--- title: "Equivalent circuit and half-load efficiency of 50 kVA transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:50kva-2500-250-test-secondary-half" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/50kva-2500-250-test-secondary-half site: BCT Engineering --- # Equivalent circuit and half-load efficiency of 50 kVA transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2069 Poush, 8 marks - Video: [#3 Equivalent circuit, Voltage Regulation and efficiency of single phase transformer #transformer](https://www.youtube.com/watch?v=L94UpXXINe8) — Dr. B. M. Manjunath - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/50kva-2500-250-test-secondary-half - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question A 50 kVA, 2500V/250V, 50 Hz single phase transformer draws a current of 0.3 Amp at no load and consumes 300 Watts. When the primary winding is supplied by 100 V with secondary winding short circuited, the primary draws a current of 20 Amp and consumes 500 watts. Calculate the equivalent circuit parameters of the transformer referred to secondary side. Also calculate the efficiency of the transformer at half load. ## Short answer From open-circuit test on HV: R0=20833 Ω, X0=9092 Ω, referred to secondary become 208.33 Ω and 90.92 Ω. Short-circuit test gives R01=1.25 Ω, X01=4.841 Ω, referred to secondary: R02=0.0125 Ω, X02=0.0484 Ω. At half load, copper loss=125 W, iron loss=300 W, efficiency (upf)=98.33%. ## Answer **Interpretation:** the no-load test (0.3 A, 300 W) is at rated primary voltage, 2500 V, and the SC test is on the primary (HV) with the secondary shorted. Rated primary current $= 50000/2500 = 20$ A, so the SC test is at full load. Factor HV to LV: $\left(\frac{250}{2500}\right)^2 = 0.01$. #### Shunt branch (no-load test on HV: 2500 V, 0.3 A, 300 W) $$ \begin{aligned} \cos\phi_0 &= \frac{300}{2500 \times 0.3} = 0.4, \quad \sin\phi_0 = 0.9165 \\ I_w &= 0.3 \times 0.4 = 0.12\ \text{A}, \quad I_\mu = 0.3 \times 0.9165 = 0.275\ \text{A} \\ R_0 &= \frac{2500}{0.12} = 20833\ \Omega, \quad X_0 = \frac{2500}{0.275} = 9092\ \Omega \ (\text{HV side}) \end{aligned} $$ Referred to secondary: $$ R_0' = 20833 \times 0.01 = 208.33\ \Omega, \qquad X_0' = 9092 \times 0.01 = 90.92\ \Omega $$ #### Series branch (SC test on HV: 100 V, 20 A, 500 W) $$ \begin{aligned} Z_{01} &= \frac{100}{20} = 5\ \Omega, \quad R_{01} = \frac{500}{20^2} = 1.25\ \Omega \\ X_{01} &= \sqrt{5^2 - 1.25^2} = 4.841\ \Omega \end{aligned} $$ Referred to secondary: $$ \begin{aligned} R_{02} &= 1.25 \times 0.01 = 0.0125\ \Omega \\ X_{02} &= 4.841 \times 0.01 = 0.0484\ \Omega \\ Z_{02} &= 5 \times 0.01 = 0.05\ \Omega \end{aligned} $$ #### Equivalent circuit referred to secondary ```text I1 -> I0 I2 o-----+--------+---[ R02 ]---[ jX02 ]---o + | | + [ R0 ] [ jX0 ] load V1' | | V2 - | | - o-----+--------+------------------------o All values referred to secondary (250 V): R0 = 208.33 ohm, X0 = 90.92 ohm R02 = 0.0125 ohm, X02 = 0.0484 ohm ``` #### Efficiency at half load Iron loss $P_i = 300$ W; full-load copper loss $= 500$ W (test at rated current). Assuming unity pf: $$ \begin{aligned} \text{Output} &= 0.5 \times 50000 \times 1 = 25000\ \text{W} \\ P_{cu} &= 0.5^2 \times 500 = 125\ \text{W} \\ \eta &= \frac{25000}{25000 + 300 + 125}\times 100 = 98.33\% \end{aligned} $$ **Answer:** Referred to secondary: $R_0 = 208.33\ \Omega$, $X_0 = 90.92\ \Omega$, $R_{02} = 0.0125\ \Omega$, $X_{02} = 0.0484\ \Omega$; half-load efficiency (upf) = **98.33%**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To refer HV-side impedances of a 2500/250 V transformer to the secondary, they are multiplied by: A. 10 B. 100 C. 0.1 D. 0.01 Answer: D. Impedance scales by the square of the turns ratio (250/2500)^2 = 0.01. (Key point: Referring to LV side) 2. For the SC test (100 V, 20 A, 500 W), R01 and X01 are: A. R01 = 1.25 ohm, X01 = 4.841 ohm B. R01 = 5 ohm, X01 = 1.25 ohm C. R01 = 25 ohm, X01 = 4.841 ohm D. R01 = 1.25 ohm, X01 = 5 ohm Answer: A. R01 = W/I^2 = 500/400 = 1.25 ohm, Z01 = 100/20 = 5 ohm, X01 = sqrt(25 - 1.5625) = 4.841 ohm. (Key point: Series branch from SC test) 3. In the OC test (2500 V, 0.3 A, 300 W), the working component Iw and R0 (HV side) are: A. 0.12 A and 20833 ohm B. 0.275 A and 9092 ohm C. 0.3 A and 8333 ohm D. 0.12 A and 9092 ohm Answer: A. cos phi0 = 0.4, so Iw = 0.12 A and R0 = 2500/0.12 = 20833 ohm. (Key point: Shunt branch from OC test) 4. Which losses are used for efficiency at half load (unity pf) in this problem? A. 300 W iron and 500 W copper B. 300 W iron and 125 W copper C. 150 W iron and 125 W copper D. 300 W iron and 250 W copper Answer: B. Iron loss is constant (300 W); copper loss is 0.5^2 x 500 = 125 W, giving eta = 25000/25425 = 98.33%. 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