--- title: "230/2300 V transformer with load impedance" subject: "Electrical Machine" chapter: 2 key: "semester_4/electrical_machine:transformer-230-2300-load-impedance" source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-230-2300-load-impedance site: BCT Engineering --- # 230/2300 V transformer with load impedance Electrical Machine · Chapter 2: Transformer - Asked in: 2070 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-230-2300-load-impedance - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2 ## Question A 230 V / 2300 V single-phase transformer is excited by 230 V ac voltage. The equivalent resistance and reactance referred to primary side are 0.1 Ω and 0.4 Ω respectively. Given that Ro = 500 Ω and Xo = 200 Ω. The load impedance is (400 + j600) Ω. Calculate: (i) Primary current and input power factor (ii) Secondary terminal voltage. ## Short answer Load referred to primary: 4+j6 Ω. Primary current I1=31.5 A, input pf=0.533 lagging. Secondary terminal voltage=2182 V. ## Answer The approximate equivalent circuit referred to the primary is used: the shunt branch (R₀ ∥ jX₀) is placed across the supply, and the series branch carries the load current. #### Equivalent circuit (referred to primary) ```text I1 Re=0.1 Xe=j0.4 I2' o----->-------[===]----[~~~]--+---->----+ | | | V1=230V I0 ---> +------+ | Zl'=4+j6 | | | | | | Ro=500 jXo=200 | | | | | | o-------------------+------+--+---------+ ``` Turns ratio a = 2300/230 = 10. #### Step 1: refer load to primary Zl' = Zl / a² = (400 + j600) / 100 = 4 + j6 Ω #### Step 2: load current referred to primary Z_total = (0.1 + j0.4) + (4 + j6) = 4.1 + j6.4 Ω, |Z| = 7.601 Ω I₂' = 230 / (4.1 + j6.4) = 30.26 ∠−57.36° A = (16.32 − j25.48) A #### Step 3: no-load (exciting) current - Core-loss component I_c = 230 / 500 = 0.46 A - Magnetising component I_m = 230 / 200 = 1.15 A (lagging) - I₀ = 0.46 − j1.15 = 1.239 A #### (i) Primary current and input power factor I₁ = I₂' + I₀ = (16.32 + 0.46) − j(25.48 + 1.15) = 16.78 − j26.63 A |I₁| = 31.48 A at −57.78°, so pf = cos 57.78° = **0.533 lagging** #### (ii) Secondary terminal voltage Voltage drop in the series impedance = I₂' × (0.1 + j0.4) = 30.26 × 0.412 = 12.47 V V₂' = 230 − I₂'(0.1 + j0.4) = 218.2 V (check: I₂' × |Zl'| = 30.26 × 7.211 = 218.2 V) V₂ = a × V₂' = 10 × 218.2 = **2182 V** **Answer: I₁ = 31.5 A, input pf = 0.533 lagging, secondary terminal voltage = 2182 V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load impedance (400 + j600) Ω referred to the primary for the 230/2300 V transformer? A. Multiplied by 100 B. Divided by 10 C. Multiplied by 10 D. Divided by 100 Answer: D. Zl' = Zl/a² with a = 10, giving 4 + j6 Ω. (Key point: Refer load) 2. How is the no-load current I0 found in the approximate circuit? A. From V1/Ro and V1/Xo placed across the supply B. From the series impedance C. From the load D. From V2 Answer: A. Ic = 230/500 = 0.46 A and Im = 230/200 = 1.15 A. (Key point: Exciting current) 3. How is the primary current I1 found? A. I1 = I2' only B. I1 = I0 only C. I1 = I2' + I0 (phasor sum) D. I1 = I2'/a Answer: C. Primary current is the phasor sum of load current and exciting current. (Key point: Primary current) 4. How is the secondary terminal voltage obtained? A. V2 = V2'/10 B. V2 = V1 C. V2 = V1 × a² D. V2' = V1 − I2' (Re + jXe), then V2 = 10 V2' Answer: D. V2' = 218.2 V and V2 = 2182 V. 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