### rushabh project physics 1.docx

• 1. transformer Transformer is a device used for converting a low alternating voltage to a high alternating voltage or a high alternating voltage into a low alternating voltage. It is a static electrical device that transfers energy by inductive coupling between its winding circuits. Transformers range in size from a thumbnail-sized coupling transformer hidden inside a stage microphone to huge units weighing hundreds of tons used in power plant substations or to interconnect portions of the power grid. All operate on the same basic principles, although
• 2. therange of designs is wide. While new technologies have eliminated the need for transformers in some electronic circuits, transformers are still found in many electronic devices. Transformers are essential for high-voltage electric power transmission, which makes long- distance transmission economically practical. A transformer is most widely used devicein both low and high current circuit. In a transformer, the electrical energy transfer from onecircuit 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. Transformer is, therefore, an essentialpiece of apparatus both for high and low current circuits. transformer.A transformer which decreases the A.C. voltages is called a step-down transformer.
• 3. Transformer is, therefore, an essentialpiece of apparatus both for high and low current circuits.  PRINCIPLE It is based on the principle of mutual induction thatis if a varying current is set-up in a circuit then induced e.m.f. is produced in the neighboring circuit. The varying current in a circuit produce varying magnetic flux which induces e.m.f. in the neighboring circuit.
• 4. CONTRUCTION A transformer consists of a rectangular shaft iron core made of laminated sheets, well insulated from one another. Two coils p1 & p2 and s1 & s2 are wound on the same core, but are well insulated with each other. Note that the both the coils are insulated from the core, the source of alternating e.m.f is connected to p1 p2, the primary coil and a load resistance R is connected to s1 s2, the secondary coil through an open switch S. Thus, no current can be drawn through the secondary coil aslong as the switch is open For an ideal transformer, we assume that the resistance of the primary & secondary windings is negligible. Further, the energy loses due to magnetic flux andiron core is also negligible. For operation at low frequency, we may use a soft iron. The soft iron core is insulating by joining thin iron strips coated with varnish to insulate them to reduce energy losses by eddy currents. The input circuit is
• 5. calledprimary and the output circuit is called secondary. An ideal voltage step-down transformer. The secondary current arises from the action of the secondary EMF on the (not shown) load impedance.
• 6. The ideal transformer as a circuit element
• 7.  THEORY AND WORKING 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 thesecondary, 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 induced e.m.f in the primary and the secondary coil and Np and Ns are the no. of turns of the primary secondary coils of the transformer and, dфB/dt = rate of change of flux in each turn of the coil at this instant, we have Ep = -Np dфB/dt …. (1) Es = -Ns dфB/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
• 8. primary coil p1, so the instantaneouscurrent 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 of p1 p2 coil, then the instantaneous current Ip in the primary coil is given by, I =E – Ep / Rp E – Ep = Ip Rp 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.
• 9. In a step down transformer Es < E so K < 1, hence Ns < Np If Ip=value of primary current at the same instant tand Is =value of sec. current at this instant, then Input power at the instant t = Ep Ip and Output power at the same instant t = 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 In a step up transformer As, k > 1, so Ip > Is or Is < Ip i.e. current in sec. 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
• 10. transformer in reality steps down the current & astep down transformer steps up the current. Efficiency 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 are no power losses, η = 1. But in actual practice, there are many power losses; therefore, the efficiency oftransformer is less than one.
• 11. Energy losses In practice, the output energy of a transformer is always less than the input energy, because energylosses occur due to a number of reasons as explained below, Loss of Magnetic Flux: The coupling betweenthe coils is seldom perfect. So, whole of the magnetic flux produced by the primary coil is not linked up with the secondary coil. Iron Loss: In actual iron cores in spite of lamination, Eddy currents are produced. The magnitude of eddy current may, however be small. And a part of energy is lost as the heat produced in the iron core.
• 12. Copper Loss: In practice, the coils of the transformer possess resistance. So a part of theenergy is lost due to the heat produced in the resistance of the coil. Hysteresis Loss: The alternating current in thecoil tapes the iron core through complete cycle of magnetization. So Energy is lost due to hysteresis. Magneto restriction: The alternating current in the Transformer may be set its parts in to vibrations and sound may be produced. It is called humming. Thus, a part of energy may be lost due to humming. A Big Transformer USES OF TRANSFORMER
• 13. A transformer is used in almost all a.c. operations  In voltage regulator for T.V., refrigerator, computer,air conditioner etc.  In the induction furnaces.  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.