Question

a) Transformers work by the Principle of Electromagnetic Induction. Briefly explain this to a layman. b) i. [2 marks] [1 mark] State the two (2) types of losses in a transformer and also state where in the [2 marks] State two (2) advantages of an auto-transformer over an ordinary [2 marks] State why DC voltages do not work in transformers. 11. transformer these losses can be found. 111. transformer. c) Explain briefly why electric power is transmitted at very high voltages such as 700 kV as against very low voltages such as 240 V required for domestic use. [3 marks] d) An ideal transformer has a turns ratio of 12:1 and is supplied at 180 V when the primary current is 4 A. Calculate the secondary voltage and current. [3 marks] e) A 250 kVA, 11 000 V/400 V, 50 Hz single-phase transformer has 80 turns on the secondary. Calculate: i. the approximate values of the primary and secondary currents; [2 marks] 11. the approximate number of primary turns; [2 marks] f) A 50 kVA single-phase transformer has a turns ratio of 300/20. The primary winding is connected to a 2200 V, 50 Hz supply. Calculate: the secondary voltage on no load; i. [2 marks] the approximate values of the primary and secondary currents on full [2 marks] 11. load.

Answer 1

a) Flecho magnete indectron. FA o > lieu When mutually coupled coile ar placed to geather and it supplied by an Ac source, the flux produced by one coil interacts with nearby coil and induces the voltage in it. This emf produced in another coil is cexd for working principle of electromos nebic induchor. b (1) When oc is given there is no change in flux linkage and hence no emf 18 induced. So transformer dognot work. Corc lorry (11) Then ar too losses - copper losjes corc losses occur in core magnche medium Copper losses occur in winding material (in) Auto transfor uses single coil. This can transfer the more porexy compared to dwinding kronebermer of Same rating, copper is saved greatly.

g for the gruan power rating → current will be lower of high trandmission voltages → Lower in current magnitude reduces the ohmic losses during transmission → Due to these woltage regulation and efficiency will be impraed. het = is volts] W I = V. IL 180* 4 = 15% IL Il = 48 Amp C) S=250 KU A vilu = 11000/400 V (0) = 250x10 $ = a paid = google = ( 7= 22.7.2 prop Va I 2 = 250x10 T = 250 x 10 20x Iz = 625 Amp

IN = 2200 ) . Nil M= 300/20 (il Secondary voltage on no load = V2 = Via Mz) = 2200 * () 120 = 2200 + 1 300 TV2 = 146.67 / volts (1 5= = S 2200 --> 5= 22.727 Amp In some 5 » 340.9) Amp 146.67