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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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137
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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143
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
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
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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
145
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
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
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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
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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
P = e i= E0 I0 sin ω t
2ωsin
t = - E0 I0 sin ω t cos ω t =
2
2sin00
ωtIE
L
Mains
CA ..
153
153
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
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
Therefore, average power dissipated in a pure capacitive ac circuit = Pav = 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
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
A16. The average power dissipated across the capacitor is Pav = T
PdtT 0
1 =
Tt
otT
IEωt2sin
2
00 dt = 0
Therefore, average power dissipated in a pure capacitive ac circuit = Pav = 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
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
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
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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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182
(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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185
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
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