Question 1. parameters: A single-phase, 10 kVA, 4400/1100 V, 50 Hz transformer has the following High voltage (HV) side winding parameters: RƒV = R₁ = 5N, Low voltage (LV) side winding parameters: RLV = R2 = 0.35 2, Core loss resistance and magnetizing inductance: Rc(HV) = LHV = L₁₁ = 0.09 H LLV = L12 = 0.0032 H 150 kN, Lm(HV) = 180 H (a) Obtain an equivalent circuit of the transformer referred to the HV side. (b) Determine the voltage regulation at full load with 0.8 power factor (PF) leading. (c) Determine the no-load power loss (core loss) based on the circuit model in (a).

The direct electrical connection of the windings allows transient over voltages to pass through the auto transfonner more easily, and that is an important disadvantage of the autotransformer. (a) True (b) False

The symbol shown is a(n) a. iron core transformer. b. auto transformer. c. current transformer. d. air core transformer.

A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.

Consider an ideal transformer with N1=3000andN2=1000 turns. Let winding 1 be connected to a source whose voltage is e1(t)=100(1| t |)volts for 1t1ande1(t)=0 for | t |1 second. A2- farad capacitor is connected across winding 2. Sketch e1(t),e2(t),i1(t),andi2(t) versus time t.

The ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.

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Single phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R2= 36 mQ Ro= 3000 Q X₁₁ = 0.60 X12= 0.0024 02 X= 6000 Input voltage= 1440 V (also known as rms voltage) Load impedance of secondary winding is Z₂= 0.12 + j0.3 a) Copper losses on primary and secondary windings b) Core loss c) Transformer efficiency V₁ Ī₁ Ro ww m Xu R₁ R'2 ww m ww I₂ X12 m

Single phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R₂= 36 mQ Ro= 3000 Q X₁₁ = 0.6Q X12= 0.0024 Q X0= 6000 Q Input voltage-1440 V (also known as rms voltage) Load impedance of secondary winding is z₂= 0.12 + j0.3 Q a) Secondary winding current phasor 12 (NOT to be confused for I'₂) b) Load voltage phasor V2 (NOT to be confused for V'2) c) Primary winding current phasor ī₁ → X11 X12 R₁ R'2 www V₂ V₁ ī₁ R m ×°