Text Book Back Questions and Answers - Chapter 4 Electromagnetic Induction and Alternating Current 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Electromagnetic Induction and Alternating Current
MultipleChoice Questions
Question 1.
An electron moves on a straight, line path $\mathrm{XY}$ as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?

(a) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced Electron
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil
Question 2.
A thin semi-circular conducting ring (PQR) of radius $r$ is falling with its plane vertical in a horizontal magnetic field $\mathrm{B}$, as shown in the figure. The potential difference developed across the ring when its speed $\mathrm{v}$, is- 

(a) Zero
(b) $\frac{B v \pi r^2}{2}$
(c) $\pi \mathrm{rBv}$ and $\mathrm{R}$ is at higher potential
(d) $2 \mathrm{rBv}$ and $\mathrm{R}$ is at higher potential
Answer:
(d) $2 \mathrm{rBv}$ and $\mathrm{R}$ is at higher potential
Question 3.
The flux linked with a coil at any instant $\mathrm{t}$ is given by $\Phi_{\mathrm{B}}=10 \mathrm{t}^2-50 \mathrm{t}+250$. The induced emf at $\mathrm{t}=$ $3 \mathrm{~s}$ is-
(a) $-190 \mathrm{~V}$
(b) $-10 \mathrm{~V}$
(c) $10 \mathrm{~V}$
(d) $190 \mathrm{~V}$
Answer:
(b) $-10 \mathrm{~V}$
Question 4.
When the current changes from $+2 \mathrm{~A}$ to $-2 \mathrm{~A}$ in $0.05 \mathrm{~s}$, an emf of $8 \mathrm{~V}$ is induced in a coil. The coefficient of self-induction of the coil is-
(a) $0.2 \mathrm{H}$
(b) $0.4 \mathrm{H}$
(c) $0.8 \mathrm{H}$
(d) $0.1 \mathrm{H}$
Answer:
(d) $0.1 \mathrm{H}$
Question 5.
The current $i$ flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be-

Answer

Question 6.
A circular coil with a cross-sectional area of $4 \mathrm{~cm}^2$ has 10 turns. It is placed at the centre of a long solenoid that has 15 turns $/ \mathrm{cm}$ and a cross-sectional area of $10 \mathrm{~cm}^2$. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?
(a) $7.54 \mu \mathrm{H}$
(b) $8.54 \mu \mathrm{H}$
(c) $9.54 \mu \mathrm{H}$
(d) $10.54 \mu \mathrm{H}$
Answer:
(a) $7.54 \mu \mathrm{H}$
Question 7.
In a transformer, the number of turns in the primary and the secondary are 410 and 1230 , respectively. If the current in primary is $6 \mathrm{~A}$, then that in the secondary coil is-
(a) $2 \mathrm{~A}$
(b) $18 \mathrm{~A}$
(c) $12 \mathrm{~A}$
(d) $1 \mathrm{~A}$
Answer:
(a) $2 \mathrm{~A}$
Question 8.
A step-down transformer reduces the supply voltage from $220 \mathrm{~V}$ to $11 \mathrm{~V}$ and increase the current from $6 \mathrm{~A}$ to $100 \mathrm{~A}$. Then its efficiency is-
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 9.
In an electrical circuit, $\mathrm{R}, \mathrm{L}, \mathrm{C}$ and $\mathrm{AC}$ voltage source are all connected in series. When $\mathrm{L}$ is removed from the circuit, the phase difference between the voltage and current in the circuit is $\frac{1}{2}$. Instead, if $\mathrm{C}$ is removed from the circuit, the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is- (NEET 2012)
(a) $\frac{1}{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) 1
(d) $\frac{\sqrt{ } 3}{2}$
Answer:
(c) 1
Question 10.
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is-
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{6}$
(d) zero
Answer:
(a) $\frac{\pi}{4}$
Question 11.
In a series resonant RLC circuit, the voltage across $100 \Omega$ resistor is $40 \mathrm{~V}$. The resonant frequency co is $250 \mathrm{rad} / \mathrm{s}$. If the value of $\mathrm{C}$ is $4 \mu \mathrm{F}$, then the voltage across $\mathrm{L}$ is-
(a) $600 \mathrm{~V}$
(b) $4000 \mathrm{~V}$
(c) $400 \mathrm{~V}$
(d) $1 \mathrm{~V}$
Answer:
(c) $400 \mathrm{~V}$
Question 12.
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $\mathrm{v}=10 \sin 340 \mathrm{t}$. The power loss in AC circuit is-
(a) $0.76 \mathrm{~W}$
(b) $0.89 \mathrm{~W}$
(c) $0.46 \mathrm{~W}$
(d) $0.67 \mathrm{~W}$
Answer:
(c) $0.46 \mathrm{~W}$

Questions 13.
The instantaneous values of alternating current and voltage in a circuit are $i=\frac{1}{\sqrt{ } 2}=\sin (100 \pi t) \mathrm{A}$ and $\mathrm{v}=\frac{1}{\sqrt{ } 2} \sin \left(100 \pi t+\frac{\pi}{3}\right) \mathrm{V}$. The average power in watts consumed in the circuit is-
(IIT Main 2012)
(a) $\frac{1}{4}$
(b) $\frac{\sqrt{ } 3}{4}$
(c) $\frac{1}{2}$
(d) $\frac{1}{8}$
Answer:
(d) $\frac{1}{8}$
Question 14.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is-
(a) $\frac{Q}{2}$
(b) $\frac{Q}{\sqrt{ } 3}$
(c) $\frac{Q}{\sqrt{ } 2}$
(d) $\frac{Q}{2}$
Answer:
(c) $\frac{Q}{\sqrt{ } 2}$
Question 15.
$\frac{20}{\pi^2} \mathrm{H}$ inductor is connected to a capacitor of capacitance $\mathrm{C}$. The value of $\mathrm{C}$ in order to impart maximum power at $50 \mathrm{~Hz}$ is-
(a) $50 \mu \mathrm{F}$
(b) $0.5 \mu \mathrm{F}$
(c) $500 \mu \mathrm{F}$
(d) $5 \mu \mathrm{F}$
Answer:
(d) $5 \mu \mathrm{F}$
ShortAnswer Questions

Question 1.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.
Question 2.
State Faraday's laws of electromagnetic induction.
Answer:
First law:
Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.
Second law:
The magnitude of induced emf in a closed circuit is equal to the time rate of change of magnetic flux linked with the circuit.
Question 3.
State Lenz's law.
Answer:
State Lenz's law:
Lenz's law states that the direction of the induced current is such that it always opposes the cause responsible for its production.
Question 4.
State Fleming's right hand rule.
Answer:
The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the Electromagnetic Induction and Alternating Current thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.
Question 5.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 6.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:
1. By changing the magnetic field $B$
2. By changing the area $\mathrm{A}$ of the coil and
3. By changing the relative orientation 0 of the coil with magnetic field
Question 7.
What for an inductor is used? Give some examples.
Answer:
Inductor is a device used to store energy in a magnetic field when an electric current flows through it. The typical examples are coils, solenoids and toroids.
Question 8.
What do you mean by self-induction?
Answer:
If the magnetic flux is changed by changing the current, an emf is induced in that same coil. This phenomenon is known as self-induction.
Question 9.
What is meant by mutual induction?
Answer:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction.
Question 10.
Give the principle of $\mathrm{AC}$ generator.
Answer:
Alternators work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. The magnitude of the induced emf is given by Faraday's law of electromagnetic induction and its direction by Fleming's right hand rule.
Question 11.
List out the advantages of stationary armature-rotating field system of $\mathrm{AC}$ generator.
Answer:

1. The current is drawn directly from fixed terminals on the stator without the use of brush contacts.
2. The insulation of stationary armature winding is easier.
3. The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.
4. Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.
Question 12.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.
Question 13.
Define average value of an alternating current.
Answer:
The average value of alternating current is defined as the average of all values of current over a positive half-cycle or negative half-cycle.
Question 14.
How will you define RMS value of an alternating current?
Answer:
RMS value of alternating current is defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time.
Question 15.
What are phasors?
Answer:
A sinusoidal alternating voltage (or current) can be represented by a vector which rotates about the origin in anti-clockwise direction at a constant angular velocity $\omega$. Such a rotating vector is called a phasor.

Question 16.
Define electric resonance.
Answer:
When the frequency of the applied alternating source is equal to the natural frequency of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance.
Question 17.
What do you mean by resonant frequency?
Answer:
When the frequency of the applied alternating source $\left(\omega_{\mathrm{r}}\right)$ is equal to the natural frequency $\left[\frac{1}{\sqrt{L C}}\right]$ of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance. The frequency at which resonance takes place is called resonant frequency. Resonant angular frequency, $\omega_{\mathrm{r}}=\frac{1}{\sqrt{L C}}$
Question 18.
How will you define Q-factor?
Answer:
It is defined as the ratio of voltage across $\mathrm{L}$ or $\mathrm{C}$ to the applied voltage.
Q-factor $=\frac{\text { Voltage across } \mathrm{L} \text { or } \mathrm{C}}{\text { Applied voltage }}$
Question 19.
What is meant by wattles current?
Answer:
The component of current $\left(\mathrm{I}_{\mathrm{RMS}} \sin \varphi\right)$, which has a phase angle of $\frac{\pi}{2}$ with the voltage is called reactive component. The power consumed is zero. So that it is also known as 'Wattless' current.
Question 20.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit. It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.
$
\cos \phi=\frac{\text { True power }}{\text { Apparent power }}=\frac{P_{\mathrm{av}}}{\mathrm{V}_{\mathrm{ms}} \mathrm{I}_{\mathrm{rms}}}
$

Question 21.
What are LC oscillations?
Answer:
Whenever energy is given to a LC circuit, the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations. During LC oscillations, the total energy
remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
LongAnswer Questions
Question 1.
Establish the fact that the relative motion between the coil and the magnet induces an emf in the coil of a closed circuit.
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.
The relative motion between the coil and the magnet induces:
In the first experiment, when a bar magnet is placed close to a coil, some of the magnetic field lines of the bar magnet pass through the coil i.e., the magnetic flux is linked with the coil. When the bar magnet and the coil approach each other, the magnetic flux linked with the coil increases. So this increase in magnetic flux induces an emf and hence a transient electric current flows in the circuit in one direction (Figure(a)).

At the same time, when they recede away from one another, the magnetic flux linked with the coil decreases. The decrease in magnetic flux again induces an emf in opposite direction and hence an electric current flows in opposite direction (Figure (b)). So there is deflection in the galvanometer when there is a relative motion between the coil and the magnet.

In the second experiment, when the primary coil $\mathrm{P}$ carries an electric current, a magnetic field is established around it. The magnetic lines of this field pass through itself and the neighbouring secondary coil $\mathrm{S}$.

When the primary circuit is open, no electric current flows in it and hence the magnetic flux linked with the secondary coil is zero (Figure(a)).
However, when the primary circuit is closed, the increasing current builds up a magnetic field around the primary coil. Therefore, the magnetic flux linked with the secondary coil increases. This increasing flux linked induces a transient electric current in the secondary coil (Figure(b)).
When the electric current in the primary coil reaches a steady value, the magnetic flux linked with the secondary coil does not change and the electric current in the secondary coil will disappear. Similarly, when the primary circuit is broken, the decreasing primary current induces an electric current in the secondary coil, but in the opposite direction (Figure (c)). So there is deflection in the galvanometer whenever there is a change in the primary current
Question 2.
Give an illustration of determining direction of induced current by using Lenz's law.
Answer:
Illustration 1:
Consider a uniform magnetic field, with its field lines perpendicular to the plane of the paper and pointing inwards. These field lines are represented by crosses ( $\mathrm{x}$ ) as shown in figure (a). A rectangular metallic frame $\mathrm{ABCD}$ is placed in this magnetic field, with its plane perpendicular to the field. The $\operatorname{arm} \mathrm{AB}$ is movable so that it can slide towards right or left.

If the arm $\mathrm{AB}$ slides to our right side, the number of field lines (magnetic flux) passing through the frame $\mathrm{ABCD}$ increases and a current is induced. As suggested by Lenz's law, the induced current opposes this flux increase and it tries to reduce it by producing another magnetic field pointing outwards i.e., opposite to the existing magnetic field.
The magnetic lines of this induced field are represented by circles in the figure (b). From the direction of the magnetic field thus produced, the direction of the induced current is found to be anti-clockwise by using right-hand thumb rule.

The leftward motion of arm $\mathrm{AB}$ decreases magnetic flux. The induced current, this time, produces a magnetic field in the inward direction i.e., in the direction of the existing magnetic field (figure (c)). Therefore, the flux decrease is opposed by the flow of induced current. From this, it is found that induced current flows in clockwise direction.
Illustration 2:
Let us move a bar magnet towards the solenoid, with its north pole pointing the solenoid as shown in figure (b). This motion increases the magnetic flux of the coil which in turn, induces an electric current. Due to the flow of induced current, the coil becomes a magnetic dipole whose two magnetic poles are on either end of the coil.

In this case, the cause producing the induced current is the movement of the magnet. According to Lenz's law, the induced current should flow in such a way that it opposes the movement of the north pole towards coil. It is possible if the end nearer to the magnet becomes north pole (figure (b)).
Then it repels the north pole of the bar magnet and opposes the movement of the magnet. Once pole ends are known, the direction of the induced current could be found by using right hand thumb rule. When the bar magnet is withdrawn, the nearer end becomes south pole which attracts north pole of the bar magnet, opposing the receding motion of the magnet (figure (c)). Thus the direction of the induced current can be found from Lenz's law.
Question 3.
Show that Lenz's law is in accordance with the law of conservation of energy.
Answer:
Conservation of energy:
The truth of Lenz's law can be established on the basis of the law of conservation of energy.
According to Lenz's law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion. As a result, there will always be a resisting force on the moving magnet.

Work has to be done by some external agency to move the magnet against this resisting force. Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.
Question 4.
Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force:
Consider a straight conducting rod $\mathrm{AB}$ of length 1 in a uniform magnetic field $\vec{B}$ which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity $\vec{v}$ towards right side.
When the rod moves, the free electrons present in it also move with same velocity $\vec{v}$ in $\vec{B}$. As a result, the Lorentz force acts on free electrons in the direction from $\mathrm{B}$ to $\mathrm{A}$ and is given by the relation $\vec{F}_{\mathrm{B}}=-\mathrm{e}(\vec{v} \times \vec{B})$

The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field $\mathrm{E}$ directed along BA. Due to the electric field E, the coulomb force starts acting on the free electrons along $A B$ and is given by $\vec{F}_{\mathrm{E}}=-\mathrm{e} \vec{E}$
The magnitude of the electric field $\vec{E}$ keeps on increasing as long as accumulation of electrons at the end A continues. The force $\vec{F}_{\mathrm{E}}$ also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force $\vec{F}_{\mathrm{B}}$ and the coulomb force $\vec{F}_{\mathrm{E}}$ balance each other and no further accumulation of free electrons at the end $A$ takes place, i.e.,
$
\begin{aligned}
& \left|\vec{F}_B\right|=\left|\vec{F}_E\right| \\
& |-e(\vec{v} \times \overrightarrow{\mathrm{B}})|=|-e \overrightarrow{\mathrm{E}}| \\
& \text { vB } \sin 90^{\circ}=\mathrm{E} \\
& \text { vB }=\mathrm{E} \ldots \ldots . \text { (3) }
\end{aligned}
$
The potential difference between two ends of the rod is

Figure: Motional emf from Lorentz force
$
\begin{aligned}
& \mathrm{V}=\mathrm{El} \\
& \mathrm{V}=\mathrm{vB} 1
\end{aligned}
$
Thus the Lorentz force on the free electrons is responsible to maintain this . potential difference and hence produces an emf $\varepsilon=\mathrm{Blv} \ldots .(4)$
As this emf is produced due to the movement of the rod, it is often called as motional emf.
Question 5.
Using Faraday's law of electromagnetic induction, derive an equation for motional emf.
Answer:
Motional emf from Faraday's law:
Let us consider a rectangular conducting loop of width 1 in a uniform magnetic field $\vec{B}$ which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field.

Figure: Motional emf from Faraday's law
When the loop is pulled with a constant velocity $\vec{v}$ to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday's law, an electric current is induced in the loop which flow's in a direction so as to oppose the pull of the loop.
Let $\mathrm{x}$ be the length of the loop which is still within the magnetic field, then its area is $1 \mathrm{x}$. The magnetic flux linked with the loop is
$
\begin{aligned}
\Phi_{\mathrm{B}} & =\int_{\mathrm{A}} \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos \theta \\
& =\mathrm{B} \mathrm{A} \\
\Phi_{\mathrm{B}} & =\mathrm{B} l x
\end{aligned}
$
Here $\theta=0^{\circ}$ and $\cos 0^{\circ}=1$
As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
$
\varepsilon=\frac{d \Phi_B}{d t}=\frac{d}{d t}(\mathrm{Blx})
$
Here, both $\mathrm{B}$ and 1 are constants. Therefore,
$
\varepsilon=\mathrm{Bl} \frac{d x}{d l}=\mathrm{Blv}
$
where $\mathrm{v}=\frac{d x}{d t}$ is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.
Question 6.
Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are
1. Induction stove
2. Eddy current brake
3. Eddy current testing
4. Electromagnetic damping

1. Induction stove:
Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone.

When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.
2. Eddy current brake:
This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.
In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.
3. Eddy current testing:
It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field.
When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.
4. Electro magnetic damping:
The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 7.
Define self-inductance of a coil interns of (i) magnetic flux and (ii) induced emf.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when $1 \mathrm{~A}$ current flows through it.
When the current $i$ changes with time, an emf is induced in it. From Faraday's law of electromagnetic induction, this self-induced emf is given by
$
\begin{aligned}
& \varepsilon=-\frac{d\left(\mathrm{~N}_{\mathrm{B}}\right)}{d t}=-\frac{d(L i)}{d t} \\
& \therefore \varepsilon=-\mathrm{L} \frac{d i}{d t} \text { or } \mathrm{L}=\frac{-\varepsilon}{d i / d t}
\end{aligned}
$
The negative sign in the above equation means that the self-induced emf always opposes the change
in current with respect to time. If $\frac{d i}{d t}=1 \mathrm{As}^{-1}$, then $\mathrm{L}=-\varepsilon$. Inductance of a coil is also defined as the opposing emf induced in the coil when the rate of change of current through the coil is $1 \mathrm{~A} \mathrm{~s}^{-1}$.
Question 8.
How will you define the unit of inductance?
Answer:
Unit of inductance: Inductance is a scalar and its unit is $\mathrm{Wb} \mathrm{A}^{-1}$ or $\mathrm{V} \mathrm{s} \mathrm{A}^{-1}$. It is also measured in henry $(\mathrm{H})$.
$
1 \mathrm{H}=1 \mathrm{~Wb} \mathrm{~A}^{-1}=1 \mathrm{VsA}^{-1}
$
The dimensional formula of inductance is $\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}$.
If $\mathrm{i}=1 \mathrm{~A}$ and $\mathrm{N} \Phi_{\mathrm{B}}=1 \mathrm{~Wb}$ turns, then $\mathrm{L}=1 \mathrm{H}$.
Therefore, the inductance of the coil is said to be one henry if a current of 1 A produces unit flux linkage in the coil.
If $\frac{d i}{d t}=1 \mathrm{As}^{-1}$ and $\varepsilon=-1 \mathrm{~V}$, then $\mathrm{L}=1 \mathrm{H}$.
Therefore, the inductance of the coil is one henry if a current changing at the rate of $1 \mathrm{~A} \mathrm{~s}^{-1}$ induces an opposing emf of $1 \mathrm{~V}$ in it.

Question 9.
What do you understand by self-inductance of a coil? Give its physical significance.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when $1 \mathrm{~A}$ current flows through it.
When the current $i$ changes with time, an emf is induced in it. From Faraday's law of electromagnetic induction, this self-induced emf is given by
$
\begin{aligned}
\varepsilon & =-\frac{d\left(\mathrm{~N} \Phi_{\mathrm{B}}\right)}{d t}=-\frac{d(\mathrm{~L} i)}{d t} \\
\therefore \quad \varepsilon & =-\mathrm{L} \frac{d i}{d t} \quad \text { or } \quad \mathrm{L}=\frac{-\varepsilon}{d i / d t}
\end{aligned}
$
Physical significance of inductance:
When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit. Likewise, when circuit is broken, the decreasing current induces an emf in the reverse direction. This emf now opposes the decay of current.

Figure: Induced emf $\varepsilon$ opposes the changing current $\mathrm{i}$
Thus, inductance of the coil opposes any change in current and tries to maintain the original state.
Question 10.
Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.
Answer:
Self-inductance of a long solenoid:
Consider a long solenoid of length 1 and cross-sectional area $\mathrm{A}$. Let $\mathrm{n}$ be the number of turns per unit length (or turn density) of the solenoid. When an electric current $i$ is passed through the solenoid, a magnetic field is produced by it which is almost uniform and is directed along the axis of the solenoid. The magnetic field at any point inside the solenoid is given by
$
\mathrm{B}=\mu_0 \mathrm{ni}
$
As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is
$
\begin{array}{ll}
\Phi_{\mathrm{B}}=\int_{\mathrm{A}} \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos \theta=\mathrm{BA} \\
\Phi_{\mathrm{B}}=\left(\mu_0 n i\right) \mathrm{A} & \left(\text { since } \theta=0^{\circ}\right)
\end{array}
$
The total magnetic flux linked or flux linkage of the solenoid with $\mathrm{N}$ turns (the total number of turns $\mathrm{N}$ is given by $\mathrm{N}=\mathrm{nl}$ ) is

$
\begin{aligned}
& \mathrm{N} \Phi_{\mathrm{B}}=\mathrm{n}(\mathrm{nl})\left(\mu_0 \mathrm{ni}\right) \mathrm{A} \\
& \mathrm{N} \Phi_{\mathrm{B}}=\left(\mu_0 \mathrm{n}^2 \mathrm{Al}\right) \mathrm{i} \ldots . .(1)
\end{aligned}
$
From the self induction
$
\mathrm{N} \Phi_{\mathrm{B}}=\mathrm{LI} \text {.... (2) }
$
Comparing equations (1) and (2), we have $\mathrm{L}=\mu_0 \mathrm{n}^2 \mathrm{Al}$
From the above equation, it is clear that inductance depends on the geometry of the solenoid (turn density $\mathrm{n}$, cross-sectional area $\mathrm{A}$, length 1 ) and the medium present inside the solenoid. If the solenoid is filled with a dielectric medium of relative permeability $\mu_{\mathrm{r}}$, then
$
\begin{aligned}
& \mathrm{L}=\mu_0 \\
& \mathrm{~L}=\mu \mathrm{n}_0 \mu_{\mathrm{r}} \mathrm{n}^2 \mathrm{Al}
\end{aligned}
$
Question 11.
An inductor of inductance $\mathrm{L}$ carries an electric current i. How much energy is stored while establishing the current in it?
Answer:
Energy stored in an inductor:
Whenever a current is established in the circuit, the inductance opposes the growth of the current. In order to establish a current in the circuit, work is done against this opposition by some external agency. This work done is stored as magnetic potential energy.

Let us assume that electrical resistance of the inductor is negligible and inductor effect alone is considered. The induced emf $e$ at any instant $\mathrm{t}$ is
$
\varepsilon=-\mathrm{L} \frac{d i}{d t}
$
Let $\mathrm{dW}$ be work done in moving a charge dq in a time dt against the opposition, then $\mathrm{dW}=-\varepsilon \mathrm{dq}=-\varepsilon \mathrm{idi}=$ iidi $\quad[\because \mathrm{dq}=\mathrm{idt}]$
Substituting for $s$ value from equation (1) $=-\left(-\mathrm{L} \frac{d i}{d t}\right)$ idt $\mathrm{dW}=\operatorname{Lid} \ldots . .(2)$
Total work done in establishing the current $i$ is

$
\begin{aligned}
& \mathrm{W}=\int d \mathrm{~W}=\int_0^i \mathrm{~L} i d i=\mathrm{L}\left[\frac{i^2}{2}\right]_0^i \\
& \mathrm{~W}=\frac{1}{2} \mathrm{~L} i^2
\end{aligned}
$
This work done is stored as magnetic potential energy.
$
\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{Li}^2
$
Question 12.
Show that the mutual inductance between a pair of coils is same $\left(M_{12}=M_{21}\right)$.
Answer:
Mutual induction:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.

Consider two coils which are placed close to each other. If an electric current $i_1$ is sent through coil 1 , the magnetic field produced by it is also linked with coil 2. Let $\Phi_{21}$ be the magnetic flux linked with each turn of the coil 2 of $\mathrm{N}_2$ turns due to coil 1, then the total flux linked with coil $2\left(\mathrm{~N}_2 \Phi_{21}\right)$ is proportional to the current $i_1$ in the coil 1.
$\mathrm{N}_2 \Phi_{21} \propto i_1$
$\mathrm{N}_2 \Phi_{21}=\mathrm{M}_{21} i_1$
or $\quad \mathrm{M}_{21}=\frac{\mathrm{N}_2 \Phi_{21}}{i_1}$
The constant of proportionality $\mathrm{M}_{21}$ is the mutual inductance of the coil 2 with respect to coil 1 . It is also called as coefficient of mutual induction. If $i_1=1 \mathrm{~A}$, then $\mathrm{M}_{21}=\mathrm{N}_2 \Phi_{21}$.
Therefore, the mutual inductance $\mathrm{M}_{21}$ is defined as the flux linkage of the coil 2 when $1 \mathrm{~A}$ current flows through coil 1. When the current changes with time, an emf $\varepsilon_2$ is induced in coil 2 . From Faraday's law of electromagnetic induction, this mutually induced emf $\varepsilon_2$ is given by

$
\begin{aligned}
& \varepsilon_2=-\frac{d\left(\mathrm{~N}_2 \Phi_{21}\right)}{d t}=-\frac{d\left(\mathrm{M}_{21} i_1\right)}{d t} \\
& \varepsilon_2=-\mathrm{M}_{21} \frac{d i_1}{d t} \quad \text { or } \quad \mathrm{M}_{21}=\frac{-\varepsilon_2}{d i_1 / d t}
\end{aligned}
$
The negative sign in the above equation shows that the mutually induced emf always opposes the change in current $\mathrm{i}$, with respect to time. If $\frac{d i}{d t}=1 \mathrm{As}^{-1}$, then $\mathrm{M}_{21}=-\varepsilon_2$. Mutual inductance $\mathrm{M}_{21}$, is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil 1 is $1 \mathrm{As}^{-1}$. Similarly, if an electric current $i_2$ through coil 2 changes with time, then emf $\varepsilon_1$ is induced in coil 1 . Therefore,
$
\mathrm{M}_{12}=\frac{\mathrm{N}_1 \Phi_{12}}{i_2} \text { and } \mathrm{M}_{12}=\frac{-\varepsilon_1}{d i_2 / d t}
$
where $\mathrm{M}_{12}$ is the mutual inductance of the coil 1 with respect to coil 2 . It can be shown that for a given pair of coils, the mutual inductance is same, i.e., $M_{21}=M_{12}=M$.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.
Question 13.
How will you induce an emf by changing the area enclosed by the coil?
Answer:
Induction of emf by changing the area of the coil:
Consider a conducting rod of length 1 moving with a velocity $\mathrm{v}$ towards left on a rectangular metallic framework. The whole arrangement is placed in a uniform magnetic field $\vec{B}$ whose magnetic lines are perpendicularly directed into the plane of the paper. As the rod moves from $\mathrm{AB}$ to $\mathrm{DC}$ in a time $\mathrm{dt}$, the area enclosed by the loop and hence the magnetic flux through the loop decreases.

The change in magnetic flux in time dt is $\mathrm{d} \Phi_{\mathrm{B}}=\mathrm{B} \times$ change in area
$\mathrm{B} \times$ Area $\mathrm{ABCD}$
$=\mathrm{Blvdt}$ since Area $\mathrm{ABCD}=1($ vdt $)$ or $\frac{d \Phi_B}{d t}=\mathrm{Blv}$
As a result of change in flux, an emf is generated in the loop. The magnitude of the induced emf is $\varepsilon=\frac{d \Phi_B}{d t}=\mathrm{Blv}$
This emf is called motional emf. The direction of induced current is found to be clockwise from Fleming's right hand rule.
Question 14.
Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.
Answer:
Induction of emf by changing relative orientation of the coil with the magnetic field:
Consider a rectangular coil of $\mathrm{N}$ turns kept in a uniform magnetic field $\vec{B}$ figure (a). The coil rotates in anti-clockwise direction with an angular velocity $\omega$ about an axis, perpendicular to the field. At time $=0$, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value $\Phi_{\mathrm{m}}=$ BA (where A is the area of the coil).

In a time $t$ seconds, the coil is rotated through an angle $\theta(=\omega \mathrm{t})$ in anti-clockwise direction. In this position, the flux linked is $\Phi_{\mathrm{m}} \cos \omega t$, a component of $\Phi_{\mathrm{m}}$ normal to the plane of the coil (figure (b)). The component parallel to the plane $\left(\Phi_{\mathrm{m}} \sin \omega t\right)$ has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is $N \Phi_{\mathrm{B}}=N \Phi_{\mathrm{m}} \cos \omega t$. According to Faraday's law, the emf induced at that instant is

$
\begin{aligned}
& \varepsilon=\frac{d}{d t}\left(\mathrm{~N} \Phi_{\mathrm{B}}\right)=\frac{d}{d t}\left(\mathrm{~N} \Phi_{\mathrm{m}} \cos \omega \mathrm{t}\right) \\
& =-\mathrm{N} \Phi_{\mathrm{m}}(-\sin \omega \mathrm{t}) \omega=\mathrm{N} \Phi_{\mathrm{m}} \omega \sin \omega t
\end{aligned}
$
When the coil is rotated through $90^{\circ}$ from initial position, $\sin \omega t=1$, Then the maximum value of induced emf is
$
\varepsilon_{\mathrm{m}}=\mathrm{N} \Phi_{\mathrm{m}} \omega=\mathrm{NBA} \omega \text { since } \Phi_{\mathrm{m}}=\mathrm{BA}
$
Therefore, the value of induced emf at that instant is then given by
$
\varepsilon=\varepsilon_{\mathrm{m}} \sin \omega \mathrm{t}
$
It is seen that the induced emf varies as sine function of the time angle $\omega t$. The graph between induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf.
Question 15.
Elaborate the standard construction details of $\mathrm{AC}$ generator.
Answer:
Construction:
lternator consists of two major parts, namely stator and rotor. As their names suggest, stator is stationary while rotor rotates inside the stator. In any standard construction of commercial alternators, the armature winding is mounted on stator and the field magnet on rotor. The construction details of stator, rotor and various other components involved in them are given below.
(i) Stator:
The stationary part which has armature windings mounted in it is called stator. It has three components, namely stator frame, stator core and armature winding.
Stator frame:
This is the outer frame used for holding stator core and armature windings in proper position. Stator frame provides best ventilation with the help of holes provided in the frame itself.

Stator core:
Stator core or armature core is made up of iron or steel alloy. It is a hollow cylinder and is laminated to minimize eddy current loss. The slots are cut on inner surface of the core to accommodate armature
windings.

Armature winding:
Armature winding is the coil, wound on slots provided in the armature core. One or more than one coil may be employed, depending on the type of alternator. Two types of windings are commonly used. They are (i) single-layer winding and (ii) double-layer winding. In single-layer winding, a slot is occupied by a coil as a single layer. But in double-layer winding, the coils are split into two layers such as top and bottom layers.
(ii) Rotor:
Rotor contains magnetic field windings. The magnetic poles are magnetized by DC source. The ends of field windings are connected to a pair of slip rings, attached to a common shaft about which rotor rotates. Slip rings rotate along with rotor. To maintain connection between the DC source and field windings, two brushes are used which . continuously slide over the slip rings.
There are 2 types of rotors used in alternators:
1. salient pole rotor
2. cylindrical pole rotor.
1. Salient pole rotor:
The word salient means projecting. This rotor has a number of projecting poles having their bases riveted to the rotor. It is mainly used in low-speed alternators.
2. Cylindrical pole rotor:
This rotor consists of a smooth solid cylinder. The slots are cut on the outer surface of the cylinder along its length. It is suitable for very high speed alternators.
The frequency of alternating emf induced is directly proportional to the rotor speed. In order to maintain the frequency constant, the rotor must run at a constant speed. These are standard construction details of alternators.
Question 16.
Explain the working of a single-phase $\mathrm{AC}$ generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase $\mathrm{AC}$ generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed here. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.
Working:
The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin $\mathrm{O}$ in the graph between induced emf and time angle.

When field magnet rotates through $90^{\circ}$, magnetic field becomes parallel to PQRS. The induced $\mathrm{cmfs}$ across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming's right hand rule. Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field.
For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Hence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper.
For PQ, it is downwards B and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of $180^{\circ}$ from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point $B$. The field magnet becomes again parallel to PQRS for $270^{\circ}$ rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP This is represented by point $\mathrm{C}$.

On completion of $360^{\circ}$, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature. Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency
of the induced emf depends on the speed at which the field magnet rotates.

Question 17.
How are the three different emfs generated in a three-phase $\mathrm{AC}$ generator? Show the graphical representation of these three emfs.
Answer:
Three-phase $\mathrm{AC}$ generator:
Some AC generators may have more than one coil in the armature core and each coil produces an alternating emf. In these generators, more than one emf is produced. Thus they are called poly-phase generators. If there are two alternating emfs produced in a generator, it is called two-phase generator. In some $\mathrm{AC}$ generators, there are three separate coils, which would give three separate emfs. Hence they are called three-phase $\mathrm{AC}$ generators.

In the simplified construction of three-phase $\mathrm{AC}$ generator, the armature core has 6 slots, cut on its inner rim. Each slot is $60^{\circ}$ away from one another. Six armature conductors are mounted in these slots. The conductors 1 and 4 are joined in series to form coil 1 . The conductors 3 and 6 form coil 2 while the conductors 5 and 2 form coil 3. So, these coils are rectangular in shape and are $120^{\circ}$ apart from one another.

The initial position of the field magnet is horizontal and field direction is perpendicular to the plane of the coil 1 . As it is seen in single phase $\mathrm{AC}$ generator, when field magnet is rotated from that position in clockwise direction, alternating emf $\varepsilon_1$ in coil 1 begins a cycle from origin $O$.

The corresponding cycle for alternating emf $\varepsilon_2$ in coil 2 starts at point $\mathrm{A}$ after field magnet has rotated through $120^{\circ}$. Therefore, the phase difference between $\varepsilon_1$ and $\varepsilon_2$ is $120^{\circ}$. Similarly, emf $\varepsilon_3$ in coil 3 would begin its cycle at point $\mathrm{B}$ after $240^{\circ}$ rotation of field magnet from initial position. Thus these emfs produced in the three phase $\mathrm{AC}$ generator have $120^{\circ}$ phase difference between one another.
Question 18.
Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric
current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction:
In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.
The coil across which alternating voltage is applied is called primary coil $\mathrm{P}$ and the coil from which output power is drawn out is called secondary coil $\mathrm{S}$. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.
Working:
If the primary' coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coif This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils. As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil $\varepsilon_{\mathrm{p}}$ is almost equal and opposite to the applied voltage $v_p$ and is given by
$
v_{\mathrm{p}}=\varepsilon_{\mathrm{p}}=-\mathrm{N}_{\mathrm{p}} \frac{d \Phi_B}{d t} \ldots \cdots(1)
$
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil eg is given by
$
\varepsilon_{\mathrm{s}}=-\mathrm{N}_{\mathrm{s}} \frac{d \Phi_B}{d t}
$
where $\mathrm{N}_{\mathrm{p}}$ and $\mathrm{N}_{\mathrm{s}}$ are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then $\varepsilon_{\mathrm{S}}=v_{\mathrm{S}}$ where $v_{\mathrm{S}}$ is the voltage across secondary coil.
$
v_{\mathrm{s}} \varepsilon_{\mathrm{s}}=-\mathrm{N}_{\mathrm{s}} \frac{d \Phi_B}{d t} \ldots \ldots \ldots(2)
$
From equation (1) and (2),
$
\frac{v_s}{\varepsilon s}=\frac{N_s}{N_p}=\mathrm{K} \ldots \ldots . .(3)
$
This constant $\mathrm{K}$ is known as voltage transformation ratio. For an ideal transformer, Input power $v_p i_p=$ Output power $v_s i_s$

where $i v_p$ and $i_s$ are the currents in the primary and secondary coil respectively. Therefore, $\frac{v_s}{v p}=\frac{N_s}{N p}=\frac{i_p}{i s}$
Equation (4) is written in terms of amplitude of corresponding quantities, $\frac{V_s}{V p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}=\mathrm{K}$
(i) If $\mathrm{N}_{\mathrm{s}}>\mathrm{N}_{\mathrm{p}}($ or $\mathrm{K}>1)$
$\therefore \mathrm{V}_{\mathrm{s}}>\mathrm{V}_{\mathrm{p}}$ and $\mathrm{I}_{\mathrm{s}}<\mathrm{I}_{\mathrm{p}}$
This is sthe case of step-up transformer in which voltage is decreased and the corresponding current is decreased.
(ii) If $\mathrm{N}_{\mathrm{s}}<\mathrm{N}_{\mathrm{p}}$ (or $\mathrm{K}<1$ )
$\therefore \mathrm{V}_{\mathrm{s}}<\mathrm{V}_{\mathrm{p}}$ and $\mathrm{I}_{\mathrm{s}}>\mathrm{I}_{\mathrm{p}}$
This is step-down transformer where voltage is decreased and the current is increased.
Question 19.
Mention the various energy losses in a transformer.
Answer:
Energy losses in a transformer: Transformers do not have any moving parts so that its efficiency is much higher than that of rotating machines like generators and motors. But there are many factors which lead to energy loss in a transformer.
(i) Core loss or Iron loss:
This loss takes place in transformer core. Hysteresis loss and eddy current loss are known as core loss or Iron loss. When transformer core is magnetized and demagnetized repeatedly by the alternating voltage applied across primary coil, hysteresis takes place due to which some energy is lost in the form of heat.
Hysteresis loss is minimized by using steel of high silicon content in making transformer core. Alternating magnetic flux in the core induces eddy currents in it. Therefore there is energy loss due to the flow of eddy current, called eddy current loss which is minimized by using very thin laminations of transformer core.
(ii) Copper loss:
Transformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss which is minimized by using wires of larger diameter.
(iii) Flux leakage:
Flux leakage happens when the magnetic lines of primary coil arc not completely linked with secondary coil. Energy loss due to this flux leakage is minimized by winding coils one over the other.

Question 20.
Give the advantage of $\mathrm{AC}$ in long distance power transmission with an example.
Answer:
Advantages of $\mathrm{AC}$ in long distance power transmission:
Electric power is produced in a t large scale at electric power stations with the help of $\mathrm{AC}$ generators.
These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places.

Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission. But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating $\left(i^2 \mathrm{R}\right)$ in the transmission lines which are hundreds of kilometer long.
This power loss can be tackled either by reducing current $\mathrm{i}$ or by reducing resistance $\mathrm{R}$ of the transmission lines. The resistance $\mathrm{R}$ can be reduced with thick wires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent.
At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss.

At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.
Illustration:
An electric power of $2 \mathrm{MW}$ is transmitted to a place through transmission lines of total resistance, say $\mathrm{R}=40 \Omega$, at two different voltages. One is lower voltage $(10 \mathrm{kV})$ and the other is higher $(100 \mathrm{kV})$. Let us now calculate and compare power losses in these two cases.
Case I:
$\mathrm{P}=2 \mathrm{MW} ; \mathrm{R}==40 \Omega ; \mathrm{V}=10 \mathrm{kV}$ Power,
Power, $\mathrm{P}=\mathrm{VI}$
$\therefore$ Current, $\mathrm{I}=\frac{P}{V}=\frac{2 \times 10^6}{10 \times 10^3} 200 \mathrm{~A}$
Power loss $=$ Heat produced $=\mathrm{I}^2 \mathrm{R}=(200)^2 \times 40=1.6 \times 10^6 \mathrm{~W}$
$\%$ of power loss $=\frac{1.6 \times 10^6}{2 \times 10^6} \times 100 \%=0.8 \times 100 \%=80 \%$
Case II:
$\mathrm{P}=2 \mathrm{MW} ; \mathrm{R}=40 \Omega ; \mathrm{V}=100 \mathrm{kV}$
$\therefore$ Current, $\mathrm{I}=\frac{P}{V}=\frac{2 \times 10^6}{100 \times 10^3} 20 \mathrm{~A}$
Power loss $=$ Heat produced $=\mathrm{I}^2 \mathrm{R}=(20)^2 \times 40=0.016 \times 10^6 \mathrm{~W}$
$\%$ of power loss $=\frac{0.6 \times 10^6}{2 \times 10^6} \times 100 \%=0.008 \times 100 \%=0.8 \%$

Question 21.
Find out the phase relationship between voltage and current in a pure inductive circuit.
Answer:
AC circuit containing only an inductor:
Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source. The alternating voltage is given by the equation. $v=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} \ldots(1)$
The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by
Back emf, $\varepsilon-\mathrm{L} \frac{d i}{d t}$
By applying Kirchoff's loop rule to the purely inductive circuit, we get

$
\begin{aligned}
& \mathrm{v}+\varepsilon=0 \\
& \mathrm{~V}_{\mathrm{m}} \sin \omega \mathrm{t}=\mathrm{L} \frac{\mathrm{di}}{d t} \\
& \mathrm{di}=\mathrm{L} \frac{V_m}{L} \sin \omega \mathrm{t} \mathrm{dt} \\
& \mathrm{i}=\frac{V_m}{L} \int \sin \omega \mathrm{tdt}=\frac{V_m}{L \omega}(-\cos \omega \mathrm{t})+\text { constant }
\end{aligned}
$
The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.
$
\begin{aligned}
& {\left[\begin{array}{l}
\cos \omega t=\sin \left(\frac{\pi}{2}-\omega t\right) \\
-\sin \left(\frac{\pi}{2}-\omega t\right)=\sin \left(\omega t-\frac{\pi}{2}\right)
\end{array}\right]} \\
& i=\frac{\mathrm{V}_m}{\mathrm{~L} \omega} \sin \left(\omega t-\frac{\pi}{2}\right) \quad \text { or } \quad i=\mathrm{I}_{\mathrm{m}} \sin \left(\omega t-\frac{\pi}{2}\right) \\
&
\end{aligned}
$
where $\frac{V_m}{L \omega}=\mathrm{I}_{\mathrm{m}}$, the peak value of the alternating current in the circuit. From equation (1) and (2), it is evident that current lags behind the applied voltage by $\frac{\pi}{2}$ in an inductive circuit.
This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by $90^{\circ}$.
Inductive reactance $\mathrm{X}_{\mathrm{L}}$ :
The peak value of current $\mathrm{I}_{\mathrm{m}}$ is given by $\mathrm{I}_{\mathrm{m}}=\frac{V_m}{L \omega}$. Let us compare this equation with $\mathrm{I}_{\mathrm{m}}=\frac{V_m}{R}$ from resistive circuit. The equantity $\omega_{\mathrm{L}}$ Plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)$. It is measured in ohm.
$
\mathrm{X}_{\mathrm{L}}=\omega_{\mathrm{L}}
$
The inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)$ varies directly as the frequency.
$
\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL} \ldots \ldots .(3)
$
where $f$ is the frequency of the alternating current. For a steady current, $f=0$. Therefore, $\mathrm{X}_{\mathrm{L}}=0$.

Thus an ideal inductor offers no resistance to steady DC current.

Question 22.
Derive an expression for phase angle between the applied voltage and current in a series RLC circuit.

Answer:
AC circuit containing a resistor, an inductor and a capacitor in series - Series RLC circuit:
Consider a circuit containing a resistor of resistance $\mathrm{R}$, a inductor of inductance $\mathrm{L}$ and a capacitor of capacitance $\mathrm{C}$ connected across an alternating voltage source. The applied alternating voltage is given by the equation.
$v=V_m \sin \omega t$

Let $\mathrm{i}$ be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed
across $\mathrm{R}, \mathrm{L}$ and $\mathrm{C}$.
We know that voltage across $R\left(V_R\right)$ is in phase with $i$, voltage across $L\left(V_L\right)$ leads $i$ by $\pi / 2$ and voltage across $\mathrm{C}\left(\mathrm{V}_{\mathrm{C}}\right)$ lags $\mathrm{i}$ by $\pi / 2$.
The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor
$\overrightarrow{O I}, \mathrm{~V}_{\mathrm{R}}$ by $\overrightarrow{O A} ; \mathrm{V}_{\mathrm{L}}$ by $\overrightarrow{O B}$ and $\mathrm{V}_{\mathrm{C}}$ by $\overrightarrow{O C}$.
The length of these phasors are $\mathrm{OI}=\mathrm{I}_{\mathrm{m}}, \mathrm{OA}=\mathrm{I}_{\mathrm{m}} \mathrm{R}, \mathrm{OB}=\mathrm{I}_{\mathrm{m}}, \mathrm{X}_{\mathrm{L}} ; \mathrm{OC}=\mathrm{I}_{\mathrm{m}} \mathrm{X}_{\mathrm{c}}$
The circuit is cither effectively inductive or capacitive or resistive that depends on the value of $V_1$ or $\mathrm{V}_{\mathrm{c}}$ Let us assume that $\mathrm{V}_{\mathrm{L}}>\mathrm{V}_{\mathrm{C}}$. so that net voltage drop across $\mathrm{L}-\mathrm{C}$ combination is $\mathrm{V}_{\mathrm{L}}<\mathrm{V}_{\mathrm{C}}$ which is represented by a phasor $\overrightarrow{A D}$.
By parallelogram law, the diagonal $\overrightarrow{O E}$ gives the resultant voltage $\mathrm{u}$ of $\mathrm{VR}$ and $\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)$ and its length $\mathrm{OE}$ is equal to $\mathrm{V}_{\mathrm{m}}$. Therefore,
$
\begin{aligned}
\mathrm{V}_m^2 & =\mathrm{V}_{\mathrm{R}}^2+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2 \\
& =\sqrt{\left(\mathrm{I}_m \mathrm{R}\right)^2+\left(\mathrm{I}_m \mathrm{X}_{\mathrm{L}}-\mathrm{I}_m \mathrm{X}_{\mathrm{C}}\right)^2}=\mathrm{I}_m \sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}
\end{aligned}
$
or $\quad I_m=\frac{\mathrm{V}_m}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}}$
or $\quad \mathrm{I}_m=\frac{\mathrm{V}_m}{\mathrm{Z}}$
Where $\quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}$
$\mathrm{Z}$ is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the graphical figure.

From phasor diagram, the phase angle between $\mathrm{n}$ and $\mathrm{i}$ is found out from the following relation
$
\tan \phi=\frac{\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{R}}}=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{R}}}
$
Special cases Figure: Phasor diagram for a series
(i) If $\mathrm{X}_{\mathrm{L}}>\mathrm{X}_{\mathrm{C}},\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)$ is positive and phase angle $\varphi$
is also positive. It means that the applied voltage leads the current by $\varphi$ (or current lags behind voltage by $\varphi)$. The circuit is inductive.
$
\therefore v=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} ; \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin (\omega \mathrm{t}+\varphi)
$
(ii) If $\mathrm{X}_{\mathrm{L}}<\mathrm{X}_{\mathrm{C}},\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)$ is negative and $\varphi$ is also negative. Therefore current leads voltage by $\varphi$ and the circuit is capacitive.
$\therefore v=\mathrm{Vm} \sin \omega \mathrm{t} ; \mathrm{i}=\operatorname{Im} \sin (\omega \mathrm{t}+\varphi)$
(iii) If $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}, \varphi$ is zero. Therefore current and voltage are in the same phase and the circuit is resistive.
$\therefore v=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} ; \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$
Question 23.
Define inductive and capacitive reactance. Give their units.
Answer:
Inductive reactance $\mathrm{X}_{\mathrm{L}}$ :
The peak value of current $\mathrm{I}_{\mathrm{m}}$ is given by $\mathrm{I}_{\mathrm{m}}=\frac{V_m}{L \omega}$. Let us compare
this equation with $\mathrm{I}_{\mathrm{m}} \frac{V_m}{R}$ from resistive circuit. The quantity $\omega \mathrm{L}$ plays the same role as the resistance $\mathrm{R}$ in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)$.
It measured in ohm.
$
\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}
$
Capacitive reactance $\mathrm{X}_{\mathrm{C}}$ :
The peak value of current $\mathrm{I}$ is given by $\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{v}_{\mathrm{m}}}{1 / \mathrm{c} \omega}$. Let us compare this equation with $\mathrm{I}_{\mathrm{m}}=\frac{V_m}{R}$ from resistive circuit. The quantity $\frac{1}{\omega C}$ plays the same role as the resistance $\mathrm{R}$ in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)$. It measured in ohm.

$
\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega C}
$
Question 24.
Obtain an expression for average power of $\mathrm{AC}$ over a cycle. Discuss its special cases. Power of a circuit is defined as the rate of consumption of electric energy in that circuit.
Answer:
It is given by the product of the voltage and current. In an $\mathrm{AC}$ circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle.
The alternating voltage and alternating current in the series RLC circuit at an instant are given by $v=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$ and $\mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin (\omega \mathrm{t}+4>)$
where $\varphi$ is the phase angle between $v$ and i. The instantaneous power is then written as $\mathrm{P}=v \mathrm{i}=\mathrm{V}_{\mathrm{m}} \mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t} \sin (\omega \mathrm{t}+\varphi)$

$=V_m I_m \sin \omega t(\sin \omega t \cos \varphi-\cos \omega t \sin \varphi)$
$\mathrm{P}=\mathrm{V}_{\mathrm{m}} \mathrm{I}_{\mathrm{m}}\left(\cos \varphi \sin ^2 \omega \mathrm{t}-\sin \omega \mathrm{t} \cos \omega \mathrm{t} \sin \varphi\right) \ldots \ldots$ (1)
Here the average of $\sin ^2 \omega t$ over a cycle is $\frac{1}{2}$ and that of $\sin \omega t \cos \omega t$ is zero. Substituting these values, we obtain average power over a cycle.
$\mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{m}} \mathrm{I}_{\mathrm{m}} \cos \varphi \mathrm{x} \frac{1}{2}=\frac{V_m}{\sqrt{ } 2} \frac{I_m}{\sqrt{ } 2} \cos \varphi$
$P_{a v}=V_{R M S} I_{R M S} \cos \varphi \ldots \ldots$ (2)
where $V_{R M S} I_{R M S}$ is called apparent power and $\cos \varphi$ is power factor. The average power of an $A C$ circuit is also known as the true power of the circuit.
Special Cases:
(i) For a purely resistive circuit, the phase angle between voltage and current is zero and cos $\varphi=1$
$\therefore \mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{RMS}} \mathrm{I}_{\mathrm{RMS}}$
(ii) For a purely inductive or capacitive circuit, the phase angle is $\pm \frac{\pi}{2}$ and $\cos \left( \pm \frac{\pi}{2}\right)=0$ $\therefore \mathrm{P}_{\mathrm{av}}=0$
(iii) For series RLC circuit, the phase angle $\varphi=\tan ^{-1}\left(\frac{\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{c}}}{\mathrm{R}}\right)$
$\therefore \mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{RMS}} \mathrm{I}_{\mathrm{RMS}} \cos \varphi$
(iv) For series RLC circuit at resonance, the phase angle is zero and $\cos \varphi=1$.
$\therefore \mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{RMS}} \mathrm{I}_{\mathrm{RMS}}$
Question 25.
Show that the total energy is conserved during LC oscillations.
Answer:
Conservation of energy in LC oscillations: During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor.
Although, these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy,
$
\mathrm{U}=\mathrm{U}_{\mathrm{E}}+\mathrm{U}_{\mathrm{B}}=\frac{q^2}{2 \mathrm{C}}+\frac{1}{2} \mathrm{~L} i^2
$
Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.

Case I:
When the charge in the capacitor, $q=Q_m$ and the current through the inductor, $i=0$, the total energy is given by
$
\mathrm{U}=\frac{\mathrm{Q}_m^2}{2 \mathrm{C}}+0=\frac{\mathrm{Q}_m^2}{2 \mathrm{C}}
$
The total energy is wholly electrical.
Case II:
When charge $=0$; current $=\mathrm{I}_{\mathrm{m}}$, the total energy is
$
\begin{aligned}
\mathrm{U} & =0+\frac{1}{2} \mathrm{LI}_m^2=\frac{1}{2} \mathrm{LI}_m^2=\frac{\mathrm{L}}{2} \times\left(\frac{\mathrm{Q}_m^2}{\mathrm{LC}}\right) \quad \text { since } \mathrm{I}_m=\mathrm{Q}_m \omega=\frac{\mathrm{Q}_m}{\sqrt{\mathrm{LC}}} \\
& =\frac{\mathrm{Q}_m^2}{2 \mathrm{C}}
\end{aligned}
$
The total energy is wholly electrical.
Case III:
When charge $=\mathrm{q}$; current $=\mathrm{i}$, the total energy is
$
\mathrm{U}=\frac{q^2}{2 \mathrm{C}}+\frac{1}{2} \mathrm{Li}^2
$
Since $\mathrm{q}=\mathrm{Q}_{\mathrm{m}} \cos \omega \mathrm{t}, \mathrm{i}=\frac{b q}{d t}=\mathrm{Q}_{\mathrm{m}} \omega \sin \omega \mathrm{t}$. The negative sign in current indicates that the charge in the capacitor in the capacitor decreases with time.
$
\begin{aligned}
& \mathrm{U}=\frac{\mathrm{Q}_m^2 \cos ^2 \omega t}{2 \mathrm{C}}+\frac{L \omega^2 Q_m^2 \sin ^2 \omega t}{2} \\
& =\frac{\mathrm{Q}_m^2 \cos ^2 \omega t}{2 \mathrm{C}}+\frac{\mathrm{L}^2 \mathrm{Q}_m^2 \sin ^2 \omega t}{2 \mathrm{LC}} \quad \text { since } \omega^2=\frac{1}{\mathrm{LC}} \\
& \mathrm{U}=\frac{\mathrm{Q}_m^2}{2 \mathrm{C}}\left(\cos ^2 \omega t+\sin ^2 \omega t\right)=\frac{\mathrm{Q}_m^2}{2 \mathrm{C}} \\
&
\end{aligned}
$
From above three cases, it is clear that the total energy of the system remains constant.

Question 26.
Prove that energy is conserved during electromagnetic induction.
Answer:
The mechanical energy of the spring-mass system is given by
$
\mathrm{E}=\frac{1}{2} m v^2+\frac{1}{2} k x^2
$
The energy E remains constant for varying values of $x$ and $v$. Differentiating $E$ with respect to time, we get
$
\frac{d \mathrm{E}}{d t}=\frac{1}{2} m\left(2 v \frac{d v}{d t}\right)+\frac{1}{2} k\left(2 x \frac{d x}{d t}\right)=0
$
or $\quad m \frac{d^2 x}{d t^2}+k x=0$
Since $\frac{d x}{d t}=v$ and $\frac{d v}{d t}=\frac{d^2 x}{d t^2}$
This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form $x(t)=X_m \cos (\omega t+\varphi) \ldots \ldots(3)$
where $X_m$ is the maximum value of $x(t), \omega$, the angular frequency and $\varphi$, the phase constant.
Similarly, the electromagnetic energy of the $\mathrm{LC}$ system is given by
$
\mathrm{U}=\frac{1}{2} \mathrm{~L} i^2+\frac{1}{2}\left(\frac{1}{\mathrm{C}}\right) q^2=\text { constant }
$
Differentiating $U$ with respect to time, we get
$
\frac{d \mathrm{U}}{d t}=\frac{1}{2} \mathrm{~L}\left(2 i \frac{d i}{d t}\right)+\frac{1}{\mathrm{C}}\left(2 q \frac{d q}{\mathrm{dt}}\right)=0
$
or $\quad \mathrm{L} \frac{d^2 q}{d t^2}+\frac{1}{\mathrm{C}} q=0$
since $i=\frac{d q}{d t}$ and $\frac{d i}{d t}=\frac{d^2 q}{d t^2}$
Equation (2) and (5) are proved the energy of electromagnetic induction is conserved.

Question 27.
Compare the electromagnetic oscillations of LC circuit with the mechanical oscillations of block spring system to find the expression for angular frequency of LC oscillators mathematically.
Answer:
The mechanical energy of the spring-mass system is given by
$
\mathrm{E}=\frac{1}{2} m v^2+\frac{1}{2} k x^2
$
The energy $E$ remains constant for varying values of $\mathrm{x}$ and $\mathrm{v}$. Differentiating $\mathrm{E}$ with respect to time, we get
$
\frac{d \mathrm{E}}{d t}=\frac{1}{2} m\left(2 v \frac{d v}{d t}\right)+\frac{1}{2} k\left(2 x \frac{d x}{d t}\right)=0
$
or
$
m \frac{d^2 x}{d t^2}+k x=0
$
Since $\frac{d x}{d t}=v$ and $\frac{d v}{d t}=\frac{d^2 x}{d t^2}$
This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form $\mathrm{x}(\mathrm{t})=\mathrm{X}_{\mathrm{m}} \cos (\omega t+\varphi) \ldots \ldots(3)$
where $X_m$ is the maximum value of $\mathrm{x}(\mathrm{t}), \omega$, the angular frequency and $\varphi$, the phase constant. Similarly, the electromagnetic energy of the $\mathrm{LC}$ system is given by
$
\mathrm{U}=\frac{1}{2} \mathrm{~L} i^2+\frac{1}{2}\left(\frac{1}{\mathrm{C}}\right) q^2=\text { constant }
$
Differentiating $\mathrm{U}$ with respect to time, we get

$
\frac{d \mathrm{U}}{d t}=\frac{1}{2} \mathrm{~L}\left(2 i \frac{d i}{d t}\right)+\frac{1}{\mathrm{C}}\left(2 q \frac{d q}{\mathrm{dt}}\right)=0
$
or
$
\mathrm{L} \frac{d^2 q}{d t^2}+\frac{1}{\mathrm{C}} q=0
$
since $i=\frac{d q}{d t}$ and $\frac{d i}{d t}=\frac{d^2 q}{d t^2}$
Equation (2) and (5) are proved the energy of electromagnetic induction is conserved. $\mathrm{q}(\mathrm{t})=\mathrm{Q}_{\mathrm{m}} \cos (\omega \mathrm{t}+\varphi)$.....(6)
where $\mathrm{Q}_{\mathrm{m}}$ is the maximum value of $\mathrm{q}(\mathrm{t}), \omega$, the angular frequency and $\varphi$, the phase constant.
Current in the LC circuit:
The current flowing in the LC circuit is obtained by differentiating $q(t)$ with respect to time. $\mathrm{i}(\mathrm{t})=\frac{d q}{d t}=\frac{d}{d t}\left[\mathrm{Q}_{\mathrm{m}} \cos (\omega \mathrm{t}+\varphi)\right]=\mathrm{Q}_{\mathrm{m}} \omega \sin (\omega \mathrm{t}+\varphi)$ since $\mathrm{I}_{\mathrm{m}}=\mathrm{Q}_{\mathrm{m}} \omega$
(or)
$\mathrm{i}(\mathrm{t})-\mathrm{I}_{\mathrm{m}} \sin (\omega \mathrm{t}+\varphi)$
The equation (7) clearly shows that current varies as a function of time $t$. In fact, it is a sinusoidally varying alternating current with angular frequency $\omega$.
Angular frequency of LC oscillations:
By differentiating equation (6) twice, we get $\frac{d^2 q}{d t}=-\mathrm{Q}_{\mathrm{m}} \omega^2 \cos (\omega \mathrm{t}+\varphi) \ldots \ldots .$. (8)
Substituting equations (6) and (8) in equation (5),
we obtain $\mathrm{L}\left[-\mathrm{Q}_{\mathrm{m}} \omega^2 \cos (\omega \mathrm{t}+\varphi)\right]+\frac{1}{C} \mathrm{Q}_{\mathrm{m}} \cos (\omega \mathrm{t}+\varphi)=0$
Rearranging the terms, the angular frequency of LC oscillations is given by $\omega=\frac{1}{\sqrt{L C}} \cdots \cdots(9)$
This equation is the same as that obtained from qualitative analogy.