The Transformer 1 π π X X X X L X B field into page L ππ½ ππ© β π³π π π π =β =β ππ ππ Case A: If B and L and orientation, π , constant ππ© β π³π π π π =β =π ππ No voltage on terminals 2 π π X X X X L X B field into page L ππ½ ππ© β π³π π π π =β =β ππ ππ Case B: If B changes as a function of time with L and orientation, π , constant π΅ = βπ΅πππ₯ sin(ππ‘)π π π = ππ³π π©πππ ππ¨π¬(ππ) Voltage on terminals due to a changing magnetic field 3 X X X X X π π L B field into page ππ½ ππ© β π³π π π π =β =β ππ ππ L Case C: If B and orientation, π , constant with L increasing as a function of time and π π X X X X M X X ππ΄ π π = ππ ππ L π£ Voltage on terminals βLinear motorβ βRail Gunβ 4 X X X X π π X L B field into page ππ½ ππ© β π³π π π π =β =β ππ ππ L Case D: If B and L constant with orientation, π , changing as a function of time and π΅ ππ‘ π π΅ β π = π΅πππ ππ‘ π π = ππ³π ππππ ππ Voltage on terminals βRotational motorβ βGeneratorβ 5 Loop on load side βSecondaryβ Loop on source side βPrimaryβ Zg π π π΅(π‘) πβ² π Zload I(t) Transformer N1 Primary N2 core Secondary Transformer optimize coupling, perform transformation N1 Primary N2 core Secondary Core: Designed such that as much π© produced in the primary passes to the secondary Iron core Only a few field lines pass through secondary Magnetic circuit guides field lines from primary to secondary ON PRIMARY SIDE One loop Area A ο=BA Three loops Area 3A ο=3BA X Two loops Area 2A ο=2BA X N1 loops Area N1A X X ο=N1BA 8 ο1=BA Flux for each loop on primary N1 loops π π Area N1A X ο=N1BA ππ½ ππ©π¨ π π =β = π΅π β ππ ππ ON PRIMARY SIDE ππ©π¨ ππ π = π΅π β ππ Voltage produced for N1 loops Voltage produced for one loop 9 ON SECONDARY SIDE One loop Two loops Area A X οβ=BA X Three loops X Area 2A οβ=2BA N2 loops Area 3A X Area N2A οβ=3BA 10 οβ=N2BA N2 loops ο1β=BA X Flux for each loop on secondary Area N2A πβ² π 11 οβ=N2BA ππ½β² ππ©π¨ πβ² π = β = π΅π β ππ ππ ON SECONDARY SIDE ππ©π¨ ππ π = π΅π β ππ Voltage produced for N2 loops Voltage produced for one loop ππ π Area N1A Prefect flux coupling X ο=N1BA X Area N2A ππ π = π΅π ππ©π¨ β ππ ππ π ππ π = π΅π π΅π ππ π 12 οβ=N2BA ππ π = π΅π ππ©π¨ β ππ Voltage transformation 12 ππ (π) IDEAL TRANSFORMER No power loss X ππ π N1 Voltage transformation X N2 ππ π ππ π = π΅π π΅π ππ π 13 ππ (π) Power (IN) Power (OUT) π·ππ = ππ π ππ π π·πππ = ππ π ππ π ππ π ππ π = π΅π π΅π Current transformation ππ (π) IDEAL TRANSFORMER No power loss X ππ π N1 ππ π ππ π = π΅π π΅π X N2 ππ π combine 14 ππ (π) Zeq looking into transformer ππ π ππ π = π΅π π΅π πππ = π΅π π΅π π πππππ Zload ππ (π) IDEAL TRANSFORMER No power loss X ππ π N1 X N2 ππ π 15 ππ (π) ππ (π) ππ π Zeq N1 πππ = Remove transformer π΅π π΅π π πππππ Zload ο¨ A 10-kVA; 6600/220 V/V; 50 Hz transformer is rated at 2.5 V/Turn of the winding coils. Assume that the transformer is ideal. Calculate the following: A) step up transformer ratio ο‘ B) step down transformer ratio ο‘ C) total number of turns in the high voltage and low voltage coils ο‘ D) Primary current as a step up transformer ο‘ E) Secondary current as a step down transformer. ο‘ SOLUTION PROVIDED IN CLASS 16 N1 βPrimaryβ N2 βSecondaryβ 2β¦ π π πβ² π 32Ξ© I(t) ο¨ Find N1/N2 ratio such that maximum power transfer to the load is observed. SOLUTION PROVIDED IN CLASS πππ = π΅π π΅π π πππππ 17 Work in phasor domain Zs π π Zload I(t) In general: Then π§π = π π + πππ π π πΌ= = π ππ + πππππ π§ππππ = π ππππ + ππππππ π πΌ= π π + π ππππ + π ππ + πππππ 2 Time average power to load π = πΌ π ππππ 2 Find maximum 18 π π= π π + π ππππ 2 2 + ππ + πππππ 2 Step 1: Make ( ) as small as possible with respect to the complex βreactanceβ part π= π 2 π π + π ππππ 2 π ππππ 2 2 Find maximum π ππππ 2 2 πππππ = βππ Find maximum 19 π π= 2 π π + π ππππ 2 π ππππ 2 2 Rs= 50β¦ ππ Rload= 50β¦ π§π = π π + πππ 2 = 200 Find maximum π§ππππ = π π β πππ 20 Autotransformer Ip N1 Primary core Is N2 Secondary Starting motors ELEC 4602 Single phase Ip N1 Primary N2 core Is Secondary Single phase Maximum power transfer to the load Center tapped Ip Ns1 Is1 Vs1 N1 Ns2 Primary core Is2 Secondary Vs2 ground With ground Vs1 180o out of phase with Vs2. Two phase household Center tapped Ip Ns1 Is1 Vs1 N1 Ns2 Primary core Is2 Secondary Vs2 ground With ground Vs1 180o out of phase with Vs2. Two phase household Center tapped A B C Three phase NA1 NAβ2 NB1 NBβ2 NC1 NCβ2 Aβ Bβ Cβ Interconnection n turns ratio Topic of ELEC 3508: Power Electronics 35 36 LabVolt Module used in ELEC 3508 37
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