NCERT Solutions Class 12th Physics by Study Innovations. 468 Pages

For
For

A neutron is removed from a nucleus. The corresponding nuclear reaction

can be written as:

It is given that:

Mass = 39.962591 u

Mass ) = 40.962278 u

Mass = 1.008665 u

The mass defect of this reaction is given as:

Δm =

∴Δm = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:

For , the neutron removal reaction can be written as:
It is given that:

Mass = 26.981541 u

Mass = 25.986895 u

The mass defect of this reaction is given as:

Hence, the energy required for neutron removal is calculated as:

Question 13.25:
A source contains two phosphorous radio nuclides (T1/2 = 14.3d) and (T1/2 =
25.3d). Initially, 10% of the decays come from . How long one must wait until 90%
do so?
Answer

Half life of , T1/2 = 14.3 days
Half life of , T’1/2 = 25.3 days

nucleus decay is 10% of the total amount of decay.
The source has initially 10% of nucleus and 90% of nucleus.
Suppose after t days, the source has 10% of nucleus and 90% of nucleus.

Initially:

Number of nucleus = N

Number of nucleus = 9 N
Finally:

Number of

Number of

For nucleus, we can write the number ratio as:

For , we can write the number ratio as:
On dividing equation (1) by equation (2), we get:

Hence, it will take about 208.5 days for 90% decay of .

Question 13.26:

Under certain circumstances, a nucleus can decay by emitting a particle more massive
than an α-particle. Consider the following decay processes:

Calculate the Q-values for these decays and determine that both are energetically allowed.
Answer

Take a emission nuclear reaction:

We know that:

Mass of m1 = 223.01850 u
Mass of m2 = 208.98107 u

Mass of , m3 = 14.00324 u
Hence, the Q-value of the reaction is given as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 208.98107 − 14.00324) c2
= (0.03419 c2) u
But 1 u = 931.5 MeV/c2

∴Q = 0.03419 × 931.5
= 31.848 MeV
Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the
reaction is energetically allowed.

Now take a emission nuclear reaction:

We know that:

Mass of m1 = 223.01850

Mass of m2 = 219.00948

Mass of , m3 = 4.00260
Q-value of this nuclear reaction is given as:
Q = (m1 − m2 − m3) c2
= (223.01850 − 219.00948 − 4.00260) C2
= (0.00642 c2) u

= 0.00642 × 931.5 = 5.98 MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is
positive, the reaction is energetically allowed.

Question 13.27:

Consider the fission of by fast neutrons. In one fission event, no neutrons are

emitted and the final end products, after the beta decay of the primary fragments, are

and . Calculate Q for this fission process. The relevant atomic and particle
masses are

m =238.05079 u

m =139.90543 u

m = 98.90594 u
Answer

In the fission of , 10 β− particles decay from the parent nucleus. The nuclear reaction
can be written as:

It is given that:

Mass of a nucleus m1 = 238.05079 u

Mass of a nucleus m2 = 139.90543 u

Mass of a nucleus , m3 = 98.90594 u

Mass of a neutron m4 = 1.008665 u
Q-value of the above equation,

Where,
m’ = Represents the corresponding atomic masses of the nuclei

= m1 − 92me
= m2 − 58me
= m3 − 44me
= m4

Hence, the Q-value of the fission process is 231.007 MeV.

Question 13.28:
Consider the D−T reaction (deuterium−tritium fusion)

Calculate the energy released in MeV in this reaction from the data:
= 2.014102 u
= 3.016049 u
(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is
the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To
what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy
required for one fusion event =average thermal kinetic energy available with the
interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
Answer

Take the D-T nuclear reaction:
It is given that:

Mass of , m1= 2.014102 u
Mass of , m2 = 3.016049 u
Mass of m3 = 4.002603 u
Mass of , m4 = 1.008665 u
Q-value of the given D-T reaction is:
Q = [m1 + m2− m3 − m4] c2
= [2.014102 + 3.016049 − 4.002603 − 1.008665] c2
= [0.018883 c2] u
But 1 u = 931.5 MeV/c2
∴Q = 0.018883 × 931.5 = 17.59 MeV

Radius of deuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m
Distance between the two nuclei at the moment when they touch each other, d = r + r = 4
× 10−15 m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy between the two nuclei is given as:

Where,
∈0 = Permittivity of free space

Hence, 5.76 × 10−14 J or of kinetic energy (KE) is needed to overcome the

Coulomb repulsion between the two nuclei.

However, it is given that:

KE
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1
T = Temperature required for triggering the reaction

Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.

Question 13.29:
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ
decays in the decay scheme shown in Fig. 13.6. You are given that
m (198Au) = 197.968233 u
m (198Hg) =197.966760 u

Answer

It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV
energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decay is given as:
E1 = 1.088 − 0 = 1.088 MeV
hν1= 1.088 × 1.6 × 10−19 × 106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiation radiated by γ1-decay

It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV
energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decay is given as:
E2 = 0.412 − 0 = 0.412 MeV
hν2= 0.412 × 1.6 × 10−19 × 106 J
Where,
ν2 = Frequency of radiation radiated by γ2-decay

It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV
energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decay is given as:
E3 = 1.088 − 0.412 = 0.676 MeV

hν3= 0.676 × 10−19 × 106 J
Where,
ν3 = Frequency of radiation radiated by γ3-decay

Mass of = 197.968233 u

Mass of = 197.966760 u

1 u = 931.5 MeV/c2

Energy of the highest level is given as:

β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
∴Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
∴Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV

Question 13.30:
Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within
Sun and b) the fission of 1.0 kg of 235U in a fission reactor.
Answer

Amount of hydrogen, m = 1 kg = 1000 g

1 mole, i.e., 1 g of hydrogen ( ) contains 6.023 × 1023 atoms.

∴1000 g of contains 6.023 × 1023 × 1000 atoms.

Within the sun, four nuclei combine and form one nucleus. In this process 26
MeV of energy is released.

Hence, the energy released from the fusion of 1 kg is:

Amount of = 1 kg = 1000 g

1 mole, i.e., 235 g of contains 6.023 × 1023 atoms.

∴1000 g of contains

It is known that the amount of energy released in the fission of one atom of is 200
MeV.

Hence, energy released from the fission of 1 kg of is:

∴

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the
energy released in the fission of 1 kg of uranium.

Question 13.31:
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten
percent of which was to be obtained from nuclear power plants. Suppose we are given
that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of
thermal energy produced in a reactor was 25%. How much amount of fissionable uranium
would our country need per year by 2020? Take the heat energy per fission of 235U to be
about 200MeV.
Answer

Amount of electric power to be generated, P = 2 × 105 MW
10% of this amount has to be obtained from nuclear power plants.

∴Amount of nuclear power,
= 2 × 104 MW
= 2 × 104 × 106 J/s
= 2 × 1010 × 60 × 60 × 24 × 365 J/y
Heat energy released per fission of a 235U nucleus, E = 200 MeV
Efficiency of a reactor = 25%
Hence, the amount of energy converted into the electrical energy per fission is calculated
as:

Number of atoms required for fission per year:

1 mole, i.e., 235 g of U235 contains 6.023 × 1023 atoms.

∴Mass of 6.023 × 1023 atoms of U235 = 235 g = 235 × 10−3 kg
∴Mass of 78840 × 1024 atoms of U235

Hence, the mass of uranium needed per year is 3.076 × 104 kg.

Question 14.1:
In an n-type silicon, which of the following statement is true:
Electrons are majority carriers and trivalent atoms are the dopants.
Electrons are minority carriers and pentavalent atoms are the dopants.
Holes are minority carriers and pentavalent atoms are the dopants.
Holes are majority carriers and trivalent atoms are the dopants.
Answer

The correct statement is (c).
In an n-type silicon, the electrons are the majority carriers, while the holes are the
minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as
phosphorus, are doped in silicon atoms.

Question 14.2:
Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Answer

The correct statement is (d).
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the
minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as
aluminium, are doped in silicon atoms.

Question 14.3:

Carbon, silicon and germanium have four valence electrons each. These are characterised
by valence and conduction bands separated by energy band gap respectively equal to
(Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true?
(Eg)Si < (Eg)Ge < (Eg)C
(Eg)C < (Eg)Ge > (Eg)Si
(Eg)C > (Eg)Si > (Eg)Ge
(Eg)C = (Eg)Si = (Eg)Ge
Answer

The correct statement is (c).
Of the three given elements, the energy band gap of carbon is the maximum and that of
germanium is the least.
The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge

Question 14.4:
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
free electrons in the n-region attract them.
they move across the junction by the potential difference.
hole concentration in p-region is more as compared to n-region.
All the above.
Answer

The correct statement is (c).

The diffusion of charge carriers across a junction takes place from the region of higher
concentration to the region of lower concentration. In this case, the p-region has greater
concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse
from the p-region to the n-region.

Question 14.5:
When a forward bias is applied to a p-n junction, it
raises the potential barrier.
reduces the majority carrier current to zero.
lowers the potential barrier.
None of the above.
Answer

The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value of potential barrier.
In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the
potential barrier across the junction gets reduced.

Question 14.6:
For transistor action, which of the following statements are correct:
Base, emitter and collector regions should have similar size and doping concentrations.
The base region must be very thin and lightly doped.
The emitter junction is forward biased and collector junction is reverse biased.
Both the emitter junction as well as the collector junction are forward biased.
Answer

The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very
thin. Also, the emitter junction must be forward-biased and collector junction should be
reverse-biased.

Question 14.7:
For a transistor amplifier, the voltage gain
remains constant for all frequencies.
is high at high and low frequencies and constant in the middle frequency range.
is low at high and low frequencies and constant at mid frequencies.
None of the above.
Answer

The correct statement is (c).
The voltage gain of a transistor amplifier is constant at mid frequency range only. It is
low at high and low frequencies.

Question 14.8:
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz.
What is the output frequency of a full-wave rectifier for the same input frequency.
Answer

Input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
∴Output frequency = 2 × 50 = 100 Hz

Question 14.9:
For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2
kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input
signal voltage and base current, if the base resistance is 1 kΩ.
Answer

Collector resistance, RC = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, RB = 1 kΩ = 1000 Ω
Input signal voltage = Vi
Base current = IB
We have the amplification relation as:

Voltage amplification

Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation:

Therefore, the base current of the amplifier is 10 μA.

Question 14.10:
Two amplifiers are connected one after the other in series (cascaded). The first amplifier
has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01
volt, calculate the output ac signal.
Answer

Voltage gain of the first amplifier, V1 = 10
Voltage gain of the second amplifier, V2 = 20
Input signal voltage, Vi = 0.01 V
Output AC signal voltage = Vo
The total voltage gain of a two-stage cascaded amplifier is given by the product of
voltage gains of both the stages, i.e.,
V = V1 × V2
= 10 × 20 = 200
We have the relation:

V0 = V × Vi
= 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.

Question 14.11:
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it
detect a wavelength of 6000 nm?
Answer

Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energy of a signal is given by the relation:

E=
Where,
h = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s

E
= 3.313 × 10−20 J

But 1.6 × 10−19 J = 1 eV
∴E = 3.313 × 10−20 J

The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV −
the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.

Question 14.12:
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 ×
1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number
of electrons and holes. Given that ni= 1.5 × 1016 m−3. Is the material n-type or p-type?
Answer

Number of silicon atoms, N = 5 × 1028 atoms/m3
Number of arsenic atoms, nAs = 5 × 1022 atoms/m3
Number of indium atoms, nIn = 5 × 1020 atoms/m3
Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3
Number of electrons, ne = 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022
Number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are
related as:
nenh = ni2

Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes
is about 4.51 × 109. Since the number of electrons is more than the number of holes, the
material is an n-type semiconductor.

Question 14.13:
In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much
smaller than electron mobility and independent of temperature. What is the ratio between
conductivity at 600K and that at 300K? Assume that the temperature dependence of
intrinsic carrier concentration niis given by

where n0 is a constant.
Answer

Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier-concentration is written as:

Where,
kB = Boltzmann constant = 8.62 × 10−5 eV/K
T = Temperature
n0 = Constant

Initial temperature, T1 = 300 K
The intrinsic carrier-concentration at this temperature can be written as:

… (1)
Final temperature, T2 = 600 K
The intrinsic carrier-concentration at this temperature can be written as:

… (2)
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between
the respective intrinsic carrier-concentrations at these temperatures.

Therefore, the ratio between the conductivities is 1.09 × 105.
Question 14.14:
In a p-n junction diode, the current I can be expressed as

where I0 is called the reverse saturation current, V is the voltage across the diode and is
positive for forward bias and negative for reverse bias, and I is the current through the
diode, kBis the Boltzmann constant (8.6×10−5 eV/K) and T is the absolute temperature. If
for a given diode I0 = 5 × 10−12 A and T = 300 K, then
What will be the forward current at a forward voltage of 0.6 V?

What will be the increase in the current if the voltage across the diode is increased to 0.7
V?
What is the dynamic resistance?
What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer

In a p-n junction diode, the expression for current is given as:

Where,
I0 = Reverse saturation current = 5 × 10−12 A
T = Absolute temperature = 300 K
kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1
V = Voltage across the diode
Forward voltage, V = 0.6 V

∴Current, I

Therefore, the forward current is about 0.0256 A.
For forward voltage, V’ = 0.7 V, we can write:

Hence, the increase in current, ΔI = I' − I
= 1.257 − 0.0256 = 1.23 A
Dynamic resistance

If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost
remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will
be infinite.
Question 14.15:
You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR
gate while the circuit (b) acts as AND gate.

Answer

A and B are the inputs and Y is the output of the given circuit. The left half of the given
figure acts as the NOR Gate, while the right half acts as the NOT Gate. This is shown in
the following figure.

Hence, the output of the NOR Gate =

This will be the input for the NOT Gate. Its output will be =A+B

∴Y = A + B

Hence, this circuit functions as an OR Gate.

A and B are the inputs and Y is the output of the given circuit. It can be observed from the
following figure that the inputs of the right half NOR Gate are the outputs of the two
NOT Gates.

Hence, the output of the given circuit can be written as:

Hence, this circuit functions as an AND Gate.
Question 14.16:
Write the truth table for a NAND gate connected as given in Fig. 14.45.

Hence identify the exact logic operation carried out by this circuit.
Answer

A acts as the two inputs of the NAND gate and Y is the output, as shown in the following
figure.

Hence, the output can be written as:
The truth table for equation (i) can be drawn as:

A
Y

01
10
This circuit functions as a NOT gate. The symbol for this logic circuit is shown as:

Question 14.17:
You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify
the logic operation carried out by the two circuits.

Answer

In both the given circuits, A and B are the inputs and Y is the output.
The output of the left NAND gate will be , as shown in the following figure.

Hence, the output of the combination of the two NAND gates is given as:

Hence, this circuit functions as an AND gate.
is the output of the upper left of the NAND gate and is the output of the lower half

of the NAND gate, as shown in the following figure.

Hence, the output of the combination of the NAND gates will be given as:

Hence, this circuit functions as an OR gate.

Question 14.18:
Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and
identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1.
Similarly work out the values of Y for other combinations of A and B. Compare with the
truth table of OR, AND, NOT gates and find the correct one.)
Answer

A and B are the inputs of the given circuit. The output of the first NOR gate is . It

can be observed from the following figure that the inputs of the second NOR gate become

the out put of the first one.

Hence, the output of the combination is given as:

The truth table for this operation is given as:

A B Y (=A + B)

00 0

01 1

10 1

11 1

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

Question 14.19:

Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only.
Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Answer

A acts as the two inputs of the NOR gate and Y is the output, as shown in the following

figure. Hence, the output of the circuit is .

The truth table for the same is given as:

A
Y

01

10

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained

in solution (a), we can infer that the outputs of the first two NOR gates are as
shown in the following figure.

are the inputs for the last NOR gate. Hence, the output for the circuit can be
written as:

The truth table for the same can be written as:

A B Y (=A⋅B)

00 0

01 0

10 0

11 1

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

Question 15.1:
Which of the following frequencies will be suitable for beyond-the-horizon
communication using sky waves?
10 kHz
10 MHz
1 GHz
1000 GHz
Answer

(b) Answer:
10 MHz
For beyond-the-horizon communication, it is necessary for the signal waves to travel a
large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size.
The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz
frequencies get reflected easily from the ionosphere. Hence, signal waves of such
frequencies are suitable for beyond-the-horizon communication.

Question 15.2:
Frequencies in the UHF range normally propagate by means of:
Ground waves.
Sky waves.
Surface waves.
Space waves.
Answer

Answer:
Space waves
Owing to its high frequency, an ultra high frequency (UHF) wave can neither travel along
the trajectory of the ground nor can it get reflected by the ionosphere. The signals having
UHF are propagated through line-of-sight communication, which is nothing but space
wave propagation.

Question 15.3:
Digital signals
Do not provide a continuous set of values,
Represent values as discrete steps,
Can utilize binary system, and
Can utilize decimal as well as binary systems.
Which of the above statements are true?

(i) and (ii) only
(ii) and (iii) only
(i), (ii) and (iii) but not (iv)
All of (i), (ii), (iii) and (iv).
Answer

(c) Answer:
A digital signal uses the binary (0 and 1) system for transferring message signals. Such a
system cannot utilise the decimal system (which corresponds to analogue signals). Digital
signals represent discontinuous values.

Question 15.4:
Is it necessary for a transmitting antenna to be at the same height as that of the receiving
antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How
much service area can it cover if the receiving antenna is at the ground level?
Answer

Line-of-sight communication means that there is no physical obstruction between the
transmitter and the receiver. In such communications it is not necessary for the
transmitting and receiving antennas to be at the same height.
Height of the given antenna, h = 81 m
Radius of earth, R = 6.4 × 106 m
For range, d = 2Rh, the service area of the antenna is given by the relation:
A = πd2
= π (2Rh)
= 3.14 × 2 × 6.4 × 106 × 81
= 3255.55 × 106 m2
= 3255.55
∼ 3256 km2

Question 15.5:
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be
the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer

Amplitude of the carrier wave, Ac = 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = Am
Using the relation for modulation index:

Question 15.6:
A modulating signal is a square wave, as shown in Fig. 15.14.

The carrier wave is given by
Sketch the amplitude modulated waveform
What is the modulation index?
Answer

It can be observed from the given modulating signal that the amplitude of the modulating
signal, Am = 1 V
It is given that the carrier wave c (t) = 2 sin (8πt)
∴Amplitude of the carrier wave, Ac = 2 V

Time period of the modulating signal Tm = 1 s
The angular frequency of the modulating signal is calculated as:

The angular frequency of the carrier signal is calculated as:
From equations (i) and (ii), we get:
The amplitude modulated waveform of the modulating signal is shown in the following
figure.

(ii)Modulation index,

Question 15.7:
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the
minimum amplitude is found to be 2 V. Determine the modulation index μ. What would
be the value of μ if the minimum amplitude is zero volt?
Answer

Maximum amplitude, Amax = 10 V
Minimum amplitude, Amin = 2 V
Modulation index μ, is given by the relation:

Question 15.8:
Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at
the receiving station, there is a facility for generating the carrier. Show that if a device is
available which can multiply two signals, then it is possible to recover the modulating
signal at the receiver station.
Answer

Let ωc and ωs be the respective frequencies of the carrier and signal waves.
Signal received at the receiving station, V = V1 cos (ωc + ωs)t
Instantaneous voltage of the carrier wave, Vin = Vc cos ωct

At the receiving station, the low-pass filter allows only high frequency signals to pass
through it. It obstructs the low frequency signal ωs. Thus, at the receiving station, one can

record the modulating signal , which is the signal frequency.

Class XII Chapter 7 – Alternating Current Physics

Question 7.1:
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer
Resistance of the resistor, R = 100 Ω
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
(a) The rms value of current in the circuit is given as:

(b) The net power consumed over a full cycle is given as:
P = VI
= 220 × 2.2 = 484 W

Question 7.2:
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer
(a) Peak voltage of the ac supply, V0 = 300 V
Rms voltage is given as:

(b) Therms value of current is given as:
I = 10 A
Now, peak current is given as:

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Class XII Chapter 7 – Alternating Current Physics

Question 7.3:
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of
the current in the circuit.
Answer
Inductance of inductor, L = 44 mH = 44 × 10−3 H
Supply voltage, V = 220 V
Frequency, ν = 50 Hz

Angular frequency, ω=

Inductive reactance, XL = ω L
Rms value of current is given as:

Hence, the rms value of current in the circuit is 15.92 A.

Question 7.4:
A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of
the current in the circuit.
Answer
Capacitance of capacitor, C = 60 µF = 60 × 10−6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω=

Capacitive reactance

Rms value of current is given as:

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Class XII Chapter 7 – Alternating Current Physics

Hence, the rms value of current is 2.49 A.

Question 7.5:
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete
cycle. Explain your answer.
Answer
In the inductive circuit,
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where,
Φ = Phase difference between V and I
For a pure inductive circuit, the phase difference between alternating voltage and current
is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can ve obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and
current is 90° i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.

Question 7.6:
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 µF and R
= 10 Ω. What is the Q-value of this circuit?
Answer

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Class XII Chapter 7 – Alternating Current Physics

Inductance, L = 2.0 H
Capacitance, C = 32 µF = 32 × 10−6 F
Resistance, R = 10 Ω
Resonant frequency is given by the relation,

Now, Q-value of the circuit is given as:

Hence, the Q-Value of this circuit is 25.

Question 7.7:
A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular
frequency of free oscillations of the circuit?
Answer
Capacitance, C = 30µF = 30×10−6F
Inductance, L = 27 mH = 27 × 10−3 H
Angular frequency is given as:

Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.

Question 7.8:
Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total
energy stored in the circuit initially? What is the total energy at later time?
Answer

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Class XII Chapter 7 – Alternating Current Physics

Capacitance of the capacitor, C = 30 µF = 30×10−6 F
Inductance of the inductor, L = 27 mH = 27 × 10−3 H
Charge on the capacitor, Q = 6 mC = 6 × 10−3 C
Total energy stored in the capacitor can be calculated by the relation,

Total energy at a later time will remain the same because energy is shared between the
capacitor and the inductor.

Question 7.9:
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µF is connected to a variable-
frequency 200 V ac supply. When the frequency of the supply equals the natural
frequency of the circuit, what is the average power transferred to the circuit in one
complete cycle?
Answer
At resonance, the frequency of the supply power equals the natural frequency of the
given LCR circuit.
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 µF = 30 × 10−6 F
AC supply voltage to the LCR circuit, V = 200 V
Impedance of the circuit is given by the relation,

At resonance,

Current in the circuit can be calculated as:

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Class XII Chapter 7 – Alternating Current Physics

Hence, the average power transferred to the circuit in one complete cycle= VI
= 200 × 10 = 2000 W.

Question 7.10:
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz
to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the
range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC
circuit should be equal to the frequency of the radiowave.]
Answer
The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz
Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz
Effective inductance of circuit L = 200 µH = 200 × 10−6 H
Capacitance of variable capacitor for ν1 is given as:

C1
Where,
ω1 = Angular frequency for capacitor C1

Capacitance of variable capacitor for ν2,

C2
Where,

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Class XII Chapter 7 – Alternating Current Physics

ω2 = Angular frequency for capacitor C2

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

Question 7.11:
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L
= 5.0 H, C = 80µF, R = 40 Ω

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating
frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show
that the potential drop across the LC combination is zero at the resonating frequency.
Answer
Inductance of the inductor, L = 5.0 H
Capacitance of the capacitor, C = 80 µH = 80 × 10−6 F
Resistance of the resistor, R = 40 Ω
Potential of the variable voltage source, V = 230 V
(a) Resonance angular frequency is given as:

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

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Class XII Chapter 7 – Alternating Current Physics

(b) Impedance of the circuit is given by the relation,

At resonance,

Amplitude of the current at the resonating frequency is given as:
Where,
V0 = Peak voltage

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the
current is 8.13 A.
(c) Rms potential drop across the inductor,
(VL)rms = I × ωRL
Where,
I = rms current

Potential drop across the capacitor,

Potential drop across the resistor,

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Class XII Chapter 7 – Alternating Current Physics

(VR)rms = IR

= × 40 = 230 V
Potential drop across the LC combination,

At resonance,
∴VLC= 0

Hence, it is proved that the potential drop across the LC combination is zero at
resonating frequency.

Question 7.12:
An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10
mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t =
0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e.,
stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the
capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as
heat?
Answer
Inductance of the inductor, L = 20 mH = 20 × 10−3 H
Capacitance of the capacitor, C = 50 µF = 50 × 10−6 F

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Class XII Chapter 7 – Alternating Current Physics

Initial charge on the capacitor, Q = 10 mC = 10 × 10−3 C
(a) Total energy stored initially in the circuit is given as:

Hence, the total energy stored in the LC circuit will be conserved because there is no
resistor connected in the circuit.
(b)Natural frequency of the circuit is given by the relation,

Natural angular frequency,

Hence, the natural frequency of the circuit is 103 rad/s.

(c) (i) For time period (T ), total charge on the capacitor at time

t,
For energy stored is electrical, we can write Q’ = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical

at time, t =
(ii) Magnetic energy is the maximum when electrical energy, Q′ is equal to 0.

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Class XII Chapter 7 – Alternating Current Physics

Hence, it can be inferred that the energy stored in the capacitor is completely magnetic

at time,
(d) Q1 = Charge on the capacitor when total energy is equally shared between the
capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy

stored in the capacitor = (maximum energy).

Hence, total energy is equally shared between the inductor and the capacity at time,

(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat
energy in the circuit. The resistance damps out the LC oscillation.

Question 7.13:
A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac
supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer

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