm{X}_{\mathrm{c}})=25 \Omega\), Inductance Reactance \(\mathrm({X}_{\mathrm{L}})=31 \Omega\) and AC Source \(=110 \mathrm{~V}\)

For a series LCR circuit, Impedance (Z) of the circuit is given by:

\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

\(=\sqrt{8^{2}+(31-25)^{2}}\)

\(=\sqrt{64+36}\)

\(=10 \Omega\)

The power factor (P) of a series LCR-circuit is given by:

\(\cos \Phi=\frac{R}{Z}\)

\(=\frac{8}{10}\)

\(=0.8\)

Hence, the correct option is (C).

The Lenz law states that the induced emf in a coil due to a changing magnetic flux is such that the magnetic field created by the induced emf opposes the change in a magnetic field.

When the north pole of a magnet is moved towards a coil that is connected to a circuit, the distance between the magnet and the coil will reduce, and magnetic flux associated with the coil is increased.

Due to this change in magnetic flux, an emf will induce in the coil and the direction of the induced emf will be such that it tries to stop the change of magnetic flux.

Therefore, the direction of current will be such that it stops the motion of the magnet and it is only possible when the north pole is formed on the magnet side of the coil so that the coil can repel the magnet.

We know that if the current in the coil is clockwise, the face of the coil towards the observer behaves as the south pole and if the current in the coil is anti-clockwise, the face of the coil towards the observer behaves as the north pole.

So, for the formation of the north pole on the magnet side, the current in the coil will be anti-clockwise when the coil is seen from the magnet side.

Hence, the correct option is (B).

Given that:

Potential \((\mathrm{v})=v_{0} \sin (\omega \mathrm{t})\)

Current \((i)=i_{0} \sin \left(\omega t+\frac{\pi}{3}\right)\)

Phase difference \((\theta)=\frac{\pi}{3}\)

Peak current \((I)=i_{0}\)

Peak voltage \((\mathrm{V})=v_{0}\)

rms current \(\left(I_{r m s}\right)=\frac{I}{\sqrt{2}}=\frac{i_{0}}{\sqrt{2}}\)

rms voltage \(\left(V_{r m s}\right)=\frac{V}{\sqrt{2}}=\frac{v_{0}}{\sqrt{2}}\)

We know that:

\(P_{\text {avg }}=V_{\text {rms }} I_{r m s} \operatorname{Cos} \theta\)

Average power \(\left(P_{\text {avg }}\right)=\frac{v_{0}}{\sqrt{2}} \times \frac{i_{0}}{\sqrt{2}} \times \operatorname{Cos} \frac{\pi}{3}=\frac{v_{0} i_{0}}{2} \times \frac{1}{2}=\frac{v_{0} i_{0}}{4}\)

Hence, the correct option is (B).

According to the second experiment of Faraday and Henry, we can say that when a current-carrying coil is moved towards or away from another coil then an emf gets induced in the coil.

If the circuit is closed in which the coil is connected then a current will also get induced in the coil. The relative motion between the two coils is required to induce a current in the coil.

In the given case when the primary coil is rotated about its axis there will be no relative motion between the primary and the secondary coil.

Since, there is no relative motion between the primary and the secondary coil so no current will induce in the primary coil.

Hence, the correct option is (B).

Given: Number of turns (\(\mathrm{N})=1\) turns, Size\((\mathrm{A})=1 \mathrm{~m} \times 0.5 \mathrm{~m}=5 \times 10^{-1} \mathrm{~m}^{2}\), Initial magnetic field \(\mathrm({~B}_{1})=1 \mathrm{~T}\), and Final magnetic field (\(\mathrm{B}_{2})=0.5 \mathrm{~T}\)

Cange in magnetic field,

\(\Delta \mathrm{B}=1-0.5=0.5 \mathrm{~T}\)

Change in Magnatic flux,

\(\Delta \phi=A \Delta B\)

\(\Rightarrow \Delta \phi=5 \times 10^{-1} \times 0.5=2.5 \times 10^{-1}=0.25 \mathrm{~Wb}\)

Hence, the correct option is (B).

Strong electromagnets are situated above the rails in some electrically powered trains.

When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train.

When a changing magnetic flux is applied to a bulk piece of conducting material then circulating currents called eddy currents are induced in the material.

So, by the above explanation, we can understand that the generation of eddy current depends on the magnetic field but it is independent of the direction of the magnetic field.

Therefore, the eddy current will also generate when the north and the south poles are replaced with each other and that eddy currents induced in the rails oppose the motion of the train. So, the velocity of the train will decrease.

Hence, the correct option is (B).

Given: Self-inductance of first coil \(\left(L_{1}\right)=108 \mathrm{mH}\), Radius of both coils are same i.e., \(r_{1}=r_{2}\) \(=r\), Number of turns of the first coil \(\left(\mathrm{N}_{1}\right)=200\), and Number of turns of the second coil \(\left(\mathrm{N}_{2}\right)=500\)

The Self-induction for the first coil is:

\(L_{1}=\frac{\mu_{0} N_{1}^{2} \pi r^{2}}{2} \quad \cdots\) (1)

The Self-induction for the second coil is:

\(L_{2}=\frac{\mu_{0} N_{2}^{2} \pi r^{2}}{2} \quad \cdots\) (2)

On dividing equation (1) and (2), we get:

\(\frac{L_{2}}{L_{1}}=\left(\frac{N_{2}}{N_{1}}\right)^{2}\)

\(L_{2}=L_{1}\left(\frac{N_{2}}{N_{1}}\right)^{2}\)

\(=108 \times\left(\frac{500}{200}\right)^{2}\)

\(=108 \times 6.25\)

\(=675 \mathrm{~mH}\)

Hence, the correct option is (C).

We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.

The mutual inductance of both the solenoids is given as,

\(M_{12}=M_{21}=\mu_{o} n_{1} n_{2} \pi r_{1}^{2} l \quad \cdots\) (1)

Where, \(n_{1}=\) number of turns per unit length of solenoid \(1, n_{2}=\) number of turns per unit length of solenoid \(2, r_{1}=\) radius of the inner solenoid, and \(l=\) length of both the solenoids

By equation (1) it is clear that the mutual inductance of both the solenoids does not depend on the current in the solenoids.

Therefore, when the current in both the solenoids is doubled, the mutual inductance of both the solenoids \({S}_{1}\) and \({S}_{2}\) will remain unchanged.

Hence, the correct option is (C).

Given: \(\mathrm{A}=0.7 \mathrm{~m}^{2}, \mathrm{~N}=100\) turns, \(\mathrm{f}=60\) rev/minute \(=1 \mathrm{rev} / \mathrm{sec}\), and \(\mathrm{B}=0.02 \mathrm{~T}\)

Where, \(A=\) area of the coil, \(N=\) number of turns, \(f=\) frequency of rotation, \(\omega=2 \pi f=\) angular speed of coil, and \(B=\) magnetic field

We know that the maximum emf induced in the AC generator is given as,

\(\epsilon_{\max }=\mathrm{NBA} \omega\)

\(=100 \times 0.02 \times 0.7 \times 2 \pi \times 1\)

\(=\frac{100 \times 0.02 \times 0.7 \times 2 \times 22}{7}\)

\(=8.8 \mathrm{~volt}\)

Hence, the correct option is (A).

Given: \(\mathrm{C}_{1}=\mathrm{C}_{2}=8 \mu \mathrm{F}=8 \times 10^{-6} \mathrm{~F}\) and \(\mathrm{L}=0.4 \mathrm{mH}=0.4 \times 10^{-3} \mathrm{H}\)

Where, \(L=\) self-inductance and \(C=\) capacitance

In the given figure, the two capacitors are in series so the equivalent capacitance \(\mathrm{C}\) is given as,

\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\)

\(\Rightarrow \frac{1}{C}=\frac{1}{8 \times 10^{-6}}+\frac{1}{8 \times 10^{-6}}\)

\(\Rightarrow \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)

We know that, the angular frequency of the LC oscillation circuit is given as,

\(\omega_{o}=\frac{1}{\sqrt{L C}}\)

\(=\frac{1}{\sqrt{0.4 \times 10^{-3} \times 4 \times 10^{-6}}}\)

\(=25 \times 10^{3} \mathrm{rad} / \mathrm{sec}\)

Hence, the correct option is (C).

Given: DC voltage EP = 120 V (Primary coil)

• The transformer works on the principle of mutual inductance.
• To induce emf in the secondary coil of the transformer, the magnetic flux associated with the secondary coil must change with respect to time.
• When the DC voltage is applied at the primary coil of the transformer, the magnetic flux associated with the coil will remain constant with respect to time. So the emf will not induce at the secondary coil.
• Therefore, when the DC voltage is applied at the primary coil, the induced emf in the secondary coil will be zero.

Hence, the correct option is (D).

When the magnet oscillates in and out of the spring, it induces an EMF, the direction in which EMF is getting induced will be different.

Due to the induced EMF, a current will be set up in the coil which will deflect the pointer in the galvanometer in the opposite directions, as the magnet oscillates in and out of the spring an eddy current is set up in it decreases the amplitude of oscillation.

In short, the EMF will be induced in opposite directions, i.e., Left and Right, as the magnet oscillates in and out of the spring also the eddy current will reduce the amplitude of oscillation as the time goes on (Damping).

Hence, the correct option is (D).

Given: Number of turns in the primary coil \(\left(\mathrm{N}_{\mathrm{s}}\right)=100\), Number of turns in the secondary coil \(\left(\mathrm{N}_{\mathrm{p}}\right)=250\), the peak input voltage \(\left(\mathrm{V}_{\mathrm{p}}\right)=28 \mathrm{~V}\)

In a transformer, the voltage in the secondary coil is calculated by,

\(\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}} \quad \cdots\) (1)

Putting the value in equation (1),

\(\frac{250}{100}=\frac{V_{s}}{\frac{28}{\sqrt{2}}}\)

\(V_{s}=50 V\)

Hence, the correct option is (A).

The Lenz law states that the induced emf in a coil due to a changing magnetic flux is such that the magnetic field created by the induced emf opposes the change in a magnetic field.

When the north pole of a magnet is moved away from a coil that is connected to a circuit, the distance between the magnet and the coil will increase, and magnetic flux associated with the coil is decreased.

We know that if the current in the coil is clockwise, the face of the coil towards the observer behaves as the south pole and if the current in the coil is anti-clockwise, the face of the coil towards the observer behaves as the north pole.

So, for the formation of the south pole on the magnet side, the current in the coil will be clockwise when the coil is seen from the magnet side.

Hence, the correct option is (A).

In the given figure, when the metallic plate A is allowed to swing like a simple pendulum between the magnetic poles, the magnetic flux associated with the plate keeps on changing as the plate moves in and out of the region between magnetic poles.

Due to this changing magnetic flux, the eddy currents induced in the plate oppose the motion of the plate. If a slot is cut in plate A area then the area available to the flow of eddy currents is less. Thus, the pendulum plate with holes or slots reduces electromagnetic damping, and the plate swings more freely.

Therefore, the times taken to by the plate to come to rest will increase when a slot is cut in the plate.

Hence, the correct option is (A).

Given: \(\eta=80 \%, V_{s}=440 V, I_{s}=2 A\), and \(V_{P}=220 V\)

The output and input power of the transformer is given by,

Input Power \(\left(P_{i}\right)=V_{P} \times I_{P}=220 \times I_{P}\)

Output power \(\left(P_{0}\right)=V_{s} \times I_{s}=440 \times 2=880 \mathrm{~W}\)

The efficiency (\(\eta\)) of the transformer is given by,

\(\eta=\frac{\text { Output power }}{\text { Input power }}\)

\(\Rightarrow \eta=\frac{P_{0}}{P_{i}}\)

\(\Rightarrow 80=\frac{880}{220 \times I_{p}} \times 100\)

\(\Rightarrow I_{P}=5\) Ampere

Hence, the correct option is (A).

Given: Initial magnetic flux \(\left(\phi_{1}\right)=5.5 \times 10^{-4} \mathrm{~Wb}\), Final magnetic flux \(\left(\phi_{2}\right)=5 \times 10^{-5} \mathrm{~Wb}\), Resistance \((\mathrm{R})=10 \Omega\), number of turns \((\mathrm{N})=1000\), and change in time \((\Delta t)=0.1~\mathrm{sec}\)

Change in flux:

\(\mathrm{d} \phi=\left(\phi_{2}-\phi_{1}\right)=\left(5 \times 10^{-5}-5.5 \times 10^{-4}\right)=-5 \times 10^{-4} \mathrm{Wb}\)

Induced emf in coils,

\(e=-N \frac{d \phi}{d t}=-1000 \frac{\left(-5 \times 10^{-4}\right)}{0.1}=5 V\)

Induced current in the coil

\(i=\frac{e}{R}=\frac{5}{10}=0.5 \mathrm{~A}\)

Hence, the correct option is (B).

Given: \(A=2.5 \mathrm{~m}^{2}, N=800\) turns, \(\omega=50 \mathrm{rad} / \mathrm{sec}, \mathrm{B}=0.05 \mathrm{~T}\), and \(\mathrm{R}=200 \Omega\)

Where, \(A=\) area of the coil, \(N=\) number of turns, \(R=\) resistance, \(\omega=\) angular speed of coil, and \(B=\) magnetic field

We know that the maximum current induced in the AC generator is given as,

\(I_{\max }=\frac{N B A \omega}{R}\)

\(\Rightarrow I_{\max }=\frac{800 \times 0.05 \times 2.5 \times 50}{200}\)

\(\Rightarrow I_{\max }=25 \mathrm{~A}\)

Hence, the correct option is (A).

Voltage in the AC circuit, \(\mathrm{V}_{\mathrm{rms}}=\mathrm{E}\)

We know that:

The power dissipated in an AC circuit,

\(P=V_{r m s} I_{r m s} \cos \phi\)

Where, rms current is denoted by \(\left(I_{r m s}\right)\)

rms voltage is denoted by \(\left(V_{r m s}\right)\)

\(P=E \times \frac{E}{|Z|} \times \cos \phi\)

\(\Rightarrow P=E \times \frac{E}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \times \frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\)

\(\therefore\) Power, \(P=\frac{E^{2} R}{\left[R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}\right]}\)

Where, R is the resistance, and ω is the angular frequency, L is inductance, C is capacitance, Z is impedence  and E is emf.

Hence, the correct option is (C).