Pmt Class Xii Physics EPMT_Class_Xmi & Ac

MEDICAL

ELECTROMAGNETIC INDUCTION& ALTERNATING CURRENT

P H Y S I C SS T U D Y M A T E R I A L

NARAYANA INSTITUTE OF CORRESPONDENCE COURSESF N S H O U S E , 6 3 K A L U S A R A I M A R K E TS A R V A P R I Y A V I H A R , N E W D E L H I - 1 1 0 0 1 6PH.: (011) 32001131/32/50 • FAX : (011) 41828320Websi te : w w w . n a r a y a n a i c c . c o mE-mai l : i n f o @ n a r a y a n a i c c . c o m

2004 NARAYANA GROUP

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CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS


Theory

Solved Examples

Exercises

• True and False Statements

• Fill in the Blanks

• Assertion-Reason Type Questions

• Multiple Choice Questions

• MCQ’s Asked in Competitive Examinations

• Subjective Questions

Answers

CO

NT

EN

TS

CO

NT

EN

TS

CO

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EN

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Physics : Electromagnetic Induction & Alternating Current

FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320

1

INSTITUTE OF CORRESPONDENCE COURSES

NARAYANA

CONTENTS

!!!!! Introduction

!!!!! Magnetic Flux

!!!!! Elecromagnetic Induction

!!!!! Fleming’s Right Hand Rule

!!!!! Self Induction

!!!!! Mutual Induction

!!!!! Growth and Decay of Current inL-R Circuit (Transient current)

!!!!! Average Value of AC over Half Cycle

!!!!! AC Through LR Circuit

!!!!! AC Through CR Circuit

!!!!! AC Through LCR Circuit

!!!!! Power in LCR Circuit

!!!!! Choke Coil

!!!!! Transformer

!!!!! Solved Examples

!!!!! Exercise

!!!!! Answers

A credit card’s number, expiration date, and

cardholder name are coded into a magnetized pattern

in a stripe on the back. When the card is “swiped”

through a card reader, the moving stripe bathes the

reader’s circuitry in a varying magnetic field that

induces current in the circuits. These currents transmit

the information in the stripe to the cardholder’s bank.

“What would happen if the card were not swiped, but

just sat motionless in the reader’s slot?”




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MAGNETIC FLUX

The magnetic flux streaming through an area is also referred to as the flux linked with that area. It also representsthe number of magnetic lines of induction passing through that area.

Let A!

= a small plane area, represented as a vector normal to its plane,

B!

= magnetic induction of the magnetic field at the site of A!

(considered to be uniform over A!

)then φ∆ = the magnetic flux linked with A

!

Where θ = angle between A!

and B!

= A B⋅! ! (dot product of A

!and B!

)= (A) (B) (cos θ )

Note : Magnetic flux is a scalar quantity. If the area is not in one plane, we can still determine the total fluxthrough the area, considering continuous small area elements d A

!which are reasonably plane and then

summing up or integrating the flux elements through these area.=φ total flux through the area A

BA!!

⋅Σ∆= or φ = B. dA∫!!

ELECTROMAGNETIC INDUCTION

Whenever the magnetic flux linked with an electric circuit or loop changes, an emf is induced in the circuit .Thisphenomenon is called electro-magnetic induction.There are three laws concerning this phenomenon. These are Faradays 1st and IInd Laws and Lenzs Law.The three laws can be summarized in one equation given by

Instantaneous induced emf in the loop = (the rate of change of magnetic flux φ)

⇒ Emf Induced = ( )φ−dtd

The negative sign only shows that the induced emf tends to oppose the very cause of its generation i.e. change ofmagnetic flux inducing the emf.If the circuit is closed, a current starts flowing through it. Such current, in turn, produces a magnetic field whosedirection is such that it opposes the very change of original magnetic field or flux. However, if the circuit is notclosed, the induced emf cannot set up a current in the circuit. Even then the current that would have resulted dueto the change of flux is in same direction as for the closed loop.

INTRODUCTION

W e know that electric current produces magnetic field. But the mutual relationship ofelectricity and magnetism does not stop here. We shall see that under certain conditionsa magnetic field can be used to produce an electric current. No current is induced in

a loop of wire that is stationary in a constant magnetic field. However, if the magnetic fieldchanges with time, or if the wire loop moves across or is rotated in the field, a current is inducedin the wire.The use of this interrelationship of electricity and magnetism are legion. One of the familiarexample is in the playing of a casette tape. The music you hear was encoded as tiny variations ina magnetic field. These variations produce electrical impulses, which are amplified and drivethe speakers. The speakers, in turn, utilize electromagnetic interactions to translate the electricalimpulses back into audible sound.



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FLEMING’S RIGHT HAND RULE

This simple rule helps to determine the direction of the induced emf in a straight conductor moving with a velocityperpendicular to the length of the conductor and also perpendicular to the magnetic field. The rule is as givenbelow:(a) Stretch the thumb and the first finger of the right hand, perpendicular to each other, keeping both in the plane

of the palm.(b) Stretch the second finger (middle finger) perpendicular to the palm.(c) Orient the outstretched thumb such that it represents the direction of the velocity (motion).(d) Orient the first finger so that it represents the direction of the magnetic field.(e) Then the second finger will represent the direction of the induced emf

i.e. First finger → field, thumb → motion, second finger → Current (induced)

INDUCED CHARGE DUE TO FLUX CHANGE

When the magnetic flux lined with a closed circuit of known resistance changes, the induced emf tends to set upa current in the circuit i.e. the charges flow in directed motion to set up the current. Let the total charge flow be

q∆ .

Then q∆ = tI∆ Rt

tt

Remf ∆

∆φ∆=∆=

Rφ∆=

It may be noted that the flow of charge does not depend on the rate at which magnetic flux is changed. It ratherdepends on the total flux change.

ROTATION OF A COIL IN A MAGNETIC FIELD—EMF INDUCED

When a plane coil (having N turns) of area A each, rotates in a uniform magnetic field (B) with a constant angularvelocity ( ω) about an axis coinciding with the plane of the coil and perpendicular to the uniform magnetic field(B) the emf induced (e) in the coil as a function of time t is given by e = NAB sin (ωt)

The induced emf in the coil is sinusoidal. When the plane of the coil is parallel to B!

, the induced emf is maximum.When the plane is perpendicular to B

!, the induced emf is zero.

Example 1 :

Determine the magnitude of the emf generated between the ends of the axle of a railway carriage, 1 m in length,when it is moving with a velocity of 36 km/hr along a horizontal track, given horizontal component of Earthsmagnetic field BH = 4 × 105 Tesla and angle of dip δ = 60o.

Solution :

Note : When the carriage runs over the horizontal track, the axle sweeps a horizontal area and hence it sweepsmagnetic flux created by vertical component of Earths magnetic induction (BV).

BV = BH tan θ= (0.4 × 104) (tan 60o) = 0.6928 × 104 Tl = 1 mv = 36 km/hr = 10 m/s

Then, the induced emf across the ends of the axlee = l BVv

= (1) ( )4-100.6928 325 ×

= 5.773 × 104 Volt



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NARAYANA

Example 2 :

A metallic disc of radius 25 cm is kept rotating at a constant speed of 1800 r.p.m. about its own axis, directedparallel to a uniform magnetic field of induction B = 0.1 Tesla. If the rotation is anticlockwise when seen in thedirection of the field, find the magnitude and direction of the emf induced in the disc.

Solution :

Consider small length element dl. emf dV is developed across itdV = Bvdl

Then dv Bvd∫ ∫ l

= B dω" " = 2

2lBω

Length of the rod l = 0.25 m

Number of revolutions per sec, N 3060

1800 == revolution/s

B = 0.1 tesla

Induced emf e = 2l B.2 N.

2

2Βω π=2

l = ( ) ( ) ( )2l N Bπ

( )( ) ( )( )0.1 30 1025 22-×π= = 0.589 V

By Flemings R.H. Rule, this emf is such that end K is positive & end L is negative.

Example 3 :

A circular coil of 200 turns and mean radius 30 cm is placed in a uniform magnetic field of induction 0.02 Teslaand is free to rotate about an axis coinciding with its own plane, but perpendicular to the uniform magnetic field.The coil is in closed circuit having a total resistance of 50 Ω . If the plane of the coil is initially perpendicular to thefield, find the charge flown through the circuit, when the coil is rotated through (i) 60o (ii) 180o.

Solution :

Initially angle 00 =θ ( )as A||B! !

Intially flux NAB0 =φ cos 0o = NAB

Final flux oo

60NAB cos NAB cos 60φ = θ= NAB

21=

NAB180 cos NAB o180o −==φ

(i) 1φ∆ = change in flux during rotation for 0 to 60o

NABNAB21 −= NAB

21−=

∴ =∆ 1q (induced charge in the circuit)

RNAB

21

R1 −=φ∆=

( ) ( ) ( ) ( ) ( )2-2

1

1/2 200 30 10 0.02 q

50π ×

∆ = C1031.11 3−×=



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NARAYANA

(ii) =φ∆ 2 change in the magntic flux during rotation from o o0 to 180θ = θ == (NAB) (NAB)= 2NAB

∴ 2q∆ = induced charge

= RNAB2

= 4.52 × 103C

Example 4 :

A small conducting loop of area 1 mm2 is placed in the plane of a long straight wire at a distance of 20 cm fromit. Current in the straight wire changes from 100 A to zero in 0.01 second. Find the average emf induced duringthis period.

Solution :

Average emf tte 21

∆φ−φ=

∆φ∆−=

Here 1φ linked with the loop = B.A. = Ad2I0

πµ

Weber10)10(2.02100104 116

7−−

−

=××π

××π=

1φ (= final flux) = 0.

So, volt1001.0

010e 811

−−

=−=

Example 5:

Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edgeenters the 20 cm wide magnetic field at t = 0. Find the emf induced in the loop at (a) t = 2s, (b) t = 10 s, (c)t = 22 s and (d) t = 30s.

×

××

××

×

××

××

×

××

××

×

××

××

20 cm

5 cm

B = 0.6 T

Solution:

(a) At t = 2 s, the loop is partly inside the magnetic field. Hence emf = Bvl = 0.6T × 0.01 m/s × 0.05 m = 3 × 104 volt.

(b) At t = 10 s, the loop is fully inside the magnetic field. As the field is uniform therefore no emf is generated.

(c) At t = 20 s, the loop just starts coming out of the magnetic field. So at t = 22 s, it is partly inside and partlyoutside the field & also it is moving. Hence

emf = Bvl = 0.6 T × 0.01 m/s × 0.05 m = 3 × 104 volt

(d) At t = 30 s, the loop is fully outside the magnetic field. Therefore, no emf is generated.

It may be used that the current generated in the loop at t = 2s is anticlock wise, whereas t = 22 s, the currentis clockwise.



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NARAYANA

Example 6 :

A uniform magnetic field B exists in a cylindrical region of radius 10 cm as shown in figure. A uniform conductingwire of length 80 cm and resistance 4.0 Ω is bent into a square frame abcd and is placed with one side along adiameter of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s, find the currentinduced in the frame.

××

×× ×

×× ×

××

××

a

b c

d

Solution :

Flux linked with the square loop is only due to flux change in one half of the circular area. so

B.Aφ =!!

(where A = πr2/2) = BA cos 0º = BA

In the loop, emf dtd E φ= =

d (BA)dt

2dB r dBE Adt 2 dt

−π⇒ = = ×

= 2π− (0.1)2 (0.01) = 1.57 × 104 volt

Current in the loop I = 4E 1.57 10

R 4×= Amp.

= 3.9 × 105 Amp.Current in the loop will be from a to b, so that it can create a magnetic field opposite to the existing magnetic fieldbecause the magnetic field is increasing. Had the magnetic field been decreasing, the current would be from b to a.

Example 7 :

A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure.A uniform magnetic field B exists in the perpendicular direction. Find the emf induced (a) in the loop abc, (b) inthe segment bc, (c) in the segment ac and (d) in the segment ab. Given length of bc = l.

a

b

c

v

B

θ

Solution :

(a) As the loop moves in a uniform magnetic field, net emf induced in the loop is zero.(b) Part bc of the loop is a straight wire and its length is perpendicular to the velocity. Hence emf. Vbc = Bvl.

Here Vb < Vc, so that end b is negative compared to end c i.e. Vb Vc = Bvl.(c) Part ac has length parallel to velocity so, net emf Vac = 0.(d) Part ab moves such that its length makes θ with its velocity v.

So Vab = Bv(l1) sin θ (where l1 = l length of ab = l sin θ )

= Bv(l/sin θ ) sin θ

= Bvl. Here Va > Vb



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NARAYANA

Example 8 :

A rectangular frame of wire abcd has dimensions 32 cm × 8.0 cm and a total resistance of 2.0 Ω . It is pulled outof a magnetic field B = 0.020 T by applying a force of 3.2 × 105 N. It is found that the frame moves with constantspeed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the pointsa and b and (d) the potential difference between the points c and d.

×

××

×

×

××

×

×

××

×

×

××

×a

b c

dF

Solution :

(a) Due to change in magnetic flux, emf generated is e = Bvl.

RBvl

Re I == . This current causes magnetic force Fm (= BIl) on the frame towards the left. As the loop

moves with constant speed, so F must equal Fm.So F = Fm.

⇒2 2Bv BF BI B. . v

R Rl l= = =" "

⇒5

2 2 2 2

FR 3.2 10 2 v 25m /sB (0.02) (0.08)

× ×= = =×"

(b) Emf induced E = Bv" = 5

2 2FR FR 3.2 10 2B . 0.04 V

B 0.02 0.08B

−× ×= = =×

lll

(c) E = FRBv B

=""

because all the emf is produced across the element ab only.

However due to current flowing through ab, there is a potential drop across abNet Vab = E Ir where r is resistance of part ab.

Also E 0.04 V 8 cm 8I 0.02 A and r R. 2 =0.2 R 2 40 cm 40

= = = = = × ΩΩ

So Vab = 0.04 0.02 × 0.2 = 0.036 volt(d) No emf is generated across cd. However due to current flow there is potential drop between c and d.

Vcd = IR = 0.02 A × 0.2 Ω = 0.04 volt.

Example 9:

Figure shows a square frame of wire placed coplanar with a long, straight wire. The wire carries a current i givenby i = i0 sin .tω Find (a) the flux of the magnetic field through the square frame, (b) the emf induced in the frame.

a

b

i

Solution:

(a) Current i through the straight wire creates magnetic field B within the loop. As this B is not constant, we have



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NARAYANA

to adopt method of integration. Let us consider a small area (shown shaded). Flux φd through shaded part

= B.dA###!!

= 0i adx2 xµπ

so total flux φ through the loop, is ai

dxx

+

πµ=

πµ=φ=φ ∫∫

+=

=b

abln2

ai x

dx 2

ai d 0ba x

bx

0

(b) φ is directly proportional to i. Therefore when i changes w.r.t. time (as per i = i0 sin tω ), φ also changes andtherefore emf is generated in the loop.Here i = i0 sin tω

So, tidtdi ωω= cos0

Induced emf. dtdi

babln

2a

dtde 0

+

πµ

=φ−=

= t cos i b

abln 2

a 00 ωω

+

πµ

Example 10 :

Region I shown in the figure is free of magnetic field and the region II has a uniform magnetic field B!

, directedinto the plane of the paper. ACD is a semi circular conducting loop of radius r with centre at O, the plane of theloop coinciding with the plane of the paper. The loop is now made to rotate with a constant angular velocity ωabout an axis passing through O and perpendicular to the plane of the paper. The effective resistance of the loopis R.

XXXXXXXXXXX

XXXXXXXXXXX

X X X

X X X

X X X

X X XX X XX X XX X XX X XX X XX X XX X X

X X X

X X X

X X X


B

D

O

A

C

ωRegion I Region II

(i) Obtain an expression for the magnitude of the induced current in the loop.(ii) Show the direction of the current when the loop is entering the region II.(iii) Plot a graph between the induced emf and the time of rotation for one period of revolution.

Solution :

(i) At any instant t, when the circular loop has entered the region II and has its diameter inclined to the borderof region II. Let θ be the angle between the diameter and the boundary.

A (area of the loop inside the magnetic field of induction 21B) r2

= θ!

φ = magnetic flux linked with the loop

θ= r

21 2

(B)

e (induced emf in the loop) ( )φ−= dtd

21 d r B2 dt

θ= −

XXXXXXXXXXX

XXXXXXXXXXX

X X X

X X XX X X


X X X

X X XX X X


Region I Region II

θ

ω−= Br 21 2



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NARAYANA

Since ω = constant, the above equation indicates the emf is steady when the loops is just partly or justcompletely inside the magnetic field

(ii) The induced emf reverses its direction after when the loop enters into or emerges out of the magnetic fieldi.e., the direction reverses every half revolution.When area inside the field is increasing, the current will be anitclockwise so as to produce a magnetic fieldupwards.By Lenzs Law, this upward magnetic field attempts to neutralise. But when the loop is coming outof the field, area is decreasing, the current direction is oppsoite i.e. clockwiseTherefore we can write as below, a single expression for induced emf.

e = (1)n

ωBr21 2 where n indicates the number of half time periods spent.

(iii) Graph of e → t is

T/2 Tt

e

SELF INDUCTION

When electric current flowing through a coil (or a solenoid) changes, the magnetic field inside the coil changesproportionally. Due to this, flux linked with the coil also changes. This change of flux causes an induced emf in thesame coil. This induced emf has a tendency to oppose the change of flux and thereby oppose the change ofcurrent.Let =φ flux linked with the coil

I = current in the coil

Then I∝φ

and LI=φ

Here L is a constant of proportionality and is called coefficient of self induction or self inductance. Its S.I. unit is Henry.

So, 1 Henry = Ampere 1 Weber1

Also, dtdIL

dtd =φ

⇒ Induced emf e = dtdI L−

Here sign appears because induced emf is in such a direction that it opposes the change of current. Forexample, if the current in the circuit is increasing, the induced emf tends to produce current in opposite direction,so that the growth of current is effectively opposed.

Self inductance of a long solenoid

Let a solenoid be quite long (practically infinite). Let its cross section be A, length be l and no of turns per unitlength be n. On passing current i through the solenoid,

nIB 0µ=

Then flux through each turn is inAB 0µ=⋅=φ A



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NARAYANA

Emf induced in one turn is 0d de nAdt dtφ=− =−µ i

Total emf in nl turns is E = Ne = nle = dtdAn2

0ilµ

Comparing with E = L dtdi

, we get

L = lAn20µ

Alternative, we can use the term total flux linked with N turns ( )lφ , whereas flux with one turn is φ.Then φ1 = N φ

Then total emf E = N.(e) = ( )d NdN

dt dt− φφ − =

( )dt

d lφ−=

Therefore total emf induced is the rate of change of total linked flux (which, is N times the flux linked with a single turn)

Energy Stored in a SolenoidLet emf = e be induced emf at any instant and dq a very small charge being sent against this opposing emf.Let external work needed be dW to push this charge dq through the solenoid circuit.Then dW = e . dq = e I dt

⇒ dW = dtdIL Idt = L . I . dI

⇒ Total work 0

dII

W LI= ∫ = 2LI21

Energy Density Inside a SolenoidEnergy density is the magnetic energy contained within a unit volume of the magnetic field.Let Solenoid length = L and cross sectional area = A.

Energy stored in the solenoid is 2I L21U = and volume of solenoid = AL

⇒ 220 I ALn

21U µ⋅=

( ) ( )0

2

0

20 volume.B

21VolumenI

21

µ=⋅

µµ=

Energy density VolumeUUd =

⇒0

2

d 2BUµ

=

Self inductance of a circular coilLet a coil have N turns, each of radius r.

Then due to current i in the coil, B = N × field due to one turn = N. 0

2ir

µ= 0

2Nir

µ

2 20iN. N.BA N . r2rµ

φ = φ= = πl

20N r i

2µ π

=

E (induced emf) = 2

0

2ld N r di

dt dt− φ µ π= ⋅

Also E = L dtdi

, so by comparison, we can write

L = 2

rN20 πµ



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NARAYANA

MUTUAL INDUCTION

If two electric circuits are arranged in such a manner that the magnetic field due to the current in one circuit ispartly or wholly linked with the other circuit, the two circuits are said to be electro-magnetically coupled circuits.Then, if the current in one circuit changes, the magnetic field inside the circuit changes. Therefore, magnetic fluxlinked with the other circuit also changes. Hence the latter opposes the change of magnetic flux within itself byinducing an emf in itself. Thus change of current in one circuit causes induced emf in the other circuit. Thisphenomenon is called Mutual Induction.The circuit in which the current is made to change is called the primary circuit, whereas the circuit, whichopposes the change of current in the primary is called the secondary circuit.The induced emf in the secondary, due to the change of current in the primary is called mutually induced emf.The coefficient of mutual induction between two coupled circuits is the magnetic flux linked with the secondary,per unit current in the primary.It is also called Mutual inductance. Let

i1 = current in the primary

2φ = the magnetic flux linked with the secondary due to i1 current in the primary.Then 2 1MIφ =

2e (induced emf in the secondary) 2d dt

=− φ

⇒ e2 = M dtdI1

Similarly if current I2 in the secondary circuit changes, it causes a change in flux linked with the primary, andthereby an emf is induced in the primary. Then,

21 MI=φ

and e1 = dtd 2φ−

⇒ e1 = M dtdI2

Note : For a given alignment (i.e., electromagnetic coupling) of the pair of circuits, M is a constant.The coefficient of mutual indcution between two given elecro-magnetically coupled circuit can also be defined asthe induced emf in one circuit per unit rate of change of current in other circuit.SI (MKS) unit of the coefficients of self-induction (L) and mutual induction (M) is the same i.e., 1 henry or 1 H.

Example 11 :

Calculate the coefficient of self induction of a solenoid coil of 2000 turns, length 0.5 m and radius 5 cm, when thecore is filled with(i) air

(ii) soft iron ( =µr for soft iron = 1000)

Solution :

N = 2000 turns, l = 0.5 m, A = ( )22-105 ×π

(i) For air core ( )1r =µ

( )( ) ( )( )l A n L 20air µ=

( ) ( )( )( )0.5 105 0.5

2000 104 2-2

7 ×π

×π= −

= 1.579 H



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NARAYANA

(ii) For soft iron core

( ) 01000 µ= µ , Therefore

( )( )airiron soft L 1000L =

= 1579 H

Example 12 :

An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction tothe same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.

Solution :

Here ∆ I = I2 I1 = 2.5 (2.5) = 5.0 A

∆ t = 0.1 second

Average emf = tIL

∆∆

⇒ L = avge . t 20 volt 0.1 sec 0.4 Henry

I 5.0A∆ ×= =

∆

Example 13 :

The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of0.8 A/s. Find the emf induced in it.

Solution :

N = 240, l = 0.12 m, 2 2A r (.02)=π =π ×

2 7 2 20 N A 4 10 (240) (0.02)L

0.12lµ π × × × π×

= =

= 7.58 × 104 H

emf induced A/s 0.8 H 10 7.58 dtdi L |e| 4 ××==

= 6.06 × 104 volt

GROWTH AND DECAY OF CURRENT IN L-R CIRCUIT (TRANSIENT STAGE)

(a) Growth of current in LR circuit :

Consider a coil of a conductor having a self inductance only, a source of constantpotential difference, a pure resistor and an open key in a closed series circuit. LetV0 = the steady p.d. of the sourceR = pure resistance in the circuitL = pure self inductance in the circuit

2R L

V0

Key

When the key K is open, there is no current in the circuit. Let at time t = 0, the key be closed (i.e., switched on).The source tends to set up a current in the circuit, which the self-inductor opposes, by self inducing an oppositeemf in the circuit. The steady source maintains its p.d. and gradually overcomes the opposition of self inductance.During this interval the inductors opposition weakens, the current grows and finally attains a steady value. Letthis steady value be I0.



13


NARAYANA

Ifi = the current at time t, during the growth,iR = the p.d. across the pure resistor,

dtdiL = the self induced emf across the self inductor

Then by Kirchoffs law to the closed circuit

0dt

LdiRiV0 =−−

Separating the variables i and t we have

dtL1

RiVdi

0=

− ..... (i)

Integrating both sides, and putting initial condition i.e. i = 0 at t = 0, we get

−=

− tLR

0 e1 RVi ..... (ii)

Graphically this is shown herex

i

OO t

i0The current grows as time progresses and gradually becomes steady.Equation (ii) shows that(i) When t → ∞

00

Vi iR

→ =

(ii) RL

has the dimensions of time. RL

is called time constant of the LR circuit and denoted by τ

When t = τ = RL

then 0 01i i 1 0.632Ie

= − =

(b) Decay of Current in LR circuit

Consider an electric circuit as shown below

(1)

(2)K

R

L

V

(i) the source of emf sets up a current in the L.R. circuit, which eventually becomes steady = i0 (say) After thesteady state is set in, let at time t = 0,

(ii) the L.R. circuit is therefore, cut off from its source and becomes a closed circuit. The induced emf acrossthe self inductor spends itself in setting a current in the circuit, and in the process potential difference acrossthe inductor grade and the current in the circuit finally vanishes.At any instant t after the switch off of the source, let i = current in the circuit

Then by Kirchoffs loop theorem, 0RidtdiL =+



14


NARAYANA

Separating the variables i and t and integrating and putting the initial condition i.e., i = i0 when t = 0

We get i = τ/t0ei − where ==

RLτ time constant

i.e., the current decreases exponentially.Graphically :The current decays exponentially

Xi0

i

OO t

RL

= time constant

= time required for the current to decay to value of ( )00 i 368.0.e.iei

starting from maximum current i0.

Example 14:

A circuit has L = 0.1 H and R = 10.0 Ω . Find the value of time after which the current in this LR circuit buildsupto (a) 90% (b) 99.9% of the steady state value.

Solution :

(a) Here I = 0.9 I0. Substituting this value in current equation, we get 0.9 I0 = I0 (1 τ− /te )

⇒ 1.0e /t =τ−

⇒ τ/te = 10

⇒ /t τ = loge 10 = 2.303

⇒Lt 2.303 2.303R

= τ=

0.1H2.30310

= ×Ω

= 0.023 second

(b) Here I = 0.999 I0.

Therefore, 0.999 I0 = I0 (1 τ− /te )

⇒ t /e 0.001− τ =

⇒ τ/te = loge 103 = 6.9

⇒ RL 6.9 6.9 t =τ=

second. 0.069 second 100.1 6.9 =×=

Example 15:

A coil of resistance Ω40 is connected across a 4.0 V battery 0.10 s after the battery is connected, the current inthe coil is 63 mA. Find the inductance of the coil.

Solution :

Here E = 4.0 V, R = Ω40 , t = 0.1 sec, I = 6.31 × 102 A.

)e1(RE I /t τ−−=

6.3 × 102 = 404

− τ

− 1.0

e1



15


NARAYANA

⇒ τ− 1.0

e = 0.37

e0.1 1log

0.37 = τ

= 1.01 e

0.37 ∴ =

⇒ 0.1τ = ⇒ 0.1LR

=

⇒ L = 0.1 × 40 = 4.0 H.

Example 16 :

A R-L circuit has L = 1.0 H and R = Ω20 . It is connected across a current source having an emf of 2.0 V at t= 0. Find rate of change of current w.r.t. time at t = 100 ms.

Solution :

Here )e1(II /t0

τ−−=Differentiating both sides

( )τ−−

τ−= /t

0 etIdtdI

= t/0I t e− τ

τ

Here 0E 2 L 1I = = 0.1A, τ= = = 0.05sR 20 R 20

=

So 05.01.0

e05.0

1.01.dtdI −×0= = 0.2 (e2) = 0.027 Amp/second

Example 17 :

A coil has an inductance 50 H and a resistance 30 Ω . If it is connected across a battery of constant emf 100 V,at time t = 0, find(i) time constant of the circuit(ii) the maximum steady current in the circuit(iii) current in the circuit at t = 0.1 sec.(iv) time t. when the current is one half of the steady value(v) electromagnetic energy stored in the circuit at the steady state(vi) power dissipated in the circuit at time t = 0.1 sec.

Solution :

(i) Time constant τ ΩΩ==

3050

RL

(ii) i0 (the steady value of the current) = A333.3A30

100RV0 ==

(iii) i0.1 (current in the circuit at t = 0.1 sec) = ( )τ/t0 e1i −−

( )3 / 5 0 .10i 1 e − = − = 0.1496 A

(iv) Let t1/2 denote the time for the current to be 1/2 i0.R tL

0 01 i i 1 e2

− = −

t

LR

e-1 21 −

= ⇒ 2et

LR

=−

Taking logarithim on both sides,



16


NARAYANA

⇒ 693.0elog2logt

53

e

e == [ ]elog e 1∴ =

⇒ ( ) sec 155.10.693 35t =

=

(v) UL = Electrical energy stored in the circuit at steady state= energy stored in the inductor at steady state (as resistor does not store energy)

= 2Li21

∞ = 21

(50) 2

310

= 277.8 J

(v) P = power dissipated in the circuit at t = 0.1 sec = rate of heat generated in the resistor (as inductor does not dissipate power)

= 21.0i R = (0.1946)2 (30) = 1.136 watt

Example 18 :

A coil of inductance 20 H and of resistance 100 Ω is connected across the terminals of a source of constant emf100 volt. After the steady state is reached and at time t = 0, the source is cut off and the LR circuit is closed. Attime t = 0.1 sec, calculate(i) the current in the circuit,(ii) the power dissipated across the resistance(iii) their rate at which energy is depleted from the inductor

Solution :

(i) R = 100 Ω , L = 20 H, V0 = 100 Volt

Then

i0 (Steady current) 100 1100

= = ampere

τ =LR 100

20= = 0.2 sec

i0.1 (current at t = 0.1 sec) = τ/t0ei − = (1) (e5×0.1))

649.11

e1

5.0 == = 0.607 A

(ii) P0.1 (Power dissipated across resistance)

( ) ( ) ( )100 0.607R i 221.0 == = 36.85 Watt

(iii) The energy stored in the inductor is gradually reduced and the same amount of energy appears as heatacross the resistor. Therefore, the rate of depletion of energy in the inductor is equal to the rate of heatgeneration across the resistor i.e., the power dissipated in R.

= (1) ( )0.607 20

100

= 3.035 A/s

Therefore, rate of energy dissipated from inductance = 36.845 Watt



17


NARAYANA

ALTERNATING CURRENT

AVERAGE VALUE OF AC OVER HALF CYCLE

Average or Mean value of alternating current over half cycle is that steady current which will send the sameamount of charge through the circuit in the time interval of one half cycle, as is sent actually by the alternatingcurrent.

00

2/T

00

2/T

0av I636.0I2dttsinI

T2

2/T

IdtI =

π=ω== ∫

∫

ROOT MEAN SQUARE VALUE OF AC

Root mean Square value (rms value) or Virtual value of AC is that steady current which would produce the sameheat in given resistance in one cycle as is done by the alternating current when passed through the same resistancefor one cycle.

T2

T2 20 0

rms 00

I RdtI1I RT (I sin t) dt

T T 2= = ω =∫

∫

AC THROUGH A RESISTOR

(i) Let E (= E0 sin tω ) be the alternating emf applied across a pure resistance R and I be the current resultingfrom this emf

Then I = RtsinE

RE 0 ω= R

E0= sin tω

R

⇒ I = I0 sin tω where I0 RE0=

The variation of E and I with tω is represented graphically and in the phasor diagram as follows

:

I = I sin t0 ω

ωt

E = E sin t0 ω

(I,E)O

ππ2

E

I I0

E0

Fig. Phasor diagram

Power in a Purely Resistive Circuit

(i) P = EvIv where Ev and Iv are virtual or rms values of emf across the element (resistor) and current throughthe element.



18


NARAYANA

⇒0 0 0 0.

22 2E I E IP = =

P

t

(ii) The variation of instantaneous power with time isgraphically represented as shown in figure.

AC THROUGH AN INDUCTOR

(i) Let E = E0 sin tω be the emf applied across the pure inductor of self inductance L.Let I be the current resulting from this emf. Then

0EI sin ( t / 2)L

= ω −πω

I = I0 sin t2π ω −

where

L

00

EIL

=ω

(ii) The variation of E and I with time and the phasor diagram are shown below :

ωt

(E, I)

0

E I

2ππ

I0

E0

90º

Y

I

wt

E

Inductive Reactance (XL)

(i) XL = ωL = fL2π where f is frequency of AC. Unit of reactance is ohm.

(ii) The variation of XL with f is shown in figure

Power in a Purely Inductive Circuit

P = EvIv cos φ where φ is phase difference between E and I. Here φ = 90º so P = EvIv cos 90º = 0. Therefore ina pure inductive circuit, the average power consumed over any time interval as well as over one full cycle is zero.

AC ACROSS A CAPACITOR

(i) If E = E0 sin tω then C

0I E C sin ( t / 2)= ω ω + π

⇒ 0I I sin t2π = ω +

where 0 0

0C

E EI1/ C X

= =ω where XC = Cω is capacitative reactance.

Here current leads the applied voltage by 90º.

E=E sin t0 ω

I = I sin( t + /2)0 ω π

ωtE, I

Y

X

(iii) The phasor diagram is shown in figure.

f

XL

L = const t



19


NARAYANA

I

E E0

90oωt

x

I0

O

Power in a Purely Capacitative CircuitP = EvIv cos φ

= EvIv cos 90º = 0 because phase difference φ between E and I is 90ºSo in a pure capacitor circuit power consumed over one cycle or any part of the cycle is zero.

AC THROUGH LR CIRCUIT

Let I = I0 sin tω be the AC current resulting in a L-R circuit.then, applied emf. E = E0 sin ( tω + φ) where E0 = I0 Z and E = IZ and Z is Impedance of the circuit. Impedanceis the term used for effective opposition to current, whose dimensions are similar to resistance and reactance andwhose unit is Ω . In a series L-R circuit applied emf leads the circuit current by φ.

L R

V

VL VR

I

φ

VL

VR

V

Inductive Impedance2 2 2 2 2

LZ R (X ) R L= + = + ωThe reciprocal of Impedence is called Admittance.

RL

RX

VVtan L

R

L ω===φZR

VVcos R ==φ

Power in LR Circuit

Average power = EvIv cos φ = 222 LR

RIEω+

νν

AC THROUGH CR CIRCUIT

If E = E0 sin tω ,

R

C

Iφ

VC

VR

V



20


NARAYANA

then I = I0 sin ( tω + φ) where ZEI 0

0 = and ,ZEI= Z being the impedance

Here R

X~XRXtan CL==φ

Here current leads emf by φ

22CL

22 R)X~X(RXZ +=+=

Capacitive Impedance

22 2

1Z RC

= +ω

CR1

RC/1

VVtan

R

C

ω=ω==φ

AC THROUGH LCR CIRCUIT

If I = I0 sin tω ,

then E = E0 ( tω + φ)

2 20 0 L CE I R (X X )= + −

Z = 2 20

L C0

E R (X X )I

= + −

tan φ = L CX XR−

Power in LCR Circuit

Average power over complete cycle,

0 0 0a

E I IEP E I cos cos 2 2 2ν ν ν= φ = φ = ×

a 22

RAlso, P E I1R L

C

ν ν ν= + ω − ω

22

C1LR

ω−ω+

φR

L ω 1 C ω

2

av 22

E RAlso, P

1R LC

ν= + ω − ω

Average power is also known as true power. The quantity EvIv is called the apparent power or virtualpower. It is customary to express true power in kW and apparent power in kVA.cos φ is called the power factor of LCR circuit. Its value varies from zero to 1. For purely resistive circuit

φcos = 1 and for pure inductive or pure capacitative or inductive cum capacitative circuit, φcos = 0.Power factor is defined as the ratio of true power to apparent power.

LR C

φ

VL

VC

I0

XVR

E0



21


NARAYANA

Power factor 2

2

R Rcos =Z 1R + Lω-

Cω

= φ=

SUMMARISED RESULTS FOR VARIOUS AC CIRCUITS

S. AC circuit Current for E=E0sin tω equivalent opposition Power Power factor =φ)(cosNo. containing to current

1. R Only I = I0 sin tω R RI2ν 1

2. L Only I = I0 sin ( tω -90o) XL = Lω Zero 0

3. C Only I = I0 sin ( tω +90o) C1XC ω

= Zero 0

4. R & L I = I0 sin ( tω - φ) 2L

2C XRZ += φνν cosIE 2 2

L

R

R X+

5. R & C I = I0 sin ( tω + φ) 2 C

2C XRZ += φνν cosIE

2 2C

RR X+

6. LCR I = I0 sin ( tω ± φ) ( )2CL2 XXRZ −+= φνν cosIE ( )2CL

2 XXR

R

−+

Wattless Current or Idle Current

If the voltage and current differ in phase by π/2, the power factor is cos 90o = 0. Although current flow throughthe circuit, but it carries no power. Such a current is called wattless current. The energy metre cannot registerenergy charges in such cases although an ammeter will show such current.

CHOKE COIL

A choke coil is an inductance coil having a large number of turns of thick insulated copper wire. It has practicallyzero resistance. It plays the same role in AC circuit as does a resistance in DC circuit.When alternating current is passed in a choke coil, a self induced emf is set up in the coil by virtue of self-induction.

R

By Lenzs law, this induced emf opposes the flow of current and thus works as a resistance in AC circuit [Lenzslaw].Instead of choke coil, we may use resistance to control alternating current. But then there will be a lot of wastageof electrical power on account of Joule heating. On other hand, the power consumed by choke coil is nearly zero.Thus choke coil is preferred over resistance to control alternating current.The chokes are also extensively used in power supplies, battery eliminators, radios, TV sets etc. Radio frequencychokes i.e. high frequency chokes have air core while audio frequency chokes i.e. low-frequency chokes have aniron core.



22


NARAYANA

TRANSFORMER

A transforer is a device, for stepping up or down an alternating voltage. Its works on the principle of mutualinduction. It consists of a primary coil wound over a soft iron cylindrical core, with the secondary coil wound overother part of the same soft iron core and insulated from each other as well as from the soft iron core. The soft ironcore has a permeability µ ,nearly 103 times the permeability of free space. The soft iron core helps in increasingeffiiciency of the transformer.

In an ideal transformer, the magnetic field B due to the curernt in the primary circuit is completley linked with thesecondary. Then

B = the common magnetic induction in the two coils.

A = area of either coils ( as They are wound on the same core)

N1 = number of turns in the primary

N2 = number of terms in the secondary

i1 = current in the primary

i2 = current in the secondary.

then

1φ = (Linked magnetic flux in the primary) = N1AB

2φ = (Linked magnetic flux in the secondary) = N2AB

When the curent i1 varies, there is self induced emf in secondary which is e2

then

1 1

2 2

e Ne N

= ratio of emf

i.e. Ratio of emf = Ratio of number of turns

Again in an ideal transformer, power is transfered from primary to the secondary without any loss. Then

e1i1 = e2i2

∴1 2 1

2 1 2

e i Ne i N

= =

If e2 > e1, it is a set up transformer i.e., voltage is increased. then Ns > Np and is < ip



23


NARAYANA

Example 1 :

An alternating current varies as per equation I = 6 sin 30 t Amp. What is the rms value of this current shown by(i) an ac ammeter (ii) an ordinary moving coil galvanometer.

Solution :

(i) The equation is similar to I = I0 sin tω where I0 represents peak value of current. Then rms value = 2/I0 .Here I0 = 6 Amp.

Hence rms current = .Amp24.42/62/I0 ==

So the a.c. ammeter reads 4.24 Ampere.(ii) A moving coil galvanometer can measure only dc current, because the direction of coils rotation depends

upon the direction of current. As net current or average current is zero, so the galvanometer shows zerovalue for an a.c. current flow.

Example 2 :

An alternating emf is represented by

E = 100 sin ( )4/t120 π+π volt. Calculate

(i) average or mean value of emf(ii) RMS value of emf(iii) frequency of alternating emf(iv) the shortest time interval after start at which emf is zero.

Solution :

(i)π

= 0av

E2E = 0.636 E0 = 0.636 × 100 volt = 63.65 volt

(ii) 0rms

E 100E2 2

= = = 70.7 volt

(iii) On comparing with E = E0 sin )t( φ+ω , we get 120 ,ω = π

⇒ 2 120 60Hzf fπ = π ⇒ =

(iv) Substituting zero for instantaneous value of emf

0 = )4/t120(sin001 π+π

⇒ sin 120 t 04π π + =

For value of t to be positive, π=π+π4

t120

⇒ π×π=

1201

43t sec = 6.25 × 103 sec.

SOLVED EXAMPLES



24


NARAYANA

Example 3 :

A coil of inductance 1.0 H and resistance 100 Ω is connected to an ac supply designated as 230 V, 50 Hz.Calculate(a) maximum current in the coil(b) TIme lag between the peak voltage and the peak current.

Solution :(a) Inductive reactance

Ωπ=××π=π= 1000.1502fL2XL

0 rmsE 2E 2 230 volt 325 volt= = × =Impedance

2 2 2 2LZ X R (100 ) (100) 330= + = π + = Ω

A39.1330

V460Z

EI 00 =

Ω==

(b) As the circuit is inductive, the voltage leads the current by angle φ

tan π=π==φ100

100R

XL

⇒ φ = 72.34º.

Then time lag3

o1 72.34ºt T. 4 10 sec

2 50 360−φ ∆ = = = × π

Example 4 :

The electric current in a circuit is given by

τ= tii 0 for a period t = 0 to t = τ . Calculate the rms value of the

current for this period.Solution :

2 2rms

0(I ) I dt

ττ=∫

⇒2 2 2 3 2

2 20 0 0rms 0 2 2

0 0

i i it(I ) i dt t dt33

τ τ τ τ τ = = = =∫ ∫ τ τ τ

⇒2 2

2 0 0rms

i i(I )3 3

τ= =τ

⇒0

rmsiI3

=

Example 5 :When 100 volt dc is applied across an inductor, a current of 2.0 A flows through it. If 100 V 50 Hz ac is appliedacross the same inductor, the current reduces to 1.25 A. Calculate(i) resistance of the inductor(ii) inductance of the inductor.

Solution:(i) It is clear that the inductor has both resistance & inductance because it offers unequal resistance to dc and

ac emfs.In case of dc voltage

E E 100I = = 0 R = = 50R I 2.0

⇒ = Ω

(ii) In case of ac, ZEI=



25


NARAYANA

E 100Z 80I 1.25

⇒ = = = Ω 2 2X R 80⇒ + = , where R = 50.

2 2X 80 50 62.45⇒ = − = Ω45.62fL2 =π⇒

H2.0502

45.62L =×π

=⇒

Example 6:A pure inductor and a pure resistor are connected in series and an ac supply is connected across this combination.Ideal ac volt meters V1 and V2 show 120 volt and 160 volt respectively. Calculate

L R

~

V1 V2

(i) phase difference between V1 and V2.(ii) applied voltage V and its phase differences w.r.t. current

Solution:(i) As current is common through both the elements, so in the inductor, voltage (V1) is ahead of current I while,

in the resistor, voltage (V2) is in same phase as current. Hence phase difference between V1 and V2 is 90º.(ii) Here applied emf 2 2

1 2V V V= +

= 22 )160()120( +

φ

V

I

V1

V2

= 200 volt1

2

V 120 3tanV 160 4

φ = = =

⇒ φ= 36.9º. Therefore applied voltage leads the circuitcurrent by 36.9º.

Example 7:An inductance of 2.0 H, a capacitance of 18 Fµ and a resistance of 10 Ωk are connected to an ac source of 20 Vwith adjustable frequency.(i) At what frequency, will the current be maximum in the circuit ?(ii) What is this maximum current?

Solution:

(i) Current will be maximum at resonant frequency fr.

rr 6

1 1f 26.5Hz2 2 LC 2 2 18 10−

ω= = = =π π π × ×

(ii) At resonance R = Z.

At resonant frequency, max maxmax

E E 20 2I 2.83mAZ R 10000

= = = =Ω

Example 8:

A Ω50 electric iron is connected to an ac supply of 200 V, 50 Hz. Calculate (i) average power delivered to iron(ii) peak power and (iii) energy spent in one minute.



26


NARAYANA

Solution:

(i) Average power is based on rms value of emf.

Average power = 2 2

V(E ) 200= =800wattR 50

(ii) Peak power = watt160050

)2200(RE 22

0 ==

(iii) Energy spent is based on rms power.So energy spent = P × t = 800 watt × 60 sec = 48000 Joule.

Example 9 :

The coefficient of mutal inductance betwen the primary and secondary of a transformer is 5 H. Calculate theinduced emf in the secondary when 3 A current in the primary is cut off to zero in 25 × 106 sec.

Solution :

e2 = (induced emf in the secondary) = 1diMdt ( ) 5

6

35 6 10 V25 10−

= − = − × ×

Example 10 :

A power transformer is used to step up an alternating emf of 200 V to 4 KV and to transmit 5 KW power.If the primary is of 1000 turns, calculate, assuming(i) the number of turns in the secondary and the transformer being ideal(ii) the current rating of the secondary

Solution :

(i) For an ideal transformer 2 2

1 1

e Ne N

=

1000N

2004000 s= ⇒ turns102N 4

s−×=

(ii) P = (power to be transferred) = e1i1

⇒3

11

P 5 10ie 200

×= = = 25 A

then3

2 14

1 2

i N 10 1i N 2 10 20

= = =× ∴ ( )2 1

1 25i i 1.25A20 30

= = = or

Alternatively, 32 2e i 5 10 W= ×

∴3

2 3

5 10i 1.25A4 10×= =×



27


NARAYANA

EXERCISES

Section A : (i) True and False Statements

(ii) Fill in the Blanks

(iii) Assertion-Reason Type Questions

Section B : (i) Multiple Choice Questions

(ii) MCQ’s asked in Competitive Examinations

Section C : Subjective Questions



28


NARAYANA

1. If the direction of current passing through a d.c. motor be reversed, the direction of motors rotation is alsoreversed.

2. Electromagnetic induction is analogous to mechanical inertia.3. An observer is facing magnetic north. An electron in front of him flies horizontally towards the north and deflects

towards his left. From there observations, we can in infer that the observer is in southern hemisphere4. A coil of metal will is stationary in a non-uniform magnetic field. And emf is induced in the coil5. Induced emf tries to keep the energy conserved.6. The greater the value of L, the faster is the decay.7. Alternating current does cause electrolysis.8. A transformer cant work on d.c. supply9. 1/WCR is dimensionless quantity.10. The potential difference and current an alternating circuit are represented by E=100 sin (100t), I=100 sin(100t+π).

The electrical power dissipated in the circuit is 2.5 watt.

FILL IN THE BLANKS1. A.C generator is based on the principle of _________2. The quantity which is increased in a step down transformer is ________.3. The average emf induced in a coil in which the current changes from 2A to 4A in 0.05 sec. is 8V. The self

inductance of the coil is ________ .4. A coil of copper wire is being pulled with a constant velocity in a magnetic field. If its ohmic resistance is

increased, it will be ______ to pull it.5. If a coil is removed from a magnet field (a) rapidly (b) slowly. work done will be more when the coil removed

from the magnetic field.6. The time required by the current in an inductor of 0.1 µ and whose resistance is 20 Ω which is connected to a

12 V battery to reach 63% of its steady maximum value is about _______ sec.7. The device that controls the alternating a.c without much loss of energy is called as ________8. When a fluorescent tube is used in a.c. circuit, a choke is connected in_____9. If an a.c. of 50Hz is flowing through a conducting wire the current becomes zero____ times in one second.10. A uniformly wound solenoid coil of self inductance 41.8 10−× henry and resistance 6Ω is broken up into two

identical coils. These identical coils are then connected in parallel across a 12 volt battery of negligible resistance.The time constant for the current in the circuit is ________ sec and the steady state current through the batteryis ________ amp.

EXERCISESSECTION - A

TRUE AND FALSE STATEMENTS



29


NARAYANA

(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.(b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion.(c) If Assertion is true but the Reason is false.(d) If both Assertion and Reason are false.

1. Assertion : Self-inductance is called the inertia of electricity.Reason : Self inductance is the phenomenon, according to which an opposing induced emf is produced

in a coil as a result of change in current or magnetic flux linked in the coil.2. Assertion : An aircraft flies along the meridian, the potential at the ends of its wings will be the same.

Reason : whenever there is change in the magnetic flux emf induces.3. Assertion : An inductor cannot have zero resistance.

Reason : This because inductor has to be made up of some material, which must have some resistance.4. Assertion : When a magnetic is made to fall freely through a closed coil, its acceleration is always less

than acceleration due to gravity.Reason : Current induced in the coil opposes the motion of the magnet, as per Lenzs law.

5. Assertion : A step up transformer can also be used as a step down transformer

Reason : This is because s s

p p

E nE n

= .

6. Assertion : A glowing bulb becomes dim when an iron bar is put in the inductor in the a.c circuit.Reason : Resistance of the circuit increases.

7. Assertion : A spark occurs sometime when an electric iron is switched off.Reason : Sparking is due to large self induced emf in the circuit, during make

8. Assertion : Use is made of eddy current in induction brakes.Reason : As eddy currents always oppose the relative motion.

9. Assertion : The presence of large magnetic flux through a coil maintains a current in the coil if thecircuit is continuous.

Reason : Only a change in magnetic flux will maintain an induced current in the coil.10. Assertion : In an anti-resonance circuit or rejector circuit, alternating voltage and current have a phase

difference of 180°.Reason : In a resonance circuit, alternating voltage and current are in the same phase

ASSERTION AND REASON TYPE QUESTIONS



30


NARAYANA

1. When a magnet is moved with its north polaritytowards a coil placed in a closed circuit, then thenearer face of the coil

(1) shows south polarity

(2) shows north polarity

(3) shows sometimes south polarity

(4) shows sometimes north and some times southpolarity

2. A horizontal straight conductor when placed alongsouth-north direction falls under gravity, there is

(1) an induced current from south to north direction

(2) an induced current from north to south direction

(3) no induced e.m.f. along the length of theconductor

(4) an induced e.m.f. along the length of theconductor

3. When two inductors L1 and L2 are connected inparallel, the equivalent inductance is

(1) L1 + L2

(2) between L1 and L2

(3) less than both L1 and L2

(4) None of the above

4. An inductor may store energy in its

(1) electric field

(2) coils

(3) magnetic field

(4) electric and magnetic fields

5. A coil has 500 turns and flux of 10-4 weber ispassing through it. Calculate e.m.f producedwhen the flux falls to 52 10−× webers in 0.1second is

(1) 0.04 volt (2) 0.4 volt

(3) 4 volt (4) 40 volt

6. Lenz law is a consequence of the law ofconservation of

(1) Charge (2) Mass

(3) Energy (4) Momentum

7. The current passing through a choke coil of 5henry is decreasing at the rate of 2 amp/sec. TheEMF developed across the coil is

(1) 10 volt (2) - 10 volt

(3) 2.5 volt (4) - 2.5 volt

8. If L and R represent inductance and resistancerespectively, then the dimensions of

LR will be

(1) 0 0 1M L T − (2) 0M L T

(3) 0 0M L T(4) Cant be represented in terms of M, L and T

9. When the current changes from 2A to 4A in 0.05second, an e.m.f. of 8 volt is induced in a coil. Thecoefficient of self induction of the coil is

(1) 0.1 H (2) 0.2 H

(3) 0.4 H (4) 0.8 H

10. The time constant of a C-R circuit is

(1)1

CR (2)CR

(3) CR (4)RC

11. A coil is wound on a frame of rectangular crosssection. If all the linear dimensions of the frameare increased by a factor of 2 and the number ofturns per unit length of the coil remains the same,the self inductance increases by a factor of

(1) 4 (2) 8

(3) 16 (4) 32

12. A current carrying coil is subjected to a uniformmagnetic field. The coil will orient so that its planebecomes

SECTION - BMULTIPLE CHOICE QUESTIONS



31


NARAYANA

(1) inclined at 45o to magnetic field

(2) inclined at any arbitrary angle to the magneticfield

(3) parallel to the magnetic field

(4) perpendicular to the magnetic field

13. L,C,R represent the physical quantities, inductance,capacitance and resistance respectively. Thecombination which have the dimensions offrequency are

(1)1

RC (2)RLC

(3)R

LC (4)CL

14. Faradays law of electromagnetic induction is

(1) related to law of conservation of charge

(2) related to law of conservation of energy

(3) related to third law of Newton

(4) related to law of conservation of angularmomentum

15. As shown in the figure, a magnet is moved with afast speed towards a coil at rest. Due to this inducedelectromotive force, induced current and inducedcharge in the coil are E, I and Q respectively. If thespeed of magnet is doubled, the incorrect statementis

S N

GGalvanometer

(1) E increases

(2) I increases

(3) Q remains the same

(4) Q increases

16. In general in an alternating current circuit

(1) the average value of current is zero

(2) the average value of square of the current iszero

(3) average power dissipation is zero

(4) the phase difference between voltage andcurrent is zero

17. A transformer is employed to

(1) convert A.C. to D.C.

(2) convert D.C. to A.C.

(3) obtain a suitable A.C. voltage

(4) obtain a suitable D.C. voltage

18. If the instantaneous current in a circuit is given by

( )2cosi tω φ= − amperes, the r.m.s value of thecurrent is

(1) 2 ampere (2) 2 ampere

(3) 2 2 ampere (4) Zero ampere

19. The potential difference V across and the currentflowing through an instrument in an A.C. circuitare given by

V 5cos tω= volt

I 2sin tω=

The power dissipated in the instrument is

(1) Zero watt (2) 5 watt

(3) 10 watt (4) 2.5 watt

20. If a current I given by 0 sin2

I t πω − flows in

an A.C. circuit across which an A.C. potential of

0 sinE E tω= has been applied, then the powerconsumption P in the circuit will be

(1)0 0

2E IP = (2) 0 0

2E IP =

(3) 2EIP = (4) P = Zero



32


NARAYANA

21. Using an A.C. voltmeter the potential differencein the electrical line in a house is 234 volt. If theline frequency is known to be 50 cycles persecond, the equation for the line voltage is

(1) ( )165sin 100V tπ=

(2) ( )331sin 100V tπ=

(3) ( )220sin 100V tπ=

(4) ( )440sin 100V tπ=

22. In an A.C. circuit, the instantaneous values ofe.m.f. and current are 200sin 314e = t volt and

sin 3143

i t π = + ampere. The average power

consumed in watt is

(1) 200 (2) 100

(3) 50 (4) 25

23. In an LCR circuit having L = 8.0 henry,0.5C Fµ= and R = 100 ohm in series, the

resonant frequency (in per second) is

(1) 600 radian (2) 600 hertz

(3) 500 radian (4) 500 hertz

24. In a circuit containing an inductance of zeroresistance, the current lags behind the appliedA.C. voltage by a phase angle of

(1) 90o (2) 45o

(3) 30o (4) 0o

25. The power factor in a circuit connected to anA.C. power supply has a value which is

(1) unity when the circuit contains an idealinductance only

(2) unity when the circuit contains ideal resistanceonly

(3) zero when the circuit contains an ideal resistanceonly

(4) unity when the circuit contains an idealcapacitance only

26. A power transformer with an 8 : 1 turns ratio has60 Hz, 120 V across primary, the load in secondaryis 104 ohm. The current in the secondary is

(1) 96 A (2) 0.96 A

(3) 9.6 A (4) 96 mA

27. In a region of uniform magnetic induction210−=B tesla, a circular coil of radius 30 cm and

resistance 2π ohm is rotated about an axis whichis perpendicular to the direction of B and whichforms a diameter of the coil. If the coil rotates at200 r.p.m. the amplitude of the alternating currentinduced in the coil is

(1) 24 mAπ (2) 30mA

(3) 6 mA (4) 200 mA

28. In an L-R circuit, the A.C. source has voltage 220volt. The potential difference across theinductance is 176 volt. The potential differenceacross the resistance will be

(1) (220 176− ) volt (2) ( )220 176+ volt

(3) 220 176× volt (4) 2 2220 176− volt

29. In an L-C-R series A.C. circuit, the current is

(1) always in phase with voltage

(2) always lags the generator voltage

(3) always leads the generator voltage

(4) none of the above

30. The angular frequency of a resonant LCcombination ( )10 mH and 1.0µFL C= = is

(1) 41.0 10× radian per sec

(2) 41.0 10× Hertz

(3) 41.0 10−× radian per sec

(4) 41.0 10−× Hertz

31. A step-up transformer(1) converts low voltage, low amperage current into

high voltage and high amperage current(2) converts high amperage to low amperage

current(3) converts high voltage to low voltage current(4) generates electricity



33


NARAYANA

32. A resistor, an inductor and a capacitor are connectedin series to an a.c. power supply. When measuredwith the help of an a.c. voltmeter, the voltage acrossthem are found to be 80 V, 30 V and 90 Vrespectively. What is the supply voltage?

(1) 200 V (2) 100 V

(3) 140 V (4) 200 2 V

33. A choke coil is used to control

(1) alternating current

(2) direct current

(3) both a.c. and d.c.

(4) neither a.c. nor d.c.

34. A 2000 ohm resistor and a 1 Fµ capacitor areconnected in series across a 120 volt rms, 60Hertz line. The impedance of the circuit is

(1) 3322 ohm (2) 2001 ohm

(3) 2233 ohm (4) 2322 ohm

35. The instantaneous values of current and voltage inan A.C. circuit are 4sinI tω= and

100cos3

E t πω = + respectively, the phase

difference between voltage and current is

(1) 3π

(2)23π

(3)56π

(4)76π

36. A wire coil carries the current i. The potentialenergy of the coil does not depend upon

(1) the value of i

(2) the number of turns in the coil

(3) whether the coil has an iron core or not

(4) the resistance of the coil

37. The inductive reactance XL due to a pure inductorin a D.C. circuit is

(1) L (2)1L

(3) 0 (4) ∞

38. Choke used to limit high frequency A.C. has(1) air core(2) iron core(3) a paramagnetic core(4) a diamagnetic core

39. In a step up transformer, the turn ratio is 1 : 20. ALeclanche cell (emf 1.5 V) has been connectedacross the primary. The voltage across thesecondary is

(1) 30 V (2) 0.75 V

(3) 75 V (4) Zero

40. A power transformer is used to step up an alternatingemf of 220 V to 11 kV to transmit 4.4 kW of power.If the primary coil has 1000 turns, what is the currentrating of the secondary? Assume 100% efficiencyfor the transformer

(1) 4 A (2) 0.4 A

(3) 0.04 A (4) 0.2 A

41. An alternating current circuit consists of aninductance and a resistance in series. In this circuit

(1) the potential difference across and current inresistance leads the potential difference acrossinductance

(2) the potential difference across and current inresistance lags behind the potential difference

across inductance by an angle 2π

(3) the potential difference across and current inresistance lags behind the potential differenceacross inductance by an angle π

(4) the potential difference across resistance lagsbehind the potential difference across

inductance by an angle 2π

, while the current

in resistance leads the potential difference

across inductance by an angle 2π

42. In the circuit shown here, the voltages across L andC are

~



34


NARAYANA

(1) in phase(2) out of phase by 90o

(3) out of phase by 180o

(4) bear a phase difference which depends on thevalues of L and C

43. If φ is phase difference between current andvoltage, the wattles component of current is

(1) cosvI φ (2) sinvI φ

(3) tanvI φ (4) 2cosvI φ

44. The core used in transformers and otherelectromagnetic devices is laminated(1) to increase the magnetic field(2) to increase the level of magnetic saturation of

the core(3) to reduce the magnetism in the core(4) to reduce eddy current losses in the core

45. An inductive circuit has zero resistance. When ACvoltage is applied across this circuit, then the currentlags behind the applied ac voltage by an angle(1) 30o (2) 45o

(3) 90o (4) 0o

46. Which one of the following curves representsvariation of capacitive reactance with frequency

(1)

t

xC

(2)

t

xC

(3)

t

xC

(4)

t

xC

47. In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should bethe change in inductance, so that circuit remains inresonance?

(1) 4 times (2)14

times

(3) 8 times (4) 2 times

48. In an AC circuit, the potential difference acrossinductance and resistance joined in series are 16V and 20 V respectively. The net potentialdifference across the circuit is

(1) 20 V (2) 25.6 V

(3) 31.9 V (4) 36 V

49. What quantity is increased in step-downtransformer

(1) current (2) voltage

(3) power (4) frequency

50. A normal domestic electric supply is an alternatingcurrent whose average value is

(1) zero

(2) half the peak value

(3) peak value multiplied by 2π

(4) peak value divided by 2π



35


NARAYANA

1. The coefficient of mutual induction of two coils is 6mH. If the current flowing in one coil is 2 ampere,then the induced emf in the second coil will be(1) 12 mV (2) 3 mV(3) 3 V (4) Zero

[Pb.CET 1996]2. Whenever magnetic lines of force move across a

conductor, there is produced an electric field. This isreferred to as [DPMT 1997](1) Lenzs law (2) Faradays law(3) Ohms law (4) Gausss law

3. A coil of area 5 cm2 and having 20 turns is placed ina uniform magnetic field of 103 gauss. The normalto the plane of the coil makes an angle of 60o withthe magnetic field. The flux in maxwell through thecoil is [CPMT 1998](1) 105 (2) 5 × 104

(3) 2 × 104 (4) 5 × 103

4. A coil having an area A0 is placed in the magneticfield which changes from B0 to 4 B0 in time interval t.The emf induced in the coil will be [MNR 2002](1) 3 A0 B0/t (2) 4 A0B0/t(3) 3 B0/A0t (4) 4 B0/A0t

5. An emf of 100 millivolt is induced in a coil when thecurrent in another nearby coil becomes 10 amperefrom zero in 0.1 second. The coefficient of mutualinduction between the two coils will be(1) 1 millihenry (2) 10 millihenry(3) 100 millihenry (4) 1000 millihenry

[MP PET 1996]6. Which of the following statements is correct ?

(1) A circular ring is rotated about its own axis in auniform magnetic field. No emf is induced in it.

(2) A circular ring is rotated about its own axis in anon-uniform magnetic field. An emf is inducedin it

(3) A circular ring is rotated about one of its diameterin a uniform magnetic field and induction of themagnetic field is increasing at a constant rate.Emf, induced in the ring may be equal to zerofor an elemental time interval

(4) None of these [CMC LDH 2002]

7. Which of the following statements, about a solenoid,is correct ? [Manipal 2002](1) Self inductance per unit length of solenoid is

greater near its ends than at centre(2) Energy density in a solenoid is greater near its

ends than that at the centre(3) Self inductance per unit length and energy

density, both are uniform over entire length ofthe solenoid

(4) None of these8. A coil of wire of a certain radius has 600 turns and

a self inductance of 108 mH. The self inductanceof another similar coil of 500 turns will be(1) 74 mH (2) 75 mH(3) 76 mH (4) 77 mH

[MP PMT 1990]9. A coil of diameter 0.40 m is in a variable magnetic

field. As the magnetic induction of the field changesby 127.4 T during 2s, an emf of 200 V is induced inthe coil. Then the number of turns in the coil isapproximately(1) 20 (2) 25(3) 30 (4) 50 [AIIMS 1999]

10. A 50 turns circular coil has a radius of 3 cm. It iskept in a magnetic field acting normal to the area ofthe coil. The magnetic field B is increased from0.10 T to 0.35 T in 2 m s. The average induced emfin the coil is [MP PET 1994](1) 1.77 V (2) 17.7 V(3) 177 V (4) 0.177 V

11. A horizontal telegraph wire 2.5 km long runningbetween east and west is a part of a closed circuitwhose resistance is 35 Ω . The wire falls to theground from a height of 10 m. If g = 9.8 m s2 andBH = 2 × 105 T, the maximum current induced inthe circuit is [AFMC 1996](1) 0.7 A (2) 0.02 A(3) 0.01 A (4) 0.04 A

12. An emf of 6 V is induced in a given coil when thecurrent in it changes at the rate of 30 A/min. The selfinductance of the coil is [All India PM/PD 1999](1) 0.2 H (2) 5 H(3) 12 H (4) 180 H

MCQ’S ASKED IN COMPETITIVE EXAMINATIONS



36


NARAYANA

13. An alternating emf of angular frequency ωis appliedacross an inductance. The instantaneous currentdeveloped in the circuit has an angular frequency

(1) 4ω

(2) 2ω

(3) ω2 (4) ω4[EAMCET 2002]

14. The magnetic field energy in an inductor changesfrom maximum value to minimum value in 5.0 mswhen connected to an AC source. The frequencyof the source is [AIIMS 2001](1) 20 Hz (2) 50 Hz(3) 200 Hz (4) 500 Hz

15. An inductive coil has a resistance of 100 Ω . Whenan AC signal of frequency 1000 Hz is fed to the coil,the applied voltage leads the current by 45o. What isthe inductance of the coil ? [EMACET 1999](1) 10 mH (2) 12 mH(3) 16 mH (4) 20 mH

16. In a LCR circuit, if the rms voltage of the AC supplyremains constant when the frequency is variedbetween 50 Hz and 50 kHz, which of the graphsbelow best illustrates the variation of current throughthe ammeter with frequency ?

[London School of Examination]

(1) I

f

(2)

(3) I

f

(4) I

f

17. The reactance of a coil when used in the domesticAC power supply (220 volt 50 cycles per second) is50 ohm. The inductance of the coil is nearly

[MP PMT 2000](1) 2.2 H (2) 0.22 H(3) 1.6 H (4) 0.16 H

18. The primary and secondary coils of a transformerhave 50 and 1500 turns respectively. If the magneticflux φ linked with the primary coil is given by

0 4tφ= φ + , where φ is in webers, t is time in

seconds and 0φ is a constant, the output voltageacross the secondary coil is

[CBSE-PMT 2007](1) 30 volts (2) 90 volts(3) 120 volts (4) 220 volts

19. What is the value of inductance, L for which thecurrent is a maximum in a series LCR circuit with C= 10µF and ω = 1000 s1 ?

[CBSE-PMT 2007](1) 10 mH(2) 100 mH(3) l mH(4) cannot be calculated unless R is known

20. A transformer is used to light a 100 W and 110 Vlamp from a 220 V mains. If the main current is 0.5amp, the efficiency of the transformer isapproximately

[CBSE-PMT 2007](1) 10 % (2) 30 %(3) 50 % (4) 90 %



37


NARAYANA

1. A magnetic flux of 8 × 104 weber is linked with each turn of a 200-turn coil and there is a current of 4 A in thecoil. Calculate the self-inductance (L) of the coil.

2. At what frequency would the reactance of 10.0 µF capacitor equal that of a 1.0 mH inductor ?

3. Two inductors 1L and 2L sufficient distance apart are connected (i) in series (ii) in parallel. What is theirequivalent inductance?

4. A closed loop of wire is being moved so that it remains in a uniform magnetic field. What is the current induced?

5. What is power dissipation in an a.c. circuit in which voltage and current are given by V 300sin t2π = ω +

and

I 5sin t= ω ?6. At very high frequency of a.c., a capacitor behaves as a pure conductor. Why?7. A conducting loop of face-area A and resistance R is placed perpendicular to a magnetic field B. The loop is

withdrawn completely from the field. Find the charge which flows through any cross-section of the wire in theprocess. Note that the flown charge is independent of the shape of the loop as well as the way it is withdrawn.

8. A coil has radius 10 cm, resistance 40 Ω and 1000 turns. It is placed with its plane vertical and its axis parallelto the magnetic meridian. The coil is connected to a galvanometer and is rotated about its vertical diameterthrough an angle of 180º. Find the charge which flows through the galvanometer if the horizontal component ofthe earths magnetic field is BH = 3.0 × 105 T.

9. Figure shows a wire of resistance R sliding on two parallel, conducting rails placed at a separation l. The rails arejoined at one end. A magnetic field B exists in a direction perpendicular to the plane of the rails. What force isnecessary to keep the wire moving at a constant velocity v ?

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

vl

10. A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v anda uniform magnetic field of induction B exists in the direction perpendicular to the plane of the wire. Find the emfinduced between the ends of the wire in the following cases(a) the velocity is perpendicular to the diameter joining free ends(b) the velocity is parallel to this diameter.

11. A series LCR circuit with L = 10 mH, C = 2 µF, and R = 5 Ω is driven by a generator of peak emf100 V and variable angular frequency ω. Calculate

(a) the resonance frequency 0ω and(b) Irms at resonance. When ω= 8000 rad/s, find(c) XC and XL(d) Z and Irms,

(e) the phase angle φ.12. A coil having inductance 2.0 H and resistance 20 Ω is connected to a battery of emf 4.0 V. Find

(a) the current at an instant t = 0.20 s after the connection is made and(b) the magnetic field energy at this instant.

SECTION - CSUBJECTIVE QUESTIONS



38


NARAYANA

13. The mutual inductance between two coils is 2.5 H. If the current in one coil is changed at the rate of1 A/s, what will be the emf induced in the other coil ?

14. Figure shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slidon the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length.At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram,showing the induced emf as a battery. Calculate the current in the circuit.

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X X

X X

X X

vl

15. A LR circuit contains an inductor of inductance 0.5 mH, a resistor of 25.0 Ω and a cell of emf 5.00 V in series.Find the potential difference across the resistor at t =(a) 20.0 ms (b) 1.00 s

16. A conducting loop of area 5.0 cm2 is placed in a magnetic field which varies sinusoidally with time as B =B0 sintω where B0= 0.20 T and ω= 300 s1, such that the normal to the coil makes an angle of 60º with the field. Find

(a) the maximum emf induced in the coil(b) the emf induced at t= (π/900)s

(c) the emf induced at s)600/(π=t .

17. The magnetic field in a region is given by yLBkB 0!!

= where L is a fixed length. A conducting rod of length L lies

along the Y-axis between the origin and the point (0, L, 0). The rod is made to move with a velocity 0v v i.=!

Findthe emf induced between the two ends of the rod.

18. A system containing a pair of rails and a wire is kept vertically in a uniform horizontal magnetic field B that isperpendicular to the plane of the rails. It is found that the wire stays in equilibrium. If the wire ab is replaced byanother wire of double its mass, how long will it take in falling through a distance equal to its length ? Note thata constant current I is being sent through the wire in each case......

.....

.....

.....

I

ab

19. A rectangular metallic loop of length l and width b is placed in the plane of a long wire carrying a current i. Theloop is moved perpendicular to the wire with a speed v in its own plane as shown. Calculate the emf induced inthe loop at the instant shown when the rear end of the loop is at a distance a from the wire.

v

a

i bl

20. Consider the circuit shown in figure. The switch is in closed position for a long time.(a) Find the current through the battery(b) Suppose the switch is again opened at t = 0. What is the time constant of the discharging circuit ?(c) Find the current through the inductor after one time constant.

R1

R2 L

E S



39


NARAYANA

ANSWERSSECTION - A

(True and False Statements)

1. False 2. True3. True 4. False5. True 6. False7. False 8. True9. True 10. True

(Fill in the Blanks)

1. Electromagnetic induction 2. Current3. 0.2 µ 4. Easier5. Rapidly 6. 0.0057. Choke coil 8. Series with the tube9. 100 10. 3×10-5s, 8A

(Assertion-Reason Type Questions)

1. (2) 2. (2)3. (4) 4. (1)5. (1) 6. (3)7. (3) 8. (1)9. (4) 10. (4)

SECTION - B(Multiple Choice Questions)

1. (2) 2. (3) 3. (3) 4. (3) 5. (2)6. (3) 7. (1) 8. (3) 9. (2) 10. (3)

11. (2) 12. (4) 13. (1) 14. (2) 15. (4)16. (1) 17. (3) 18. (2) 19. (1) 20. (4)21. (2) 22. (3) 23. (3) 24. (1) 25. (2)26. (4) 27. (3) 28. (4) 29. (4) 30. (1)31. (2) 32. (2) 33. (1) 34. (1) 35. (3)36. (4) 37. (3) 38. (1) 39. (4) 40. (2)41. (2) 42. (3) 43. (2) 44. (4) 45. (3)46. (1) 47. (1) 48. (2) 49. (1) 50. (1)

(MCQ’s asked in Competitive Examinations)

1. (4) 2. (2) 3. (2) 4. (1) 5. (1)6. (1) 7. (4) 8. (2) 9. (2) 10. (2)

11. (2) 12. (3) 13. (3) 14. (2) 15. (1)16. (1) 17. (3) 18. (3) 19. (2) 20. (4)



40


NARAYANA

1. 0.04 H2. 1591 Hz

3. (i) 1 2L L+ (ii) 1 2

1 2

L LL L+

4. Zero5. Zero6. AT high frequency e

1X 0Wc

= $ , so VIXc

= is very high.7. BA/R8. 57 µC

9.R

vlB 22

10. (a) 2 Brv (b) zero11. (a) 7070 rad/s (b) 14.1 A (c) XC = 62.5 Ω , XL = 80 Ω

(d) Z = 8.2 Ω , Irms = 3.89 A (e) φ = 74º

12. (a) 0.17 A (b) 0.03 J13. 2.50 V14. (Blv)/(2r(l + vt)15. (a) 3.16 V (b) 4.97 V (c) 5.00 V16. (a) 15 mV (b) 7.5 mV (c) zero

17.2

00 lVB

18. )/(2 gl

19.)(2 ala

ibvl+π

µ

20. (a) 0.45 (b) 0.45 (c) 1.0 W (d) 0.64 W

SECTION - C(Subjective Questions)

Documents

Pmt Class Xii Physics EPMT_Class_Xmi & Ac