SOLVED MODEL QUESTION PAPER-1 (2024-25)

(ISSUED BY PUE BOARD)

II PUC - PHYSICS (33)
Time: 3 hours. Maximum Marks: 70
No of questions: 45
General Instructions:
1. All parts A to D are compulsory. Part-E is only for visually challenged students.
2. For Part – A questions, first written-answer will be considered for awarding marks.
3. Answers without relevant diagram / figure / circuit wherever necessary will not carry any marks.
4. Direct answers to numerical problems without relevant formula and detailed solutions will not carry
any marks.
PART – A
I. Pick the correct option among the four given options for ALL of the following
questions: [15 × 1 = 15]
1.The S.I. unit of electric charge is:
(A) coulomb metre (B) coulomb per metre (C) coulomb (D) per coulomb
Answer: (C) coulomb
2 The angle between equipotential surface and electric field is:
(A) 90º (B) 0º (C) 180º (D) 45º
Answer: (A) 90º
3 Statement-I: The resistivity of metals increases with increase in temperature.
Statement-II: Increasing the temperature of metals causes more frequent collisions of electrons.
(A) both I and II are true and II is the correct explanation of I.
(B) both I and II are true but II is not the correct explanation of I.
(C) I is true but II is false.
(D) both I and II are false.
Answer: (A) both I and II are true and II is the correct explanation of I.
4 A moving coil galvanometer can be converted into a voltmeter by connecting:
(A) a low resistance in parallel with galvanometer.
(B) a low resistance in series with galvanometer.
(C) a high resistance in parallel with galvanometer.
(D) a high resistance in series with galvanometer.
Answer: (D) a high resistance in series with galvanometer
5 When a bar magnet is suspended freely, it points in the direction of:
(A) east-west (B) north-south
(C) northeast-southeast (D) northwest-southwest
Answer: (B) north-south
6 The energy stored in an inductor of inductance L in establishing the current I in it is
1 (B) LI 2
1
(A) LI (C) LI (D) LI 2
2 2
1
Answer: (D) LI 2
2
7 The direction of current induced in the loop ‘abc’ shown in the figure is:
(A) along ‘abc’ if I is increasing
(B) along ‘abc’ if I is decreasing
(C) along ‘acb’ if I is increasing
(D) along ‘acb’ if I is constant
Answer: (B) and (C)

1
8 An ideal step-up transformer decreases _.
(A) current (B) voltage (C) power (D) frequency
Answer: (A) current
9 The displacement current is due to:
(A) flow of electrons (B) flow of protons
(C) changing electric field (D) changing magnetic field
Answer: (C) changing electric field
10 An object of finite height is placed in front of a concave mirror within its focus. It forms
(A) a real enlarged image (B) a real diminished image
(C) a virtual enlarged image (D) a virtual diminished image
Answer: (C) a virtual enlarged image
11A beam of unpolarised light of intensity I0 is passed through a pair of polaroids with their pass-axes
inclined at an angle of θ. The intensity of emergent light is equal to:
Answer: ( D)

I0 I0
(A) I 0 cos 2  (B) I 0 cos  (C) cos (D) cos2 
2 2
I0
Answer: (D) cos2 
2
12 Emission of electrons from a metal surface by heating it is called:
(A) photoelectric emission (B) thermionic emission
(C) field emission (D) secondary emission
Answer: (B) thermionic emission
13 When alpha particles are passed through a thin gold foil, most of them go undeviated because:
(A) most of the region in an atom is empty space
(B) alpha particles are positively charged particles
(C) alpha particles are heavier particles
(D) alpha particles move with high energy
Answer: (A) most of the region in an atom is empty space
14 Nuclei with same atomic number are called:
(A) isotopes (B) isobars (C) isomers (D) isotones
Answer: (A) isotopes
15 The column-I is the list of materials and the column-II, the list of energy band gaps Eg.
Identify the correct match.

2
 (A) (i) - (a), (ii) - (b), (iii) - (c) (B) (i) - (b), (ii) - (a), (iii) - (c)
(C) (i) - (c), (ii) - (a), (iii) - (b) (D) (i) - (b), (ii) - (c), (iii) - (a)
Answer: (D) (i) –(b), (ii)-(c), (iii)- (a )

II. Fill in the blanks by choosing appropriate answer given in the bracket for ALL
the following questions: [5 × 1 = 5]
(photon, polar, zero, infinite, phase, phasor)
16. A molecule possessing permanent dipole moment is called molecule.
Answer: polar
17. The net magnetic flux through any closed surface is .
Answer: zero
18. A rotating vector used to represent alternating quantities is called .
Answer: phasor
19. A wavefront is a surface of constant .
Answer: phase
20. In interaction with matter, light behaves as if it is made up of packet of energy called .
Answer: photon

PART – B
III. Answer any FIVE of the following questions: [5 × 2 = 10]
21. State and explain Gauss’ law in electrostatics.
Answer:
Gauss’ law in magnetism: It states that the net magnetic flux through any closed surface is always zero.
According to Gauss’ theorem in magnetism, the magnetic flux  B is given by:

B   B  d S  0
S

Where:
 
 B =magnetic flux , B = uniform magnetic field, dS = small area vector of Gaussian surface with a unit
vector n̂ along the normal drawn to the plane of the element and  = angle made by the normal to the
element with the direction of the magnetic field.

22. Define drift velocity and mobility of free electrons in conductors.

Answer:
Drift velocity (Vd) :The average velocity with which the free electrons in a conductor are drifted in a
direction opposite to the direction of electric field is called drift velocity.

Mobility of free electron: The mobility (  ) is defined as the magnitude of drift velocity (vd) per unit
electric field ( E).

vd

E

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23. A long air-core solenoid of 1000 turns per unit length carries a current of 2 A. Calculate the
magnetic field at the mid-point on its axis.
Answer:
Given: n  1000 , I  2 A and  0  4  10 7 Tm  12 .56Tm
To calculate: B
Solution:
Magnetic field at the mid-point of a current carrying solenoid is:
B   0 nI
B  12.56 107 1000  2
B  2.5 103 T
24. Give the principle of AC generator. Why is a current induced in an AC generator called
alternating current?
Answer: AC generator works on the principle of electromagnetic induction.
As the induced current changes its magnitude and direction after regular time intervals, so it is called
alternating current.
25. Write any two uses of ultraviolet radiations.
Answer: Highly focused UV-rays are used: in eye surgery and to kill germs in water purifiers.
26. Name the objective used in
a) refracting type telescope and
b) reflecting type telescope.
Answer:
a) Convex lens of large aperture & large focal length is used as objective in refracting type telescope.
b) Concave mirror of large aperture & large focal length is used as objective in reflecting type telescope.
27. Write the two conditions for the total internal reflection to occur.
Answer:
Conditions for total internal reflection:
I.A ray of light must travel from denser to rarer medium.
ii.The angle of incidence must be greater than critical angle.
28. Name the majority and the minority charge carriers in n-type semiconductor.
Answer: The free electrons are majority and holes are minority charge carriers in n-type semiconductor.

PART – C

IV. Answer any FIVE of the following questions: [5 × 3 = 15]

29. Write any three properties of electric field lines.
Answer:
i. The electric field lines start from a positive charge and end at a negative charge and do not form
closed loops
ii. Two electric field lines never intersect each other.
iii. Electric field lines do not pass through a conductor.

4
30. Obtain the expression for the effective capacitance of two capacitors connected in parallel.
Answer: Consider two capacitors connected in parallel.

Let :
C1 and C2 = capacitances of two capacitors connected in parallel.
V = potential difference across each capacitor
q1 and q2 = charges on the two capacitors .
Total charge of the combination is:
q = q1 + q2 ……………….(1)
But:q1 = C1V, and q2 = C2V
Equation (1) becomes:
q = C1 V + C2 V
q = V(C1 +C2) ……………(2)
If Cp is the equivalent capacitance of the combination, then
q
Cp =
V
q = Cp V
Equation (2) leads to :
Cp V = V ( C1 + C2 )
Cp = C1 + C2 ……………………..(3)
Thus, equivalent capacitance of the parallel combination of two capacitors is equal to the sum of the
individual capacitances.
31. What is Lorentz force? Write its expression and explain the terms.
Answer: The force experienced by a charged particle moving in space where both electric and magnetic
field exist, is called Lorentz force.

Expression for Lorentz force:


 



 
 

F  q v  B  qE  q v  B  E

Where: F= Lorentz force, q=charge on the particle, v=velocity of the particle, B=magnetic field and
E=electric field.

32. Write any three differences between diamagnetic and paramagnetic materials.
Answer:
Diamagnetic materials Paramagnetic materials
i. Diamagnetic materials are weakly i.The paramagnetic materials are weakly
magnetized in a direction to opposite to the magnetized in the direction of the applied
direction of the applied magnetic field. magnetic field.

ii.Relative Permeability is less than 1. ii.Relative Permeability is more than 1.

iii.Susceptibility has small negative value. iii.The susceptibility of a ferromagnetic
materials has a very high positive value

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33. Describe an experiment to demonstrate the phenomenon of electromagnetic induction using a bar
magnet and a coil.
Answer: Coil and magnet experiment:

When the magnet NS is moved towards or away from the coil C1, the magnetic flux linked with the coil
changes. As a result, emf is induced and current flows through the coil which causes a momentary
deflection in the galvanometer G. If the magnet is kept stationary and the coil is moved, similar results
are obtained. It is also observed that faster the magnet, greater is the deflection. When the coil and
magnet are kept stationary (no relative motion between the coil and the magnet), then no deflection is
observed.
34. Give any three results of experimental study of photoelectric effect.
Answer: (Any three of the following)
i..As soon as the light of particular frequency falls on the metal it emits photoelectrons without any
delay. Therefore, the process of photoelectric emission is instantaneous.
ii.For a given photometal , there is a particular frequency below which no photoelectric emission
occurs. This minimum frequency is called threshold frequency of the metal  o . 
iii.If the frequency of incident radiation is greater than the threshold frequency of metal then electrons
are emitted even if the intensity of incident radiation is low.
iv.For a given frequency of radiation greater than the threshold frequency of a metal, the photoelectric
current is directly proportional to the intensity of incident radiation.
v.The kinetic energy of emitted photoelectrons is independent of the intensity of radiation.
35. Write the three postulates of Bohr’s atom model.
Answer: Bohr’s postulates:
i.First postulate: It states that an electron in an atom could revolve in certain stable orbits without the
emission of radiant energy.
ii.Second postulate: It states that the electron revolves around the nucleus only in those orbits for
h
which the angular momentum is some integral multiple of . Where: h is Planck’s constant.
2
Therefore, the angular momentum L of the electron of mass m revolving with a speed v in a stationary
orbit of radius r is quantized.
 h 
L  mvr  n  Where: n=positive integer=1,2,3……
 2 
This postulate is known as quantization of angular momentum of an electron.
iii.Third postulate: It states that an electron might make the transition from one of its specified non-
radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to
the energy difference between the initial and final states.
The frequency of the emitted photon (   due to the transition of an electron from the state of higher
energy Efto that of lower energy Eiis given by:
h  E 2  E1  E f  Ei
Where: E1=Ei= energy of the initial state and E2=Ef=energy of the final state.
This postulate is called Bohr’s frequency condition.
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36. Find the energy equivalent of one atomic mass unit, first in joule and then in
MeV. Given: 1𝑢 = 1.6605 × 10–27 𝑘𝑔, 𝑒 = 1.602 × 10–19 𝐶 and 𝑐 = 2.9979 × 108 𝑚 𝑠 –1 .
Answer: The mass( m), energy ( E) and speed of light ( c) are related by the relation
E  mc 2
Substituting, m  1u  1.6605 1027 kg and c  2.9979 108 m / s in the above equation:
E  mc 2

E  1.6605  10  27  2.9979  10 8 2

E  1.49 1010 J
This is the energy equivalent to 1u in J.

Energy in MeV :
1.49 10 10
E
1.6 10 19
E  0.931109 eV 1eV  1.6 1019 J
E  931106 eV
E  931MeV
Therefore, the energy equivalent to 1u is 931MeV.

PART – D
V. Answer any THREE of the following questions: [3 × 5 = 15]
37. Derive the expression for the electric field at a point on the axis of an electric dipole.
Answer: Consider an electric dipole of moment p having charges -q and +q separated by a distance 2a.
Let P be a point on the axis of the dipole at which electric field E is to be calculated and r be the distance
of P from the centre O the dipole on the side of +q.

Electric field at P due to +q is :

1 q
E1  ...................................(1) It is along BP(+x direction).
4 0 r  a 2
Electric field at P due to -q is :
1 q
E2  ...................................(2) It is along PA( -x direction).
4 0 r  a 2
Since E1>E2, resultant electric field at P is: E  E1  E2
1 q 1 q
 
4 0 r  a  4 0 r  a 2
2

q  1 1 
   
4 0  r  a  r  a 2 
2

q  r  a   r  a  
2 2
  
4 0  r  a 2 r  a 2 


  
q  r 2  a 2  2ra  r 2  a 2  2ra  
 
4 0  r  a 2 r  a 2 

7
 q  4ra 
  
 
4 0  r 2  a 2 2 

1  2(q 2a)r 
    q2a  p  dipole.moment
 
4 0  r 2  a 2 2 

1  2 pr 
E   ……….................…(3)
 
4 0  r 2  a 2 2 
This gives the expression for electric field at a point on the axial line of an electric dipole.
The vector form of equation(3) is:
 1 2 pr 
E= p …….................…(4)

4 0 r 2  a 2 2 

Where: p = unit vector in the direction of dipole moment p (from negative charge to the positive charge).

38. Two cells of different emfs and different internal resistances are connected in series. Derive the
expression for effective emf and effective internal resistance of the combination.
Answer: Consider two cells of emfs  1 &  2 with internal resistances r1 and r2in series connection.

Let:
I = current flowing through the combination.
 eq =equivalent emf of the series combination.
req = equivalent internal resistance of the series combination.
VA, VB and VC = potentials at A , B and C respectively.
Potential difference between A and C is:
VAC = VA- VC =  1 - Ir1------------------------(1)
Potential difference between C and B is:
VCB = VC- VB =  2 - Ir2------------------------(2)
Potential difference between A and B is:
VAB = VA- VB
=(VA- VC)+(VC- VB)
=  1 - Ir1 +  2 - Ir2
=  1   2 - Ir1 - Ir2
VAB =  1   2 –I(r1 +r2)-------------------------(3)
If the series combination of two cells is replaced by a single cell of emf  eq and internal resistance req,
then the potential difference between A and B is:
VAB =  eq –Ireq---------------------------------(4)
Comparing equations (3) and (4):
 eq   1   2 ------------------------------------(5)
 The equivalent emf of a series combination of two cells is equal to the sum of the emfs of the
two cells.
req  r1  r2 ------------------------------------(6)
 The equivalent internal resistance a series combination of n cells is equal to the sum of the
internal resistances of the individual cells.
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39. Derive the expression for the magnetic field at a point on the axis of a circular current loop.
Answer: Consider a circular coil of radius R carrying a current I. Let P be a point on the axis of the coil
at a distance x from the centre of the coil at which magnetic field is to be calculated. To calculate the
magnetic field at P, consider two small current elements each of length dl located diametrically opposite
to each other at C and D. Let r be the distance between the point and the element.

According to Biot- Savart’s law, magnetic field at P due to the element dl at C is:
 Idl sin 
dB  0 . -------------------------------------------(1)
4 r2
As the line joining the point and the element is perpendicular to dB,   90 .
0

Equation (1) becomes:

 Idl
dB  0 2 ---------------------------------------------------(2)
4 r
It acts long PM.
Similarly , magnetic field at P due to the element dl at D is:
 Idl
dB  0 2 -----------------------------------------------------(3)
4 r
It acts along PN.
The magnetic field dB is resolved into two components: dBcos  and dBsin  . The components dBcos
 along PY and PY1 are equal and opposite. Hence, they cancel each other. The components dBsin 
along PX add up. Therefore, net magnetic field at P due to the elements at C and D is = 2dBSin .
Resultant magnetic field at P due to one turn of the coil is :
B   2dBSin
   2I
  0  2  dlSin
 4  r
   2I R Circumference R
  0  2  .R    dl   R & sin  
 4  r r 2 r
   2 .I .R 2
B 0  -------------------------------------(3)
 4  r
3

From the figure :

r 2  x2  R2

r  x2  R2  1
2


r3  x2  R2 2  3

Equation (3) leads to the form:

   2 .IR 2
B 0  -----------------------------------(4)….It acts along PX.
 4  2
3
(x  R )
2 2

This gives the expression for magnetic field at an axial point due to a current carrying circular
coil.
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40. a) Two coherent waves of a constant phase difference undergo interference. Obtain the expression
for the resultant displacement. (3)
b) Write the conditions for constructive and destructive interference in terms of phase difference.
(2)
Answer:
(a) Answer: Consider two light waves of same frequency travelling in a medium in the same direction.
The displacements y1 and y2 of the two waves of each of amplitude ao at any instant t having a phase
difference  and angular frequency  are given by :
y1 = aosin  t …………………………………….....(1)
and
y2 = aosin (  t +  ) ……..…………………………....(2)
Applying the principle of superposition, the resultant displacement is:
y = y1 + y2
= aosin  t + aosin (  t +)
= aosin  t + ao[sin  t cos + cos  tsin]
= aosin  t + aocos. sin  t + aosincos  t
y =ao(1 + cos) sin  t + (aosin)cos  t ……………(3)
Let :
A cos  = ao( 1 + cos) …………….....…....…….(4)
A sin  = aosin .…...…………………................(5)
Where : A = amplitude of the resultant wave and  = phase difference between the resultant wave & the
first wave.
Equation (1) becomes: y = A cos  sin  t + Asin  cos  t
= A [sin  t cos  + cos  t sin  ]
y=A sin (  t +  )…………………………………..(6)
This equation represents that resultant wave is also simple harmonic wave of same frequency.
b)
Condition for constructive interference in terms of phase difference: For constructive interference,
the phase difference between the two waves should be an even multiple of  .   2n
Where n = 0, 1, 2, 3 …….
Condition for destructive interference in terms of phase difference:
For destructive interference, the phase difference between the two waves should be an odd multiple of 
  (2n  1) Where: n =1, 2, 3 …………
41. What is a rectifier? Explain the working of a full-wave rectifier using a neat circuit diagram.
Draw its input-output waveforms.
Answer: A device which ac voltage into dc voltage is called rectifier.
Working of p-n junction diode as a full wave rectifier:

T D1

S
RL
AC P D
C
Output

D2

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 The AC voltage to be converted into dc is applied across the primary P. Between the points C and D, a load
resistance is placed. The point C on the secondary is called centre tapped point. The diodes D1 and D2 are
connected as shown in the circuit diagram. During positive half of ac, D 1 becomes forward biased and D2
gets reverse biased. Therefore, positive half of ac is converted into dc by D 1 .During negative half of ac, D2
becomes forward biased and D1 becomes reverse biased. Therefore, negative half of ac is converted into dc by
D2.Thus, there is a full wave rectification.

VI. Answer any TWO of the following questions: [2 × 5 = 10]

42. a) Calculate the potential at point P due to a charge of 400nC located 9 cm away.
b) Obtain the work done in moving a charge of 2nC from infinity to the point P. Does the
answer depend on the path along which the charge is moved?
Answer:
Given:
1
q  400nC  400 10 9 C , q0  2nC  2 10 9 C , r  9cm  9 10 2 m and  9 109 Nm 2 C 2
4 0
To Calculate: a) V and b) W.
Solution:
a) Potential at point P due to a charge q located at a distance r is:
1 q
V
4 0 r
400 10 9
V  9 109 
9 10 2
V  4104 V
b) Work done in moving a charge q0 from infinity to the point P is:
W  q0 V
W  2 109  4 104
W  8 105 J
No, the work done in moving a charge from infinity to a point P is independent of the path followed by
the charge.

43. In the following network, find the current I3.

Answer:
Given: I  1A, P  10  , Q 10 , R  20 , S  5 and G  5 .
To calculate: I3

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 Solution:

Kirchhoff’s loop rule to ABDA:

I 1 P  I 3G  I 2 R  0
I1 10  I 3  5  1  I1  20  0  I1  I 2  1  I 2  1  I 1
10I1  5I 3  20  20I1  0
30 I1  5 I 3  20 ………………………………(1)
Multiplying both sides of this equation by 4:
120 I1  20 I 3  80 ……………………………(2)

Kirchhoff’s loop rule to BCDB:

I1  I 3 Q  I 2  I 3 S  I 3G  0
I1  I 3 10  I 2  I 3  5  I 3  5  0
I1  I 3 10  1  I1  I 3  5  I 3  5  0  I 2  1  I1
I1  I 3 10  1  I1  I 3  5  I 3  5  0
10I1  10I 3  5  5I1  5I 3  5I 3  0
15I1  20I 3  5 ……………………………….(3)

Adding equations (2) and (3):

135I1  85
85
I1 
135
I1  0.63A

Substituting this value of I1 in equation (1):

30  0.63  5I 3  20
30  0.63  5I 3  20
18.9  5I 3  20
5I 3  20  18.9
5I 3  1.1
1 .1
I3 
5
I 3  0.22 A

44. An AC source of frequency 50Hz is connected in series with an inductor of 1H, a capacitor of
90µF and a resistor of 100Ω. Does the current leads or lags the voltage? Calculate the phase
difference between the current and the voltage.
Answer: Answer:
Given: f  50Hz , L  1H , C  90 F  90 106 F and R  100.
12
 To calculate: 
Solution:
Inductive reactance is:
X L  2 f L
X L  2 3.14  50 1
X L  314
Capacitive reactance is:
1
XC 
2 f C
1
XC 
2  3.14  50  90 10 6
1
XC 
2  3.14  50  9 10 5
10 5
XC 
2826
100000
XC 
2826
X C  35.4 
XL  XC
tan 
R
314  35.4
tan 
100
278.6
tan  
100
tan  2.786
  tan 1 2.786
  70012
The current lags behind the voltage by:   70012 , as capacitive reactance is less than inductive
reactance.
45. An equilateral prism is made of glass of unknown refractive index. A parallel beam of light is
incident on a face of the prism. The angle of minimum deviation is 40º. Find the refractive index of
the material of the prism. If the prism is placed in water of refractive index 1.33, find the new
angle of minimum deviation of a parallel beam of light.
Answer:
Given: A  60 0 , D  40 0 and n w  1.33 .
To calculate: n g and Dnew .
Solution:
i) Calculation of refractive index of material of prism:
 A D
sin  
ng   2 
A
sin
2
 60 0  40 0 
sin  
ng   
2
0
60
sin
2
13
 100 0
sin
ng  2
sin 30 0
sin 50 0
ng 
sin 30 0
0.766
ng 
0.500

ng  1.532

ii) Calculation of new angle of minimum deviation: When the prism of RI ng is immersed in water of
RI nw, then:
 A  Dnew 
sin  
ng
  2 
nw A
sin
2
 60 0  Dnew 
sin  
1.532
  2 
0
1.33 60
sin
2
 60  Dnew 
0
sin  
1.152   2 
0
sin 30
 60 0  Dnew 
sin  
1.152   2 
0.5
 60 0  Dnew 
sin    0.5  1.152

 2 
 60 0  Dnew 
sin    0.576

 2 
 60  Dnew 
0
   sin 1 0.576
 
 2 
 60 0  Dnew 
   35 010 //
 
 2 
60  Dnew
0
 35010 /
2
60  Dnew  70 0 20 /
0

Dnew  70 0 20 /  60 0

Dnew  10 0 20 /

14
 PART – E
(For Visually Challenged Students only)
7) A circular conducting loop is placed in the plane of the paper to the right of a long straight
conductor carrying current I in the upward direction. The direction of current induced in the
loop is:
(A) clockwise, if I is increasing (B) clockwise, if I is decreasing
(C) anticlockwise, if I is increasing (D) anticlockwise if I is constant
Answer: Both (B) and (C)
21) Name the shape of a Gaussian surface suitable to find electric flux due to a point charge. What
is the net electric flux through a closed surface enclosing two equal and opposite charges?
Answer: Spherical shape and Zero.
43) In a Wheatstone bridge, AB = 10Ω , BC = 10Ω , CD = 5Ω and DA = 20Ω are connected in cyclic
order. A galvanometer of 2Ω is connected between B and D. A current of 1A enters at A and
leaves the network at C. Find the current through the galvanometer.
Answer:
Given: I  1A, P  10  , Q 10 , R  20 , S  5 and G  2 .
To calculate: I3
Solution:

Kirchhoff’s loop rule to ABDA:

I 1 P  I 3G  I 2 R  0
I1 10  I 3  2  1  I1  20  0  I1  I 2  1  I 2  1  I 1
10I1  2I 3  20  20I1  0
30 I 1  2 I 3  20
15I1  I 3  10 ……………………………….…(1)
Kirchhoff’s loop rule to BCDB:
I1  I 3 Q  I 2  I 3 S  I 3G  0
I1  I 3 10  I 2  I 3  5  I 3  2  0
I1  I 3 10  1  I1  I 3  5  I 3  2  0  I 2  1  I1
I1  I 3 10  1  I1  I 3  5  I 3  2  0
10I1  10I 3  5  5I1  5I 3  2I 3  0
15I1  17 I 3  5 ……………………………….(2)
Subtracting equation (2) from equation (1):
18I 3  5
5
I3 
18
I 3  0.28 A
This is the current through the galvanometer.
15