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# Lecture 1 transformer

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### Lecture 1 transformer

1. 1. Single Phase Transformer
2. 2. • The efficiency of electrical power transmission has been improved by the use of higher voltages. • This is one of the main reasons that alternating current (AC) has nearly entirely replaced direct current (DC) for power transmission and distribution. • While it is true that the AC generator is better than the DC generator for producing higher voltages. • The transformer is the device or machine primarily responsible for the wide use of AC today.
3. 3. • Basically, the transformer is a device for transferring electrical energy from one circuit to another circuit without a change in frequency. • The transformer accomplishes the change in voltage without use of moving parts, and therein lies its great advantage.
4. 4. • The cost per kilowatt is comparatively low, and the efficiency is high. • As a matter of fact, the transformer is the most efficient piece of electrical machinery, and efficiencies of 98 and 99% are not at all uncommon. • Since there are no moving parts, maintenance is simpler and cheaper, and the required insulation for the extremely high voltages obtained can more easily be constructed.
5. 5. • A transformer is a static (or stationary) piece of apparatus by means of which electrical power in one circuit is transformed into electric power of the same frequency in another circuit. • It can raise (step up) or lower (step down) the voltage in a circuit but with a corresponding decrease or increase in current. • Physically, a transformer is mutual induction between two circuits linked by a common magnetic flux.
6. 6. Voltage Transformation Ratio (K) • If N2>N1 i.e. K>1, then transformer is called step-up • If N2< N1 i.e. K<1, then transformer is called step-down transformer (We will discuss about EMF in next class)
7. 7. • For Ideal Transformer, input VA= output VA V1 x I1= V2 x I2= 1/k
8. 8. Solve the problem: The maximum flux density in the core of a 250/3000 volts,50 Hz single phase transformer is 1.2 Wb/m2. If the e.m.f. per turn is 8 volt, determine- • Primary and secondary turns
9. 9. Sol: E1= N1 x e.m.f. induced/ turn • N1= 250/8= 32 • N2= 3000/8= 375