Single phase transformers Construction Faraday’s laws on Induction 1. Any change in the magnetic field of a coil of wire will cause an emf to be induced in the coil. This emf is called induced emf and if the conductor circuit is closed, the current will also circulate through the circuit and this current is called induced current 2. The magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. The flux linkage of the coil is the product of number of turns in the coil and flux associated with the coil. ๐ธ = 4.44๐๐ะค๐๐๐ฅ E= Effective Voltage Induced (V) f= Frequency(Hz) N= No of turns in the coil ะคmax = Peak Value of the flux (Wb) Question 01 A coil has 3000 turns and linked with an ac flux having a peak value of 2mWb. If the frequency is 50Hz, Calculate the effective value and the frequency of the induced voltage Transformer types • Power transformers • Distribution transformers • Substation transformers • Generator transformers • Instrument transformers • Voltage/ Potential transformers (PT) • Current transformers (CT) • Auto transformers • Isolating transformers Ideal Transformers • Magnetic circuit (iron path) has infinite permeability • No magnetic leakage • Iron losses of the core are zero • Copper losses are zero Voltage ratio According to the Faraday’s Law, Where ratios, ๐ธ1 = 4.44๐1 ะค๐๐๐ฅ ๐ ๐ธ2 = 4.44๐2 ะค๐๐๐ฅ ๐ E1 and E2 are primary and secondary voltages, N1 and N2 are primary and secondary turns ะคmax is the magnetic flux and f is the frequency Therefore, ๐ธ1 ๐1 = =๐ ๐ธ2 ๐2 a = Turns ratio Question 02 A not quite Ideal transformer having 90 turns at the primary and 2250 turns at secondary, is connected to 120V and 50Hz source. The coupling between the primary and the secondary is perfect, but the magnetizing current is 4A. Calculate, a. The effective voltage across the secondary terminals b. Peak Voltage across the secondary terminals c. The instantaneous voltage across the secondary terminal when the voltage across primary is 30V Current ratio Input power = Output power ๐ธ1 ๐ผ1 = ๐ธ2 ๐ผ2 Therefore, ๐1 ๐ผ1 = ๐2 ๐ผ2 Let I1= rms supply current I2= rms secondary current Net flux in the core is constant ๐ผ1 ๐2 1 = = ๐ผ2 ๐1 ๐ a = Turns ratio Question 03 An Ideal transformer has 90 turns on the primary and 2250 turns on the secondary and is connected to a 200V, 50Hz source. The load across the secondary draws a current of 2A at a power factor of 80 percent lagging Calculate, a. Effective value of the primary current b. Instantaneous current in the primary when the instantaneous current in the secondary is 100mA c. The peak flux linked by the secondary winding d. Draw the phasor diagram Impedance ratio ๐1 ๐ธ2 . ๐ธ1 ๐1 ๐2 ๐๐ฅ = = = ๐2 ๐ผ1 ๐ 2 ๐ผ2 . ๐1 2 ๐ธ2 . ๐ผ2 ๐๐ฅ = ๐2 ๐ Real Transformers • Resistance in the winding • Magnetization current • Losses in the core • Flux leakage from windings
