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SUBMITTED TO: MR. FAISAL SHAREEF
SUBMITTED BY: CH.MURLIDHAR
( XIITH ‘A’)
1. Input and output voltage.
2. Number of turnings in the
secondary coil and primary coil of a
self made transformer.
CH.Murlidhar
This is to certifythat CH MURLIDHAR of class 12th A has
satisfactorilycompletedhis physicsInvestigatory Project as
prescribedby the CBSE courseduringthe academic year
2014-2015
I furthercertifiesthat thisis ownworkand not submitted anywhere
else.
SIGNATURE (TEACHER) SIGNATURE (EXAMINER)
SIGNATURE (PRINCIPAL)
CH.Murlidhar
I hereby acknowledge my
deep sense of gratitude
and indebtedness to my
principal Mr. D. ASHOK
and my physics teacher
Mr. FAISAL SHAREEF
PGT (physics) whose
immense help, genius
guidance encouragement,
necessary suggestions
initiations, enthusiasm
and inspiration made this
work a master art and a
joint enterprise.
CH.Murlidhar
CH.Murlidhar
 1. Input and output voltage.
 2. Number of turnings in the
secondary coil and primary coil of a
self made transformer.
CH.Murlidhar
Transformer is a device which increase or decrease the
voltage. It is based on the Principle of mutual induction.
According to this principle, the amount of magnetic flux
linked with a coil changing, an e.m.f is induced in the
neighboring coil.
A transformer is an electrical device which is used for
changing the A.C. voltages. A transformer is most widely
used device in both low and high current circuit. As such
transformers are built in an amazing strength of sizes. In
electronic, measurement and control circuits, transformer
size may be so small that it weight only a few tens of grams
where as in high voltage power circuits, it may weight
hundred of tones.
In a transformer, the electrical energy transfer from one
circuit to another circuit takes place without the use of
moving parts.
A transformer which increases the voltages is called a
step-up transformer. A transformer which decreases the
A.C. voltages is called a step-down transformer. an
essential piece of apparatus both for high and low current
circuits.
CH.Murlidhar
 When an altering e.m.f. is supplied to the primary coil
p1p2, an alternating current starts falling in it. The
altering current in the primary produces a changing
magnetic flux, which induces altering voltage in the
primary as well as in the secondary. In a good-
transformer, whole of the magnetic flux linked with
primary is also linked with the secondary, and then the
induced e.m.f. induced in each turn of the secondary is
equal to that induced in each turn of the primary. Thus if
Ep and Es be the instantaneous values of the e.m.f.’s
induced in the primary and the secondary and Np and
Ns are the no. of turns of the primary secondary coils of
the transformer and
dф / dt = rate of change of flux in each,
turnoff the coil at this instant
we have,
Ep = -Np dф/dt _______________ (1) and
CH.Murlidhar
Ep = -Np dф/dt ……………………………..
(1)
and,
Es = -Ns dф/dt
……………………………….. (2)
Since the above relations are true at every
instant, so by dividing 2 by 1,
we get,
Es / Ep = - Ns / Np
………………………………… (3)
As Ep is the instantaneous value of back
e.m.f induced in the primary coil p1, so the
instananeous current in primary coil is due to
the difference (E – Ep ) in the instantaneous
values of the applied and back e.m.f. further if
Rp is the resistance o, p1p2 coil, then the
instantaneous current Ip in the primary coil is
given by.
Ip = E – Ep / Rp
E – Ep = Ip Rp
CH.Murlidhar
When the resistance of the primary is
small, Rp Ip can be neglected so therefore.
E – Ep = 0 or Ep = E
Thus back e.m.f = input e.m.f
Hence equation 3 can be written as
Es / Ep = Es / E = output e.m.f / input
e.m.f = Ns / Np = K
Where K is constant, called turn or
transformation ratio.
CH.Murlidhar
CH.Murlidhar
Es < E so K < 1, hence Ns < Np
If Ip = value of primary current at the same
instant.
And,
Is = value of secondary current at this
instant,
then,
Input power at the instant = Ep Ip
and,
Output power at the same instant = Es
Is
If there are no losses of power in the
transformer.
then,
Input power = output power
Or
Ep Ip = Es Is Or
Es / Ep = Ip / Is = K
CH.Murlidhar
CH.Murlidhar
Es > E so K > 1, hence Ns > Np
As, k > 1, so Ip > Is or Is < Ip
i.e. current in secondary is weaker when
secondary voltage is higher.
Hence, whatever we gain in voltage, we lose
in current in the same ratio.
Similarly it can be shown, that in a step down
transformer, whatever we lose in voltage, we
gain in current in the same ratio.
Thus a step up transformer in reality steps
down the current & a step down transformer
steps up the current.
CH.Murlidhar
Efficiency of a transformer is defined as the
ratio of output power to the input power.
i.e.
η = output power / input power = Es Is / Ep
Ip
Thus in an ideal transformer, where there is
no power losses, η = 1. But in actual practice,
there are many power losses; therefore the
efficiency of transformer is less than one.
CH.Murlidhar
 Following are the major sources of energy loss
in a transformer:-
 Copper loss:- Is the energy loss in the form
of heat in the copper coils of a transformer.
This is due to joule heating of conducting
wires.
 Iron loss :-Is the energy loss in the form of
heat in the iron core of the transformer. This is
due to formation of eddy currents in iron core.
It is minimized by taking laminated cores.
 Hysteresis loss :- Is the loss of energy
due to repeated magnetization and
demagnetization of the iron core when A.C.
is fed to it.
CH.Murlidhar
 Leakage of magnetic flux:-
Occurs inspite of best insulations. Therefore,
rate of change of magnetic flux linked with
each turn of S1S2 is less than the rate of
change of magnetic flux linked with each turn
of P1P2.
 Magneto striation:- i.e. humming
noise of a transformer.
CH.Murlidhar
CH.Murlidhar
CH.Murlidhar
Take thick iron rod and cover it with a thick paper and
wind a large number of turns of thin Cu wire on thick
paper (say 60). This constitutes primary coil of the
transformer.
Cover the primary coil with a sheet of paper and wound
relatively smaller number of turns (say 20) of thick
copper wire on it. This constitutes the secondary coil. It
is a step down transformer.
Connect p1, p2 to A.C main and measure the input
voltage and current using A.C voltmeter and ammeter
respectively.
Similarly, measure the output voltage and current
through s1and s2.
Now connect s1and s2to A.C main and again measure
voltage and current through primary and secondary coil
of step up transformer.
Repeat all steps for other self made transformers by
changing number of turns in primary and secondary coil.
CH.Murlidhar
 In voltage regulator for T.V., refrigerator,
computer, air conditioner, etc.
 A step down transformer is used for welding
purposes.
 A step down transformer is used for obtaining
large current.
 A step up transformer is used for the
production of X-Rays and NEON
advertisement.
 Transformers are used in voltage regulators
and stabilized power supplies.
 Transformers are used in the transmissions of
a.c. over long distances.
 Small transformers are used in Radio sets,
telephones, loud speakers and electric bells
etc
CH.Murlidhar
SOURCES OF ERROR
 Values of current can be changed
due to heating effect.
 Eddy current can change the
readings.
CH.Murlidhar
 1. The output voltage of the
transformer across the
secondary coil depends upon
the ratio (Ns/Np) with respect to
the input voltage.
 2. The output voltage of the
transformer across the
secondary coil depends upon
the ratio (Ns/N p) with respect to
the input voltage.
 3. There is a loss of power
between input and output coil of
a transformer.
CH.Murlidhar
PRECAUTIONS
Keep safe yourself from high
voltage.
While taking the readings of
current and voltage the A.C should
remain constant.
CH.Murlidhar
BIBLIOGRAPHY
NCERT
TEXTBOOK
PHYSICS LAB
MANNUAL
www.wikipedia.com
CH.Murlidhar

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TOPIC-To investigate the relation between the ratio of :- 1. Input and output voltage. 2. Number of turnings in the secondary coil and primary coil of a self made transformer.

  • 1. SUBMITTED TO: MR. FAISAL SHAREEF SUBMITTED BY: CH.MURLIDHAR ( XIITH ‘A’) 1. Input and output voltage. 2. Number of turnings in the secondary coil and primary coil of a self made transformer. CH.Murlidhar
  • 2. This is to certifythat CH MURLIDHAR of class 12th A has satisfactorilycompletedhis physicsInvestigatory Project as prescribedby the CBSE courseduringthe academic year 2014-2015 I furthercertifiesthat thisis ownworkand not submitted anywhere else. SIGNATURE (TEACHER) SIGNATURE (EXAMINER) SIGNATURE (PRINCIPAL) CH.Murlidhar
  • 3. I hereby acknowledge my deep sense of gratitude and indebtedness to my principal Mr. D. ASHOK and my physics teacher Mr. FAISAL SHAREEF PGT (physics) whose immense help, genius guidance encouragement, necessary suggestions initiations, enthusiasm and inspiration made this work a master art and a joint enterprise. CH.Murlidhar
  • 5.  1. Input and output voltage.  2. Number of turnings in the secondary coil and primary coil of a self made transformer. CH.Murlidhar
  • 6. Transformer is a device which increase or decrease the voltage. It is based on the Principle of mutual induction. According to this principle, the amount of magnetic flux linked with a coil changing, an e.m.f is induced in the neighboring coil. A transformer is an electrical device which is used for changing the A.C. voltages. A transformer is most widely used device in both low and high current circuit. As such transformers are built in an amazing strength of sizes. In electronic, measurement and control circuits, transformer size may be so small that it weight only a few tens of grams where as in high voltage power circuits, it may weight hundred of tones. In a transformer, the electrical energy transfer from one circuit to another circuit takes place without the use of moving parts. A transformer which increases the voltages is called a step-up transformer. A transformer which decreases the A.C. voltages is called a step-down transformer. an essential piece of apparatus both for high and low current circuits. CH.Murlidhar
  • 7.  When an altering e.m.f. is supplied to the primary coil p1p2, an alternating current starts falling in it. The altering current in the primary produces a changing magnetic flux, which induces altering voltage in the primary as well as in the secondary. In a good- transformer, whole of the magnetic flux linked with primary is also linked with the secondary, and then the induced e.m.f. induced in each turn of the secondary is equal to that induced in each turn of the primary. Thus if Ep and Es be the instantaneous values of the e.m.f.’s induced in the primary and the secondary and Np and Ns are the no. of turns of the primary secondary coils of the transformer and dф / dt = rate of change of flux in each, turnoff the coil at this instant we have, Ep = -Np dф/dt _______________ (1) and CH.Murlidhar
  • 8. Ep = -Np dф/dt …………………………….. (1) and, Es = -Ns dф/dt ……………………………….. (2) Since the above relations are true at every instant, so by dividing 2 by 1, we get, Es / Ep = - Ns / Np ………………………………… (3) As Ep is the instantaneous value of back e.m.f induced in the primary coil p1, so the instananeous current in primary coil is due to the difference (E – Ep ) in the instantaneous values of the applied and back e.m.f. further if Rp is the resistance o, p1p2 coil, then the instantaneous current Ip in the primary coil is given by. Ip = E – Ep / Rp E – Ep = Ip Rp CH.Murlidhar
  • 9. When the resistance of the primary is small, Rp Ip can be neglected so therefore. E – Ep = 0 or Ep = E Thus back e.m.f = input e.m.f Hence equation 3 can be written as Es / Ep = Es / E = output e.m.f / input e.m.f = Ns / Np = K Where K is constant, called turn or transformation ratio. CH.Murlidhar
  • 11. Es < E so K < 1, hence Ns < Np If Ip = value of primary current at the same instant. And, Is = value of secondary current at this instant, then, Input power at the instant = Ep Ip and, Output power at the same instant = Es Is If there are no losses of power in the transformer. then, Input power = output power Or Ep Ip = Es Is Or Es / Ep = Ip / Is = K CH.Murlidhar
  • 13. Es > E so K > 1, hence Ns > Np As, k > 1, so Ip > Is or Is < Ip i.e. current in secondary is weaker when secondary voltage is higher. Hence, whatever we gain in voltage, we lose in current in the same ratio. Similarly it can be shown, that in a step down transformer, whatever we lose in voltage, we gain in current in the same ratio. Thus a step up transformer in reality steps down the current & a step down transformer steps up the current. CH.Murlidhar
  • 14. Efficiency of a transformer is defined as the ratio of output power to the input power. i.e. η = output power / input power = Es Is / Ep Ip Thus in an ideal transformer, where there is no power losses, η = 1. But in actual practice, there are many power losses; therefore the efficiency of transformer is less than one. CH.Murlidhar
  • 15.  Following are the major sources of energy loss in a transformer:-  Copper loss:- Is the energy loss in the form of heat in the copper coils of a transformer. This is due to joule heating of conducting wires.  Iron loss :-Is the energy loss in the form of heat in the iron core of the transformer. This is due to formation of eddy currents in iron core. It is minimized by taking laminated cores.  Hysteresis loss :- Is the loss of energy due to repeated magnetization and demagnetization of the iron core when A.C. is fed to it. CH.Murlidhar
  • 16.  Leakage of magnetic flux:- Occurs inspite of best insulations. Therefore, rate of change of magnetic flux linked with each turn of S1S2 is less than the rate of change of magnetic flux linked with each turn of P1P2.  Magneto striation:- i.e. humming noise of a transformer. CH.Murlidhar
  • 19. Take thick iron rod and cover it with a thick paper and wind a large number of turns of thin Cu wire on thick paper (say 60). This constitutes primary coil of the transformer. Cover the primary coil with a sheet of paper and wound relatively smaller number of turns (say 20) of thick copper wire on it. This constitutes the secondary coil. It is a step down transformer. Connect p1, p2 to A.C main and measure the input voltage and current using A.C voltmeter and ammeter respectively. Similarly, measure the output voltage and current through s1and s2. Now connect s1and s2to A.C main and again measure voltage and current through primary and secondary coil of step up transformer. Repeat all steps for other self made transformers by changing number of turns in primary and secondary coil. CH.Murlidhar
  • 20.  In voltage regulator for T.V., refrigerator, computer, air conditioner, etc.  A step down transformer is used for welding purposes.  A step down transformer is used for obtaining large current.  A step up transformer is used for the production of X-Rays and NEON advertisement.  Transformers are used in voltage regulators and stabilized power supplies.  Transformers are used in the transmissions of a.c. over long distances.  Small transformers are used in Radio sets, telephones, loud speakers and electric bells etc CH.Murlidhar
  • 21. SOURCES OF ERROR  Values of current can be changed due to heating effect.  Eddy current can change the readings. CH.Murlidhar
  • 22.  1. The output voltage of the transformer across the secondary coil depends upon the ratio (Ns/Np) with respect to the input voltage.  2. The output voltage of the transformer across the secondary coil depends upon the ratio (Ns/N p) with respect to the input voltage.  3. There is a loss of power between input and output coil of a transformer. CH.Murlidhar
  • 23. PRECAUTIONS Keep safe yourself from high voltage. While taking the readings of current and voltage the A.C should remain constant. CH.Murlidhar