UNIT IV ELECTROMAGNETIC INDUCTION & A...128 128 15.Principle of AC Generator:- Electromagnetic...

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UNIT IV—ELECTROMAGNETIC INDUCTION & A.C 1. Gist of the lesson (including basic concepts and important formulae)

1. Magnetic Flux:- The no. of magnetic field lines passing through an area inside

a magnetic field region is known as magnetic flux passing through that area.

The magnetic flux through a plane surface placed inside a uniform magnetic

field is given by

ϕ = cosBAAB

The magnetic flus passing any surface placed inside a magnetic field (uniform

of non-uniform) is given by

ϕ = AdB

The SI unit of magnetic flux is Tm2 or Wb. It is a Scalar quantity.

The dimension of magnetic flux is [ML2T

-2A

-1].

2. Electromagnetic Induction:- “Whenever there is change in magnetic flux

linked with a conductor or conducting coil, an emf is induced in the conductor

or coil. The emf induced lasts so long as the magnetic flux linked with

conductor or coil changes. This phenomenon is called EMI.”

3. Faraday’s Law of Electromagnetic Induction:- “The magnitude of the induced

emf in a circuit is proportional to the time rate of change of magnetic flux

through the circuit.”

dt

d

4. Lenz’s Law:-“ The polarity of induced emf is such that it tends to tends to

produce a current which opposes the change in magnetic flux that produces

it.” It is based of law of conservation of energy.

Mathematically, Faraday’s & Lenz’s laws are combined in the following

expression

dt

d

5. Induced current and induced charge:- If a coil is closed and has resistance R,

then current induced in the coil,

dt

d

R

N

Ri

ampere,

and the induced charge,

q = i∆t = R

N =

Resistance

linkangeflux Total

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6. Motional emf:-

(a) The emf induced in a straight conductor moving inside a uniform magnetic

field with a velocity perpendicular to its length as well as the magnetic field

induction

ɛ = Bvl

(b) The emf induced in a disc rotating inside a uniform magnetic field directed

parallel to the axis of rotation of disc/ the emf induced in a straight

conductor rotating about its one end inside a uniform magnetic field

directed parallel to the axis of rotation

ɛ = 22

2BwlBvl

7. Eddy Currents:- “When bulk pieces of conductors are subjected to changing

magnetic flux, induced currents are produced in them due to electromagnetic

induction. The flow pattern of these currents resemble swirling water, so these

are known as eddy currents or whirlpool currents.”

These currents are also known as Foucault’s currents after the name of the

discoverer of eddy currents. Eddy currents are undesirable in many devices

such as transformers, electric motors etc. since they heat up the core and

dissipate electrical energy in the form of heat.

Eddy currents are minimised by using laminations of metal to make a metal

core. The laminations are separated by an insulating material like lacquer. The

plane of the laminations must be arranged parallel to the magnetic field, so

that they cut across the eddy current paths. This arrangement reduces the

strength of the eddy currents.

8. Applications of eddy currents:-

(i) Magnetic braking in trains: (ii) Electromagnetic damping:

(iii) Induction furnace: (iv) Electric power meters:

9. Self-Induction:- “When the current in a coil is changed, a back emf is induced

in the coil that opposes the change in the current. This phenomenon is known

as self induction or electrical inertia.”

10. Self-Inductance L :-

This quantity is the measure of self-induction of a coil. It is a scalar quantity.

SI unit of this quantity is Henry and the dimension is [ML2T

-2A

-2]. This

quantity is also known as ‘coefficient of self-induction’.

Self-inductance of a coil is defined numerically equal to, “the back emf

induced in the coil when the current flowing through its turns changes at the

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rate of 1 A/s.” OR “the magnetic flux linked with the coil when the current

flowing through its turns is unity.”

The formula for the self-inductance of any coil is

IL or

dtdI

L

11. Self-inductance of a long solenoid is

L = μr μ0 n2 A l

12. Mutual-Induction:- “When two coils are placed nearby and the current in one

coil is changed, an emf is induced in the neighbouring coil due to the change

in magnetic flux linked with it. This phenomenon is known as mutual

induction.”

13. Mutual-Inductance M12:- the mutual induction between two coils is given

mathematically by quantity ‘Mutual-Inductance’ or ‘coefficient of mutual

induction’.

Mutual inductance of two coils is defined as the “Magnetic flux linked with

one coil due to the unit amount of current flowing in the neighbouring coil”.

OR “the back emf induced in one coil due the unit rate of change of current in

the neighbouring coil”.

1

212 I

M

or

dtdI

M1

2

12

Mutual-Inductance of two coils depends on the linkage of magnetic field lines

between them apart from other factors. If the magnetic fields lines of one coil

are completely linked with the neighbouring coil then they are called perfectly

coupled.

For two perfectly coupled co-axial solenoids, mutual-inductance is given by

M12 = μ0 n1 n2 πr2 l where r is the radius of inner coil.

OR,

M12 = √³L1L2´ where L1 and L2 are the self-inductances of the two coils.

14. The magnetic energy stored in a current carrying solenoid:-

U = ½ LI2

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15. Principle of AC Generator:- Electromagnetic induction i.e. when a coil is

rotated about its axis perpendicular the direction of a uniform magnetic field

across it then an induced emf is produced in it.

16. Alternating emf and alternating current:- The emf/ current whose polarity/

direction reverses after a regular interval of time periods is called an

alternating emf/ alternating current.

The alternating emf / alternating current produced by an a.c. generator is a

sinusoidally varying alternating emf/ current.

The instantaneous magnetic flux associated with coil is )cos( tNBA or

)sin( tNBA

The instantaneous emf and instantaneous current is

e =E0 sin (ɷt+ϕ) or e = E0 cos (ɷt+ϕ)

i = I0 sin (ɷt+ϕ) or i = I0 cos (ɷt+ϕ)

The peak value or amplitude of the emf/ current is

0E = NBA and 0I = RNBA /

17. The average value of alternating emf / alternating current is ‘zero’ for full

cycle.

18. The average value of alternating emf / alternating current for the ‘half’ cycle is

02V

or

02I

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19. RMS value / Effective value / Virtual value of the alternating current:- It is the

value of alternating emf/ current that is measured by a.c. metres which are

based on heating effect of current.

RMS value of the current is numerically equal to that value of constant (D.C.)

current which when flows through a resistor for a certain time period

produces the same amount of heat as is produced by the alternating current in

the same time period for same resistor.

Irms = Io/√2 = 0.707 I0 and Erms = Eo/√2 = 0.707 E0

20. Phase difference between the alternating current and alternating voltage:-

The potential difference across resistor remains in phase with the alternating

current.

The potential difference across inductor leads the alternating current with a

phase angle π/2.

The potential difference across capacitor lags behind the alternating current

by a phase angle π/2.

(a) Phasor diagram and a wave diagram for a resistor v =Vm sin ɷt1 and i = Im

sin ɷt1

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(b) Phasor diagram and a wave diagram for a inductor v =Vm sin ɷt1 and i = Im

sin (ɷt1-π/2)

(c) Phasor diagram and a wave diagram for a capacitor v =Vm sin ɷt1 and i =

Im sin (ɷt1+π/2)

21. L-C-R series circuit:-

22. Impedance and reactance:- “The obstruction offered by the pure inductance

or pure capacitance to the flow of a.c. which is frequency dependent and has

the dimension of resistance but is not a source of power dissipation is called

reactance.”

“The obstruction offered by a circuit to the flow of a.c. that comprises of a

frequency dependent component as well as frequency independent component

is called impedance of the circuit. It has a dimension of resistance.”

The SI unit of reactance and impedance is Ω.

Reactance of an Inductor (or Inductive Reactance) XL = ɷL

Reactance of a Capacitor (or Capacitive Reactance) XC = 1/ɷC

Impedance of a series L-C-R circuit Z = √{(XL – XC)2 + R

2} and voltage V =

√{(VL – VC)2 + VR

2}

23. Phasor diagram for a series L-C-R circuit with an a.c. source:-

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24. Phase difference between voltage and current for a series L-C-R circuit:-

When the source frequency f > fr

R

XX CL1tan or

R

CL

V

VV1tan

When the source frequency f<fr

R

XX LC1tan or

R

LC

V

VV1tan

25. Graphs of Reactance Vs frequency and Impedance Vs frequency for a.c.

circuits:-

26. Graph of current Vs frequency for a series L-C-R circuit:-

27. Resonance in a series L-C-R circuit:-

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“The a.c. current flowing in the series L-C-R circuit is maximum for a certain

frequency of the a.c. source when the impedance of the circuit is minimum.

This phenomenon is called resonance. And the corresponding frequency when

the impedance is minimum is called the resonance frequency.”

28. The resonance frequency is:-

At resonance XL = XC

Angular resonance frequency LC

r

1 and resonance frequency

LCf r

2

1

29. Quality factor of series L-C-R circuit:-

The mathematical factor that is measure the sharpness of resonance of L-C-R

circuit is called the quality factor of the series L-C-R circuit.

The Q-factor of L-C-R circuit is defined as

“the ratio of voltage drop across inductor (or capacitor) to the voltage drop

resistor at resonance.” OR

“the ratio of resonance frequency to the frequency band width of the resonant

curve.”

Q = C

L

R

1=

2

r = R

Lr

30. Power dissipation in a.c. circuit:-

The average power dissipation in a.c. circuit depends upon the phase

difference between current and voltage.

For pure inductive or capacitive circuit, where the phase difference between

current and voltage is π/2, the average power dissipation is zero as for one half

of the cycle the electrical energy is transformed into magnetic/ electrostatic

energy and in the next half cycle the magnetic energy/ electrostatic energy is

retransformed into electrical energy.

The power dissipation in an a.c. circuit is <P> = ErmsIrms cosϕ

The factor ‘cosϕ’ is called power factor. Z

Rcos

The power factor is maximum at resonance and is zero for pure capacitive or

pure inductive circuit.

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31. Watt-less current:- the component of current phasor that is at π/2 phase

difference with the voltage is called watt-less current as it is not source of any

power dissipation.

The watt-less current is given by expression Irmscos ϕ

32. LC oscillations:-

A circuit consisting of an inductance L and a capacitor C in parallel is called

an LC resonant or tank circuit.

In this circuit if the capacitance is charged initially and the source is then

removed, then electrostatic energy of capacitor q02/2C is converted into

magnetic energy of inductor 2

2

1LI and vice versa periodically. Such

oscillations of energy are called LC oscillations and form the basis of the

oscillators.

The frequency of oscillations is given by

LC

1 ,

LCf

2

1

Practically, the LC oscillations do not sustain for a long time due to various

losses of energy.

The two main sources of energy loss in LC oscillations are:-

a) The resistance of the inductor

b) The radiation loss in the form of EM waves

33. Transformer:-

It is a device used for converting low alternating voltage at high current into

high voltage at low current and vice-versa.

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Principle: It works on the principle of mutual induction i.e. if two coils are

inductively coupled and when current or magnetic flux is changed through

one of the two coils, then induced emf is produced in the other coil.

Transformers are of two types:-

1. Step – up transformer The transformer having more number of turns in the

secondary coil than primary coil (i.e. NS > NP) and used to convert low voltage at

high current to high voltage at low current.

2. Step – down transformer The transformer having more number of turns in the

primary coil than secondary coil (i.e. NS < NP) and used to convert high voltage at

low current to low voltage at high current.

Transformation Ratio:-

“The output to input voltage ratio of transformer is equal to the ratio of no. of

turns in the secondary to the primary of transformer. This ratio is called the

transformation ratio of the of the transformer.”

P

S

v

v=

P

S

N

N

Efficiency of Transformer:- η = ipp

ss

P

P

Iv

Iv 0

100% efficient transformer is called an ideal transformer as there is no energy

loss in this transformer

For an ideal transformer;-

a) The windings of transformer should have no resistance.

b) The output of transformer should be in open circuit.

c) There should be no flux leakage.

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d) There should be no hysteresis loss or any other loss.

The various sources of energy loss in transformer are:-

1. Flux losses:- The coupling of the two coils of transformer is not perfect. As a

result, whole of the magnetic flux linked to the primary coil can not be linked to

the secondary coil.

2. Copper losses:- Some electrical energy is always converted into heat energy in

the resistance of the copper wire used in the winding of the coil.

3. Iron losses:- The changing magnetic flux leads to production of induced emf in

the iron core of the transformer which also leads to loss of some electric energy

in the eddy current produced in the iron core. It is minimized using laminated

iron core. It is prepared by joining two similar iron strips together after coating

with varnish. As a single iron strip is very thin so its resistance becomes large

which leads to production of very small eddy currents in it and in this way only a

small amount of heat is produced in the core.

4. Hysteresis losses:- Due to alternating current flowing through the coil, the iron

core is magnetised and demagnetised again and again. During each cycle of

magnetisation and demagnetisation, some energy is lost due to Hysteresis. It can

be minimized by selecting a material for iron core whose area of hysteresis loop

is very small.

5. Humming losses:- Due to passage of AC current, the core of the transformer

starts vibrating and produces humming sound in which some part of the electrical

energy is lost in the form of sound.

Uses of transformer:-

1. Transformers are used for voltage regulators and stabilized power supplies.

2. Small transformers are used in radio sets, televisions, telephones, loud speakers

etc.

3. A step-up transformer is used in the production of X-rays.

4. A step-down transformer is used for obtaining large current for electric welding

or in the induction furnace for melting metals.

5. The most important use of the transformer is the long distance transmission of

electric energy. Due to the large distance between the electric power generating

station and the electric power consuming places, there is a considerable loss of

electric power due to the large resistance of the electric power transmission lines

as well as it also results in the considerable voltage loss across them. Hence the

power and the voltage drop obtained at the destination will be much smaller than

the actual values of the power and the voltage supplied to the transmission lines.

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2. Formulae at a glance

Physical Quantity Formula SI unit Dimension

Magnetic flux (ϕ) cosBAAB

= AdB

Wb = Tm2 [ML

2T

-2A

-1]

Induced emf (e)

dt

d

Induced current dt

d

R

N

Ri

Induced charge q = i∆t = R

N

Motional emf induced in a straight

conductor

(i) Linear motion = Blv

(ii) Rotation about one end = Bl2ɷ/2

Volt [ML2T

-1A

-1]

Self-inductance I

L and

dtdI

L

Self-inductance of a long solenoid

L = μr μ0 n2 A l

Henry [ML2T

-2A

-2]

Mutual inductance

1

212 I

M

and

dtdI

M1

2

12

Mutual-inductance of two long co-axial

solenoids

M12 = μ0 n1 n2 πr2 l, M12 = √³L1L2´

Henry [ML2T

-2A

-2]

Magnetostatic

energy stored U = ½ LI

2 Joule [ML

2T

-2]

Alternating

current and

voltage

e =E0 sin (ɷt+ϕ) or e = E0 cos

(ɷt+ϕ)

i = I0 sin (ɷt+ϕ) or i = I0 cos

(ɷt+ϕ)

Irms = Io/√2 = 0.707 I0 and Erms =

Eo/√2 = 0.707 E0

Phase relationship

For R : No phase difference betn V and

I

For L: Voltage leads the current by π/2

For C: Current leads the voltage by π/2

For LCR circuit: if f > fr

Unit less Dimensionless

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R

XX CL1tan or

R

CL

V

VV1tan

If f<fr

R

XX LC1tan or

R

LC

V

VV1tan

Reactance and

impedance

Inductive reactance XL = ɷL

Capacitive reactance XC = 1/ɷC

Impedance of LR circuit Z = √{XL2 +

R2}

Impedance of RC circuit Z = √{XC2 +

R2}

Impedance of LCR circuit Z = √{(XL –

XC)2 + R

2}

Ohm [ML2T

-1A

-2]

Resonance

frequency

LCf r

2

1 , angular frequency

LCr

1

Hertz, rad/s [T-1

]

Quality factor Q = C

L

R

1=

2

r =R

Lr=

CRr

1 Unit less Dimensionless

Power dissipated

in ac circuit

In pure inductor and capacitor: Zero

In pure resistive circuit: I2R/2

In a combination of L,C and R:

VrmsIrmscosϕ

Watt [ML2T

-3]

Power factor cos ϕ = R/Z Unit less Dimensionless

Wattles current Irms cos ϕ Ampere [A]

Frequency of LC

oscillations LCf r

2

1 Hertz [T

-1]

Energy of ideal

LC oscillator ½ Q

2/C + ½ LI

2 = ½ Q0

2/C Joule [ML

2T

-2]

Transformation

ratio and

efficiency of

transformer

P

S

v

v=

P

S

N

N

Efficiency:- η = ipp

ss

P

P

Iv

Iv 0

Unit less Dimensionless

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3. Mind maps of formulae -- EMI

Mind maps of formulae of chapter AC

Alternating current

and alternating emf

i = I0 sin (ɷt+ϕ)

e = E0 sin (ɷt+θ)

rms value of ac

Irms = I0/√2 Average value for half

cycle = 2I0/π

Phase difference

between alternating

voltage and current

For L: +π/2

For C: -π/2

For R: No phase diff.

Average power dissipation in a pure

inductive or pure capacitive circuit = 0

Average power dissipation in a resistive

circuit = I02R/2

Average power dissipation in a series

LCR circuit = V0I0cosϕ/2

Power factor = Z

Rcos

Transformer steps up or steps down ac

voltage

Transformation ratio

P

S

v

v=

P

S

N

N

Efficiency η =

ipp

ss

P

P

Iv

Iv 0

Q factor of LCR circuit

Q =

C

L

R

1=

2

r=

R

Lr

Inductive reactance XL = ɷL

Capacitive reactance XC = 1/ɷC

Impedance of Series LCR circuit

Z = √{(XL – XC)2 + R2}

Phase difference between current and

voltage for series LCR circuit

R

XX LC1tan

At resonance XL = XC

Resonance frequency

LCf r

2

1

Induced emf

e = - N dt

d

Magnetic

flux

Φ = AB

Static charge

B = 0, Φ = 0,

Induced emf e = 0

Charge in uniform

motion

B = constant (μ0nI),

Φ = constant, e = 0

Accelerating charge

B = varying:, Φ =

varying:,

e = non zero

Mutual inductance

M = ϕ2/I1 =

dtdI

e

1

2

SI unit : Henry

Mutual inductance of two perfectly

coupled long co axial solenoids

L = μ0n2lA

Emf induced in the armature of an

ac generator

e = NBAɷ sin ɷt or e = NBAɷ cos ɷt

maximum emf = NBAɷ

average emf = 0

Self inductance

L = ϕ/I =

dtdI

e

SI unit : Henry

Self inductance

of long solenoid

L = μ0n2lA

Induced emf in moving conductor

e = N dt

d = Blv

for rotating disc or conductor

e = Bl2ɷ/2

Power produced due to induced emf

P = r

vlB 222

, F = r

vlB 22

, I =

r

Blv

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4. Important Topics for slow learners in UNIT – EMI and AC

1. Statement of Faraday’s and Lenz’s law

2. Simple questions of identification of direction of induced current using Lenz’s

law

3. Questions based on formula V= Bvl

4. Eddy currents and its applications

5. Unit and definition of self-inductance and self-induction

6. Numerical Questions based on formula L = ϕ/I

7. Diagram and working principle of a.c. generator

8. Numerical Questions based on the rms, average and peak value of a.c.

9. Deriving phase relationship between current and voltage for an inductor and

capacitor.

10. Graph of XL Vs f and XC Vs f

11. Phasor diagram for an L-C-R diagram

12. Derivation of impedance and phase relationship between current and voltage

of a series L-C-R circuit using phasor diagram

13. Graph of Impedance Vs f and Current Vs f for a series L-C-R circuit

14. Definition of Quality factor and its significance

15. Simple conceptual questions based on the variation of frequency, inductance,

capacitance etc. of a.c. circuit

16. Transformer- principle, working, diagram and energy losses

17. Numerical Questions based on transformer

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5. Important Derivation Questions Unit 4-- EMI and A-C

1. Prove that the magnetic energy stored in a long solenoid of cross-section A,

length l, no of turns per unit length n, magnetic permeability inside the solenoid

to be 0

will be lAB2

02

1

, where B is the uniform magnetic field along the axis of

the solenoid.

Ans. If ‘I’ is the instantaneous current in the circuit then at any instant ‘t’ then

rate of work done will be

dt

dW = e I = L I dt

dI , ignoring the resistive effects and considering only the

inductive effects.

Total amount of work done will be W = dW = I

LIdI0

= 2

1 L I

2

This is an expression for the energy stored in the inductor.

U = 2

1 L I

2 =

2

10

n2lA

2

0

n

B

= lAB

2

02

1

2. Obtain an expression for average ac current passing through a circuit over half a

cycle.

Ans. The instantaneous value of ac in a circuit is given by i = I0 sin ω t

Total charge that will pass through the circuit in time T/2 through a resistor will

be

q = 2

0

T

dq = 2

0

T

idt = 2

0

0 sin

T

tI dt = 2

0

0 sin

T

tI dt = 2

0

0

cosT

tI

=

TI o

If Im is mean value of ac for half cycle then q = 2

TI m . So,

02II m = 0.636 I0

3. What is meant by the root mean square or effective value of alternating current?

Derive a relation between it and its peak value.

Ans. RMS value of the current is numerically equal to that value of constant

(D.C.) current which when flows through a resistor for a certain time period

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produces the same amount of heat as is produced by the alternating current in the

same time period for same resistor.

Let R

ei =

R

tE ωsin0 = I0 sin ω t

The instantaneous power dissipated in the resistor is p = i2R = i0

2 R sin

2

ωt

The heat dissipated over a cycle is H = T

Pdt0

= T

tdtRi0

22

0 ωsin H =

T

dtt

Ri0

2

02

cos2ω-1 = RTi2

02

1

If the rms value of current is Irms then the heat produced over time period T is H =

Irms2RT

Irms2RT = RTi2

02

1 Irms=I0/√2

4. Show that the alternating voltage leads the alternating current by π/2 in phase

through a pure inductor.

Ans. Consider a pure inductor with inductance ‘L’ (no resistance) is connected to

an ac source which applies the instantaneous value of emf given by e = E0

sin ω t

By the energy conservation, for an ideal inductor,

The instantaneous value of applied emf = induced emf in the inductor

Therefore, E0 sin ω t = Ldt

dI , L is self-inductance of the inductor

tL

EdI ωsin0 dt tdt

L

Edii ωsin0 = t

L

Eωcos

ω

0 + Integration constant

Integration constant has the dimensions of the current and is time-independent. As

the source has an emf which oscillates symmetrically about zero, and hence the

current also oscillates symmetrically about zero so that no constant or time-

independent component of current exists. So, Integration constant = 0.

So, tii ωcos0 =

2

πωsin0 ti , where

L

Ei

ω

00 is the amplitude of current.

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Clearly, current phasor I is π/2 behind the voltage phasor V.

5. Prove that the average power dissipated in an inductor in an a.c. circuit is zero.

Ans. The instantaneous power in the circuit is

P = e i= E0 I0 sin ω t

2ωsin

t = - E0 I0 sin ω t cos ω t =

2

2sin00

ωtIE

The average dissipated over a cycle is Pav = T

PdtT 0

1 =

Tt

otT

IEωt2sin

2

00 dt = 0

Therefore, average power dissipated in a pure inductive ac circuit = Pav = 0

6. A sinusoidal emf is applied to a circuit containing a capacitor only. Show that

current leads the voltage by π/2.

Ans. Consider a capacitor of capacitance ‘C’ connected to an ac source of voltage

e = E0 sin ω t.

At any instant, if the current in the circuit is i, the charge on capacitor is q then the

instantaneous potential difference across the capacitor will be C

q .

By the energy conservation,

The instantaneous value of applied emf = instantaneous potential difference across

the capacitor

E0 sin ω t = C

q q = C E0 sin ω t

Therefore, current in circuit

i = dt

d

dt

dq (C E0 sin ω t) i = C E0 ω cos ω t =

C

E

ω1

0 cos ω t i = C

E

ω1

0 sin

2ω

t

where C

EI

ω1

00 = peak value of alternating current through capacitor

Clearly, alternating current through pure capacitor leads the emf by π/2 in

phase.

7. Prove that the average power dissipated in a capacitor in ac circuit is zero.

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Ans. The instantaneous power in the circuit is


2ωsin

t = E0 I0 sin ω t cos ω t =

2

2sin00

ωtIE


PdtT 0

1 =

Tt

otT

IEωt2sin

2

00dt = 0

Therefore, average power dissipated in a pure capacitive ac circuit = Pav = 0

8. Explain, with the help of a necessary diagram, the principle and the working of a

step up transformer and obtain the expression for the transformer

equationp

s

p

s

s

p

N

N

v

v

i

i , where the symbols have their usual meanings. What are the

two main assumptions made to derive the above relations?

Ans. It works on the principle of mutual induction i.e. if two coils are inductively

coupled and when current or magnetic flux is changed through one of the two

coils, then induced emf is produced in the other coil.

When an alternating voltage is applied to the primary, the resulting current

produces an alternating magnetic flux which links the secondary and induces an

emf in it. The value of this emf depends on the number of turns in the secondary.

We consider an ideal transformer in which the primary has negligible resistance

and all the flux in the core links both primary and secondary windings.

Let be the flux in each turn in the core at time t due to current in the primary when

a voltage vp is applied to it.

Then the induced emf in the secondary is

es = - Ns dt

dφ

The alternating flux φ also induces an emf, called back emf in the primary given

by

ep = - Np dt

dφ

But ep = vp .since the primary has zero resistance(as assumed).

If the secondary is an open circuit es = vs

So, p

s

p N

N

ese

144

144

If the transformer is assumed to be 100 % efficient (no energy losses) then the

input power is equal to the output power, i.e. Pinput = Poutput Ip ep = Is es,

p

s

ps

p

N

N

eI

I se

Two assumptions taken to derive the above relation are:-

(i) There is no flux leakage

(ii) The resistance of primary and secondary windings are small

(iii) The secondary winding is open circuited.

9. A transformer steps down an AC voltage. What appliance will you use to step

down a DC voltage?

Ans. By connecting a series high resistance

10. Explain, with the help of diagram, the principle and working of an ac generator.

Derive the expression for the emf generated in the coil in terms of its speed of

rotation.

Ans.

AC Generator

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145

Principle: It works on the phenomenon of electromagnetic induction. i.e.

when a coil is rotated about its axis perpendicular the direction of a uniform

magnetic field across it then an induced emf is produced in it.

Construction: It has the following essential parts:-

Armature: A rectangular coil ABCD having large number of turns of copper wire

wound over a soft iron core is called armature. The soft iron core is used to

increase the magnetic flux.

Field magnet: It is produced by using two cylindrical magnetic poles of strong

permanent magnet in which the coil is rotated about its axis such that the axis of

the coil remains perpendicular to the magnetic field lines.

1. Slip Rings: These are two circular conducting rinds R1 and R2 used to provide

movable contact of the armature with the external circuit terminals.

2. Brushes: The flexible metallic pieces B1 and B2, called brushes, are used to

pass on the current from armature to external load. As the slip rings rotate with the

armature, the brushes provide movable contact by keeping pressed themselves

against the rings.

Working: Consider, initially at t = 0, the armatures ABCD is vertical with arm

AB up and CD down. During the clockwise motion of the armature between the

instants t = 0 and t = T/2, the arm AB moves down and CD moves up. According

to Fleming’s right hand rule, the current in the coil flow in the direction DCBA.

During the motion of armature between t = T/2 and t = T, the arm AB moves up

and the arm CD moves down, as a result, the current in the armature moves along

the direction ABCD.

i.e. the direction of the current in the every half cycle of the rotation of the

armature is reversed. Therefore, an ac induced emf and an induced current is

generated in the generator.

ot4/T 2/T 4/3T T

e

NS

146

146

Magnitude of induced emf: Whenever the armature is vertical then momentarily

its arm AB and CD move parallel to the magnetic field and hence the rate of

change of magnetic flux through the armature becomes zero. Therefore, according

to the Faraday’s law of electromagnetic induction, the induced emf in this

situation, in the armature will be also zero. I.e. at the instant t = 0 , t = T/2 and t =

T, the induced emf in the armature is zero. Whenever the armature is horizontal,

the arms AB and CD move normally to the direction of magnetic field lines.

Hence the rate of change of magnetic flux becomes maximum. So, the induced

emf in this situation is maximum.

When the armature starts rotating from its vertical position at the angular speed of

then the angle through which it is rotated in the time interval t will be t .

Therefore, the magnetic flux linked to the coil in this situation will be

cosnBA , So, the instantaneous induced emf in the coil , at this instant will

bedt

de

= tnBA sin sinnBA t sinnBAe t _________

(5)gives the instantaneous induced emf in the armature at any instant t. This

expression can also be written as sin0ee t _________ (6)

where 0e = maximum emf induced in the coil given by 0e = nBA ________ (7)

As the value of sine function varies between + 1 and - 1, so, the polarity of the

emf changes periodically.

Since the direction of the current changes periodically and hence it is called

alternating current.

11. Show analytically that the Lenz’s law is consistent with law of conservation of

energy.

Ans.:- Lenz’s Law and Conservation of Energy

Lenz’s law states that, “The polarity of induced emf is such that it tends to

produce a current which opposes the change in magnetic flux linked to the coil

that produced it.”

If the resistance of the conductor PQ is R and the rest three arms of the

rectangular loop PQRS have negligible resistance then the overall resistance of

the loop is R. The current in the conductor PQ is R

BlI

v . Clearly, the conductor

due to this current experiences a force F = I l B, directed perpendicular inwards

opposite to v according to the Fleming’s left hand rule.

L v

P

Q

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147

The magnitude of the force F = I l B = B 2 l

2 v / R.

Therefore, the arm PQ is being pulled with a constant speed v, the power required

to do this is,

P = F v = B 2 l

2 v

2 / R ____________ (1)

This is the mechanical work done by the external agency.

This energy is dissipated in the form of heat cane verified as P = I 2 R = R

R

Blv2

= B 2 l

2 v

2 / R

Thus, mechanical energy which was needed to move the arm PQ is converted into

electrical energy and then into thermal energy.

12. A conducting rod of length l is moved in a magnetic field of magnitude B with

velocity v such that the arrangement is mutually perpendicular. Prove that the

emf induced in the rod is ׀ɛ׀= Blv.

Ans.

Consider a straight conductor moving in a uniform and time-independent

magnetic field B directed perpendicularly inwards of the plane of paper. Length of

conductor =l, conductor is moving towards the left with velocity v.

The magnetic flux ф linked with the conductor in any time ∆t ф = B l v∆t

The induced emf in the conductor will be e = - dt

d = - B l

t

tv

= - B l v

13. A metallic rod of length ‘L’ is rotated with angular frequency ‘ɷ’ with one end

hinged at the centre and the other end at the circumference of a circular metallic

ring of radius L, about an axis passing through the centre and perpendicular to the

plane of the ring. A constant and uniform magnetic field B parallel to the axis is

present everywhere. Deduce an expression for the emf induced between the

centre and the metallic ring.

lv B

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148

Ans. Let the rod PQ of length l is rotated in the plane of paper and magnetic field

is directed perpendicular inwards to the plane of paper with a uniform angular

speed ω in anticlockwise direction.

Induced emf T

lB

dt

dAB

dt

BAd

dt

de

2π

But

2T

π2π 2

lBe Therefore,

2

2

1Ble

14. What are eddy currents? Write there important applications. Why are eddy

current produced in the cores of transformers and a.c. generators

disadvantageous. How can they be minimised?

Ans. Eddy currents or Foucault Currents

When we put a bulk piece of conductor to the changing magnetic field but now

the pathway of the currents is not well defined and the flow pattern resemble the

swirling eddies in water. This effect was discovered by Foucault and hence also

called Eddy currents or Foucault currents.

The direction of the eddy currents is given by Lenz’s law.

Eddy currents are used to advantage in certain applications like:-

i. Magnetic braking in trains:

ii. Electromagnetic damping (Dead beat galvanometers): iii. Induction furnace:

iv. Electric power meters:

v. Speedometer:

Eddy currents are undesirable since they heat up the core of transformers;

electric motors etc as they dissipate electrical energy in the form of heat.

Eddy currents are minimized by using laminations of metal to make a metal

core. The laminations are separated by an insulating material. The plane of the

laminations must be arranged parallel to the magnetic field so that they cut across

the eddy current paths. This arrangement reduces the strength of the eddy

currents. Since the dissipation of electrical energy into heat depends on the square

of the strength of electrical current, heat loss is substantially reduced.

149

149

15. Define self-induction and self-inductance. Derive expression for the self-

inductance of a long solenoid.

Ans. Self-Induction

It is property of a coil by virtue of which the coil opposes the growth or decay of

the current flowing through it. Self-induction is also known as inertia of

electricity as it opposes the growth or decay of the current in the circuit.

Self-inductance

Self-inductance of a coil is defined numerically equal to, “the back emf induced

in the coil when the current flowing through its turns changes at the rate of 1

A/s.” OR “the magnetic flux linked with the coil when the current flowing

through its turns is unity.” ф = L I

L is called the self-inductance of coefficient of self-induction of the coil.

When the current is the coil changes then due to change in flux linkage, the

induced emf in the coil is given by e = dt

Nd )φ( = - L

dt

dI

Inductance is a scalar quantity and its dimensions are M L 2 T

– 2 A

– 2.

Si unit of inductance is Henry (H).

Self-inductance of a long solenoid

Consider a long solenoid of length l, area of cross-section A, number of turns per

unit length l and the permeability of the core is μ and carrying current I.

The magnetic field produced inside the solenoid is B = μ n I

Magnetic flux linkage to the solenoid is Nф = (n l) B A = (n l) μ n I A = μ n 2 I l

A

But Nф = L I, therefore L = μ n 2 l A, this is an expression for

the self-inductance of the solenoid.

16. Derive expression for the magnetic energy required to build up the current I in a

coil of self-inductance L is given by ½ LI2. Hence derive expression for the

magnetic energy density.

Ans. Since, due to self-inductance is a back emf is induced, so work is to be done

by the external agency against the back emf to establish the current in the circuit

which is stored in the coil as its magnetic potential energy.

If ‘I’ is the instantaneous current in the circuit then at any instant ‘t’ then rate of

work done will be

dt

dW = e I = L I dt


inductive effects.


LIdI0

= 2

1 L I

2

150

150

This is an expression for the energy stored in the inductor.

Magnetic energy density:- Magnetic energy stored per unit volume is called

magnetic energy density.

u = U/volume = 0

222

0

22

0

2

221

2

21

BIn

Al

lAIn

Al

LI

17. Define mutual induction. Two long co-axial solenoids of the same length but

different radii and different number of turns are wound one over the other.

Deduce the expression for the mutual inductance of this arrangement.

Ans. Mutual induction is the phenomenon of inducing emf in a coil due to the

change in magnetic flux linked to a neighboring coil coupled to it.

Consider two long coaxial solenoids each of length l. Radius of the inner solenoid

S1 is r1, number of turns is N1 and the number of turns per unit length is n1. The

corresponding quantities for the outer solenoid S2 are r2, N2 and n2, respectively.

When current I2 is set up through S2, then the flux linkage through S1 is

N 1 ф 1 = M12 I2

M12 is called the mutual inductance of solenoid S1 with respect to S2 and called

the coefficient of mutual induction.

The magnetic field due to the current I2 is S2 is μ 0 n 2 I 2. Hence the resulting

flux linkage with soil S1 is N 1 ф 1 = (n 1 l) (π 21r ) (μ 0 n 2 I 2) = μ 0 n 1 n 2 π 2

1r l I

2

Clearly, M 12 = μ 0 n 1 n 2 π 21r l Also, 12M = 21M = M

18. Obtain the expression for the mutual inductance of a pair of coaxial circular coils

of radii r and R(R >r) placed with their centres coinciding.

Ans.Two concentric circular coils, one of small radius r and the other of large

radius R, such that r<<R, are placed co axially with centres coinciding. Obtain the

mutual inductance of the arrangement.

151

151

If I1 is current flow through the outer coil then the magnetic field at its center is

B2 = μ0I1/2R

Considering B2 almost constant around its centre as r << R, the flux linkage to

inner coil is

Ф2 = π 2r B 2 = 1

2

0

2

πμI

R

r = M12I 1 M12 =

R

r

2

πμ 2

0

19. Show that the average power dissipated in a resistor connected to an a.c. source

supplying a current i = i0 sinɷt is ½ i02R.

Ans.:- AC voltage applied to a Resistor

Consider a pure resistor ‘R’ is connected to an ac source which produces

sinusoidally varying potential difference (called ac voltage) across its terminals

given by e = E0 sin ω t

e is instantaneous value of the ac voltage and E0 is its amplitude, ω is its angular

frequency.

Let the instantaneous value of emf is given by e = E0 sin ω t ______

(11)

Now, by energy conservation, instantaneous emf of the source = instantaneous

potential drop across the resistor.

If i is the current in the circuit at that instant then e = i R R

ei =

R

tE ωsin0 = I0 sin ω t

Now, R

E0 = I0 = the peak value of alternating current.

The instantaneous power dissipated in the resistor is p = i2R = i0

2 R sin

2

ωt

The average power dissipated over a cycle is Pav = T

PdtT 0

1 =

T

tdtT

Ri

0

220 ωsin

Pav =

T

dtt

T

Ri

0

20

2

cos2ω-1 = Ri 2

02

1

152

152

20. Obtain phase relationship between a.c. and alternating voltage for a pure

inductor. Also deduce the relation for the inductive reactance.

Ans:- Consider a pure inductor with inductance ‘L’ (no resistance) is connected

to an ac source which applies the instantaneous value of emf given by e

= E0 sin ω t

By the energy conservation, for an ideal inductor,

The instantaneous value of applied emf = induced emf in the inductor

Therefore, E0 sin ω t = Ldt

dI , L is self-inductance of the inductor

tL

EdI ωsin0 dt tdt

L

Edii ωsin0 = t

L

Eωcos

ω

0 + Integration constant

Integration constant has the dimensions of the current and is time-independent. As

the source has an emf which oscillates symmetrically about zero, and hence the

current also oscillates symmetrically about zero so that no constant or time-

independent component of current exists. So, Integration constant = 0.

So, tii ωcos0 =

2

πωsin0 ti , where

L

Ei

ω

00 is the amplitude of current.

Clearly, current phasor I is π/2 behind the voltage phasor V.

The quantity ω L is analogous to the resistance and is called inductive

reactance denoted by XL. XL = ω L

The dimension of inductive reactance is the same as that of resistance and its SI

unit is Ω.

21. Show that the average power dissipated in pure inductive circuit.

Ans:- he instantaneous power in the circuit is


2ωsin


2

2sin00

ωtIE

L

Mains

CA ..

153

153


PdtT 0

1 =

Tt

otT

IEωt2sin

2

00 dt = 0

Therefore, average power dissipated in a pure inductive ac circuit = Pav = 0

22. Obtain phase relationship between a.c. and alternating voltage for a pure

capacitor. Also deduce the relation for the capacitive reactance.

Ans. AC Voltage Applied to A Capacitor

Consider a capacitor of capacitance ‘C’ connected to an ac source of voltage e =

E0 sin ω t.

At any instant, if the current in the circuit is i, the charge on capacitor is q then the

instantaneous potential difference across the capacitor will be C

q .

By the energy conservation,

The instantaneous value of applied emf = instantaneous potential difference across

the capacitor

E0 sin ω t = C

q q = C E0 sin ω t

Therefore, current in circuit i = dt

d

dt

dq (C E0 sin ω t) i = C E0 ω cos ω t =

C

E

ω1

0

cos ω t

Or i = C

E

ω1

0 sin

2ω

t

where C

EI

ω1

00 = peak value of alternating current through capacitor

V

C

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154

Clearly, alternating current through pure capacitor leads the emf by π/2 in

phase.

Here, Cω

1 plays the role like resistance and it is called reactance of capacitor or

capacitive reactance and it is denoted by XC.

23. Show that the average power dissipated in pure inductive circuit.

Ans:- The average power dissipated across the capacitor is Pav = T

PdtT 0

1 =

Tt

otT

IEωt2sin

2

00 dt = 0


24. Using phasor diagram, obtain expression for the impedance of a series LCR

circuit. Also obtain an expression for the phase relationship between current and

voltage. Draw impedance Vs frequency and current Vs frequency graph.

Ans:- Consider a resistor of resistance R, an inductor of inductance ‘L’ and a

capacitor of capacitance ‘C’ are connected in series across an ac source of emf e =

E0 sin ω t

Let e and i are the instantaneous values of emf and current in the circuit. VL, VC

and VR are the instantaneous values of the potential differences across inductor L,

capacitor C and the resistor R respectively. q is the instantaneous charge across

the capacitor.

Using Kirchhoff’s loop rule, e = Ldt

di + i R + C

q or e = VL + VR + VC

LV

RV

I

φωt

155

155

As R, L and C are in series, so the instantaneous current in each of these element

is same and let it is given by i = i 0 sin (ωt + φ), where φ is the phase difference

between the instantaneous voltage of the source and the instantaneous current in

the circuit.

We know VR is in phase of I, VL leads I by π/2 while VC lags in phase by π/2 than

I.

A VL and VC are anti-parallel so they can be combined as a single phasor as VL +

VC. If |VC| > |VL| at any instant then VL + VC = |VC – VL| = VC – VL

Therefore as the VL + VC and VR are perpendicular to each other and if e is there

resultant represented by diagonal according to parallelogram law, so,

e = 22CLR VVV = 22

CL iXiXiR = i 22CL XXR

22

CL XXR

ei

= Z

ei

Here, 22CL XXRZ has the units of resistance and it is the effective

opposition offered by L, C and R to an ac and it is called impedance of the LCR-

series circuit.

Therefore, impedance Z = 22CL XXR =

22

ω

1ω

CLR

From the phasor diagram, the phase angle φ between the emf and current can be

determined as follows:- tan φ = R

XX

V

VV LC

R

LC

Special Case

1. If XC > XL, φ = + ve. Hence, in such case the current leads emf.

2. If XL > XC, φ = - ve. Hence, in such case the current lags emf.

25. Obtain an expression for the average power dissipated in a series LCR circuit.

Ans:- Let in series LCR circuit, the instantaneous current and voltage is given by

156

156

i = tI ωsin0 and e = e0 sin ωt

The instantaneous power in the circuit will be P = e i= e0 I0 sin ω t tωsin

P = e0 I0 sin ω t sin ω t cos φ + e0 I0 sin ω t cos ω t sin φ

= 2

ω2cos100

tIe

cos φ + e0 I0

2

ω2sin tsin φ

So, the average power in the ac LCR circuit over a cycle is Pav = T

PdtT 0

1

= T

Ie

2

00T

0

cos dt - T

Ie

2

00T

t0

cos2cos dt + T

Ie

2

00T

t0

sin2sin dt = 2

00 Ie cos φ = rmse rmsI cos

φ

26. Obtain an expression for frequency of LC oscillations.

Ans:-

LC oscillations

A capacitor and an inductor can store electrical and magnetic energy respectively.

When a charged capacitor is connected across an inductor then the charge and the

current exhibits the phenomenon of electrical oscillations in the circuit.

A combination of L and c is called a tank circuit.

Electrical oscillations produced by the exchange of energy between a capacitor

and an inductor are called LC oscillations.

Let at instant t = 0, a charged capacitor with initial charge q0 is connected across

an inductor then the charge on the capacitor starts decreasing, giving rise to a

current in the circuit. Let at any instant ‘t’the charge and the current in the circuit

are q and i respectively.

As there is no source of emf in the circuit, so, according to Kirchhoff’s energy

conservation law of circuits 0dt

diL

C

q 0

12

2

qLCdt

qd,

Assuming 2ω1

LC

, we can write 0ω2

2

2

qdt

qd

Let solution of this equation is given by q = A cos ω t + B sin ω t

Since, at t = 0, the capacitor was fully charged, therefore q = q0 q0 = A cos ω

(0) + B sin (0) = A

Again, the instantaneous current in the circuit will be dt

dqI = tBtA ωcosωωsinω

Since, at t = 0, I = 0, therefore, 0 = )0(cos)0(sin BA = B B = 0

Therefore, the solution will be q = q0 cos t

C

L

i

157

157

Hence, the instantaneous charge on the capacitor at any time is a harmonic

function of time oscillating with the angular frequency given by LC

1

Imoprtant about series LCR circuit

Resonance

An interesting characteristic of the series LCR circuit is the phenomenon of

resonance. It is a common phenomenon amongst the systems which oscillate at a

particular frequency called natural frequency of the system.

If such a system is driven by an external energy source of the nearly same frequency

then system starts oscillating with large amplitude.

In case of series LCR circuit, driven by an external ac source of emf amplitude e0

and angular frequency ω, the current amplitude is given by i 0 = 22

0

CL XXR

e

If the source frequency ω is varies then at a particular frequency ω0, ωL = 1/ωC i.e.

XL = XC.

In such case Z = R = minimum and the current amplitude i 0 = max. This frequency

ω0 is called resonant angular frequency and given by ω 0 = LC

1

A series LCR circuit is said to be resonant series LCR circuit when it admits

maximum current corresponding to a particular angular frequency of the ac source

and this angular frequency of the LCR-series circuit at this happen is called

resonant angular frequency.

Therefore, when resonance condition occurs in a LCR-series circuit then

1. Current and the emf are in phase of each other as φ = zero.

2. The impedance of the circuit is minimum and is equal to R.

3. The current in the series LCR circuit attains a maximum value which depends

only on the resistance of the circuit and is independent of the L and C.

Resonant circuits have a variety of applications:-

In the tuning mechanism of a radio or a TV set, the antenna of radio accepts signals

from many broadcasting stations. The signals picked up in the antenna acts as a

source in the tuning circuit of the radio so that circuit can be driven at many

frequencies. But to hear one particular radio station, we tune the radio by varying the

capacitance of a capacitor in the tuning circuit such that the resonant frequency of

the circuit becomes nearly equal to the frequency of the radio signal received. When

this happens, the amplitude of the current with the frequency of the signal of the

particular radio station in the circuit is maximum.

Quality- factor (Q-factor) and Sharpness of resonance

158

158

At resonant frequency ωr, the current in the LCR-series circuit becomes maximum

i.e. for all the values of angular frequencies ω > ωr or ω < ωr, the value of current is

less than the current at resonant angular frequency ωr.

The variation in the value of current with angular

frequency of ac source with different values of

resistance with same value of L and C can be shown

as in the given diagram.

The sharpness of the resonance is given by ω2

ω r

=

R

ω r L

The ratio R

ω r L is also called the quality factor Q of the circuit, so, Q =

R

ω r L also, Q

= CRω

1

r

Note: Clearly, if the resonance is less sharp, then not only the maximum current is

less but also the circuit is close to resonance for a larger bandwidth frequencies

giving poor tuning. So, less sharp the resonance, less is the selectivity of the circuit.

Again if quality factor Q is large, i.e. R is low or L is large the circuit is more

selective.

6. Previous years AISSCE Questions

One mark questions

Q1. When a lamp is connected to an alternating voltage supply, it lights

with the same brightness as when connected to a 12 V DC battery. What is the

peak value of alternating voltage source? (AISSCE 2005 C)

Q2. What is the power dissipated in an a.c. circuit in which voltage and

current are given by

V=230 sin (ɷt+π/3) and I = 10 sin ɷt? (AISSCE 2005)

Q3. In a series LCR circuit, the voltage across an inductor, capacitor and

resistor are 20V, 20V and 40V respectively. What is the phase difference

between the applied voltage and the current in the circuit? (AISSCE 2005)

ω

rωrωω rωω

0i1R

2R

21 RR

159

159

Q4. A bulb and a capacitor are connected in series to an a.c. source of

variable frequency. How will the brightness of the bulb change on increasing

the frequency of the a.c. Source? Give reasons. (AISSCE 2005)

Q5. Two identical loops, one of copper and another of constantan are

removed from a magnetic field within the same time interval. In which loop

will the induced current be greater? (Delhi 2005 C)

Q6. Give the direction in which the induced current flows in the coil

mounted on an insulating stand when a bar magnet is quickly moved along the

axis of the coil from one side to the other as shown in the figure. (Delhi 2005)

Q7. The power factor of an a.c. circuit is 0.5. What will be the phase

difference between voltage and current in this circuit? (Delhi 2005)

Q8. In series LCR circuit, the voltages across an inductor, a capacitor and a

resistor are 30V, 30V and 60 V respectively. What is the phase difference

between the applied voltage and the current in the circuit? (AISSCE 2007)

Q9. When current in a coil changes with time, how is the back emf induced

in the coil related to it? (AISSCE 2008)

Q10. The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin

314 t A and V = 50 sin (314t + π/2) V. What is the power of dissipation in the

circuit? (AISSCE 2008)

Q11. State the Faraday’s law of electromagnetic induction. (Foreign 2009)

Q12. Define self-inductance. Give its S.I. units. (Foreign 2009)

SS NN

160

160

Q13. A plot of magnetic flux (ϕ) versus current (I) is shown in the figure for two

inductors A and B. Which of the two has larger value of self-inductance?

(Delhi 2010)

Q14. A rectangular loop of wire is pulled to the right,

away from the long straight wire through which a

steady current I flows upwards. What is the

direction of induced current in the loop? (Foreign

2010)

Q15. Two loops of different shapes are moved in a region of uniform magnetic field

in the directions marked by arrows as shown in the figure. What is the

direction of the induced current in each loop? (Foreign 2010)

Q16. A circular loop is moved through the region of uniform magnetic field. Find

the direction of induced current (clockwise or anticlockwise) when the loop

moves: (i) into the field, and (ii) out of the field. (Foreign 2010)

ϕ

I

A

B

I

161

161

Q17. Define the term “watt-less current”. (DELHI 2011)

Q18. Two bar magnets are quickly moved towards a metallic loop connected across

a capacitor ‘C’ as shown in figure. Predict the polarity of the capacitor.

(AISSCE 2011)

Q19. What is the function of a step down transformer? (AISSCE 2011 C)

Q20. The peak value of emf in a.c. in E0. Write its (i) rms and (ii) average value

over a complete cycle. (Foreign 2011)

Q21. The current flowing through a pure inductor of inductance 4mH is i = 12 cos

300t ampere. What is the (i) rms (ii) average value of current for a complete

cycle? (Foreign 2011)

Q22. A closed loop PQRS of wire is moved into a uniform magnetic field at right

angles to the plane of the paper as shown in the figure. Predict the direction of

induced current in the loop. (Foreign 2012)

Q23. A bar magnet is moved in the direction indicated by the arrow between two

coils PQ and CD. Predict the directions of induced current in each coil.

(AISSCE 2012)

Q24. Predict the direction of induced current in

metal rings 1 and 2 when current I in the wire

is steadily decreasing? (Delhi 2012)

1

2

I

162

162

Q25. Predict the direction of induced current in a

metal ring when the ring is moved towards a

straight conductor with constant speed v. the

conductor is carrying current I in the

direction shown in the figure. (Delhi 2012)

Q26. Predict the direction of induced current in

metal rings 1 and 2 when current I in the wire

is steadily increasing? (Delhi 2012)

Q27. How does the mutual inductance between two coils change when

(i) distance between the coils is increased and

(ii) number of turns in the coils is increased?

Q28. The motion of copper plate is damped when it is allowed to oscillate between

the two poles of a magnet. What is the cause of damping?

Two mark questions

Q1. Draw the graphs showing the variations of (i) inductive reactance, and

(ii) capacitive reactance, with frequency of applied voltages in a.c. circuit.

How do the values of (i) inductive, and (ii) capacitive reactance change, when

the frequency of applied voltage is tripled? (AISSCE 2005 C)

Q2. Explain the principle on which the metal detector used at airports for

security purpose works. (AISSCE 2005 C)

Q3. Write the expression for frequency of an ideal LC circuit. In an actual

circuit, why do the oscillations ultimately die away? (AISSCE 2005 C)

Q4. Mention the factors on which the resonant frequency of a series LCR

circuit depends. Plot a graph showing variation of impedance of a series LCR

circuit with the frequency of the applied a.c. source. (AISSCE 2005)

v

I

1

I

2

163

163

Q5. State the condition under which the phenomenon of resonance occurs in

a series LCR circuit. Plot a graph showing variation of current of a series LCR

circuit with the frequency of the applied a.c. source. (AISSCE 2005)

Q6. A circular coil of radius 8cm and 20cm turns rotates about its vertical

diameter with an angular speed of 50 s-1

in a uniform horizontal magnetic field

of magnitude 3 x 10-2

T. Find the maximum and average value of emf induced

in the coil. (AISSCE 2005)

Q7. State the underling principle of an a.c. generator. Write the relationship

between the peak value and rms value of alternating voltage. (Delhi 2005 C)

Q8. The figure shows two identical

rectangular loops (1) and (2), placed on

a table along with a straight current

carrying wire between them.

(i) What will be the direction of the

induced currents in the loops when they

are pulled away from the conductor with

same velocity v?

(ii) Will the emf induced in the two

loops be equal? Justify your answer.

(Delhi 2005)

Q9. (i) Draw the graphs showing the variation of inductive reactance and

capacitive reactance with the frequency of the applied a.c. source.

(ii) Can the voltage drop across the inductor or the capacitor in a series LCR

circuit be greater than the applied voltage of the a.c. source? Justify your

answer. (Delhi 2005)

Q10. An inductor 200 mH, a capacitor C and a resistor 10 ohm are connected in

series with a 100 V, 50 s-1

a.c. source. If the current and voltage are in phase

with each other, calculate the capacitance of the capacitor. (AISSCE 2006 C)

Q11. A capacitor and a resistor are connected in series with an a.c. source. If the

potential difference across C,R are 120V, 90 V respectively and if the rms

xx

a

ab

b

v

v)1(

)2(I

164

164

current of the circuit is 3 A, calculate the (i) impedance, (ii) power factor of

the circuit. (AISSCE 2006 C)

Q12. An alternating voltage of frequency f is applied across a series LCR circuit.

Let fr be the resonant frequency for the circuit. Will the current lag, lead of

remain in phase with the applied voltage when (i) f>fr (ii) f<fr? Explain your

answer in each case. (AISSCE 2006)

Q13. A rectangular coil of area A, having no. of turns N is rotated at f revolutions

per second in a uniform magnetic field B, the field being perpendicular to the

coil. Prove that the maximum emf induced in the coil is 2πfNBA. (Delhi 2006

C)

Q14. A conducting rod of length l is moved in a magnetic field of magnitude B with

velocity v such that the arrangement is mutually perpendicular. Prove that the

emf induced in the rod is ׀ɛ׀= Blv. (Delhi 2006 C)

Q15. A perfect self-inductor when connected to an a.c. source does not produce

heating effect yet reduces the current in the circuit. Explain. (Foreign 2007)

Q16. Prove that an ideal capacitor, in an a.c. circuit does not dissipate power.

(Delhi 2008)

Q17. Derive an expression for the impedance of an a.c. circuit consisting of an

inductor and a resistor. (Delhi 2008)

Q18. Obtain the expression for the mutual inductance of a pair of coaxial circular

coils of radii r and R(R >r) placed with their centres coinciding. (Delhi 2008)

Q19. A jet plane is travelling west at 450 ms - 1

. If the horizontal component of

earth’s magnetic field at that place is 4 x10- 4

tesla and the angle of dip is 30°,

find the emf induced between the ends of wings having a span of 30 m.

Q20. The figure given below shows an

arrangement by which current flows

through the bulb (X) connected with coil

B, when a.c. is passed through coil A.

(i) Name the phenomenon involved.

A B

165

165

(ii) If a copper plate is inserted in the gap

between the two coils, explain, how the

brightness of bulb would change?

(AISSCE 2008)

Q21. Derive an expression for the self-inductance of a long air-cored solenoid of

length l and number of turns N. (Delhi 2008)

Q22. (i) How are eddy currents reduced in a metallic core?

(ii) Give two uses of eddy currents. (Foreign 2009)

Q23. An electric lamp having coil of negligible

inductance connected in series with a

capacitor and an AC source is glowing

with certain brightness. How does the

brightness of the lamp change on

reducing the (i) capacitance, and (ii) the

frequency? Justify your answer. (Delhi

2010)

Q24. A coil Q is connected to low voltage bulb B and placed near another coil P is

shown in the figure. Give reason to explain the following observations:

(a) The bulb ‘B’ lights.

(b) Bulb gets dimmer if the coil Q is moved towards left. (Delhi 2010)

Q25. Two identical loops, one of copper and the other of aluminium, are rotated

with the same angular speed in the same magnetic field. Compare (i) the

induced emf and (ii) the current produced in the two coils. Justify your

answer. (AISSCE 2010)

Lamp

Source

166

166

Q26. A rectangular loop and a circular loop are moving out of a uniform magnetic

field to a field-free region with a constant velocity ‘v’ as shown in the figure.

Explain in which loop do you expect the induced emf to be constant during the

passage out of the field region. The magnetic field is normal to the loops.

(AISSCE 2010)

Q27. Two long co-axial solenoids of the same length but different radii and

different number of turns are wound one over the other. Deduce the expression

for the mutual inductance of this arrangement. (Foreign 2010)

Q28. Current in a circuit falls from 5 A to 0 A in 0.1s. If an average emf of 200V

induced, give an estimate of the self-inductance of the circuit. (Foreign 2011)

Q29. (a) The groups (i) and (ii) represent the variation of the opposition offered by

the circuit element to the flow of alternating current with frequency of the

applied emf. Identify the circuit element corresponding to each graph.

(i)

Opposition to

current

Opposition to

current

frequency frequency

(ii)

167

167

(b) Write the expression for the impedance offered by the series combinations

of the above two elements connected across the same a.c. source. Which will

be ahead in phase in this circuit, voltage or current? (AISSCE 2011 C)

Q30. What are eddy currents? Give two applications of eddy currents. (AISSCE

2011)

Q31. A current is induced in coil C1 due to the motion of current carrying coil C2.

(a) Write any two ways by which a large deflection can be obtained in the

galvanometer G. (b) Suggest an alternative device to demonstrate the induced

current in place of a galvanometer. (Delhi 2011)

Q32. Calculate the quality factor of a series LCR circuit with L = 2 H, C = 2 μF and

R = 10 Ω. Mention the significance of quality factor in LCR circuit. (Foreign

2012)

Q33. An alternating voltage given by V = 70 sin 100πt is connected across a pure

resistor of 25 Ω. Find (i) the frequency of the source (ii) the rms current

through the resistor. (AISSCE 2012)

Q34. State the underlying principle of transformer. How is the large scale

transmission of electric energy over long distance done with the use of

transformers? (AISSCE 2012)

Q35. Define self-inductance of a coil. Show that magnetic energy required to build

up the current I in a coil of self-inductance L is given by ½ LI2. (Delhi 2012)

Q36. A metallic rod of length ‘L’ is rotated with angular frequency ‘ɷ’ with one

end hinged at the centre and the other end at the circumference of a circular

metallic ring of radius L, about an axis passing through the centre and

perpendicular to the plane of the ring. A constant and uniform magnetic field

B parallel to the axis is present everywhere. Deduce an expression for the emf

induced between the centre and the metallic ring. (Delhi 2012)

168

168

Q37. State Lenz’s law. A metal rod held horizontally along east-west direction, is

allowed to fall under gravity. Will there be an emf induced at its ends? Justify

your answer.

Answers of previous years CBSE papers

One mark Questions

A1. V0 = 12 √2 V

A2. <P> = ErmsIrms cosϕ = 230 x 10 x ½ x ½ = 575 W

A3.

R

CL

V

VV1tan = 00

A4. 22

C

rmsrmsrms

XR

e

Z

ei

and XC = 1/ɷC, therefore if the frequency is increased

then impedance of circuit decreases and current increases, So does the

brightness.

A5. Induced current is greater in copper loop as the resistance of copper loop is

lesser while the emf induced in both the loops are same. dt

d

R

N

Ri

A6. Initially clockwise at the right face and when the magnet goes on other side

the current reverses its direction in accordance with the lenz’s law.

A7. cos ϕ = ½ therefore ϕ = 600

A8.

R

CL

V

VV1tan = 00

A9. e = dt

Nd )φ( = - L

dt

dI

A10. Zero due to a phase difference of π/2 between current and voltage.

169

169

A11. The magnitude of the induced emf in a circuit is proportional to the time rate

of change of magnetic flux through the circuit.”

dt

d

A12. The magnetic flux linked with the coil when the current flowing through its

turns is unity.

SI unit - Henry

A13. ‘A’ as this curve shows more magnetic flux for same amount of current.

A14. Clockwise, because by Lenz’s law the induced current tends to oppose the

decrease in inward magnetic flux.

A15. Anticlockwise in Loop ‘abc’ and clockwise in loop ‘defg’

A16. (i) anticlockwise (ii) clockwise

A17. The component of current phasor that is at π/2 phase difference with the

voltage is called watt-less current as it is not source of any power dissipation.

The watt-less current is given by expression Irmscos ϕ

A18. A will be positive with respect to B

A19. To decrease the a.c. voltage

A20. (i) E0/√2 (ii) 0

A21. (i) 8.48 A (ii) 0

A22. Anticlockwise

A23. As the nearer faces of both coils act as S- poles, the direction of induced

current in both coils appear clockwise at the nearer face.

A24. 1-anticlockwise and 2-clockwise

A25. Clockwise

A26. 1-anticlockwise and 2-clockwise

Two marks questions

170

170

A1. The inductive reactance increases while capacitive reactance decreases with

increase in frequency.

A2. The metal detector works on the principle of resonance in ac circuits. When

you walk through a metal detector, you are, in fact, walking through a coil of

many turns. The coil is connected to a capacitor tuned so that the circuit is in

resonance. When you walk through with metal in your pocket, the impedance

of the circuit changes – resulting in significant change in current in the circuit.

This change in current is detected and the electronic circuitry causes a sound

to be emitted as an alarm.

A3. LC

f2

1 in practice the oscillations ultimately die away due to the

following two reasons:-

(i) Every inductor has some resistance. The effect of this resistance is to

introduce a damping effect on the charge and current in the circuit and the

oscillations finally die away.

(ii) Even if the resistance were zero, the total energy of the system would not

remain constant. It is radiated away from the system in the form of

electromagnetic waves.

A4. Resonance frequency depends upon

the value of inductance and

capacitance of the circuit.

A5. When the reactance of capacitor and the reactance of inductor are equalised,

resonance is obtained.

171

171

A6. Flux through each turn ϕ = cosBAAB

= B πr2 cos ɷt

For N turns ϕ = NB πr2 cos ɷt

The induced emf ɛ = NBɷ πr2 sin ɷt

Maximum emf ɛ0 = NBɷ πr2 = 0.603 V

The average emf over a cycle = 0

The maximum current = ɛ0/R = 0.0603 A

Power dissipated P = 4

00I= 0.603 x 0.0603/4 = 0.0091 W

The induced current causes a restoring torque in the coil. An external source is

responsible for the supply of energy for this torque. Thus mechanical energy is

the source of this.

A7. Elecrtomagnetic induction when a coil is rotated about its axis perpendicular

the direction of a uniform magnetic field across it then an induced emf is

produced in it.

Erms = Eo/√2

A8. (i) (1) anticlockwise (2) clockwise

(ii) no

The rate of change of area swept by loop (1) dt

dxa

dt

dA1

The rate of change of area swept by loop (2) dt

dxb

dt

dA2

Induced emf is directly proportional to the rate of change of flux, which in

turn depends upon the rate of change of area. Since, b>a, the two rates are not

equal. Hence the induced emf will not be equal.

A9. (i)

172

172

(ii) Yes, it is possible since the voltages across inductor and capacitor are out

of phase and source voltage is equal to V = √{(VL – VC)2 + VR

2}

Therefore if VL and VC = 60 V and VR = 30 V, the source voltage is 30 V

A10. Since current and voltages are in phase with each other, circuit is in resonance

and XL = XC

322 10200505014.314.34

1

4

1

Lf

C

=50μF

A11. V = √{VC2 + VR

2}=150 V

(i) Z=V/I = 50 ohm

(ii) R = VR/I =30 ohm, power factor =Z

Rcos =0.6

A12. (i) Current will lag behind because the voltage across inductor is greater than

voltage capacitor.

(ii) Current will lead the voltage because the voltage across capacitor is

greater than voltage inductor.

A13. Let the angle between the area vector of the coil and the external magnetic

field is θ = ωt at any instant ‘t’ with assumption that θ = 00 at t = 0. Hence the

flux linked to the coil is ф = B A cos ω t, where A is the area of the coil. Due

the rotational motion of the coil, the area of the coil exposed to the magnetic

field is changing with time and hence flux linked to the coil. So, the induced

emf produced in the coil is

e = - N dt

d )φ( = - N B A

dt

d )ωt(cos = N B A ω sin ωt, This is an expression for the

instantaneous induced emf in the coil.

N B A ω = N B A 2πf is the maximum value of the induced emf.

A14. Consider a straight conductor moving in a uniform and time-independent

magnetic field B directed perpendicularly inwards of the plane of paper.

Length of conductor =l, conductor is moving towards the left with velocity v.

lv B

173

173

The magnetic flux ф linked with the conductor in any time ∆t ф = B l v∆t

The induced emf in the conductor will be e = - dt

d = - B l

t

tv

= - B l v

A15. Because of its inductive reactance it reduces the current but due to phase

difference of π/2 the average power dissipation in the inductor is zero.

The instantaneous power in the circuit is


2ωsin


2

2sin00

ωtIE


PdtT 0

1 =

Tt

otT

IEωt2sin

2

00 dt = 0

A16. The average power dissipated across the capacitor is Pav = T

PdtT 0

1 =

Tt

otT

IEωt2sin

2

00 dt = 0


A17. As R, L are in series, so the instantaneous current in each of these element is

same and let it is given by i = i 0 sin (ωt + φ), where φ is the phase

difference between the instantaneous voltage of the source and the

instantaneous current in the circuit.

We know VR is in phase of I, VL leads I by π/2,

So, phasor diagram can be drawn as follows:-

powerflux

current

voltage

LV

RVI

φ

V

174

174

The lengths of these phasor is VL = i XL, and VR = i R

Therefore as the VL and VR are perpendicular to each other and if e is there resultant

represented by diagonal according to parallelogram law, so,

e = 22

LR VV = 22

LiXiR = i 22

LXR

22

LXR

ei

=

Z

ei

22

LXRZ

A18. Two concentric circular coils, one of small radius r and the other of large

radius R, such that r<<R, are placed co axially with centres coinciding. Obtain

the mutual inductance of the arrangement.

If I1 is current flow through the outer coil then the magnetic field at its

center is B2 = μ0I1/2R

Considering B2 almost constant around its centre as r << R, the flux linkage to

inner coil is

Ф2 = π 2r B 2 = 1

2

0

2

πμI

R

r = M12I 1 M12 =

R

r

2

πμ 2

0

A19. The wings of jet plane will cut the vertical component of earth’s magnetic

field, so emf is induced across the wing. The vertical component of earth’s

magnetic field BV = BH tan θ

Given that BH = 4 x 10-4

T and θ = 300

Therefore BV = 4/√3 x 10-4

T

Induced emf = BV l v = 3.12 V

R

e

L

175

175

A20. (i) Mutual induction

(ii) When a Cu sheet is inserted in the gap between the coils, the eddy currents

are induced in the coil due to which the current in coil B is reduced, so

brightness of bulb decreases.

A21. Consider a long solenoid of length l, area of cross-section A, number of turns

per unit length l and the permeability of the core is μ and carrying current I.

The magnetic field produced inside the solenoid is B = μ n I

Magnetic flux linkage to the solenoid is Nф = (n l) B A = (n l) μ n I A = μ

n 2 I l A

But Nф = L I,

Therefore L = μ n 2 l A, this is an expression for the self-inductance of

the solenoid.

A22. (i) when we put a bulk piece of conductor to the changing magnetic field but

now the pathway of the currents is not well defined and the flow pattern

resemble the swirling eddies in water. These are called eddy currents.

(ii) any two applications

A23. 22

C

rmsrmsrms

XR

e

Z

ei

and XC = 1/ɷC, therefore (i) if capacitance is decreased,

impedance of circuit is decreases, which in turns results in increase in current

and brightness of bulb. (ii) if frequency is decreased, impedance of circuit is

decreases, which in turns results in increase in current and brightness of bulb.

A24. (a) Bulb B lights up due to induced current in B because of change in flux

linked with it as a consequence of continuous variation of magnitude of

alternative current flowing A.

(b) When coil B moves upward, the magnetic flux linked with B decreases

and hence lesser current is induced in B.

A25. (i) induced emf is same in both the loops as they are identical and are rotating

with same speed.

176

176

(ii) Induced current is greater in copper loop as the resistance of copper loop is

lesser while the emf induced in both the loops are same. dt

d

R

N

Ri

A26. In rectangular coil the induced emf will remain constant because in this case

the rate of change of area in the magnetic field region remains constant, while

in circular coil the rate of change of area in the magnetic field region does not

remain constant.

A27. Consider two long coaxial solenoids each of length l. Radius of the inner

solenoid S1 is r1, number of turns is N1 and the number of turns per unit

length is n1. The corresponding quantities for the outer solenoid S2 are r2, N2

and n2, respectively.

When current I2 is set up through S2, then the flux linkage through S1 is

N 1 ф 1 = M12 I2

M12 is called the mutual inductance of solenoid S1 with respect to S2 and

called the coefficient of mutual induction.

The magnetic field due to the current I2 is S2 is μ 0 n 2 I 2. Hence the

resulting flux linkage with soil S1 is N 1 ф 1 = (n 1 l) (π 21r ) (μ 0 n 2 I 2) = μ 0

n 1 n 2 π 21r l I 2

Clearly, M 12 = μ 0 n 1 n 2 π 21r l Also, 12M = 21M = M

A28.

dtdI

L

= )

1.0

05(

200

= 4H

A29. (a) (i) resistor (ii) inductor

177

177

(b) 22

LXR , voltage leads the current in phase.

A30. (i) when we put a bulk piece of conductor to the changing magnetic field but

now the pathway of the currents is not well defined and the flow pattern

resemble the swirling eddies in water. These are called eddy currents.

(ii) Any two applications

A31. (a) The deflection in galvanometer may be made large by (i) moving coil C2

towards C1 with high speed. (ii) by placing a soft iron laminated core at the

centre of the coil C1.

(b) The induced current can be demonstrated by connecting a torch bulb (in

place of galvanometer) in coil C1. Due to induced current the bulb begins to

glow.

A32. Q = C

L

R

1= 100

It is measure the sharpness of resonance of L-C-R circuit.

A33. (i) f = ɷ/2π = 50 Hz

(ii) Vrms = 70 x 0.707 = 49.5 V, Irms = Vrms/R = 1.98 A

A34. Mutual inductance

At the electric power generation station, a step-up transformer is used to

convert the electric power at low voltage to electric power at very high

voltage and again at the power supply places a step-down transformer is used

to convert the electric power at high voltage to electric power at low voltage.

By transmitting at high voltage, the current (I = P/V) in the transmission line

reduces thereby reducing the transmission loss (I2R).

A35. The back emf induced in the coil when the current flowing through its turns

changes at the rate of 1 A/s.

Since, self-inductance is a back emf, so work is to be done by the external

agency against the back emf to establish the current in the circuit which is

stored in the coil as its magnetic potential energy.

If ‘I’ is the instantaneous current in the circuit then at any instant ‘t’ then rate

of work done will be

178

178

dt

dW = e I = L I dt


inductive effects.


LIdI0

= 2

1 L I

2

A36. Let the rod PQ of length l is rotated in the plane of paper and magnetic field is

directed perpendicular inwards to the plane of paper with a uniform angular

speed ω in anticlockwise direction.

Induced emf T

lB

dt

dAB

dt

BAd

dt

de

2π

But

2T

π2π 2

lBe

Therefore, 2

2

1Ble

A37. The polarity of induced emf is such that it tends to tends to produce a current

which opposes the change in magnetic flux that produces it.

Yes due to horizontal component of earth’s magnetic field. ɛ = BH l v

7. Higher Order Thinking Skill (HOTS) Questions

27. A conducting rod XY slides freely on two parallel rails, A and B, with a uniform

velocity ‘V’. A galvanometer ‘G’ is connected, as shown in the figure and the

closed circuit has a total resistance ‘R’. A uniform magnetic field, perpendicular

to the plane defined by the rails A and B and the rod XY (which are mutually

perpendicular), is present over the region, as shown.

179

179

(a) With key k open:

(i) Find the nature of charges developed at the ends of the rod XY.

(ii) Why do the electrons, in the rod XY, (finally) experience no net force even

through the magnetic force is acting on them due to the motion of the rod?

(b) How much power needs to be delivered, (by an external agency), to keep the

rod moving at its uniform speed when key k is (i) closed (ii) open?

(c) With key k closed, how much power gets dissipated as heat in the circuit?

State the source of this power.

Ans. (a) (i) Y : negative , X: positive

(ii) Magnetic force, Fm, experienced by the moving electrons, gets balanced by

the electric force due to the electric field, caused by the charges developed at the

ends of the rod. Hence net force on the electrons, inside the rod, (finally) become

zero.

(b) The power, that needs to be delivered by the external agency, when key k is

closed, is P =FV = (I l B)V = R

BlVl BV =

R

VlB 222

When k is open, there is an induced emf, but no induced current. Hence power

that needs to be delivered is zero.

(c) Power, dissipated as heat = i2 R =

R

VlB 222

The source of this power is the mechanical work done by the external agency.

28. A rectangular wire frame, as shown is placed in a uniform magnetic field directed

upwards and normal to the plane of paper. The part AB is connected to a spring.

The spring is stretched and released when the wire AB has come to the position

A’B’(t = 0). How would the induced emf vary with time? Neglect damping.

A 1A

B 1B

180

180

Ans. As spring oscillates simple harmonically, so the velocity of wire AB varies

sinusoidally, hence motional induced emf will be also sinusoidal.

29. Use Lenz’ law to determine the direction of the induced current in the situations

[i] and [ii] described below:

[i] A wire of irregular shape turning into a circular shape.

[ii] A wire of rectangular shape turning into a circular shape.

(i) With same perimeter, in all plane figures, circle has maximum area. In case (1),

inward flux increases, so, according to Lenz’ law, an equivalent outward flux is

produced for which I should be along ADCB.

(ii) Outward flux increases for which an equivalent inward flux induces and hence

‘I’ is along ABCD.

4. In a given circuit, inductor L and resistor

R have identical resistance. Two similar

electric lamps B1 and B2 are connected as

shown. When switch S is closed, (a)

which one of the lamps lights up earlier,

(b) will the lamp be equally bright after

some time? Justify your answer.

(a) Inductor offers impedance Z = 22

LXR , initially, due to inductive effect of

L, there will be smaller current through L than R, so, B2 lights up earlier.

(b) Source is dc, so, after some time, I, attains its peak value and remains

constant. Inductive effect vanishes. Therefore, in both the circuits, there will be

equal opposition, so, both the bulbs will glow equally.

a

bd

c

][i

b d

c

a

][ii

L

R

1B

2B

S

181

181

5. The arm PQ of a rectangular conductor is moved from x = 0, outwards. The

uniform magnetic field is perpendicular to its plane and extends from x=0 to x =

b and is zero for x>b. only the arm PQ possess a substantial resistance r.

Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b,

and is then moved back to x = 0 with constant speed v. Obtain expressions for

the flux, the induced emf, the force necessary to pull the arm and the power

dissipated as Joule heat. Sketch the variation of these quantities with distance.

Ans.

8. Value Based Questions

EMI & AC

1. Mr. Hanuman Prasad Sharma purchased a bulb marked with 200 V – 100 W.

He wanted to operate it on 240 V – 50 Hz a.c. supply. He calculated

inductance of the choke required for operating that heater.

P

QR

S

0x bx bx 2

Blb

0x bx bx 2 bx 0x

Blv

Blv

rvlB /22

rvlB /22

rvlB /222

flux

EMF

Current

Power

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(i) Write the values shown by Mr. Hanuman Prasad Sharma.

(ii) Calculate the value of inductance required for the purpose.

Ans:- (i) Practical knowledge about the electrical devices, careful and

awareness, (ii) current rating of the bulb = 100/200 = 0.5 A, therefore

required minimum impedance of the circuit Z = V/I = 480 ohm, Resistance of

the bulb R = V2/P = 400 ohm, XL = √(Z

2-R

2) and L = XL/2πf

2. Poonam went to the market with her mother and decided to come back home

by metro. At the Metro station they were made to pass through a gateway for

security purpose. Poonam passed through it and was waiting for her mother to

come. She heard a long beep when her mother passed through the metal

detector gateway. Poonam was confused why metal detector beeped in case of

her mother. She asked the duty staff, who explained her in detail. Both were

satisfied with the security system.

(i) What values are displayed by Poonam?

(ii) What is the cause of sound through metal detector?

(iii) Write the principle on which a metal detector works.

Ans:- (i) Curiosity, power of observation. (ii) The metal detector works on the

principle of resonance in a.c. circuit. When Poonam’s mother is passed

through a metal detector, her mother, in fact, is passing through a coil of

many turns. The coil is connected to a capacitor turned so that the circuit is in

resonance. When she walks through with the metal in her pocket, the

impedance of the circuit changes – resulting in significant change in the

current in the circuit. This change in current is detected and the electronic

circuitry causes a sound to be emitted as an alarm. (iii) Resonance

3. Ambuj Tiwari, an Indian – American is working in Chicago University, USA

as an assistant professor in Computer Science. After having lived in USA for

about 18 years he returned India to meet his parents, relatives and friends. He

had a discussion with his friend Rohit Bohra on domestic power supply in US

and in India. In US domestic power supply is at 110 V, 50 Hz, whereas in

India it is at 220 V, 50 Hz. Ambuj was stressing that US supply is better than

Indian supply. But Rohit argued differently- I guess the US just likes to be

stubborn. Because one can see that the US and Britain the only countries in

the world to still be using the FPS system and not the metric system.

Both went to his physics teacher who taught them physics when they were

students of standard XII and asked his opinion on the issue. The teacher

explained that both the supplies have advantages as well as disadvantages.

(i) In your opinion what negatives values were displayed by Ambuj Tiwari

and Rohit Bohra?

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(ii) Write some advantages and disadvantages of 110V, 50 Hz and 220 V, 50

Hz AC.

(iii) What will be the peak value of 110 V, 50 Hz?

Ans:- (i) Might is write. (ii) The disadvantage of 110V as compared to 220V

a.c. is more losses due to resistance. Higher current at lower voltage (I =

P/V) creates more heat loss. The advantage of 110V is that it is less

dangerous. (iii) Peak value of 110V, 50Hz is 110√2V = 160 V

4. Mr. Nayyer an industrialist, worried about frequent power cuts for long

duration decided to install an A.C. generator of 800 kVA to cater to plant

requirement of 750 kVA. As the law did not permit such an installation in the

area, he had to get it installed 2.5 km away. He however found that the

generator failed to meet the requirement of the plant. He asked his supplier

friend to install another generator of 100 kVA. The supplier discussed his

problem and advised Mr. Nayyer that the existing generator was sufficient for

his needs. To solve his problems, he suggested installation of two transformer

220 V – 22 kV near generator and 22 kV – 220 V near the plant. The plant

now could be run smoothly.

(i) Explain how the transformers solved the problem.

(ii) In your opinion what would have been the disadvantages of installing an

additional generator?

(iii) What ethics did the power supplier show while solving the problem?

Ans:- (i) Using a step up transformer of 220V-22kV the line loss (H = i2Rt)

can be minimised during transmission and at the location of the plant another

22kV-220V transformer is used. So, the plant gets desired power. (ii) (a)

Wastage of money, (b) Higher fuel consumption, (c) Increase in pollution

level. (iii) Honest trade practices/ unbiased opinion.

5. Mr. Anil Sharma an Economics teacher in a reputed public school had gone to

visit a science exhibition with his colleague Mr. U C Gupta a science teacher

in the same school. Students were demonstrating to visitors that when a

speaker magnet is brought near a circular aluminium disc fixed on spindle of

D.C. motor which is rotating, surprisingly the speed of rotating disc decreases.

Again when the magnet is moved back the speed of rotation of disc starts

increasing. Mr. Anil Sharma was surprised to see what was happening there.

Because he knows that magnet could not attract an aluminium disc, then how

the speed of disc is changing. Mr. U C Gupta explained him that no magic

was involved there everything was happening according to the laws of

physics. The speed of disc is changing due to production of eddy currents in

the disc.

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(i) In your opinion what values were displayed by Mr. U C Gupta?

(ii) State the law Mr. U C Gupta was talking about.

(iii) What are eddy currents? Who discovered these currents?

(iv) Name the two devices in which it is used in positive sense.

Ans:- (i) (a) Sharing of knowledge, (b) Scientific temperament. (ii) Lenz’s law,

which states that induced current always opposes the cause due to which it is

created. (iii) When bulk pieces of conductors are subjected to changing

magnetic flux, induced currents are produced in them due to electromagnetic

induction. The flow pattern of these currents resemble swirling water, so these

are known as eddy currents or whirlpool currents. Focault discovered the

eddy currents. (iv) (a) Induction furnace (b) Electromagnetic Brake in electric

trains.

6. Krishnan a retired science teacher was walking with his grandson Munna by

the side of a paddy field. Munna noticed power grids carrying thick wires. He

was curious to know what the structure was and what the wires were for. So

he asked his grandfather about it. He also wanted to know if the tower could

be removed so that there would be more space for crops. Krishnan explained

in detail about the tower and the need for the wires.

a) Why is the voltage stepped up for long distance transmission?

b) What are the values exhibited by Krishnan and Munna?

ANSWERS:

a) Less power loss

b) Concern for saving energy and greenery.

Numerical Problems frequently asked in AISSCE/ important concepts

9. The instantaneous current from an ac source is I = 5 sin (314 t) ampere. What

is the average value, peak value and rms value of the current?

I0 = 5A, Iav = 0 (for full cycle), Iav = 0.636 I0 = 0.636 x 5 = 3.18 A (for half cycle),

Irms = 0.707 I0 = 3.535 A

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10. The electric mains in the house are marked 220V, 50Hz. Write down the

equation for instantaneous voltage.

erms = 220 V, = 50Hz e0 = erms 2 =311 V, = 2 = 314 rad/s

so, instantaneous value of emf is e = (311) sin (314 t) Volt

11. A capacitor blocks dc while it allows ac to pass through it. Explain.

For d.c: = 0; so, capacitive reactance XC = C2

1 =

12. What is power factor of LCR circuit in series at resonance?

At resonance, R = Z, power factor = 1

13. The magnetic flux through a coil perpendicular to its plane is varying

according to the relation Φ = (4t3 + 5t

2 + 8t + 5) Weber. Calculate the induced

current through the coil at t = 2 s if the resistance of the coil is 3.1 ohm.

e = dt

d = - (12 t

2 + 10 t + 8); at t = 2 s, e = - 76 V;

R

ei - 24.5 Volt

14. Find the maximum value of current when an inductance of one Henry is

connected to an a.c. source of 200 volts, 50 Hz.

irms =

2

2

200

LX

e

L

rms i0 = 2rmsi =

22 =0.636 x 1.414 = 0.9 A

15. A 10 mH inductor carries a current I which changes with time according to

equation: i = 2 t 2 + 3

If ‘I’ is in ampere and ‘t’ is in sec; calculate the instantaneous value of emf at t =

2.5 sec.

Represent variation of induced emf with time.

dt

diLe = 10

– 2 (4 t) = 10

– 2 x 4 x 2.5 = 0.1 V = constant

16. Find the capacitance of the capacitor that would have a reactance of 100 ohm

when used with an ac source of frequency (5 / π) kHz.

C = CX

1 = 1 μ F

17. An a.c. circuit consists of a series combination of circuit elements X and Y.

The current is ahead of the voltage in phase by π / 4. If element X is a pure

1.0

)(inVe

t

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186

resistor of 100 Ω, (i) name the circuit element Y, and (ii) calculate the rms

value of current, if rms value of voltage is 141 V.

Y is capacitor; 22

C

rmsrms

XR

ei

tan =

R

X C tan 4

=

R

X C XC = R = 100 Ω ;

22 100100

141

rmsi = 1 A

18. Determine the impedance of the circuit if reactance of C and L are 340 Ω and

300 Ω respectively and R is 30 Ω.

Z = 22 )( CL XXR = 50 Ω

19. A 15 μ F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive

reactance and the current (rms and peak) in the circuit. If the frequency is

doubled, what happens to the capacitive reactance and the current?

CXC

1 = 212 Ω;

C

rmsrms

X

ei = 1.038 A; i0 = 2rmsi =1.47 A;

If doubles then XC halves and hence i doubles.

20. Calculate the (i) impedance, (ii) wattles

current of the given ac circuit.

Vrms =22

CR VV = 50 V; Z =

252

50

rms

rms

i

V Ω

Sin rms

C

V

V = 4/5 wattles current = irms sin = 8/5 A

21. The instantaneous current and voltage of an ac circuit are given by I = 10 sin

300 t A and v = 200 sin 300 t V. What is the power dissipation in the circuit?

i and e are in phase, so, circuit is resistive only; power dissipated = erms irms = 2

00ie

= 1000 W

22. The power factor of an ac circuit is 0.5. What will be the phase difference

between the voltage and current in this circuit?

Power factor = cos = 0.5 = 600

V40 V30

rmsVA2

C R

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23. Given below is electric circuit.

Calculate the power factor of circuit.

Z = 22

LXR = R 10 ; Power factor = cos = Z

R =

10

1

24. A step-down transformer converts a voltage of 2200 V into 220 V in the

transmission line. Number of turns in primary is 5000. Efficiency of the

transformer is 90% and output power is 8 kW.

Calculate (a) the number of turns in the secondary and (b) Input power.

input

output

P

P 0.9 =

inputP

8000 Pinput = 8888.89 W;

P

S

P

S

N

N

e

e NS =

500

25. A step up transformer operates on a 220 V line and a load current of 2

Ampere. The ratio of the primary and the secondary windings is 1 : 25. What

is the current in the primary coil, the secondary voltage and power output

assuming the transformer to be an ideal one?

For ideal transformer, P

S

P

S

e

e

N

N eS = 5500 V; Poutput = eS iS = 5500 x 2 = 11 kW

S

P

P

S

i

i

N

N iP = 50 A

26. In an ideal transformer, the number of turns in the primary and the secondary

coil are 200 and 1000 respectively. If the power input to the primary is 10 kW

at 200 V, calculate [i] output voltage and [ii] current in primary.

P

S

P

S

e

e

N

N eS = 1000 V; Pinput = eP iP iP = 50 A

27. Calculate the current drawn by the primary of a transformer, which steps

down 200 V to 20 V to operate a device of resistance 20 Ω. Assume the

efficiency of the transformer to be 80%.

R

ei SS = 1 A;

input

output

P

P AiP

8

1

28. The output voltage of an ideal transformer, connected to a 240 V ac mains is

24 V. When this transformer is used to light a bulb with rating 24 V, 24 W,

calculate the current in the primary coil of the circuit.

RX L 3 R

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R = P

V 2

=24 Ω; R

Vi SS = 1A; iP = 0.1 A

29. In a step-up transformer, the transformation ratio is 100. The primary voltage

is 220 V and input power is 100 watt. The number of turns in primary is 100.

Calculate

i. the number of turns in the secondary,

ii. the current in the primary,

iii. the voltage across the secondary and

iv. the current in the secondary.

(i) 100P

S

N

N Ns = 10000 (ii) ip = Pi/vp = 5/11 A (iii)

P

S

P

S

e

e

N

N es = 22000 V

(iv)S

P

P

S

i

i

N

N is = 5/1100 A

30. An ac generator consists of a coil of 100 turns and cross-sectional area of 3

m2, rotating at a constant angular speed of 60 rad/s in a uniform magnetic

field 0.04 T. The resistance of the coil is 500 Ω. Calculate (a) maximum

current drawn from the generator, (b) maximum power dissipation in the coil.

Maximum current R

NBA

R

ei

0

0 = 1.44 A; Maximum power dissipation = Ri2

0 =

1036.8 W

UNIT IV ELECTROMAGNETIC INDUCTION & A...128 128 15.Principle of AC Generator:- Electromagnetic...

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